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Table of Contents
Chapter 1: Matter and Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
What is chemistry? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Why is chemistry relevant? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
What is matter? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
How is matter classified? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
What kinds of properties does matter have? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
An introduction to the elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
The natural states of the elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
The Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Chapter 2: Measurements in Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Chemistry is an empirical science . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Scientific notation and powers of 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Measurements and significant figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
SI units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Measurements and measured quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Derived quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Energy and heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Dimensional analysis and problem solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
Chapter 3: Atoms and the Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
A brief history of atomic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
The structure of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Atoms and protons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Atoms, neutrons, and isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Atoms and electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
The octet rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Fundamentals of electrons in atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Ground states and excited states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Electron configurations: full configurations and the periodic table . . . . . . . . . . . . . . . . 39
Electron configurations: Noble Gas configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Electron configurations: electron spin diagrams (arrow diagrams) . . . . . . . . . . . . . . . . . 48
Inner shell, outer shell, and valence electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
The electron configurations of ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Atomic properties and periodic trends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Chapter 4: Ionic Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Chemical bonds and chemical compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Lewis electron dot structures (Lewis structures) for atoms . . . . . . . . . . . . . . . . . . . . . . . 63
The names of common cations and anions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
The nomenclature of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
The names and molecular formulas of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . 70
The nomenclature of binary molecular compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
The nomenclature of acids and bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Post-script: responses to a student questions about nomenclature . . . . . . . . . . . . . . . . . 82
Chapter 5: Molecular Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
Covalent bonds and molecular compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
Covalent bonds and the Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Multiple covalent bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
Coordinate covalent bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
Molecular formulas and structural formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
The Lewis structures of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
The Lewis structures of covalent compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
Exceptions to the octet rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
The shapes of molecules and VSEPR - Valence Shell Electron-Pair Repulsion . . . . . . 101
Two Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Three Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Four Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Five Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Six Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
Polar covalent bonds and polar molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
Molecular polarity and symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
Summary of molecular polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Chapter 6: Chemical Reactions - Classification and Mass Relationships . . . . . . . . . . . . . . . . . 112
Molecular formulas and empirical formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
Molecular formulas, molecular weights, and formula weights . . . . . . . . . . . . . . . . . . . 114
The mole and Avogadro's number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Molecular weight and molar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
Chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Theoretical yield and percent yield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
Strong electrolytes, weak electrolytes, and non-electrolytes . . . . . . . . . . . . . . . . . . . . . 130
Molecular, ionic, and net ionic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Predicting the likelihood of chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
Precipitation reactions and solubility rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
Water formation in double-displacement reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
Gas formation in double-displacement reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
Electron transfer (redox) reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Post-script: responses to student questions about Chapter 6 concepts . . . . . . . . . . . . . 148
Chapter 7: Chemical Reactions - Energy, Rates, and Equilibrium . . . . . . . . . . . . . . . . . . . . . . 154
An introduction to thermodynamics and kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
Energy, work, and other thermodynamic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . 155
Internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
Gibbs free energy and the spontaneity of reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
The rates of chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Reaction rates and molecular collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
Activation energy and activated complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
The reversibility of chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
Catalysts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
The equilibrium constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
Writing equilibrium constant expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
Equilibrium constants and Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
LeChatelier's principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Chapter 8: Gases, Liquids, and Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
General gas properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
Ideal gases and the Ideal Gas Law (Universal Gas Law) . . . . . . . . . . . . . . . . . . . . . . . 190
The ideal gas law and molar mass calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
The ideal gas law and stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
The combined gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
Dalton's Law of Partial Pressures and mole fractions . . . . . . . . . . . . . . . . . . . . . . . . . . 198
States of matter, phase transitions, and enthalpies of phase transitions . . . . . . . . . . . . 201
Intermolecular forces: attractive interactions between molecules . . . . . . . . . . . . . . . . . 204
London (dispersion) forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
The magnitude of London forces increases with molecular size and surface area . . . . 208
Dipole-dipole interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Hydrogen bonds - a special case of dipole-dipole interactions . . . . . . . . . . . . . . . . . . . 209
Ion-dipole interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
Predicting the intermolecular forces in compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
Intermolecular forces and solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Intermolecular forces and physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
Chapter 9: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
General solution properties and definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
Calculating concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
Dilutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
Some properties of liquids and solutions: surface tension and capillary action . . . . . . 229
Some properties of liquids and solutions: vaporization and vapor pressure . . . . . . . . . 230
Some properties of liquids and solutions: diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
Some properties of liquids and solutions: osmosis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
Chapter 10: Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
General acid-base information: a review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
The Arrhenius acid-base theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
The Brønsted-Lowry acid-base theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
The Lewis acid-base theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
Strong and weak acids and bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
The behavior of acids and conjugate bases in water . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
The behavior of bases and conjugate acids in water . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
The acid-base behavior of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
The pH scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
Calculating [H3O+], [OH-], and pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
Buffer solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Chapter 1: Matter and Life
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as chemistry, matter, atom, molecule, element, compound,
homogeneous mixture, heterogeneous mixture, periodic table, physical property, chemical
property, physical change, chemical change, metals, nonmetals, semi-metals (or metalloids),
alkali metals, alkali earth metals, chalcogens, halogens, and noble gases.
Identify the three basic states of matter, the names of the changes between those states, and
the role of temperature in the state in which matter is found.
Describe how solids, liquids, and gases differ from each other at a molecular level.
Identify a substance as an element, compound, homogeneous mixture, or heterogeneous
mixture.
Know the natural states of the elements, especially those that occur as diatomic molecules.
Differentiate between physical and chemical properties.
Differentiate between physical and chemical changes.
Know how to determine whether an element is a metal or a nonmetal based on its location
in the periodic table.
Know whether a substance exists in the elemental state as diatomic molecules rather than as
individual atoms.
What is chemistry?
If you refer to various chemistry text books you will find a variety of definitions of
chemistry. I prefer to define chemistry as the study of things made up of atoms and molecules. It is
the study of why things made up of atoms and molecules behave the way they do. It is also the study
of how to make them behave more usefully. As everything in the world around us is composed from
atoms, by gaining a fundamental understanding of chemistry we can begin to understand many
important and interesting phenomena that affect us on a daily basis.
While one may read of a variety of different types of chemistry such as forensic chemistry,
geochemistry, food chemistry, and etc., there are really just five fundamental branches of chemistry:
Analytical chemistry is the study of what is in a substance, and how much of a particular
thing is in a substance. Most of the things in the world around us are not chemically pure but consist
of dozens or even hundreds or thousands of different chemically distinct components. Gasoline
looks like it is simply a liquid, but in fact it is composed of several hundred different chemicals. The
smell of an orange is due to the presence of nearly two hundred different compounds. And so on.
Analytical chemistry is used all around us, every day, and in a variety of ways. The study of
pollutants in the environment, the analysis of the composition of an expensive perfume, the testing
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of a patient's blood in a hospital or of an athlete's urine during a competition, and a check of the
composition of the various liquids used during the manufacture of gasoline, are a few examples of
how analytical chemistry is used.
Inorganic chemistry is the study of all of the elements and their compounds except carbon
and its compounds.
Organic chemistry is the study of carbon and its compounds. Since there are 117 known
elements, it often seems odd that an entire branch of chemistry is devoted to a single element and
its compounds while the other 116 elements and their compounds are all lumped together in a
separate discipline, but there is a very good reason for this. There are about 1.5 million known
inorganic compounds. This is a lot of compounds, and you will not be required to know them all for
this class, although it may seem like it before you're finished. The number of known organic
compounds varies, depending on the reference. These sources state that there are from 16 to 40
million or more known organic compounds. Carbon is the basic element of life, and living creatures
have developed an astonishing array of different organic compounds. By definition organic
compounds are compounds that contain carbon, i.e., compounds in which carbon is found in the
molecular formula. But there are a small number of carbon-containing compounds that are classified
as inorganic, such as the cyanides, carbonates, and bicarbonates (we’ll learn more about these in
Chapter 4). So a better working definition of organic compound is a compound that contains both
carbon and hydrogen in its molecular formula, with the exception of the bicarbonates (hydrogen
carbonates). Inorganic compounds are compounds that do not contain both carbon and hydrogen.
This is a general way of classifying organic compounds and there are numerous, notable exceptions
in the “real world,” but it will be the way by which we make the distinction between organic and
inorganic compounds in this class. You very much want to remember how to tell the difference
between the two as I always ask questions like this on the exams you’ll take in this class.
Biochemistry is the study of the chemistry of life and living things. This is a field of great
importance to those who practice in the health sciences. The ways the atoms and molecules behave
in living systems is predicated on the rules of organic chemistry.
Physical chemistry is the study of how the laws of physics affect things as small as atoms
and molecules. During the early decades of the 20th Century researchers were astonished to discover
that very small particles, like atoms and molecules, do not obey the laws of physics expounded by
Sir Isaac Newton. This spurred the development of the field known as quantum mechanics, which
explains the behavior of things that are very small.
Why is chemistry relevant?
We live in a world that grows more and more technical by the day. Even if you're not
interested in medicine or how the body works, how do you know if the claims made by someone
selling a particular health care product or beauty product are legitimate or bogus? Are you willing
to trust an industry that releases substances into the environment when it says they are safe, or will
you believe those who might be willing to distort or misrepresent the truth in order to give weight
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to their argument that the chemicals released are dangerous? Do you trust people or government
agencies that assure you that exposure to a particular chemical is not dangerous, particularly when
their arguments may be based on economic or political considerations and not on pure science?
When medicine is prescribed for a sick family member or friend, are you sure they will receive the
appropriate medication and not something that may aggravate their illness or even kill them?
Whether you're interested in science or not, whether you're interested in chemistry or not,
you need to know some of the fundamentals of the science in order to be a good consumer, a good
citizen, and a good family member and friend. You need to know a little bit about chemistry in order
to protect yourself. You need to know a little bit about chemistry in order to be able to think clearly
and correctly about many important everyday issues.
To many people chemistry is a dirty word, but that attitude is terribly naive and uninformed.
Everything around us is chemicals, from the air we breath, the water we drink, and the food we eat,
to our very bodies. All living things are made up entirely of chemicals. All dead things, and all
things that never have or never will live (such as rocks) are made up of chemicals. The earth on
which we live is nothing more, at its most fundamental, than a great mass of chemicals. The sun on
which we rely for light and heat and life itself is a giant globe of reacting chemicals. And so on. It
is true, the incorrect use of chemistry is responsible for many important health and environmental
issues. But chemistry also makes the world a better place.
Compare your life with that of someone living a year ago, ten years ago, fifty years ago, or
a hundred or more years ago. What things do you have that they did not? How are medicine and
health care different for you? personal hygiene? transportation? communication? the way you spend
your leisure time? food and nutrition? the clothes you wear? Almost inevitably, anything you have
that someone in the past did not have is attributable, either directly or indirectly, to advances made
in chemistry.
What do chemists do to make the world a better place? In order to address this question let
me tell you a little bit about the work of some of my friends who hold masters and doctoral degrees
in chemistry. They are involved in the study of:
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•
•
•
•
•
•
•
•
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The analysis and identification of DNA and large biological molecules.
The chemistry of blood clotting mechanisms.
The chemistry of cardiovascular disease.
The chemical mechanisms of infectious diseases.
New chemical methods for breast cancer detection.
The development of new and improved antibiotics and other medicinal compounds.
The development of new and improved fertilizers and pesticides.
Food chemistry.
Personal hygiene products & cosmetics.
The chemistry of surfaces, which sounds terribly arcane until one realizes that many
important reactions will only occur on certain types of surfaces.
Composite materials, such as graphite fiber products.
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The production of gasoline and other petroleum-based products.
•
Lubricants and fuel additives for automobiles.
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The detection of drugs and poisons.
This is just a small sample of the sorts of things that chemists do to make our world a better place.
What is matter?
As is the case with chemistry, there are a number of different definitions of matter. It is
common to define matter as anything that has both mass and volume. I prefer to define matter as
anything made up of atoms, which are the basic building blocks of the world around us. It is true that
there are types of matter smaller than atoms, such as protons, neutrons, electrons, and quarks, but
we leave the study of particles smaller than atoms to courses in physics.
The study of matter is the study of chemistry. This means that the study of matter is the study
of things made up of atoms and molecules. We will learn more about this in Chapter 3.
How is matter classified?
Matter can be classified either by its state or by its composition.
There are three common states of matter, solids, liquids, and gases. We do not concern
ourselves with more exotic states of matter such as plasmas or Bose-Einstein condensates when we
study elementary chemistry.
We will use the word "particle" as we discuss the three common states of matter, since
substances can be made up either of atoms or of molecules. Particle is used to generically represent
the small units of matter in the substance.
There are various ways to make distinctions between the three common states of matter. For
our purposes, the most useful comparison at the particle level is as follows:
In the solid state the particles are relatively very close together, usually well-ordered (tidily
arranged), and are held together by relatively strong attractive interactions between the solid
particles.
In the liquid state the particles are not quite as close together, not quite as well-ordered, and
the attractive interactions are not quite as strong.
In the gaseous state the particles are relatively far apart, chaotically arranged (i.e., not
ordered), and attractive interactions between the particles are either negligible or nonexistent.
There are specific names for the transitions from one state of matter to another. These transitions
are also called phase changes, or changes of state. These refer to state changes in any material, and
not just in water.
4
Matter can also be classified based on its composition.
An element is a substance in which all of the atoms are the same. In other words, an element
is atomically pure. Elements may exist as single atoms or in molecular (i.e. more than single atom)
form, such as Fe, O2, P4 and S8. Notice that the molecular formulas of elements reflect their
composition from one and only one type of atom.
Compounds are substances made from the atoms of two or more elements and which are held
together by chemical bonds. There are millions of examples of compounds, including water (H2O),
carbon dioxide (CO2), and ethyl alcohol (C2H6O). Notice that the molecular formulas of compounds
reflect their composition from two or more types of atoms. It is important to note that the different
types of atoms in compounds are held together by chemical bonds, a topic we will examine in
Chapter 4.
A mixture is a substance composed of two or more components which can be separated by
physical means, i.e., based on the different physical properties of the components. Those
components may be elements, compounds, or other mixtures. As an example, table salt (sodium
chloride) dissolved in water can be recovered by evaporating the water. The ability to separate the
components of a mixture by taking advantage of differences in physical properties means that
chemical bonds do not exist between the components. We will refine this concept more correctly in
Chapter 8 but for the time being, this will suffice for our purposes.
There are two types of mixtures. They are classified by how evenly dispersed the particles
are at the molecular level.
5
In homogeneous mixtures there is uniformity in composition, properties, and appearance
throughout the mixture down to the particle level. Examples of homogeneous mixtures include table
salt dissolved in water, red wine, gasoline, and 18 karat gold.
Heterogeneous mixtures are characterized by discontinuities in composition, properties, and
appearance at levels well before the molecular level. In other words, heterogeneous mixtures do not
have uniform composition, properties, and appearance throughout the mixture. Examples of
heterogeneous mixtures include pizza, granite, a nice fizzy soda in a glass filled with ice, and a
peanut butter and jelly sandwich. What do you see when you look at a PBJ sandwich? Does it look the
same throughout, or do you see layers? What does the presence of layers mean about its composition?
The separation of the components of a mixture can be accomplished using physical changes, i.e.,
changes based on the physical properties of the substances. The separation of the components of a
compound can only be achieved by chemical reactions. Elements are pure and cannot be separated.
What kinds of properties does matter have?
Matter has physical properties, which are properties that can be observed and measured
without changing the composition of the material. The physical properties of a substance include
its color, odor, state of matter, melting point, boiling point, heats of vaporization and fusion, density,
solubility, metallic character, electrical and thermal conductivity, magnetic properties, crystal shape,
malleability, ductility, and viscosity. Most common physical changes either involve changes of state
(phase transitions) or are the consequence of mechanical processing (e.g. grinding, crushing, slicing,
pulverizing, gluing pieces together, etc.)
The chemical properties of a material are the chemical reactivity of the substance, that is,
how a substance changes its composition when it interacts with other substances.
There is a difference between changes of state (physical changes) and chemical reactions.
Physical changes are changes involving physical properties, or, changes in which there is no change
to the chemical composition of the material(s) under observation. When chemical changes (or,
chemical reactions) occur, the starting materials are consumed and new materials are formed due
to the breaking and making of chemical bonds. Chemical reactions may be indicated by color
changes, the absorbing or release of energy (heat, light, sound, electric), or by the formation of new
materials such as gases, pure liquids, or solids (precipitates). These are general indications that a
chemical reaction has occurred but they are not absolutely correct in all instances. Use them, but use them
with care!
As an example, chemically speaking, water is known as H2O. This tells us that each water
molecule consists of two hydrogen atoms and one oxygen atom. If we take water out of our kitchen
faucet, it is in the liquid state. If we take liquid water and cool it by placing it in the freezer it will
eventually change from the liquid to the solid state. But while its state has changed, the water
remains H2O, even though we call solid water “ice.” If we take solid water (ice) and put it in a pan
on the stove, as it warms it will first change from the solid to the liquid state and then from the liquid
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to the gaseous state. However, as these changes of state occur, the water remains chemically
unchanged. It remains H2O.
If we take a glass of liquid water and place a nine volt battery in it, something interesting
occurs. This is an experiment you can safely do yourself. Should you elect to try it, I’d suggest you use a
clear drinking glass or a Mason fruit jar. It will not harm the glass. Add about one teaspoon of table salt
to one to two pints of tap water, and place the battery in the water. Streams of bubbles begin to
emanate from the battery terminals almost immediately. As you observe the formation of bubbles,
you can see that a far greater volume of bubbles are being produced at one terminal than at the other.
This is because the electrical energy of the battery is being used to break the chemical bonds that
hold the hydrogen atoms and oxygen atoms in water molecules together. Hydrogen gas is being
produced at the terminal producing the greater volume of gas, while oxygen gas is being produced
at the other terminal. Why would you expect to see more hydrogen gas produced than oxygen gas? Hint:
think about the molecular formula of water.
An introduction to the elements
As we stated earlier, elements are substances in which all of the atoms are the same. There
are 117 known elements, of which 88 are naturally occurring. The twenty most abundant elements
in the earth's crust by mass are: Note: I do not expect you to memorize this table. Also, you should know
that there is some disagreement from one source to the next as to the number of naturally-occurring versus
synthetic elements. Don’t worry about it. This is an early introduction to a reality of chemistry: there are
many areas in which chemists disagree with one another. There are sound reasons for these disagreements.
I expect you to know what I tell you in these lecture notes and to understand that as you look beyond
them, you will at times find contradictory information. It is not that these other sources are right and I am
wrong, or visa versa. It is that we respectfully disagree with one another. While these disagreements are
common, they are usually also minor. We as chemists are typically united in the larger, more important
points that we teach our students in class.
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element
percent composition
oxygen
45.5
silicon
27.2
aluminum
8.3
iron
6.2
calcium
4.7
magnesium
2.8
sodium
2.3
potassium
1.8
titanium
0.6
hydrogen
0.2
phosphorus
0.1
manganese
0.1
fluorine
0.05
barium
0.04
strontium
0.04
sulfur
0.03
carbon
0.02
zirconium
0.02
vanadium
0.01
Chlorine
0.01
all others
0.05
A table of the fifteen most abundant elements in the human body is found below. It is
interesting to note the difference in the composition of the earth’s crust and that of the human body.
Again, I do not expect you to memorize this information. It comes from a more detailed table found at
"Elemental Composition of the Human Body” http://web2.iadfw.net/uthman/elements_of_body.html should
you find yourself interested.
8
element
mass of element in a
70 kg person (kg)
percent composition
(mass percent)
oxygen
43
61.4
carbon
16
22.9
hydrogen
7.0
10.0
nitrogen
1.8
2.6
calcium
1.0
1.4
phosphorus
0.8
1.1
potassium
0.1
0.2
sulfur
0.1
0.2
sodium
0.1
0.1
chlorine
0.09
0.1
magnesium
0.02
0.02
iron
0.004
0.006
fluorine
0.003
0.004
zinc
0.002
0.003
silicon
0.001
0.001
An important source of information about the elements is found in the periodic table. We will
we will discuss it in a moment.
The natural states of the elements
How do the elements occur naturally? In other words, when we venture out into the real
world, in what state or condition might we find a sample of a particular element? Knowing the
natural states of the elements is often very important in understanding chemical reactions. You should
know the following by heart, and soon. Questions regarding this information frequently find their way into
quizzes and exams.
Most elements occur as compounds with other elements. Finding hydrogen and oxygen in
water (H2O), sodium and chlorine in sodium chloride (NaCl), and aluminum and oxygen in an
aluminum ore called bauxite (Al2O3) are three common examples.
There are some exceptions in which the elements are found in a more or less pure state.
These include the "Noble Metals," gold, silver, and platinum, although these are often found
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together as mixtures. These also include the Noble Gases (which we will discuss in a moment),
which are also often found in their pure elemental form.
Some substances occur naturally in diatomic (i.e., molecules made up of two atoms) form.
These include hydrogen, nitrogen, and oxygen (H2, N2, and O2 respectively) and the halogens
fluorine, chlorine, bromine, and iodine (F2, Cl2, Br2, and I2 respectively). You *must* remember that
these substances occur as diatomic molecules or you will come to grief in this class repeatedly!
Phosphorus and sulfur occur naturally as P4 and S8 molecules respectively.
The Periodic Table
The Periodic Table is a table of the elements arranged in rows and columns in sequence of
increasing atomic number (i.e., the number of protons in the nucleus). The rows are called periods
and the columns are called groups.
The elements are represented by symbols in the periodic table.
In most cases the relationship between the name and the symbol is obvious. For example, O-oxygen,
H-hydrogen, N-nitrogen. For a few elements the symbol is derived from a non-English language like
Latin or German, e.g., K-potassium (kalium), Fe-iron (ferrum), Au-gold (aurum), W-tungsten
(wolfram, Swedish). In these cases there is not an obvious correlation between the element's name
and its symbol.
I do expect you to know the names and symbols of the first twenty elements along with the names and
symbols of all other elements commonly used in this class. While I will provide you with a periodic table
and a list of element names and symbols on all exams, it will help you immensely and make your work go
more quickly if you have committed these names and symbols to memory. You do not need to memorize
other information such as atomic numbers and atomic weights. If pertinent they will be provided.
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The top number in the periodic table for each element is the atomic number, the number of
protons an atom has in its nucleus. The bottom number in the periodic table for each element is the
atomic weight, the weighted average of the atomic masses of all of the isotopes of a particular
element. The atomic mass is the sum of the protons and neutrons in an isotope or a particular atom.
The masses of individual atoms is expressed in amu, or atomic mass units and is equal to 1/12 the
mass of a single 12C isotope (or, 1.661 x 10-27 kg; don’t memorize this number. If you need to use it, I
will give it to you). We will discuss atomic number, atomic mass, atomic weight, and isotopes in more
detail in Chapter 3 (will you be able to wait, or will the suspense just kill you?!).
In the periodic table, groups are "chemical families," because they exhibit similar chemical
behavior. Some of the groups have special names (which you must know):
Group 1A - alkali metals
Group 2A - alkali earth metals
Group 6A - chalcogens (this is rarely used any more)
Group 7A - halogens
Group 8A - Noble Gases
The "island" by itself beneath the main body of the periodic table consists of the lanthanides
and actinides or the rare earth metals, which have special names even though they are periods rather
than groups. Interestingly there are a number of rare earth metals far more common than some of the
transition metals (the 1B-8B elements in the periodic table) such as gold, silver, platinum, and etc.
The periodic table can also be divided into three main regions, metals, nonmetals, and
metalloids.
Metals are elements characterized by the tendency to give up electrons in reactions (i.e. form
cations), good thermal and electrical conductivity, and are usually lustrous, malleable, ductile.
Metals are found to the left of the "staircase" line on the periodic table.
Nonmetals are elements characterized by the tendency to gain electrons in reactions (i.e.
form anions) and by a lack of other metallic properties, i.e. they are poor conductors, dull, and
brittle. Nonmetals are found to the right of the "staircase" line on the periodic table.
Metalloids (or, semi-metals) are elements characterized by the some of the properties of both
metals and nonmetals. Metalloids straddle the "staircase" line on the periodic table. Having
mentioned metalloids here, you can promptly forget about them as they have no relevance to us as
we study chemistry in this class. You will however, need to be able to classify each and every
element as a metal or a nonmetal, for reasons that will become apparent when we discuss chemical
bonds in Chapter 4.
Examine the periodic table in the front inside cover of your textbook. Can you find the
staircase line? Can you tell which elements are metals? nonmetals?
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Chapter 2: Measurements in Chemistry
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as scientific notation, accuracy, precision, significant figures,
density, specific heat, conversion factor, dimensional analysis.
Correctly write numbers either in regular notation or in scientific notation.
Correctly identify the number of significant figures in a number and correctly predict the
number of significant figures in the answer of a problem involving multiplication and/or
division.
Correctly identify the zeros in a number as leading, captive, or trailing zeros and assess
whether or not they contribute to the number of significant figures in a reported value.
Know the basic metric system units, the prefixes used in the metric system, and be able to
convert from one metric system unit to another, e.g., convert mg to kg.
Be able to convert between the Fahrenheit, Celsius, and Kelvin temperature scales.
Calculate the density of substances, calculate either the mass or the volume of a substance
given its density, and use density as a conversion factor in dimensional analysis problems.
Correctly set up and solve dimensional analysis problems using appropriate conversion
factors.
Chemistry is an empirical science
Chemistry is an empirical science. This means that it is based on observations and
measurements made during experiments. This chapter will address measurements and calculations
in chemistry, although we do not actually make any measurements in this course.
Scientific notation and powers of 10
It is common to use either very large or very small numbers in science. The size of most
atoms is in the trillionths of an inch (0.000000000001 inch) range. The atoms in a few ounces of a
substance may number as many as one million billion billion (1,000,000,000,000,000,000,000,000)
or more. How can we conveniently express numbers that are either very large or very small?
Scientific notation is a scientific shorthand that makes the expression of large and small
numbers easier. It is based on our use of a base ten number system. Each digit in a number is a
placeholder representing some multiple of ten. For example, a number greater than 1, such as 4321,
can also be written as 4321 = (4 x 1000) + ( 3 x 100) + (2 x 10) + (1 x 1), or, as (4 x 10 x 10 x 10)
+ (3 x 10 x 10) + (2 x 10) + (1 x 1). A number less than 1, such as 0.5678, can also be written as
0.5678 = (5 x .1) + (6 x .01) + (7 x .001) + (8 x .0001).
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4321 is written as 4.321 x 103 in scientific notation. Numbers written using scientific notation
consist of two pieces, the coefficient and the exponent. The coefficient is the part with the decimal,
in this case, 4.321. The exponent is the power to which 10 is raised, in this case, 3. In scientific
notation the coefficient is always obtained by moving the decimal until is just one place to the right
of the first non-zero digit. The magnitude of the exponent depends on how many places the decimal
is moved. The sign of the exponent depends on which direction the decimal is moved. If the number
is greater than one, the decimal will be moved to the left and the exponent will be a positive number.
If the number is less than 1 (note: less than one is a small number but not a negative number.
Negative numbers are less than 0), the decimal will be moved to the right and the exponent will be
negative. The number 0.5678 would be written as 5.678 x 10-1.
A few examples:
54405 = 5.4405 x 104
0.000006036 = 6.036 x 10-6
-201,000 = -2.01 x 105
-0.023473 = -2.3473 x 10-2
Measurements and significant figures
First, some definitions. There are many words used in science that have slightly different
meanings than in everyday use. To most people the words accuracy and precision are synonymous.
To a scientist they have similar but distinct and different meanings. Accuracy is an indication of how
close to a correct value a measurement is. Precision is a function of the closeness of the results of
a series of measurements to each other in value.
To the layman this distinction may seem absurd. It is often wondered how the correct value
of a measurement cannot be known. But in the real world, the absolute correctness of most
measurements is seldom known.
As an example, assume someone takes exactly 1.000 pound of finely powdered lead and
scatters it evenly across a 10.00 acre field. Now assume a farmer plows the lead deep into the
ground. Is there still exactly 1.000 pound of lead, or is it possible that perhaps some of the lead stuck
to the blades of the plow and the tires of the tractor? Now, assume that the field is in constant use
over a ten year period. Will the field still contain exactly 1.000 pound of lead? What processes might
be responsible for a loss of lead in the 10 acres? The action of wind and water, the integration of lead
into growing plants, and the transport of lead by either vehicles or animals moving through the field
may all contribute to a loss of lead. Lead may also move into the field, especially if there is a copper
smelter nearby, since lead is a common by-product of copper and silver refining. The finely
powdered lead that results from metal refining processes can be carried by the wind and deposited
in the field, even if the refinery is many miles distant. Also, lead may have existed in the soil before
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we added our 1.000 pound. If we are asked how much lead remains in the 10.00 acres, how accurate
will our answer be if we state 1.000 pound?
How can we determine exactly how much lead remains in the field? If we want to be exact,
we must remove all of the soil to a depth of several feet or more, over all of the ten acres, and
measure the total amount of lead. Obviously, this is not a feasible approach.
Alternatively, we may take a number of soil samples scattered across the ten acres and then
extrapolate our test results as being representative for the entire field. This may result in a loss of
accuracy, but it is more realistic in terms of time and money since testing can be expensive. The
results are treated statistically to help us estimate the accuracy of our testing. In testing the samples
we hope that the results of our tests are similar in value from one set of samples to the next. If the
results are in fact similar, we say that the results are precise. The greater the difference between the
results of various sets of samples, the more imprecise our data are.
One very important aspect of testing involves the measurement of standards. Standards are
samples in which the amount of the substance of interest is known exactly because it is added under
very carefully controlled conditions in a laboratory. Standards are always analyzed along with
samples from the real world. The results obtained from the analysis of similar standards should be
very nearly the same, i.e., they should be precise. The greater the precision in the measurement of
a series of standards, the greater the likelihood of the measurements being accurate under normal
conditions. In other words, precise measurements are also usually accurate measurements, although
this might not be entirely true for a number of reasons we will not discuss here.
Realistically, can we ever know exactly how much lead remains in the field? No. But with
careful testing we can provide a highly accurate estimation, one that is very close to the actual,
correct vaue.
Whenever measurements are made in the real world, there are limitations in the
measurements. For example, if the pound of lead was initially measured on an electronic scientific
balance before it was scattered through the field, it might be confidently said that it weighed exactly
1.000 pound. But what if the lead was initially weighed on a bathroom scale, which only indicates
in one pound increments? Or worse, what if the lead was weighed on a feed store scale which only
registers 10 pound increments? How certain might we be that the pound of lead weighed exactly
1.000 pound?
Scientists are required to know the limitations of the equipment they use when making
measurements during experiments. These limitations are reflected in the manner in which scientists
report their observations. Whenever a scientist reports a measurement, he writes the number to as
many places as he is certain are correct plus a last place, in which there is uncertainty due to the
limitations of the instrument used to make the measurements.
If we define the distance one mile by measuring it with a yardstick, we might report that 1
mile is equal to 5280 feet. However, if we measure that same distance using lasers and GPS (global
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positioning satellite) technology, we might find that 1 mile is equal to 5280.000 feet, since the
distance can be measured more exactly with the more sophisticated equipment.
The number of significant figures in a measurement is equal to all of the places in a
measurement we are certain about plus the first uncertain place. In this class we do not make
measurements. We must learn to recognize how many significant figures are in the numbers reported to
us. Nonzero digits are always significant. Zeros may or may not be significant, as described below.
As examples:
Zeros may or may not be counted as significant figures. There are three types of zeros in
numbers:
Leading zeros are found before (to the left of) the first non-zero digit in a number are never
significant. As an example, the zeros in 0.053 are leading zeros. This value has two significant
figures.
Captive zeros are found between two non-zero digits and are always significant. Given the
number 100.01, we have three captive zeros and five total significant figures.
Trailing zeros are found after (to the right of) the last non-zero digit. They may or may not
be significant, depending on the presence of a explicit decimal in a number. The zeros in 1300 are
trailing zeros but are not counted as significant figures because a decimal is not shown explicitly in
the value. But given 1300. with it’s explicit decimal shown, the two trailing zeros become significant
and the overall value has four significant figures.
Here are some examples. How many significant figures are found in each value below?
5 sf significant figures, or, sig figs as many lovingly call them; captive zeros
are always significant
73,000
2 sf trailing zeros without a decimal are not significant
73,000.
5 sf trailing zeros with a decimal are significant
0.0006149
4 sf leading zeros are never significant
0.00456000 6 sf leading zeros are never significant and trailing zeros with a decimal
are significant
0.002022300 7 sf leading zeros are never significant, captive zeros are always
significant, and trailing zeros with a decimal are significant
4321 = 4 sf
.5678 = 4 sf
0.5678 = 4 sf why isn’t the zero counted as a sig fig?
0.56078 5 sf why is the second zero counted as a sig fig when the first one isn’t?
93,567,891 = 8 sf
10101
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.123456789 = 9 sf
-23,456 = 5 sf note that the sign of a number does not affect the way its sig figs are
counted
-.456 = 3 sf
-0.456 = 3 sf why isn’t the zero counted as a sig fig?
Do not confuse the relevance of placeholder zeros with significant figures. Leading or
trailing zeros are often essential in the reporting of a number, but they may or may not be significant
(i.e., they may or may not be counted as sig figs), depending on the type of zero they are.
Numbers may need to be rounded if they are to be reported to the correct number of sig figs.
Never round any numbers until all calculations have been completed. The rules for rounding may
be found in your text if you are not familiar with them.
SI units
The International System (or SI system) is the measurement system used in science. It is also
known as the metric system. Its basic units are:
Mass - kilogram (kg)
Length - meter (m)
Time - second (s)
Temperature - Kelvin (K; note: not degrees Kelvin)
Quantity - mole
Prefixes are used to indicate how many or what part of a base unit is being described. The prefixes
and their abbreviations are as follows. Note that these abbreviations are case-sensitive. You cannot
be sloppy or casual with these and remain correct for long.
unit
abbreviation
powers of ten
examples
tera
T
1012
trillions
Every year vegetation around the world emits about 500
teragrams of isoprene, a hydrocarbon that plays an important
role in atmospheric chemistry.
giga
G
109
billions
A 4 gigabyte thumb drive can hold four billion bytes of
information.
mega
M
106
millions
A one megaton nuclear bomb delivers the same amount of
destructive energy as the detonation as one million tons of TNT.
kilo
k
103
thousands
deci
d
10-1
tenth
One kilogram is equal to exactly one thousand grams.
In sound measurements one decibel is equal to one-tenth of a
bel, a unit of sound pressure.
16
centi
c
10-2
hundredths
A centimeter is exactly one one-hundredth of a meter.
milli
m
10-3
thousandths
A milligram is exactly one one-thousandth of a gram.
micro
μ, u, or mc
10-6
millionths
Fine human hair has a diameter of about 50 micrometers, 50
one-millionths of a meter.
nano
n
10-9
billionths
A nanosecond is exactly one one-billionth of a second.
pico
p
10-12
trillionths
The diameters of a single copper atom is about 157 picometers,
or 157 trillionths of a meter.
femto
f
10-15
thousand-trillionth
atto
a
10-18
million-trillionth
There are some very fast chemical reactions that occur on a
femtosecond time scale, in thousand-trillionths of seconds.
The concentration of certain chemicals in individuals nerve cells
has been measured in the attomolar range.
Note that much femto/atto scale work is at the cutting edge of chemistry today. You must know the
base units and the prefixes from giga to pico in this class. You do not need to know femto, atto, or tera.
Measurements and measured quantities
In chemistry length is commonly reported in meters (m), centimeters (cm), and millimeters
(mm). The sizes of individual atoms and molecules are reported in the picometer (pm) and
nanometer (nm) range. A unit called the Angstrom (Å) is occasionally encountered and is equal to
0.1 nm (1 Å = 1 x 10-10m).
Mass is reported in kilograms, although the masses of individual atoms may be exceedingly
small. Note that there is a difference between mass and weight. Mass is the amount of matter present
in a substance and does not change with respect to local gravitational fields. Mass is constant.
Weight depends on gravity. As an example, if a person weighs 180 pounds on earth and travels to
the moon, the mass of the person will remain constant. How much of the person is there will not
change. However, since the gravity of the moon is about one-sixth that of earth, the person's weight
would change from 180 pounds to 30 pounds. This is why a trip to the moon might be appealing to
some who perceive themselves as overweight.
Temperature is reported using the Fahrenheit scale in the United States but the centigrade
(or Celsius) and Kelvin scales are used in science. The Fahrenheit scale is based on the boiling point
and the melting point of water at 212°F and 32°F respectively. The centigrade scale is also based
on the melting and boiling points of water, but these are stipulated as occurring at 100°C and 0°C
respectively. Note that 1°C = 1.8°F, i.e., the Celsius degree is larger by nearly a factor of 2. The
Kelvin scale is based on 0 K at absolute zero, the point at which all molecular motion stops. On this
scale we find the melting point of water at 273 K and the boiling point of water at 373 K. Note that
17
1K = 1°C, i.e., they are the same size. Also note that we refer to degrees Fahrenheit and degrees
Celsius but never to degrees Kelvin. The word degree is implied in the name Kelvin. So, we state
that the melting point of water occurs at 32 degrees Fahrenheit, 0 degrees Celsius, and at 273 Kelvin.
The equations used to convert between the temperature scales is as follows:
F to C: (F - 32)/1.8
C to F: (C x 1.8) + 32
C to K: C + 273
K to C: K - 273
K to F: convert K to C and then convert to F
F to K: convert F to C and then convert to K
Derived quantities
Two of the most important derived quantities are volume and density. Derived quantities are
based on the measurement of more than one attribute of a material or object.
Volume is the product of the length, width, and height of an object (V = l x w x h). Since the
standard unit of length is the meter, the standard unit of volume could be stated in cubic meters (m3),
but this is too large a unit to be practical unless the volume of large objects, such as houses,
mountains, or planets, is being discussed. In chemistry volume is commonly expressed in units of
cubic centimeters (cm3). Since 1 m = 100 cm, then 1 m3 = (100 cm)3 = 1 x 106 cm3. Another common
unit of volume in chemistry is the liter (L), which is equivalent to one-one thousandth of a cubic
meter, or, to one deciliter: 0.001 m3 = 1 dm3 = 1 L. There are one thousand milliliters in one liter
(1000 mL = 1 L), and a milliliter and a cubic centimeter are exactly the same size (1 mL = 1 cm3 ).
The density (D) of any material is the ratio of the mass (m) of the material to the volume (V)
occupied by that mass (D = m/V). Density is an important physical property of substances. It gives
information about a substance at both a macroscopic and a microscopic level. Given samples of two
substances, the more dense material either has more particles per unit volume, heavier particles, or
both.
Density is determined by the careful measurement of a substance's mass and volume. Mass
can be determined on a scale or balance. Volume can be determined in a variety of ways but is
commonly determined by displacement. To measure the volume of an object by displacement, water
is placed in a device in which its volume can be carefully measured. One such common device is
a thin glass cylinder with volume increments marked on its side called a graduated cylinder (see, for
example, Figure 2.3 in your text). After carefully noting the volume occupied by just the water, an
18
object may be added. It will be noticed that the level of the water in the cylinder moves upward, as
water is displaced (moved out of the way) by the object. The difference between the starting volume
and the final volume of water is equal to the volume of the object.
The following are examples of how density may be used in chemical calculations.
Throughout these lecture notes I will work many problems as examples. An essential part of your study in
this class should consist of working these same example problems on your own, along with similar
examples in the text and also the recommended end-of-chapter problems. Like it or not, chemistry is very
problem based. If you don’t practice, practice, practice then you’re going to have some real trouble with
the exams. But don’t worry. When you repeat the course because you failed to practice, maybe you’ll have
learned how important this really is to your success in this course.
•
If 10.0 g of liquid occupies a volume of 13.5 mL, what is the density of the liquid?
density = m/V = 10.0g/13.5 mL = 0.741 g/mL
•
Pure gold (24 K) has a density of 19.32 g/cm3 . A wedding band weighs 3.50 grams. How
much water must it displace (i.e., what must its volume be?) if the band is in fact 24 K gold?
if D = m/V then V = m/D
3.50 g/19.32 g/cm3 = 0.181 cm3
•
Benzene has a density of 0.880 g/mL. If you need exactly 78.12 grams of benzene, what
volume would you measure out?
if D = m/V then V = m/D
78.12 g / 0.880 g/mL = 88.77 mL
Energy and heat
In science, energy is defined as the capacity to do useful work. We speak of potential energy,
which is energy that is stored and based on position, and kinetic energy which is the energy of
moving objects. Energy can be transferred as heat, light, sound (acoustic), or electricity by various
chemical reactions. Reactions that give off energy are said to be exothermic, while reactions that
absorb energy to make them happen are said to be endothermic. We’ll discuss exothermic and
endothermic reactions in more detail in Chapter 7.
A few examples of exothermic reactions include fire, explosions, and the flow of electricity
from a battery. Examples of endothermic reactions include cooking and recharging a battery.
19
Energy is often used to increase the temperature of a substance. The common unit of energy
is the calorie (cal). One calorie is the amount of heat required to raise the temperature of 1 g of water
1°C or 1 K since these temperature units are the same relative size, they are used more or less
interchangeably in this discussion. The SI unit of energy is the Joule (J). There is a relationship between
calories and Joules: 4.184 J = 1 cal. We should note that the calories used to calculate the nutrition
value of food are actually equal to one thousand of the calories we discuss in this chapter. In other
words, in nutrition kilocalories are the basis for nutritional discussions.
Let’s do a thought experiment. Imagine taking pieces of gold and iron that are equal in mass
and immersing them in boiling water. After allowing them to reach the temperature of the water,
they are removed and dried. The gold will return to room temperature much quicker than the iron.
This is because of an important physical property called specific heat. The specific heat of a
substance is the amount of heat required to raise the temperature of one gram of the substance by
1°C. Comparatively speaking things with high specific heats heat more slowly and cool more slowly
than things with lower specific heats.
Time for another thought experiment, although this can be attempted at home. An aluminum
pan at room temperature is filled with a mass of water equal to the mass of the pan. The pan of water
is placed on a stove. Which will heat more quickly, the pan or the water? Once removed from the
burner, the water is poured out of the pan into another container at room temperature. Which will
cool more rapidly, the empty pan or the water in its new container?
The pan will heat more rapidly than the water while on the burner, not just because it is in
contact with burner, but also because aluminum has a lower specific heat than water. This is
confirmed by the second part of the experiment. The empty pan will return to room temperature
much quicker than the water. This is because aluminum in the pan has a lower specific heat than
water.
Dimensional analysis and problem solving
It is often necessary to convert from one set of units to another while working science
problems. This is made possible by a problem-solving technique called dimensional analysis or the
factor label method.
At the risk of belaboring the point I just made above, you absolutely *must* learn how to use this
technique. Chemistry is difficult, not because of the math, which is usually the sort of math most of us
learned in junior high school and high school (although we may not remember it). Chemistry is difficult
because the problems are nearly always story problems (oh damn, I hear you say) and can only be
mastered through practice, practice, and more practice. If you do not practice what follows below you will
be extremely unhappy at exam time. You have been warned!
Dimensional analysis works by using conversion factors, which make the correct conversion from
a value with one set of units to a value with a different set of units possible.
20
In mathematics there is something called the multiplicative identity, which states that any
number multiplied by 1 will result in that number i.e., 1 x y = y. A conversion factor permits the
conversion from one set of units to another without changing the true value of the number associated
with the units, because a conversion factor is always exactly equal to one. Conversion factors are
created through equivalent ways of expressing things. As examples:
•
1 mile = 5280 feet
If we divide both sides by 5280 feet we obtain
(1 mile/5280 feet) = (5280 feet/5280 feet) = 1
The right side of the equation cancels, since the numerator and denominator are equal, and
this leaves us with (1 mile/5280 feet) = 1
•
1 foot = 12 inches
If we divide both sides by 12 inches we obtain
(1 foot/ 12 inches) = (12 inches/12 inches) =1
The right side of the equation cancels, since the numerator and denominator are equal, and
this leaves us with (1 foot/12 inches) = 1
•
1 hour = 3600 seconds
If we divide both sides by 3600 seconds we obtain
(1 hour/3600 seconds) = (3600 seconds/3600 seconds) = 1
The right side of the equation cancels, since the numerator and denominator are equal, and
this leaves us with (1 hour/3600 seconds) = 1
•
1 m = 1000 mm
If we divide both sides by 1000 mm we obtain
(1 m/1000 mm) = (1000 mm/1000 mm) = 1
The right side of the equation cancels, since the numerator and denominator are equal, and
this leaves us with (1 m/1000 mm) = 1
Dimensional analysis is the process of problem solving using conversion factors. How are
conversion factors used? While if you know another approach to this technique you’re welcome to
use it, I suggest the following approach, using as an example the conversion of 150 lbs to kg:
1.
Identify what is given and how should the problem wind up. In this case we are given 150
pounds, and expect to wind up with a certain number of kg.
2.
Identify which conversion factors are needed. In this example we need to know the
relationship between pound and kilograms, which is 1 kg = 2.2 lbs or (1 kg/2.2 lbs) = 1
3.
Starting with what is given, we multiply it by our conversion factor. After using the
conversion factor do the units cancel, or is another conversion factor needed?
21
150 lbs x (1 kg/2.2 lbs) = kg; the units are correct, no additional conversion factors are
needed
4.
Do the arithmetic.
150 lbs x (1 kg/2.2 lbs) = 68.2 kg
Note: I strongly suggest that you always check to make sure that your units cancel each other out
and that you wind up with the correct units before you attempt any of the arithmetic in the
problem. I will never require you to do a dimensional analysis problem any one specific way, but
I know from experience if you take care of the units in a problem first, then the arithmetic part
nearly always turns out correct. If you rush through problems you will ultimately prove to be your
own worst enemy at exam time.
What does this mean physically? That there are 68.2 kg in 150 pounds of anything.
Some additional examples of dimensional analysis problems are as follows:
•
Convert 62.5 cm to inches
conversion factors needed: 1 inch = 2.54 cm
62.5 cm x (1 inch/2.54 cm) = 24.6 inches
•
Convert 3.62 ounces to mg
conversion factors needed: 1 lb = 16 oz; 1 lb = 453.6 g; 1 g = 1000 mg
(3.62 oz) x (1 lb/16 oz) x (453.6 g/1 lb) x (1000 mg/1 g) = 102,627 mg = 1.02 x 105 mg
Note that the solution of this problem required the use of several conversion factors. We can use as many
as we need to solve a problem. Also note that when we perform multiplication or division, the number of
significant figures we report in our answer is determined by the value with the fewest sig figs in our
calculations. Conversion factors are usually considered to be exact, and, as such, have an infinite number
of significant figures. This means that, in this class, when we work problems, we report our answers with
the correct number of sig figs, and the correct number of sig figs is determined by the information we are
given in the problem, not the conversion factors. You should also be comforted to read that I will always
provide you with all non-metric conversion factors in this class. The metric prefixes are very much your
responsibility, and I require you to know them.
22
•
Convert 125 m/s to mph (miles per hour)
conversion factors needed: 60 sec = 1 minute; 60 min = 1 hr; 1 m = 100 cm; 1 in = 2.54 cm;
1 ft = 12 in; 1 mile = 5280 ft
(125 m/s) x (60 s/1 min) x (60 min/1 hr) x (100 cm/m) x (1 in/2.54 cm) x (1 ft/12 in) x
(1 mile/5280 ft) = 279.6 mile/hr = 280. mph
•
Convert the area of an 8.5" by 11" piece of paper to square centimeters (cm2)
conversion factors needed: 1 in = 2.54 cm
step 1. 8.5 in x 11 in = 93.5 in2
step 2. (93.5 in2) x (2.54 cm/1 in)2 = 603.2 cm2 = 6.0 x 102 cm2
Note that we do not have a conversion factor that permits the conversion from square inches to square
centimeters. We can, however, create our own, simply by squaring both sides of the conversion factor given
i.e., if 1.00 in = 2.54 cm then (1.00 in)2 = (2.54 cm)2. If you need to do this you must remember to square
the number as well as the unit during the calculation. You should also note that there is more than one way
to solve this problem and come up with the correct answer. You should also note that while normally we
like to do dimensional analysis problems using one continuous string of conversion factors, sometimes it is
easier to break a problem down into pieces as we did here.
•
Convert 6 yds3 of cement to cm3
conversion factors needed: 1 yd = 3 ft; 1 ft = 12 in; 1 in = 2.54 cm
(6 yds3) x (3 ft/1 yd)3 x (12 in/1 ft)3 x (2.54 cm/1 in)3 = 4,587,329.1 cm3 = 5 x 106 cm3
Note again the use of conversion factors created, this time, by cubing both sides of a relationship, e.g.
(3 ft)3 = (1 yd)3
•
Hexane is an organic liquid used in the manufacture of gasoline. If a railroad car contains
2.5 x 104 gallons of hexane, how much does the liquid weigh in pounds? The density of
hexane is 0.6594 g/mL
Ok, this problem is a bit of a tough one but we can do it. You must know that the density is needed to act
as a bridging conversion factor between the conversion of mass to volume, or, visa versa, the conversion
of volume to mass. It is not possible to correctly convert from mass to volume, or from volume to mass,
without using the materials density as a conversion factor.
23
conversion factors needed: 1 gal = 4 qt; 1 L = 1.057 qt; 1000 mL = 1 L; 1 lb = 453.6 g
step 1. (2.5 x 104 gallons) x (4 qt/1 gal) x (1 L/1.057 qt) x (1000 mL/1 L) = 94,607,379.376
mL
step 2. (94,607,379.376 mL) x (0.6594 g/mL) x (1 lb/453.6 g) = 137,531.1 lbs = 1.4 x 105 lbs
24
Chapter 3: Atoms and the Periodic Table
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as proton, neutron, electron, ion, isotope, atomic number, atomic
mass, atomic weight, atomic mass unit (amu), cation, anion, octet rule, electromagnetic
radiation, wavelength, frequency, quantum number, shell, subshell, orbital, ground state,
excited state, electron configuration, full electron configurations, Noble gas (or valence
shell) electron configurations, electron spin diagrams, inner shell electrons, outer shell
electrons, valence electrons, atomic size, metallic character, ionization energy, electron
affinity, electronegativity.
Be familiar with Dalton's atomic theory.
Be familiar with the fundamental properties of protons, neutrons, and electrons.
Assign the correct number of protons, neutrons, and electrons to any isotope or ion.
Correctly calculate the atomic weight of any element given the masses and abundances of
its isotopes.
Predict the number of electrons an element will typically gain or lose using the octet rule and
its position on the periodic table.
Be able to assign the following for either neutral atoms or ions: full electron configurations,
Noble gas (or valence shell) electron configurations, and electron spin diagrams.
Understand the relationship between an atom's electron configuration and its position in the
periodic table.
Predict how many inner shell, outer shell, and valence electrons an atom has.
Describe in general terms the periodic trends.
A brief history of atomic theory
In the 5th Century B.C. Leucippus and his student Democritus postulated that all matter is
made of invisible, indivisible particles called "atomos," which in Greek means "indivisible." At
roughly the same time another Greek philosopher, Empedocles, proposed that everything is made
up of four basic elements: earth, air, water, and fire. This notion was later popularized by the work
of Aristotle (around 340 B.C.) and the ideas of Democritus were largely forgotten. In the 1st Century
B.C. the rejected and mostly forgotten ideas of Democritus were popularized by the Roman poet
Lucretius in his work "De Rerum Naturae."
By the time of the Renaissance the original writings of Democritus no longer existed but his
ideas were rediscovered in the late 16th Century as scholars read Lucretius. By the 17th Century
physicists such as Robert Boyle, Robert Hooke, and Isaac Newton began to reject the Aristotlean
notion of matter and to suggest that the physical properties of matter might be better explained by
forces acting upon a ultimate building block of matter. In his great work "Optiks" Newton wrote:
25
"it seems probable to me that God in the beginning formed matter in solid, massy, hard,
impenetrable, moveable particles. . . ." .
Between 1803-1807 the English scientist John Dalton proposed a theory that still serves as
the basis of our understanding of matter. He proposed:
!
!
!
!
!
!
Each element is composed of extremely small particles called atoms.
All of the atoms in an element are identical.
The atoms of an element are different from the atoms in any other element.
The atoms of two or more elements can combine to form compounds.
The law of constant composition: compounds always have the same ratios of atoms
regardless of where the compound is found.
Atoms are not created or destroyed in chemical reactions, i.e. mass is conserved in chemical
reactions.
While we now take these ideas as a basic statement of fact they were not immediately
accepted by the scientific community. Some were argued for nearly 100 years. But in the end, they
were proved and accepted. It is nothing short of amazing to realize that our current notions of matter
were initially postulated two hundred years ago.
While people began to accept the notion of atoms as the basic building blocks of matter
during the 19th Century, they still did not know what they were.
In 1897 J. J. Thomson discovered the electron. In 1909 Millikan discovered both the charge
and mass of electrons.
In 1896, at roughly the same time as Thomson was working on his electrons, Henri
Becquerel discovered the "radioactivity" of a uranium-bearing compound, pitchblende. Around
1898 the French chemists Marie and Pierre Curie discovered two additional radioactive elements,
radium and polonium.
In 1909 Ernest Rutherford, a chemist from New Zealand who had studied under Thomson,
discovered that every atom has a nucleus at its center. His work demonstrated that the nucleus of the
atom contains 99.95% of the mass of the atom but only 10-15 (one-thousand trillionth!) of its volume.
In other words, an atom is mostly just empty space. Based on this fact it has been calculated that if
the nucleus of an atom is the size of a football stadium, the atom itself would be about the size of
the earth! In 1919 Rutherford discovered the proton, a particle found in the nuclei of all atoms. This
was the first identified building block of all atomic nuclei. The name derives from the Greek word
"protos," meaning "first." In 1932 Rutherford's student, James Chadwick, discovered the neutron,
the other particle found in nearly all atomic nuclei. I do not require you to remember names and dates
in this class. However, after reading this section you will hopefully gain insight into the way that science
creeps forward, step by step, sometimes rapidly, sometimes slowly. Science rarely advances by enormous
leaps and bounds.
26
The structure of the atom
Atoms are made of three types of subatomic particles which are summarized in the table
below.
charge
relative mass
absolute mass (kg)
proton
+1
1
1.6726 x 10-27
neutron
0
1
1.6749 x 10-27
electron
-1
1/1800
9.109 x 10-31
Protons and neutrons are held together in the nucleus at the center of the atom. Under normal
circumstances particles with similar electrical charges repel each other, but protons are held together
in the nucleus despite this repulsion by the Strong Force, which is about 100 times stronger than the
electrostatic repulsion experienced by the similarly charged particles. The Strong Force, while not
a particularly clever name, is the strongest physical force in the known universe. Neutrons also help
to mitigate the electrical repulsion that arises when positively charged protons are jammed tightly
together in an atomic nucleus. Electrons orbit the nucleus relatively far away. As we stated above,
the atom is mostly empty space.
An electrostatic attraction, or, electromagnetic attraction, occurs between the positively
charged nucleus and the negatively charged electrons as they move around it. The number of
electrons, or, more correctly, the way that they are arranged around the nucleus, determines an
atom's chemical reactivity.
Atoms and protons
It is the number of protons in an atomic nucleus that determines the identity of the atom.
The number of protons in the nucleus is unique for the atoms of each element. The number of
protons in an atom is invariable in chemical reactions. If the number of protons did change in a
chemical reaction, the atom would be converted into an atom of a different element. These types of
events do occur in nature but they are nuclear events, not chemical events. The process of converting
an atom of one substance into an atom of another substance is called transmutation and will be
discussed in Chapter 11.
As we just said, and let us reiterate for emphasis: in chemical reactions, the number of protons in reacting
atoms *never* changes. In nuclear reactions, which are not chemical reactions, the number of protons in
the nuclei of reacting atoms may or may not change.
If we examine a periodic table we find three pieces of information for every element.
27
The top number is the atomic number. In the above figure we see that this number, 29, is the number
of protons in the nucleus of all copper atoms. The atomic number is sometimes referred to as the"Z"
number of an element although it is a bit archaic to do so.
Atoms, neutrons, and isotopes
While the number of protons for a given element is invariable, the number of neutrons in a
nucleus can differ. In other words, it is possible for atoms to have the same numbers of protons but
different numbers of neutrons. We call these isotopes, atoms with the same atomic number (same
number of protons) but with different numbers of neutrons. All elements have at least one isotope.
Most elements have two or more isotopes.
The sum of the protons and neutrons in any isotope is known as its atomic mass. If we want
or need to be precise we can also include the masses of the electrons when calculating atomic mass but
usually we ignore them because the are so small compared to the masses of the protons and neutrons.
We can indicate the atomic number and the atomic mass (also called the mass number) of any
isotope using it's elemental symbol:
28
The atomic mass is indicated as superscript to the left of the elemental symbol. The atomic number
of an element is indicated as a subscript to the left of the elemental symbol. It is also common to
represent isotopes using only the elemental symbol and the mass number of the isotope.
As we know the element from its symbol, we must also know its atomic number as each element has
a unique number of protons.
Isotopes are named by first stating the element and then the mass number. In this illustration
we are looking at the symbol for carbon-12.
If we know the atomic mass of an isotope we can calculate the number of neutrons found in
its nucleus. The number of neutrons in any isotope is equal to the difference between the isotope's
mass number and its atomic number (number of protons). In other words,
atomic mass - atomic number = number of neutrons
As an example carbon has three isotopes, carbon-12, carbon-13, and carbon-14. As all three of these
isotopes are of carbon ,they all have six protons. The difference between these three isotopes is that
each has a different number of neutrons from the other.
12
6
13
6
C
C
14
6
C
atomic mass
12
13
14
atomic number
6
6
6
number of neutrons
6
7
8
Students are sometimes confused as to how carbon-12 can be an isotope when it has an equal
number of protons and neutrons. Isotopes are atoms with the same atomic number but with different
numbers of neutrons. This does not preclude substances from having isotopes with the same number
of protons and neutrons. The phrase “different numbers of neutrons” means relative to other atoms
with the same numbers of protons. So carbon-12 has six protons and six neutrons. It differs from
carbon-13 in that while both have six protons, carbon-13 has seven neutrons. And carbon-12 and
carbon-13 both differ from carbon-14 in that while each isotope has six protons, carbon-14 has eight
neutrons.
29
As another example chlorine has two isotopes, chlorine-35 and chlorine-37. These isotopes
have the following numbers of protons and neutrons.
35
17
Cl
37
17
C
atomic mass
35
37
atomic number
17
17
number of neutrons
18
20
The mass of individual atoms is expressed in amu, atomic mass units. One amu is equal to
1/12 the mass of a single atom of carbon-12, and is also equal to 1.661 x 10-27 kg. In simpler terms,
we can state that protons and neutrons each have a mass of 1 amu when calculating atomic mass.
This isn’t quite exact but it’s good enough for purposes in this course. This means that the three isotopes
of carbon have atomic masses of 12 amu, 13 amu, and 14 amu respectively, while we say that the
atomic masses of the two isotopes of chlorine have masses of 35 amu and 37 amu respectively.
The bottom number in the entry for each element in the Periodic Table is the atomic weight,
which is the weighted average of the atomic masses of all of the isotopes of that element. In the
figure above the atomic weight of carbon is given as 12.011 amu. A weighted average is calculated
by taking into consideration both the mass and the abundance of each of the isotopes of an element.
We calculate the atomic weight of an element by multiplying the atomic mass of each isotope by its
fractional abundance and by then adding all of the resulting numbers for all of an element's isotopes
together.
The abundance of 12C is 98.9%, of 13C is 1.1%, and of 14C is around ~0%. This isotopic
abundance information is not found in the periodic table. We have to turn to other sources of information
to learn the abundances of the isotopes of the elements. In other words, if we take a million carbon
atoms and examine them, we will find that roughly 989,000 of them are 12C isotopes, 11,000 are 13C
isotopes, and less than 1 of every million atoms is a 14C isotope even though 14C is not very abundant
it is still useful, since it is the isotope examined when objects are carbon dated in archeological studies.
(0.989)(12) + (0.011)(13) + (0)(14) = 12.011 amu
which is the same as the atomic weight reported for carbon on your periodic table. Other examples
of atomic weight calculations:
!
The relative abundance of 35Cl is 75.78% and of 37Cl is 24.22%. The atomic weight of
chlorine is equal to:
(0.7578)(35) + (0.242)(37) = 35.48 amu
30
!
The relative abundance of 50Cr is 4.345%, of 52Cr is 83.789%, of 53Cr is 9.501%, and of 54Cr
is 2.315%. The atomic weight of chromium is equal to:
(0.04345)(50) + (0.83789)(52) + (0.09501)(53) + (0.02315)(54) = 52.03 amu
Atoms that have two or more isotopes will always have an atomic weight that is not an
integer (i.e., an integer is not a fraction, i.e. there is no decimal place in the number). Check the
periodic table in your text and determine how many elements have an atomic weight that is a whole
number. What does this suggest about the number of isotopes for each of these elements? Any
element with a whole number for its atomic weight must only have one isotope.
Atoms and electrons
In their elemental (natural) state elements are electrically neutral. In other words, in each
atom the number of electrons is equal to the number of protons. If an atom gains or loses electrons
(remember: atoms never gain or lose protons in chemistry!), an electrical imbalance will be created
and the atom will become electrically charged. Atoms that have electrical charge through the gain
or loss of electrons are called ions. We speak of two categories of ions in chemistry.
!
!
Cations have lost electrons, and since (#electrons < #protons), cations have a net positive (+)
charge.
Anions have gained electrons, and since (#electrons > #protons), anions have a net negative
(-) charge.
Charge is represented as a superscript to the right of an element's symbol for ions. The sign
tells us whether electrons have been lost or gained. Something that has lost electrons has more
protons than electrons and becomes positively (+) charged. Something that has gained electrons has
more electrons than protons and becomes negatively (-) charged. The number tells us how many
electrons have been lost or gained. This number is always a whole number. We do not lose or gain
parts of electrons. As examples:
!
Mg has 12 protons (how do we know? hint: what is its atomic number? You must learn how to
use the periodic table to help you. I don’t expect you to memorize atomic numbers, atomic masses,
or atomic weights.) and therefore 12 electrons. Mg2+ has 12 protons but only 10 electrons.
!
As has 33 protons and therefore 33 electrons. As3- has 33 protons but 36 electrons.
!
NH4+ and SO42- are examples of molecular ions. The "+" on the ammonium ion (NH4+) tells
us that the entire molecule has lost one electron, i.e., if we add up all of the protons and all
of the electrons in the molecule we will find that there is one fewer electron than protons.
The "2-" charge on the sulfate ion (SO42-) tells us that the entire molecule has gained two
electrons.
Note that by convention we write the number and then the sign when representing charge. Mg2+ is correct,
Mg+2 is incorrect. If an ion has lost or gained a single electron, we use only a "+" or "-" sign and do not
write "1+" or "1-". Na+ is correct, Na1+ is incorrect. Cl- is correct, Cl1- is incorrect. If an atom is electrically
31
neutral there is no superscript to the right of the symbol. This absence tells us that the number of protons
and electrons are equal.
The octet rule
Why would an atom want to gain or lose electrons? When two or more substances have the
same number of electrons, they are said to be isoelectronic. We will see that substances with the
same numbers of electrons also have the same arrangement of those electrons around their nucleus.
It turns out that atoms are ultimately more stable when they are isoelectronic with the nearest
Noble Gas. This is the basis for what is known as the octet rule. Atoms become isoelectronic with
the nearest Noble Gas by either gaining or losing electrons, or by forming chemical bonds (which
we will discuss later). The number of electrons an atom gains or loses depends on how far away it
is (on the Periodic Table) from the nearest Noble Gas. This behavior of becoming isoelectronic with
the nearest Noble Gas is the driving force in many chemical reactions.
For many elements we can predict how many electrons an atom may gain or lose based on
the element's position in the Periodic Table. Bear in mind that it takes energy for each electron lost
or gained, so the atoms of an element will be inclined to gain or lose the smallest possible number
of electrons necessary to become isoelectronic with the nearest Noble Gas.
All alkali metals (Group 1A elements) find it easiest to become isoelectronic with the nearest
Noble Gas by losing one electron. When lithium loses an electron, it becomes isoelectronic with
helium. When sodium loses an electron, it becomes isoelectronic with neon. When potassium loses
an electron it becomes isoelectronic with argon. The same is true for rubidium, cesium, and
francium.
All alkali earth metals (Group 2A elements) find it easiest to become isoelectronic with the
nearest Noble Gas by losing two electrons. When beryllium loses two electrons it becomes
isoelectronic with helium. When magnesium loses two electrons, it becomes isoelectronic with neon.
When calcium loses two electrons it becomes isoelectronic with argon. And so on for the rest of the
Group 2A elements.
The Group 3A elements find it easiest to become isoelectronic with the nearest Noble Gas
by losing three electrons. In other words, we often find B3+, Al3+, Ga3+, and so on.
We’re going to skip the Group 4A elements for the moment. They do obey the octet rule, but
in a way slightly more complicated than we want to worry about at the moment.
The Group 5A elements find it easiest to become isoelectronic with the nearest Noble Gas
by gaining three electrons. In other words, we often find N3-, P3-, As3-, and so on.
The chalcogens (Group 6A elements) find it easiest to become isoelectronic with the nearest
Noble Gas by gaining two electrons. In other words, we often find O2-, S2-, Se2-, and so on.
32
The halogens (Group 7A elements) find it easiest to become isoelectronic with the nearest
Noble Gas by gaining one electron. In other words, we often find F-, Cl-, Br-, and so on.
Trying to predict the charge of the transition metals and rare earth metals is not easily done
as rules other than the octet rule often govern their behavior. and will not be discussed here. You
should also know that, as a rule, metals typically lose electrons in chemical reactions and non-metals
typically gain electrons in chemical reactions.
You will find it incredibly useful to memorize the relationship between charge and position in the Periodic
Table as quickly as possible. You do not need to worry about unusual atomic or chemical behavior in this
class. If you can predict in general how things will behave, you have achieved my expectations of you.
Radiation
There are two types of radiation in the world around us.
Ionizing radiation comes from radioactive materials such as uranium, plutonium, and etc. and
is the result of the ejection of atomic nuclei, electrons, or energy by the nuclei of unstable isotopes.
Electromagnetic radiation is due to disturbance of electromagnetic fields. These
electromagnetic fields are the result of the movement of anything with charge and as a result may
be found in many places, including in very large items, such as stars, and also in very small items,
such as atoms and nuclei.
Electromagnetic radiation consists of small, massless packets of energy called photons that
travel in waves. All types of electromagnetic radiation travel through a vacuum at the same velocity,
a constant value equal to 3.00 x 1010 cm/s and represented with the symbol “c”.
Waves are defined by specific characteristics.
The amplitude of a wave is how intense (high) the wave is. The wavelength, λ, of a wave is the crest
to crest distance between waves and can range from kilometers (radiowaves) to picometers (x-rays).
33
The frequency of a wave, ν, is how many waves per second occur. The units of frequency is the
Hertz (Hz), which is equal to 1 cycle/sec.
For any wave, the product of its wavelength and its frequency is equal to the speed of light.
That is,
c = λν
I will not ask you to make any calculations using this equation. This means you don’t need to know the
value of the constant “c”.
There is also a relationship between the frequency of a wave and its energy. The energy of
a photon depends on its frequency and on a constant “h”, Planck’s constant:
E = hν
I will not ask you to make any calculations using this equation either. Having said this, you don’t need to
worry about the value of Planck’s constant. But it is useful to understand in a general way the relationship
between the frequency of electromagnetic radiation and its energy. As frequency goes up, energy goes up.
And as frequency goes down, energy also goes down.
Electromagnetic radiation occurs across a spectrum of wavelengths. Visible light, the portion
of the spectrum we can detect with our eyes, is only a very tiny fraction of the spectrum. As I am
creating all of the illustrations in these notes myself and am possessed of very limited ability in the
graphic arts, I’m not going to attempt an illustration of the electromagnetic spectrum. You will find
a very nice example by turning to the Wikipedia entry for “electromagnetic spectrum” which may
be found at http://en.wikipedia.org/wiki/Electromagnetic_spectrum. Note that the spectrum ranges
from gamma rays, which have short wavelengths and high frequencies, to radio waves, which have
long wavelengths and low frequencies. You should note that cell phones communicate using
electromagnetic radiation. Which type of electromagnetic radiation would be safer to use in cell
phones, radio waves or gamma rays? Why? As gamma rays have very high frequencies, they also have
very high energies. If you used a gamma ray cell phone, I’m afraid you’d wind up frying your little head.
Does this mean that cell phones are safe as they use radio waves, which are much lower in energy than
gamma rays? That’s an excellent question. The most accurate answer is: we’re still not certain.
Fundamentals of electrons in atoms
!
!
First, a few fundamental concepts about electrons.
In neutral atoms the number of protons is equal to the number of electrons.
Electrons do not orbit the nucleus in planetary fashion as we once believed. The position of
an electron is described in terms of statistical probability and may be calculated using
quantum mechanics (which we will not discuss in this class).
34
!
!
!
Electrons cannot exist just anywhere with respect to the nucleus. They can only be found at
certain discrete (specific) distances from the nucleus, which distances may be calculated
using quantum mechanics.
These distances correspond to energies. The further an electron is from the nucleus the
greater its energy.
Areas of high probability (90% probability) of finding an electron with a specific energy are
called orbitals.
The discrete distances at which electrons can be found from the nucleus can be broken down
into shells, subshells, and orbitals. These shells, subshells, and orbitals can be identified using
quantum numbers which help us identify where each of the electrons in an atom is likely to be
found. There are four quantum numbers, the principal quantum number “n”, the angular momentum
quantum number “l”, the magnetic quantum number “ml”, and the spin quantum number “ms“.
Every electron in an atom has its own set of four quantum numbers unique from those of any
other electron in the same atom. These four quantum numbers serve essentially as an address system
for electrons, just as your home address helps others find you. If a friend from another country
wanted to visit you, you might tell them you lived in the United States, in Utah, in Salt Lake City,
at 4600 South Redwood Road. These four bits of information identify your location to your friend.
!
!
!
!
!
The shell in which an electron is found describes its distance from the nucleus.
Shell numbers, known as principal quantum numbers (n), range from 1 to 7 in chemistry.
In other words, there are only seven average distances from the nucleus of any atom in which
electrons may be found.
The shell closest to the nucleus is n=1, the next nearest is n=2, and shell numbers increase
in integer values to the shell furthest away from the nucleus at n=7.
As n increases the distance from the nucleus increases, as does the energy of the electrons
found in that shell.
The spacing between shells is not linear. This has an important consequence, as we will
shortly see.
You should also note that there are seven rows, or periods, in the periodic table. Each period
corresponds to one of the seven principal quantum numbers. The first period corresponds to n=1,
the second period corresponds to n=2, the third period corresponds to n=3, and so on. This is
because position in which an element is found in the periodic table is directly linked not only to the
number of its protons but also to the number of its electrons and to the configuration of its electrons
with respect to the nucleus.
Each shell contains subshells, which correspond to the angular momentum quantum
numbers (l). A few important points about subshells are:
!
There are seven different types of subshells, s, p, d, f, g, h, and I subshells.
!
Each shell contains as many subshells as its principal quantum number, i.e., the shell with
n=1 has 1 subshell, the shell with n=2 has 2 subshells, the shell with n=3 has 3 subshells, and
so on to the seventh shell (n=7) which is the only shell to have all seven subshells.
35
!
!
!
While, in a given shell, all of the orbitals are at roughly the same distance from the nucleus,
and therefore at roughly the same energy, the s subshells are the lowest in energy, followed
in order by the p, d, f, g, h, and I subshells.
"
In other words, within a shell, not all of the subshells are at exactly the same distance
from the nucleus. The s subshells are the closest to the nucleus and the lowest in
energy. The I subshells are the furthest from the nucleus and the highest in energy.
"
These differences in energy and distances from the nucleus within a shell are not
great but they are nonetheless significant.
When assigning subshells to a shell, begin with the lowest energy subshell and work your
way up in terms of increasing energy.
"
The first shell (n=1) has one subshell, s. It never has a p, d, f, or any other subshell.
"
The second shell (n=2) has two subshells, s and p. There are never d, f, g, etc.
subshells in the second shell.
"
The third shell (n=3) has three subshells, s, p, and d. There are never f, g, h, or I
subshells in the third shell.
"
And so on up to the seventh subshell (n=7), which has s, p, d, f, g, h, and I subshells.
Each subshell differs from the others as to its distance from the nucleus, the number of
orbitals it contains, the shape of its orbitals, and the orientation of its orbitals with respect
to the nucleus.
subshell
#orbitals
s
1
p
3
d
5
f
7
g
9
h
11
I
13
The shapes and orientations of s, p, d, and f orbitals may be found at the following Wikipedia link:
http://en.wikipedia.org/wiki/Atomic_orbital#Orbitals_table.
Orbital shapes play an important role in the formation of chemical bonds, but I do not expect you
to memorize orbital shapes.
36
The orbitals in the various subshells have magnetic quantum numbers (ml) which describe
their orientation with respect to a Cartesian coordinate system with its origin at the nucleus of the
atom.
Any orbital, regardless of its shell and subshell, can only hold a maximum of two electrons.
Electrons within the same orbital have opposite spin quantum numbers (ms). Electrons can have one
of two spin quantum numbers, spin up or spin down.
We can summarize what we know about shells, subshells, and the total number of electrons
in a shell in a table:
shell (n)
subshells (l)
# orbitals/subshell
#electrons/
orbital
total
electrons/shell
1
s
1
2
2
2
s,p
1+3=4
2
8
3
s, p, d
1+3+5=9
2
18
4
s, p, d, f
1+3+5+7=16
2
32
5
s, p, d, f, g
1+3+5+7+9=25
2
50
6
s, p, d, f, g, h
1+3+5+7+9+11=36
2
72
7
s, p, d, f, g, h, I
1+3+5+7+9+11+13=49
2
98
There are a few interesting features in the table we should point out. Notice that the total number
of orbitals in any shell is equal to n2. The first shell holds one orbital, the second shell holds four
orbitals, the third shell holds nine orbitals, and so on. Also note that the total number of electrons
in any given shell is equal to 2n2.
Based on my years of experience as a teacher, I expect that some of you are on the verge of panic right
about now. Peace, my children. Be calm. Much of this section is commentary that goes somewhat beyond
the scope of this class. You don’t need to memorize this table. I expect you to understand what it
summarizes, which is the basic information about shells, subshells, and orbitals and how many orbitals are
in each shell and subshell. That's it. The rest is frosting as far as this class is concerned.
Ground states and excited states
Regardless of how many electrons they may actually have, the atoms of all elements have
all of the possible shells and subshells described above available. In other words, even though
hydrogen atoms only have one electron, they still have empty but available orbitals in the 2s, 2p, 3s,
3p, 3d, and etc. shells and subshells up to and including the 7i shell and subshell. However, the
37
majority of these shells and subshells are empty most of the time. Electrons always occupy orbitals
in the shells and subshells with the lowest energy first. When all of the electrons in an atom are
found in orbitals in the lowest energy shells and subshells they are said to be in the ground state, i.e.
electrons in their lowest energy state are said to be in the ground state.
When an electron absorbs energy, either thermal or electromagnetic, it will become excited
and can move to higher energy unoccupied orbitals. When an electron becomes excited and moves
to an orbital higher in energy than its ground state orbital the electron is said to be in an exited state.
We mentioned g, h, and I subshells above, and we should point out that they are virtual subshells,
only populated by excited electrons. In the ground state electrons are never found in these subshells.
The difference in energy between the orbitals of the various shells and subshells correspond
to specific amounts of energy. These, in turn, correspond to specific wavelengths of light since there
are mathematical relationships between the energy, frequency, and the wavelength of radiation.
If an atom is heated or bombarded with electromagnetic radiation (with a light or a laser for
example), electrons may become excited and jump from low-energy ground state orbitals to empty
higher energy orbitals. The electrons cannot remain excited for very long. They usually only remain
in their higher energy orbitals for a fraction of a fraction of a second. But, to return to lower energy
orbitals they must give off energy. The amount of energy given off must correspond to the exact
difference in energy between the high energy and low energy orbitals. This excess energy is usually
given off as electromagnetic radiation, and often it is emitted as light in the visible portion of the
spectrum. In other words, all of the light by which we see, regardless of source (from the sun, a light
bulb, etc.), is generated by energy given off by excited electrons returning to their ground state. The
process of returning from an excited state to the ground state is called relaxation. Sounds familiar,
doesn’t it? People and electrons are alike in this respect.
Sunlight is created by excited electrons emitting energy as they return to the ground state.
If we pass sunlight through a prism we can create a rainbow. Many elements, when heated by
themselves (i.e. in a pure state) do not emit white light, but purple or red or green or orange or blue
or yellow light, depending on the element. If their light is analyzed by passing it through a prism,
a rainbow is not observed. Rather, a series of narrow lines may be seen. These results are called line
spectra. Every element has its own unique line spectrum, its own electromagnetic fingerprint. This
makes it possible to identify individual elements in complex mixtures of substances by heating a
sample of the substance and analyzing the resulting emitted light. This also means that we can
determine the elemental composition of far away bodies that emit light, like stars, by collecting their
light, passing it through a prism, and analyzing the results. Examples of a few line spectra may be
seen at the web site “Spectra of Gas Discharges,” found at
http://astro.u-strasbg.fr/%7ekoppen/discharge/
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Look at the spectra at this site. How are the spectra of the various elements the same? how do they
differ? No, I do not ask you to memorize what the spectra look like. But I still expect you to take five
minutes and look the spectra at this address.
Electron configurations: full configurations and the periodic table
Of what practical value is all of this to us as we study chemistry? We are interested in being
able to predict how things will behave in chemical reactions. Believe it or not, this is a skill you will
develop in this class. One of the most important predictive tools we have is the knowledge that the
electron configuration of atoms has a direct bearing on their behavior in chemical reactions. In other
words, if we know the electron configuration of an element, we can often predict how it will react.
The electron configuration of an element is a description of the shells, subshells, and orbitals
in which its electrons are found in the ground state. While this might seem like an imposing task,
it is made easier if we remember what we said above about the relationship between shells,
subshells, orbitals, and numbers of electrons. A few other useful tidbits about electron configurations
and the ground state are as follow.
•
•
•
In the ground state electrons always occupy the lowest energy shells and subshells first. This
means that the lowest energy electron in all elements is found in the 1st shell, in its
s subshell. This is known as the Aufbau, or “building up,” principle.
In the ground state no orbital can ever hold more than two electrons. Electrons paired up in
the same orbital must have opposite spins (spin up and spin down). This is a restatement of
the Pauli exclusion principle, which tells us that in the ground state, no two electrons in the
same atom can have the same set of four quantum numbers. In other words, in the ground state
in the same atom no two electrons can have the same address. If you think about it it makes
sense, doesn’t it?.
When assigning electrons to subshells in the ground state, all orbitals must hold at least one
electron before any orbital may hold two electrons. In other words, electrons will never pair
up in an orbital together if there is an empty orbital available (can you guess why?). This is
known as Hund’s rule.
We will discuss three types of electron configurations in this class, full configurations, Noble
Gas configurations, and electron spin diagrams (also known as arrow diagrams or orbital diagrams).
We will begin with full configurations because they serve as the basis for the other two types of
configurations.
Full configurations are assigned to elements by determining how many electrons they have
and then by placing them in orbitals in shells and subshells, starting with those orbitals of the lowest
energy and working upward.
What we need is a tool to help us remember the order in which subshells are filled. You may
be amazed to learn that the periodic table is just such a device. The periodic table not only provides
39
information about the number of protons and neutrons in each of the elements, it also gives
information about the highest energy electron for each element.
The periodic table can be divided into four regions, the s block, p block, d block, and f block.
Groups 1A and 2A are known are the "s" block, because the highest energy electron of each element
in these two groups is an "s" electron. Groups 3A, 4A, 5A, 6A, 7A, and 8A are known as the "p"
block, because the highest energy electron of each element in these six groups is a"p" electron. The
transition metals (Groups 1B - 8B) are found in the "d" block, because the highest energy electron
of each element in these ten groups is a"d" electron. The lanthanides and actinides are found in the
"f" block, because the highest energy electron of each element in these 14 groups is a"f" electron.
Do you notice that the "s" block contains two groups? Did you remember that an s subshell can hold a
maximum of two electrons? The "p" block contains six groups, and a p subshell can also hold six electrons.
The "d" block holds 10 groups, and a d subshell can hold up to 10 electrons. The "f" block holds 14
groups, and a f subshell can hold 14 electrons? Coincidence? I think not. Remarkably clever? Absolutely!
How do we use the periodic table to assign electron configurations to elements? We must
find the position of the element of interest, notice which period it is in and which block it is in, and
40
observe how many groups to the right of the left-most group in the block the element is. Then we
work backwards to account for the rest of the electrons. Let's choose a few examples and see how
this works:
Hydrogen has one electron. Hydrogen’s “address” in the periodic table is in the 1st period,
in the s block, and in the first column, counting from the left within the block. Therefore the full
configuration of hydrogen is 1s1. The first number tell us the shell in which the subshell is found (1),
the letter tells us which subshell contains the orbital in which the electron is found, and the
superscripted "1" tells us that the orbital is occupied by a single electron. A handy tip: be sure to have
a periodic table handy and refer to it often when assigning electron configuration. I don’t expect you to
memorize electron configurations. But plan on seeing them on quizzes and exams, for which you will
always be provided a periodic table.
Helium has two electrons. Helium’s “address” in the periodic table is in the 1st period, in the
s block, and in the second column, counting from the left within the block. Therefore the full
configuration of helium is 1s2. In other words, there are two electrons in the single orbital of the s
subshell of the 1st shell. At this point we have filled both the 1st shell and the s subshell.
Lithium has three electrons. Lithium’s “address” in the periodic table is in the 2nd period,
in the s block, and in the first column, counting from the left within the block. Therefore the
configuration of lithium’s highest energy electron is 2s1. But this only accounts for one of lithium’s
three electrons. According to the Aufbau principle the first two electrons are assigned to orbitals in
the lowest energy shells and subshell, which as we have just seen is the orbital in the 1s shell and
subshell. This means that the two lowest energy electrons in lithium have the same configuration
as helium, 1s2. The full configuration of lithium is 1s2 2s1.
Beryllium has four electrons. Beryllium’s “address” in the periodic table is in the 2nd period,
in the s block, and in the second column, counting from the left within the block. Therefore the
configuration of beryllium’s highest energy electron is 2s2. But this only accounts for two of
beryllium’s four electrons. According to the Aufbau principle the first two electrons are assigned
to orbitals in the lowest energy shells and subshell, the 1s shell and subshell. This means that the two
lowest energy electrons in beryllium have the same configuration as helium, 1s2. The full
configuration of beryllium is 1s2 2s2. At this point we have filled the 2s subshell.
Boron has five electrons. Boron’s “address” in the periodic table is in the 2nd period, in the
p block, and in the first column, counting from the left within the block. The configuration of
boron’s highest energy electron is 2p1. But this only accounts for one of boron’s five electrons.
According to the Aufbau principle the other four electrons are assigned to orbitals in the lowest
energy shells and subshells. This means that the two lowest energy electrons in boron have the same
configuration as helium, 1s2. The next two electrons are found in the orbital in the 2s shell and
subshell and have the configuration 2s2. The full configuration of boron is 1s2 2s2 2p1.
41
Carbon has six electrons. Carbon’s “address” in the periodic table is in the 2nd period, in the
p block, and in the second column, counting from the left within the block. The configuration of
carbon’s highest energy electron is 2p2. But this only accounts for two of carbon’s six electrons.
According to the Aufbau principle the other four electrons are assigned to orbitals in the lowest
energy shells and subshells. This means that the two lowest energy electrons in carbon have the
same configuration as helium, 1s2. The next two electrons are found in the orbital in the 2s shell and
subshell and have the configuration 2s2. The full configuration of carbon is 1s2 2s2 2p2. Note how the
sum of the superscripts in the configuration tell us the total number of electrons for which the configuration
accounts. In the case of carbon, 2+2+2=6.
Nitrogen has seven electrons. Nitrogen’s “address” in the periodic table is in the 2nd period,
in the p block, and in the third column, counting from the left within the block. The configuration
of nitrogen’s highest energy electron is 2p3. But this only accounts for three of nitrogen’s seven
electrons. According to the Aufbau principle the other four electrons are assigned to orbitals in the
lowest energy shells and subshells. This means that the two lowest energy electrons in nitrogen have
the same configuration as helium, 1s2. The next two electrons are found in the orbital in the 2s shell
and subshell and have the configuration 2s2. The full configuration of nitrogen is 1s2 2s2 2p3.
Oxygen has eight electrons. Oxygen’s “address” in the periodic table is in the 2nd period,
in the p block, and in the fourth column, counting from the left within the block. The configuration
of oxygen’s highest energy electron is 2p4. But this only accounts for four of oxygen’s eight
electrons. According to the Aufbau principle the other four electrons are assigned to orbitals in the
lowest energy shells and subshells. This means that the two lowest energy electrons in oxygen have
the same configuration as helium, 1s2. The next two electrons are found in the orbital in the 2s shell
and subshell and have the configuration 2s2. The full configuration of oxygen is 1s2 2s2 2p4.
Fluorine has nine electrons. (note: fluorine is an element. Flourine - I don't know what
flourine is, perhaps an element made of ground-up wheat? Be *very* careful with spellings. Spelling
mistakes will absolutely kill you on the Chapter 4 quiz!) Fluorine’s “address” in the periodic table
is in the 2nd period, in the p block, and in the fifth column, counting from the left within the block.
The configuration of fluorine’s highest energy electron is 2p5. But this only accounts for five of
fluorine’s nine electrons. According to the Aufbau principle the other four electrons are assigned
to orbitals in the lowest energy shells and subshells. This means that the two lowest energy electrons
in fluorine have the same configuration as helium, 1s2. The next two electrons are found in the
orbital in the 2s shell and subshell and have the configuration 2s2. The full configuration of fluorine
is 1s2 2s2 2p5.
Neon has ten electrons. Neon’s “address” in the periodic table is in the 2nd period, in the p
block, and in the sixth column, counting from the left within the block. The configuration of neon’s
highest energy electron is 2p6. But this only accounts for six of neon’s ten electrons. According to
the Aufbau principle the other four electrons are assigned to orbitals in the lowest energy shells and
subshells. This means that the two lowest energy electrons in neon have the same configuration as
42
helium, 1s2. The next two electrons are found in the orbital in the 2s shell and subshell and have the
configuration 2s2. The full configuration of neon is 1s2 2s2 2p6.
At this point we've filled all of the orbitals in all of the subshells in the first two shells.
Where will we put additional electrons? If you guessed the third shell and it's subshells, you're doing
great!
Sodium is the first element in the third period (the 3rd row down). It has 11 electrons.
Sodium’s “address” in the periodic table is in the 3rd period, in the s block, and in the first column,
counting from the left within the block. The configuration of sodium’s highest energy electron is 3s1.
But this only accounts for one of sodium’s eleven electrons. According to the Aufbau principle the
other ten electrons are assigned to orbitals in the lowest energy shells and subshells. This means that
these ten electrons have the same configuration as neon, 1s2 2s2 2p6. The full configuration of
sodium is 1s2 2s2 2p6 3s1.
Magnesium has 12 electrons. Magnesium’s “address” in the periodic table is in the 3rd
period, in the s block, and in the second column, counting from the left within the block. The
configuration of magnesium’s highest energy electron is 3s2. But this only accounts for two of
magnesium’s twelve electrons. According to the Aufbau principle the other ten electrons are
assigned to orbitals in the lowest energy shells and subshells. This means that these ten electrons
have the same configuration as neon, 1s2 2s2 2p6. The full configuration of magnesium
is 1s2 2s2 2p6 3s2. At this point the 3s shell and subshell are filled.
Aluminum has 13 electrons. Aluminum’s “address” in the periodic table is in the 3rd period,
in the p block, and in the first column, counting from the left within the block. The configuration of
aluminum’s highest energy electron is 3p1. But this only accounts for one of aluminum’s thirteen
electrons. According to the Aufbau principle the other twelve electrons are assigned to orbitals in
the lowest energy shells and subshells. This means that these twelve electrons have the same
configuration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of aluminum is
1s2 2s2 2p6 3s2 3p1.
The next five elements, silicon, phosphorus, sulfur, chlorine, and argon also have electrons
that occupy the orbitals in the 3p subshell.
Silicon has 14 electrons. Silicon’s “address” in the periodic table is in the 3rd period, in the
p block, and in the second column, counting from the left within the block. The configuration of
silicon’s highest energy electron is 3p2. But this only accounts for two of silicon’s fourteen electrons.
According to the Aufbau principle the other twelve electrons are assigned to orbitals in the lowest
energy shells and subshells. This means that these twelve electrons have the same configuration as
magnesium, 1s2 2s2 2p6 3s2. The full configuration of silicon is 1s2 2s2 2p6 3s2 3p2.
Phosphorus has 15 electrons. Phosphorus’s “address” in the periodic table is in the 3rd
period, in the p block, and in the third column, counting from the left within the block. The
configuration of phosphorus’s highest energy electron is 3p3. But this only accounts for three of
43
phosphorus’s fifteen electrons. According to the Aufbau principle the other twelve electrons are
assigned to orbitals in the lowest energy shells and subshells. This means that these twelve electrons
have the same configuration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of phosphorus is
1s2 2s2 2p6 3s2 3p3.
Sulfur has 16 electrons. Sulfur’s “address” in the periodic table is in the 3rd period, in the
p block, and in the fourth column, counting from the left within the block. The configuration of
sulfur’s highest energy electron is 3p4. But this only accounts for four of sulfur’s sixteen electrons.
According to the Aufbau principle the other twelve electrons are assigned to orbitals in the lowest
energy shells and subshells. This means that these twelve electrons have the same configuration as
magnesium, 1s2 2s2 2p6 3s2. The full configuration of sulfur is 1s2 2s2 2p6 3s2 3p4.
Chlorine has 17 electrons. Chlorine’s “address” in the periodic table is in the 3rd period, in
the p block, and in the fifth column, counting from the left within the block. The configuration of
chlorine’s highest energy electron is 3p5. But this only accounts for five of chlorine’s seventeen
electrons. According to the Aufbau principle the other twelve electrons are assigned to orbitals in
the lowest energy shells and subshells. This means that these twelve electrons have the same
configuration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of chlorine is 1s2 2s2 2p6 3s2 3p5.
Argon has 18 electrons. Argon’s “address” in the periodic table is in the 3rd period, in the
p block, and in the sixth column, counting from the left within the block. The configuration of
argon’s highest energy electron is 3p6. But this only accounts for six of argon’s eighteen electrons.
According to the Aufbau principle the other twelve electrons are assigned to orbitals in the lowest
energy shells and subshells. This means that these twelve electrons have the same configuration as
magnesium, 1s2 2s2 2p6 3s2. The full configuration of argon is 1s2 2s2 2p6 3s2 3p6.
Does each of these configurations account for the correct number of electrons? How do you
know? The sum of the superscripts should equal the total number of electrons.
The next element is potassium, which has 19 electrons (how do we know it has 19
electrons?). We find potassium in the 4th period (the 4th row down). The first 18 electrons have the
same configuration as argon, 1s2 2s2 2p6 3s2 3p6 . How do we assign the 19th electron? On the one
hand we need to remember that the 3rd shell has 3 subshells, 3s, 3p, and 3d. We have yet to place
any electrons in the orbitals in the 3d subshell. On the other hand, potassium is found in the 4th
period. Is the correct full configuration of potassium 1s2 2s2 2p6 3s2 3p6 3d1 or 1s2 2s2 2p6 3s2 3p6 4s1?
We use the periodic table to help us answer this question. We find potassium located in the
4th period, and in the left-most column of the "s" block. This makes the configuration of potassium's
highest energy electron 4s1. The other eighteen electrons have the same configuration as argon, 1s2
2s2 2p6 3s2 3p6. The full configuration of potassium is 1s2 2s2 2p6 3s2 3p6 4s1.
Calcium has 20 electrons. It is located in the 4th period, second to the left in the "s" block.
Calcium's highest energy electron is 4s2. This accounts for two of calcium’s twenty electrons. The
44
other eighteen electrons have the same configuration as argon, 1s2 2s2 2p6 3s2 3p6. Calcium's full
configuration is 1s2 2s2 2p6 3s2 3p6 4s2.
Next we come to scandium, which has 21 electrons. Scandium is located in the forth period,
in the left-most group of the "d" block. Does this mean that scandium's highest energy electron has
a 4d1 configuration? Not quite. Remember that the 3rd shell has 3 subshells, s, p, and d. Recall as
well that the spacing between shells is not linear. An important consequence of this non-linearity
is that as shells move further from the nucleus, their subshells begin to overlap with those in higher
shells. The first d electrons we encounter in the periodic table are 3d electrons, meaning that
scandium's highest energy electron has a 3d1 configuration. Scandium's full configuration is
1s2 2s2 2p6 3s2 3p6 4s2 3d1.
Because of overlap, the d electrons in every period are from one shell lower than the period
in which they reside, while the s and p electrons are from the same shell as the period. In other
words, as we see in the figure near the beginning of this section, the subshells follow the order 4s
3d 4p, 5s 4d 5p, 6s 5d 6p, and 7s 6d 7p.
Where do we find elements with f electrons in the periodic table? Note the order of the
elements in the 6th period. The atomic number of cesium (Cs) is 55, the atomic number of barium
(Ba) is 56, the atomic number of lanthanum (La) is 57, and the atomic number of hafnium (Hf) is
72. Where are the elements with atomic numbers from 58 to 71? In the "f" block, in the lanthanides
and actinides, which are found in a strip below the periodic table. Why are these elements placed
separately from the main body of the periodic table rather than integrated in their proper place?
There is a reason, although it is not necessarily a very good one. By including the 14 f-block groups
in their proper place a periodic table we nearly double it's width. This means that it is difficult to fit
the entire table on an regular-sized piece of paper in print that is large enough to read clearly. A
periodic table with the f-block in its proper place looks like this:
45
To which f-block do elements 58 through 71 belong? The first shell with an f subshell is the
4th shell. Elements 58 through 71 have electrons with a 4f configuration. The f subshells overlap
even more severely with higher level subshells than the d subshells.
In the 7th period there is a gap between actinium (Z=89) and Rutherfordium (Z=104), with
the missing elements, those with atomic numbers from 90 to 103, again found in the f-block. These
are 5f elements.
To summarize, s and p electrons always have the same number as the period in which they
are found. d electrons always have a number one less than the period in which they reside, and f
electrons, 2 less.
Do you feel like you can assign full electron configurations now? Let's do a few examples.
Arsenic (As) has 33 electrons. It is found in the 4th period, three groups from the left in the
p-block. Arsenic's highest energy electron has the configuration 4p3. By using the periodic table to
remind us which subshells lie beneath 4p in energy, and their order, we assign arsenic a full electron
configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3. Note that 2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33 a
way to check that we have accounted for all of the electrons.
Palladium (Pd) has 46 electrons. It lies in the 5th period, 8 groups from the left. Palladium
has the full electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8. Again, let’s check to make
sure that we have accounted for all of the electrons: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 8 = 46
Polonium (Po) has 84 electrons. It lies in the 6th period, four groups from the left in the
p-block. Polonium has the full electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2
4f14 5d10 6p4. Note that 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 6 + 2 + 14 + 10 + 4 = 84. That’s
an awful lot of electrons, but we did it correctly.
•
•
A few final notes on full electron configurations.
As I said above, you do not need to memorize the periodic table. I will always provide you
with one on exams and permit you to use one on quizzes.
There are periodic tables available at campus bookstores which offer electron configurations
that are correct but not in the order provided by the periodic table. As an example, you might
find the electron configuration of arsenic given as 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3, rather than
as we have written it above. I require you to be able to write electron configurations as they are
found on the periodic table. You will not be permitted to use any information source that provides
you with electron configurations on exams, other than periodic table such as the one found on the
inside cover of your text. If you provide an electron configuration that is not congruent with the
order found in the periodic table, as with our arsenic example, even though it is correct, it will be
marked wrong.
46
•
The electron configurations of d and f-block elements can be tricky things. Sometimes their
full electron configurations differ slightly from those we might predict using the periodic
table. There are good reasons for these deviations, but we will not discuss them in this class
and I do not require you to know them. If I ask for the electron configuration of a d-block
element I will accept as correct the configuration derived from the periodic table, regardless
of whether it is absolutely correct or not. You will not be asked about the electron
configuration of f-block elements. In summary, my goal for you in this section is simply to be
able to provide the full configuration of any s-block, p-block, or d-block element as they appear on
any periodic table. I do not expect you to know and understand the exceptions. I do not expect you
to provide configurations for f-block elements.
Electron configurations: Noble Gas configurations
Ok, let's admit it. Writing full electron configurations, especially of elements with many
electrons, is a real pain. Let's learn a short-cut. We can substitute the symbol of the noble gas in the
previous period for the configuration of lower energy electrons in a full electron configuration. We
must place the symbol for the noble gas in square brackets to correctly indicate what we're doing.
Let's see how this works.
Element
full configuration
noble gas in
previous period
noble gas
configuration
H
1s1
-
1s1
He
1s2
-
1s2
Li
1s2 2s1
He
[He] 2s1
Be
1s2 2s2
He
[He] 2s2
B
1s2 2s2 2p1
He
[He] 2s2 2p1
C
1s2 2s2 2p2
He
[He] 2s2 2p2
N
1s2 2s2 2p3
He
[He] 2s2 2p3
O
1s2 2s2 2p4
He
[He] 2s2 2p4
F
1s2 2s2 2p5
He
[He] 2s2 2p5
Ne
1s2 2s2 2p6
He
[He] 2s2 2p6 or [Ne]
Na
1s2 2s2 2p6 3s1
Ne
[Ne] 3s1
Mg
1s2 2s2 2p6 3s2
Ne
[Ne] 3s2
Al
1s2 2s2 2p6 3s2 3p1
Ne
[Ne] 3s2 3p1
Si
1s2 2s2 2p6 3s2 3p2
Ne
[Ne] 3s2 3p2
47
P
1s2 2s2 2p6 3s2 3p3
Ne
[Ne] 3s2 3p3
S
1s2 2s2 2p6 3s2 3p4
Ne
[Ne] 3s2 3p4
Cl
1s2 2s2 2p6 3s2 3p5
Ne
[Ne] 3s2 3p5
Ar
1s2 2s2 2p6 3s2 3p6
Ne
[Ne] 3s2 3p6 or [Ar]
Note that there are not noble gas configurations for the first period elements. All of the second
period elements have 1s2 as a part of their full configuration. Since this is also the full configuration
for helium, which is the noble gas in the period before the second period, we can represent these two
electrons by using [He] in the configurations for the second period elements. And all of the third
period elements have 1s2 2s2 2p6 as a part of their configuration, which is also the full configuration
for neon which is the noble gas in the period before the third period. So we use [Ne] to represent
those electrons in their Noble gas configurations.
Let’s also look at the noble gas configurations of the three elements we did a moment ago, arsenic,
palladium, and polonium.
element
full configuration
noble gas
configuration
arsenic
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3
[Ar] 4s2 3d10 4p3
palladium
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8
[Kr] 5s2 4d8
polonium
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p4
[Xe] 6s2 4f14 5d10 6p4
We find arsenic in the fourth period. Argon is the noble gas in the third period with a full
configuration of 1s2 2s2 2p6 3s2 3p6. So we use [Ar] to represent those electrons in the noble gas
configuration for arsenic.
We find palladium in the fifth period. Krypton is the noble gas in the fourth period with a
full configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6. So we use [Kr] to represent those electrons in
the noble gas configuration for palladium.
We find polonium in the sixth period. Xenon is the noble gas in the fifth period with a full
configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6. So we use [Xe] to represent those
electrons in the noble gas configuration for polonium.
Electron configurations: electron spin diagrams (arrow diagrams)
Full electron configurations and Noble Gas configurations account for all of the electrons
in an atom, but there are times when it is helpful to display electron configurations in a slightly
different way. Electron spin diagrams, also known as arrow diagrams, orbital diagrams, or spin
48
diagrams use dashes to represent each of the orbitals in a subshell and arrows to represent electrons.
Electron spin diagrams may be drawn vertically or horizontally. The energy of electrons in the
diagram increases as we move from lower to higher levels or as we move from left to right,
depending on if the electron spin diagrams is drawn vertically or horizontally. Remember that
electrons repel each other (why?) and to minimize the repulsion, electrons that share an orbital have
different spins, spin up or spin down. Of course, this is congruent with the Pauli exclusion principle
as two electrons in the same orbital and with the same spin would have the same sets of four
quantum numbers.
Hydrogen has the full configuration 1s1. Its electron spin diagram is as follows.
Note the use of a single dash for the 1s orbital, the 2s orbital, and the 3s orbital, while the 2p and 3p
subshells are represented by three dashes because each has three orbitals. The shell and subshell
names are placed as a label beneath the appropriate dashes. There is only one dash with a single
arrow in the electron spin diagram for hydrogen because there is only one orbital in hydrogen
occupied by its lone electron. Really, we could have left the blank lines out if we had wanted to and
we usually do.
The electron spin diagram for helium with its 1s2 configuration is
Note that the 1s orbital is now filled, and the two electrons, represented by the arrows, point in
opposite directions since they have opposite spins.
49
The electron spin diagram for lithium with its 1s2 2s1 configuration is
We now have two dashes occupied, one for the orbital in the 1s shell and subshell, and one for the
2s orbital. Again, we always label the orbitals (dashes) as to which shell and subshell they belong.
The electron spin diagram for beryllium (1s2 2s2) is
At this point we have filled the 2s orbital. The next six elements will have electrons in the three 2p
orbitals. The first three elements with 2p electrons are boron (1s2 2s2 2p1 ),
carbon (1s2 2s2 2p2 ),
50
Did you notice that each 2p electron occupied an empty 2p orbital? You may have been tempted to
place both of carbon’s 2p electrons in same 2p orbital:
The above electron spin diagram for carbon is wrong because it violates Hund’s rule. All electrons
are negatively charged and are also therefore mutually repulsive. This is mitigated to a slight extent
when electrons have opposite spins, but it is better still for them to stay as far apart as possible by
occupying different orbitals and not pairing up until it becomes necessary. The spin diagram for
nitrogen is
Again, Hund’s rule requires an unpaired electron in each of the three 2p orbitals before electrons
begin to pair up in orbitals. As we look the electron spin diagrams for oxygen, fluorine, and neon
we see that it now becomes necessary to place pairs of electrons in the 2p orbitals. Here are the
electron spin diagrams for oxygen (1s2 2s2 2p4),
fluorine (1s2 2s2 2p5),
51
and neon (1s2 2s2 2p6).
And so it goes. We draw a single dash to represent the 3s orbital, and then, when necessary, three
dashes to represent the three 3p orbitals. As examples, the electron spin diagram for the first five
elements in the third period are as follow,
sodium (1s2 2s2 2p6 3s1),
magnesium (1s2 2s2 2p6 3s2),
aluminum (1s2 2s2 2p6 3s2 3p1),
52
silicon (1s2 2s2 2p6 3s2 3p2),
and phosphorus (1s2 2s2 2p6 3s2 3p3).
Have you noticed that each electron spin diagram begins with knowing the full electron
configuration for the element in question?
Inner shell, outer shell, and valence electrons
Noble gas configurations provide us with a short-hand way of writing electron
configurations. They also help us categorize an atom's electrons into one of two types, inner and
outer shell electrons.
All electrons in periods prior to the period of the element of interest, i.e., all electrons that
can be accounted for by the noble gas portion of the noble gas configuration for an element are
called inner shell electrons. Inner shell electrons never participate in chemical reactions.
All electrons found in the same period as the element of interest are known as outer shell
electrons. Outer shell electrons are often called valence electrons, but there is actually a slight
distinction between the two. Valence electrons are those outer shell electrons available to participate
in chemical reactions.
Completely filled p, d, and f subshells are extremely stable, to the extent that these electrons
behave as though they are inner shell. The electrons of filled p, d, and f subshells do not participate
in chemical reactions. This means that all of an atom's outer shell electrons may or may not be
valence electrons.
If we examine, for example, aluminum, we find its noble gas configuration to be [Ne] 3s2
3p1. This means aluminum has 13 total electrons, of which 10 are inner shell and three are outer shell
electrons, all three of which are valence electrons.
53
Arsenic has the full configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 and the noble gas
configuration [Ar]4s2 3d10 4p3. Arsenic has 33 total electrons, of which there are 18 inner shell
electrons and 15 outer shell electrons. Of the 15 outer shell electrons there are only 5 valence
electrons, the 4s and 4p electrons. The other 10 are tied up in the filled 3d subshell and are
unreactive. The following table summarizes total electrons, inner shell electrons, outer shell
electrons, and valence electrons for the first 36 elements.
element
noble gas
configuration
inner shell
electrons
outer shell
electrons
valence
electrons
commonly
formed ions
H
1s1
0
1
1
H+
He
1s2
0
2
0
none
Li
[He] 2s1
2
1
1
Li+
Be
[He] 2s2
2
2
2
Be2+
B
[He] 2s2 2p1
2
3
3
B3+
C
[He] 2s2 2p2
2
4
4
C4-
N
[He] 2s2 2p3
2
5
5
N3-
O
[He] 2s2 2p4
2
6
6
O2-
F
[He] 2s2 2p5
2
7
7
F-
Ne
[Ne]
2
8
0
none
Na
[Ne] 3s1
10
1
1
Na+
Mg
[Ne] 3s2
10
2
2
Mg2+
Al
[Ne] 3s2 3p1
10
3
3
Al3+
Si
[Ne] 3s2 3p2
10
4
4
Si4-
P
[Ne] 3s2 3p3
10
5
5
P3-
S
[Ne] 3s2 3p4
10
6
6
S2-
Cl
[Ne] 3s2 3p5
10
7
7
Sl-
Ar
[Ar]
10
8
0
none
K
[Ar] 4s1
18
1
1
K+
Ca
[Ar] 4s2
18
2
2
Ca2+
Sc
[Ar] 4s2 3d1
18
3
3
Sc3+
54
Ti
[Ar] 4s2 3d2
18
4
4
Ti4+
V
[Ar] 4s2 3d3
18
5
5
V5+
Cr
[Ar] 4s2 3d4
18
6
6
Cr6+
Mn
[Ar] 4s2 3d5
18
7
7
Mn7+
Fe
[Ar] 4s2 3d6
18
8
8
Fe3+
Co
[Ar] 4s2 3d7
18
9
9
Co2+
Ni
[Ar] 4s2 3d8
18
10
10
Ni3+
Cu
[Ar] 4s2 3d9
18
11
11
Cu2+
Zn
[Ar] 4s2 3d10
18
12
2
Zn2+
Ga
[Ar] 4s2 3d10 4p1
18
13
3
Ga3+
Ge
[Ar] 4s2 3d10 4p2
18
14
4
Ge4+
As
[Ar] 4s2 3d10 4p3
18
15
5
As3-
Se
[Ar] 4s2 3d10 4p4
18
16
6
Se2-
Br
[Ar] 4s2 3d10 4p5
18
17
7
Br-
Kr
[Kr]
18
18
0
none
Generally there is a very good correlation between the number of valence electrons an element has
and the ion it forms to satisfy the octet rule. This is especially true of the s and p block elements
found in the first three periods. The behavior of the transition metals is not well described by the
octet rule, and we will not discuss why they behave the way they behave in this class.
You should be able to accurately predict for each s and p block element whether it forms a
cation or an anion and its charge. This is based on position in the periodic table, as we have seen.
You are not responsible for remembering the various cations each of the transition metals may form.
However, you will find in Chapter 4 that when we examine transition metals compounds and their
names, there is a fairly simple way to determine the transition metal cation's charge.
The electron configurations of ions
When two substances have the same electron configuration (because they have the same
number of electrons), they are said to be isoelectronic. The Noble Gas configuration of neon is [Ne]
and of sodium is [Ne]3s1. When an atom loses electrons, it loses outer shell electrons. This means
that when sodium becomes sodium ion, Na+, it loses the 3s electron. This makes sodium ion
isoelectronic with neon. How do we know that sodium always tends to lose one electron? where is Na
55
found in the periodic table? And remember: metals tend to lose electrons in chemical reactions, nonmetals
tend to gain electrons in chemical reactions.
Magnesium has a Noble Gas configuration of [Ne] 3s2. Magnesium nearly always exists as
Mg2+ in chemical reactions. how do we know this? where is Mg found in the periodic table? Mg2+ has
lost its two 3s electrons, and has become isoelectronic with both neon and Na+ in the process.
Fluorine has the electron configuration 1s2 2s2 2p5. It tends to gain an electron in reactions.
how do we know this? where is F found in the periodic table? When an atom gains electrons, they are
always added to empty or half-filled orbitals in the outer shell. The electron configuration of F- is
1s2 2s2 2p6, which makes fluoride ion isoelectronic with neon.
Oxygen has the electron configuration 1s2 2s22p4. Oxygen tend to gain two electrons. how
do we know this? where is O found in the periodic table? The electron configuration of O2- is 1s2 2s2 2p6,
which makes it isoelectronic with neon and F-.
Note that more than two substances can be isoelectronic. In our examples in this section we
find that Na+, Mg2+, Ne, F-, and O2- are all isoelectronic. What are examples of other ions that are
isoelectronic with neon? Al3+ and N3- are two examples.
Atomic properties and periodic trends
There are four basic properties of elements, the extent of which can be predicted in a relative
sense from position of the elements in the periodic table. These are properties are called periodic
trends. They include atomic size, metallic character, ionization energy, and electronegativity.
Atomic size is the actual size of the atoms. Within a group, atoms become larger as we move
down the group. Since moving down a group involves moving to electron shells that are further
away from the nucleus, it should not surprise you that atomic size increases. As an example, iodine
is larger than bromine, which is larger than chlorine, which is larger than fluorine. Atomic size also
increases as we move from right to left within a period. This is a bit surprising but not difficult to
explain. Within a period, as we move from left to right, the number of electrons increases. Lithium
has fewer electrons (3 electrons) than beryllium (4 electrons), which has fewer electrons than boron
(5 electrons), which has fewer electrons than carbon (6 electrons), and so on. While there is a
balance between the numbers of protons and electrons in each of these elements, as the number of
protons in the nucleus increases, electrons at the same average distance from the nucleus (i.e., in the
same period) are attracted more forcefully to the nucleus. As a result, the size of the atom becomes
smaller. Atomic size increases as we move from nonmetals to metals and as we move down a group.
If we are interested in comparing the size of the atoms of two or more elements, without
actually knowing any specific information about the size of the atoms, we can still qualitatively
predict which atoms are the largest and which are the smallest. Look at a periodic table and compare
the positions of barium (Ba) and chlorine. Based on position, which element has larger atoms?
56
barium Using a periodic table compare the positions of atoms of cesium (Cs), ruthenium (Ru), and
sulfur and arrange them in order of increasing size. S (smallest) < Ru < Cs (largest)
Metallic character involves the extent to which a substance behaves as a metal. Remember
from chapter 1 that metals are elements characterized by good thermal and electrical conductivity
and are usually lustrous, malleable, and ductile. Metals are also characterized by the tendency to
give up electrons in reactions (i.e. form cations).
It is the "looseness" with which valence electrons are held that determines how metallic an
element is. The further down a group, the further the valence electrons are found from the nucleus
and the more loosely they are held. As we stated above, metallic character increases as we move
from nonmetals to metals within a period and as we move down a group.
We can qualitatively predict how metallic elements are relative to each based on this
information. While looking at a periodic table compare the atoms of cesium (Cs), ruthenium (Ru),
and sulfur and arrange them in order of increasing metallic character. S (least metallic, in fact a
non-metal) < Ru < Cs (most metallic)
Ionization energy is the energy required to remove an electron from a neutral atom, thus
making it a cation. Whenever an atom loses an electron it takes energy. The energy required to
remove an electron increases as they are held more tightly. This means that the closer electrons are
found to the nucleus (i.e., going up a group or from right to left within a period), the more energy
required to remove them from an atom.
We can qualitatively predict how ionization energies are relative to each based on this
information. While looking at a periodic table compare the atoms of cesium (Cs), ruthenium (Ru),
and sulfur and arrange them in order of increasing ionization energy. Cs (lowest ionization energy) <
Ru < S (highest ionization energy)
There is a relationship between ionization energy and ion size. When an electron is removed
from an atom and a cation is formed, an imbalance between electrons and protons results. As a
consequence, the remaining electrons are pulled nearer to the nucleus then they were in the
electrically neutral atom. In other words, cations are smaller than the neutral atom from which they
originated. It is possible to remove more than one electron from an atom and to form cations with
+2, +3 and higher charges. This loss can be thought of as occurring in a stepwise process, meaning
if an atom loses three electrons, it doesn't lose all three simultaneously, but rather, in three steps that
follow each other in rapid succession. Since each loss of an electron results in the electrons being
held more tightly by the nucleus, it should not surprise you to read that it takes more energy to
remove a second electron from an atom than it did the first, far more still to remove a third electron
than it did to remove the second, and so on. At the same time, with the loss of each electron, the
cation becomes smaller and smaller as it holds its remaining electrons more and more tightly. So,
as an example, if we compare the size of aluminum with that of its cations: Al > Al+ > Al2+ > Al3+
57
Electron affinity is the energy associated with the addition of a single electron to a neutral
atom, thus making it an anion. Whenever an atom gains an electron, it may result in the release of
energy (i.e. an exothermic process) if a stable anion is formed, or it may require an input of energy
(i.e. an endothermic process) to make it happen if an unstable anion is formed. Electron affinity does
not have a consistent periodic trend, but it is related to electronegativity which we shall soon discuss
and which does have a consistent periodicity to it.
As with cations, the size of anions differs from that of neutral atoms. When an electron is
added to an atom and an anion is formed, an imbalance between electrons and protons results. As
a consequence, the electrons are not held quite as tightly as in the neutral atom. In other words,
anions are larger than the neutral atom from which they originated. It is possible to add more than
one electron to an atom and to form anions with -2, -3 and higher charges. This addition can be
thought of as occurring in a stepwise process, meaning if an atom gains three electrons, it doesn't
gain all three simultaneously, but rather, in three steps that follow each other in rapid succession.
With the addition of each electron, the anion becomes larger and larger as it holds its electrons more
and more loosely. So, as an example, if we compare the size of nitrogen with that of its anions:
N < N– < N2- < N3Electronegativity is the tendency of an atom to attract electrons to itself. As we think about
the elements, we should recall that metals tend to lose electrons in reactions and that nonmetals tend
to gain electrons in chemical reactions. Therefore, as we move across a period from left to right,
from metals to nonmetals, we would expect that electronegativity should increase, and it does.
Electronegativity also increase as we move up a group. This is related to the tendency of smaller
atoms to cling to their electrons more tightly. The noble gases do not have electronegativity, i.e, they
have an electronegativity of zero. why?
58
We can qualitatively predict how electronegativities are relative to each based on this
information. Compare the atoms of cesium (Cs), ruthenium (Ru), and sulfur and arrange them in
order of increasing electronegativity. Cs (lowest electronegativity) < Ru < S (highest electronegativity).
Electronegativity is commonly ranked on a scale from 0 to 4. The most electronegative
element is fluorine, with an electronegativity of 4. Francium (Fr) is the least electronegative element
with a value of 0.7. The noble gases have no tendency to attract electrons to themselves and
therefore have an electronegativity of “0.” This statement about the electronegativity of noble gases that
I’ve just made is not something that all chemists agree upon. But still, we’re going to treat it this way in this
class. If you need to know the numerical of the electronegativity of a noble gas in this class, which you will
in Chapter 5, use “0".
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Chapter 4: Ionic Compounds
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as chemical bonds, metallic bond, ionic bond, covalent bond,
metallic compounds, ionic compounds, covalent compounds, electron sea model,
monatomic, polyatomic, and Lewis dot structures.
Define what a chemical bond is and describe the factors that affect bond strength.
Be able to identify the type of chemical bond found in a compound simply by looking at its
molecular formula.
Describe the three types of chemical bonds, the three types of chemical compounds, and the
basic properties of each type of compound.
Be able to provide the correct IUPAC name for an ionic compound given its molecular
formula.
Be able to provide the correct molecular formula for an ionic compound given its IUPAC
name.
Be able to provide the IUPAC name for an covalent compound given its molecular formula.
Be able to provide the correct molecular formula for an covalent compound given its IUPAC
name.
Provide a simple definition for acids and bases and be able to correctly identify the names
and molecular formulas of the common acids and bases listed in this chapter.
Ions
Let's review. Ions are substances that have gained or lost one or more valence electrons.
Metals tend to lose electrons and form cations. Nonmetals tend to gain electrons and form anions.
The number of electrons an element will gain or lose can be predicted from that element's position
on the periodic table. This is possible because of the tendency of elements to become isoelectronic
with the nearest noble gas. It takes energy to remove electrons from atoms (ionization energy). The
addition of electrons to neutral atoms may require energy (or in the case of some nonmetals, result
in a release of energy) called electron affinity.
Chemical bonds and chemical compounds
While there are substances in the world around us that consist of collections of individual
atoms, most things are made up of molecules. Molecules consist of two or more atoms that have
chemically bonded to each other.
A chemical bond occurs between two atoms. The bonding atoms may be the same types of
atoms or two atoms from different elements. Each atom donates one electron to the bond. The bond
60
is the result of the attraction between particles with opposite electrical charges. The types of
oppositely charged particles, the way the electrons are donated or shared, and the extent to which
they are shared is a function of the type of bond. The strength of the interactions between two
charged particles is described by Coulomb’s law, which is the following equation
F
kq1q2
r2
in which F is the force of attraction or repulsion between two charged particles, k is a constant, q1
and q2 are the charges on the interacting particles, and r is the distance between the interacting
particles. We use the word particle, rather than atom, because the equation describes the attractive and
repulsive interactions between all charged particles, not just atoms. While we will not use this equation
quantitatively in other words, we’re not going to plug numbers into Coulomb’s law and solve it in this class
you should understand that it proves that as the magnitude of the charges on mutually-attracting
particles increases, the force of attraction also increases.
In other words: Let’s assume we’re trying to compare the strength of the interactions
between Na+ and Cl- as compared with the interactions between Ca2+ and O2-. As Na+ is positively
charged and Cl- is negatively charged, we know that they will attract each other. The same is true
for Ca2+ and O2-. Let’s assume for a moment that these particles are the same size. They’re not but if
we make this assumption then the “r” term stays constant. Because the charges on Ca2+ and O2- are
larger than the charges on Na+ and Cl- we can safely assume that the force of attraction will be much
stronger between Ca2+ and O2-.
There are three types of chemical bonds. Each bond is found in a particular type of
compound.
Metallic bonds occur between two metal atoms (M-M). The metal nuclei remain fixed with
a specific number of electrons around them proportional to the charge of the nuclei. However, the
valence electrons are not fixed in place around the nuclei of the metal atoms but are free to move
around them, or even to flow away under appropriate conditions, just as the waters of a river flow
around the islands that lie in mid-channel. The ratio of water to land around the islands is fixed, even
though the exact water molecules adjacent to the island change from one moment to the next. This
model of metallic bonding is known as the "electron-sea model." It is commonly offered as an
analogy in which bonding metal nuclei are likened unto islands in an ocean, surrounded by a sea of
electrons. In my opinion, the "electron-river" model is a better analogy, but whatever works for you.
Ionic bonds occur between ions. They are the types of bonds that form between metal and
nonmetal atoms (M-NM). The cations and anions may be either monatomic or polyatomic ions.
Monatomic ions are composed of one type of atom while polyatomic ions are composed of two more
types of atoms. We will be discussing polyatomic ions shortly. In ionic bonds between a metal and
a nonmetal atom the bonding electrons are not shared equally. Instead, the nonmetal seizes the
bonding electrons, making itself an anion, while the metal, having lost electrons, becomes a cation.
61
This is not a perfectly correct explanation of what occurs but it is a good working description and will suffice
for our purposes in this class.
Covalent bonds occur between two nonmetals (NM-NM). In a covalent bond the two
bonding electrons are shared more or less equally. This is a very good working generalization, but one
that will require some future elaboration in Chapter 5.
Being able to recognize the type of bond between atoms is a simple but essential skill. The
type of bonds between its atoms determines the type of compound a substance is. And the names
of compounds are assigned based on the types of compounds they are. You should note that, for the
purposes of determining bond types, staircase elements are either metals or nonmetals, not
metalloids. To summarize what we have said above, a bond between two metals atoms will be a
metallic bond. A bond between a metal and a nonmetal will be an ionic bond. And a bond between
two nonmetals will be a covalent bond. This too is a very good working generalization and one you
should take care to remember, as you will soon discover.
We can determine the type of compound a substance is by looking at its molecular formula.
The molecular formula will tell us the types of bonds between the atoms in the compound. And it
is the type of bonds that determine the type of compound.
Metallic compounds are compounds in which the bonding metal atoms are held together by
metallic bonds. Pure metals and alloys are metallic compounds. We will not spend time discussing
metallic compounds further in this class.
Ionic compounds are those which are held together by ionic bonds. This includes salts and
most rocks and minerals in the world around us. Salt is not a single chemical, but rather, a family of
chemicals with hundreds of members. Ionic compounds dissociate in water, if they dissolve.
Dissociation is the process of ionizing, or, of breaking apart into ions. There is a difference between
dissociation and dissolution, the process of dissolving. We will elaborate in the future on this difference.
Solutions of ionic compounds are capable of carrying charge and of conducting electricity and are
called electrolyte solutions.
Molecular compounds (also known as covalent compounds) are held together by covalent
bonds. All organic compounds and polymers are covalent compounds. Polymers are molecular chains;
regularly repeating molecules are the links of the molecular chain. This includes all of the important
biological molecules such as proteins, carbohydrates, nucleic acids, and fats. Many molecular
compounds dissolve in water, but most of them that will dissolve do not dissociate.
The behavior of atoms as they form bonds or ions can be predicted from their position in the
Periodic Table. Are the bonding atoms metals or a nonmetals? This will determine the type of bond.
Of which group is the element a member? This will determine the number of bonds an element will
62
form. And remember the octet rule: atoms are more stable when they are isoelectronic with the
nearest noble gas.
An atom may satisfy the octet rule either through gaining or losing electrons in reactions or
through forming bonds. The alkali metals (1A) usually either form 1+ cations or one bond in
chemical reactions. The alkali earth metals (2A) usually either form 2+ cations or two bonds in
chemical reactions. The 3A elements usually either form 3+ cations or three bonds in chemical
reactions. The Group 4A elements usually form 4 bonds in chemical reactions. The Group 5A
elements usually either form 3- anions or three bonds in chemical reactions. The chalcogens (6A)
usually either form 2- anions or a two bonds in chemical reactions. And the halogens (7A) usually
either form 1- anions or a single bond in chemical reactions.
You must keep in mind that much of what we have just said are generalizations. They are very good
working generalizations, especially for this class, but there are many exceptions to these generalizations.
We will discuss a few exceptions, but they are mostly beyond the scope of this class. It is not that we are
trying to be untruthful as we teach you chemistry. It is that, in a class at this level, much of what you are
taught is generalization. You would need to major in chemistry, and perhaps do graduate work as well, to
begin to learn all of the exceptions and the reasons that they are exceptions. I hope this is not too upsetting
to you. If it is, please let me know and we’ll talk about your decision to change your major to chemistry.
Lewis electron dot structures (Lewis structures) for atoms
An important goal of chemistry is to understand why molecules behave the way they behave,
especially in biological systems. The actual three dimensional structure of molecules often
determines the behavior of molecules in chemical reactions. Although this is a first semester
chemistry class we can still learn how to predict the shapes of many small molecules and will learn
precisely how this is done in Chapter 5.
Lewis electrons dot structures, also known as Lewis structures, are a first step in this
direction. Lewis structures can be used to represent atoms, ions, and molecules. We will learn how
to draw Lewis structures for atoms in this chapter and for ionic and covalent molecules in Chapter 5.
Remember what you learned in Chapter 3 about valence electrons. Valence electrons are the
reactive outer shell s and p electrons of atoms. Inner shell electrons are unreactive. Electrons in filled
d and f subshells behave as inner shell electrons. In partially filled d subshells the d electrons are
both outer shell and valence electrons (transition metals), but to reiterate, the electrons in filled d
subshells are unreactive. Most often we will only consider elements with filled d subshells.
The Lewis structures of atoms and ions consist of the elemental symbol surrounded by one
dot for each valence electron of the substance. The Lewis structures of hydrogen and helium are as
follows:
63
The Lewis structures of the second period elements look like
and the Lewis structures of the third period elements look nearly the same, except that the element
symbols differ
Drawing Lewis structures is similar to giving the electron configuration for an element in some
respects. Note that electrons do not pair up until each orbital in a subshell is occupied by one
electron. You may be wondering what the Lewis structures for elements with d subshell valence electrons
look like. Don't bother worrying. We will only work with Lewis structures for elements with s and p valence
electrons. You may also have noticed that your text does these Lewis structures in a slightly different way.
They do not begin to pair electrons up until there is an electron on each of the four sides of the element's
symbol. Either way is acceptable and correct. What is ultimately important is not where the dots are drawn
with respect to the element symbol but that the correct number of dots is drawn.
The names of common cations and anions
First, a definition. Nomenclature is a set of rules used in the naming of chemical compounds.
To correctly name a compound we must first identify it as an ionic or a covalent compound. This
is because the rules for naming ionic and covalent compounds differ. We cannot correctly name
ionic compounds using the rules of covalent nomenclature. And it is not possible to correctly name
covalent compounds using the rules of ionic nomenclature. We will not study metallic compounds
and their names in this class. Sorry, I know how disappointed you are.
The rules of nomenclature for all compounds have been established by IUPAC, the
International Union of Pure and Applied Chemistry. However, chemistry is a very old science. Many
of the compounds we will discuss have been known for decades, some for centuries. Many were
discovered before IUPAC was organized in 1919. As a consequence many compounds are known
by more than one name. We will focus on the systematic nomenclature required by IUPAC and it
is this system that we will teach you and that you are required to learn for this class. Note: while there
are usually several ways to name most compounds you are required to use the rules outlined in this
chapter. No other names will be accepted.
Cations names are based on the names of the elements that have lost electrons. Group 1A and
2A metals and aluminum can only form one type of cation. That is, all alkali metal cations are
always 1+ cations, all alkali earth metal cations are always 2+ cations, and aluminum is always a 3+
64
cation. This is a generalization but an excellent one, especially for this class. The names of the 1A and
2A cations and of aluminum ion consists only of the element name plus the word ion.
Li+
lithium ion
Na+
sodium ion
K+
potassium ion
Rb+
rubidium ion
Cs+
cesium ion
Be2+
beryllium ion
Mg2+
magnesium ion
Ca2+
calcium ion
Sr2+
strontium ion
Ba2+
barium ion
Al3+
aluminum ion
However, most metals do not belong to Group 1A and 2A. All of these other metals, the
transition metals, the rare earth metals, and the p-block metals, are capable of forming two or more
different cations. For example, iron commonly occurs as either Fe2+ or as Fe3+. Copper can occur as
Cu+ or as Cu2+. Chromium may be found as Cr2+, Cr3+, and Cr6+. And so on. We cannot surmise the
charge on these other metal cations based on their position in the periodic table as they generally
obey rules besides the octet rule. This makes it necessary to convey charge information in the name
of the cation. This is done by naming the element, including the cation charge in Roman numerals
in parentheses, and by adding the word ion to the end of the name. As examples:
Fe2+
iron (II) ion
Fe3+
iron (III) ion
Cu+
copper (I) ion
Cu2+
copper (II) ion
Cr2+
chromium (II) ion
Cr3+
chromium (III) ion
Cr6+
chromium (VI) ion
65
Anion names depend on the type of anion. As we mentioned above there are two types of
anions, monatomic anions and polyatomic anions.
Monatomic anion are anions which consist of only one type of atom. The names of
monatomic anions consist of the element name with its ending changed to "ide."
neutral element name
anion name
symbol
fluorine
fluoride
F-
chlorine
chloride
Cl-
bromine
bromide
Br-
iodine
iodide
I-
oxygen
oxide
O2-
sulfur
sulfide
S2-
selenium
selenide
Se2-
tellurium
telluride
Te2-
nitrogen
nitride
N3-
phosphorus
phosphide
P3-
arsenic
arsenide
As3-
carbon
carbide
C4-
silicon
silicide
Si4-
boron
boride
B5-
So far we've only discussed ions consisting of a single type of atom. Many common
important ions are made up of two or more types of atoms. These are called polyatomic ions. There
are two common polyatomic cations and 15 common polyatomic anions that you must know for this
class. You must *memorize* the names, formulas, and charges of these polyatomic ions as soon as
possible. I suggest making flash cards and then drill yourself night and day until you know all of them by
heart. You will almost certainly wind up repeating this class until you do know them. So do yourself a favor
and learn them *now!* These common polyatomic cations and anions are:
66
Hg22+
mercury (I) ion
NH4+
ammonium ion
SO42-
sulfate
SO32-
sulfite
NO3-
nitrate
NO2-
nitrite
OH-
hydroxide
CO32-
carbonate
HCO3-
hydrogen carbonate
CrO42-
chromate
MnO4-
permanganate
C2H3O2-
acetate*
PO43-
phosphate
CN-
cyanide
ClO4-
perchlorate
ClO3-
chlorate
ClO2-
chlorite
ClO-
hypochlorite
The mercury (I) ion consist of two mercury (I) cations that are chemically bonded to each other. Do
not confuse the mercury (I) ion with the mercury (II) ion (Hg2+), which consists of a single mercury
cation with a 2+ charge.
*
Note: many texts, including the one for this course, will tell you that the formula for acetate ion is
CH3COO-. This is not a molecular formula. It is a condensed structural formula, something we will discuss
in Chapter 6. It certainly is not wrong, but it conveys more information than we need in our study of
nomenclature. As such, it will not be accepted when you take the online quizzes. I expect you to use
C2H3O2- to represent acetate, and if you would like credit for your work you will heed this advice.
Note that most of the polyatomic anions in our list are oxyanions, anions that contain one
or more oxygen atoms. Oxyanion is a term you should know. This is not an all-inclusive list. Many
67
polyatomic ions have not been included. But this is a good starting point, and I do hold you
accountable for *all* of these ions in this list for this class. There are a variety of ways to learn
these. I suggest you make a flash card for each of these polyatomic ions with the name on one side
and its structure on the other.
A word about the naming of ionic compounds in the past. IUPAC nomenclature replaced the
Stock system for naming ionic compounds. You do not need to know the Stock system for this class.
However, since you will almost certainly encounter cations and anions that have been named using
the system, you should be aware that many compounds are still known by their Stock system names
as well as by their systematic (IUPAC) names.
The Stock system used the suffixes "ic" and "ous" to indicate which of a pair of cations was
the higher and lower charged of the pair. The “ic” suffix was used with the higher charged cation
while the “ous” suffix was used with the lower charged cation.
Fe2+
iron (II) ion
ferrous ion
Fe3+
iron (III) ion
ferric ion
Cu+
copper (I) ion
cuprous ion
Cu2+
copper (II) ion
cupric ion
There is no intrinsic information about cation charge in this type of nomenclature. Rather, the system
requires you to know beforehand something about the cations in question and provides relative
information about cation charges.
The naming of oxyanions utilized the suffixes "ate" and "ite" to differentiate between a pair
of oxyanions with different numbers of oxygen atoms. The oxyanion with more oxygen atoms
bonded to the central atom (the first named atom in the molecular formula) received the “ate” suffix
while the oxyanion with fewer oxygen atoms bonded to the central atom received the “ite” suffix.
SO42-
sulfate
SO32-
sulfite
NO3-
nitrate
NO2-
nitrite
Again, the problem this presents to those learning chemistry is that the “ate” and “ite” suffixes do
not convey absolute information about the number of oxygen atoms bonded to the central atom. It
is presupposed that you know more than you do at this point.
68
Those series of oxyanions with the same central atom (e.g., Cl in perchlorate) but with four
different oxyanions used prefixes in combination with the suffixes. "Per" and "ate" indicated which
of the four had the most oxygen atoms, "ate" the second most, "ite" the third most, and "hypo" and
"ite" the fewest.
ClO4-
perchlorate
ClO3-
chlorate
ClO2-
chlorite
ClO-
hypochlorite
Again, remember: you do *not* need to know how to use the Stock system.
The nomenclature of ionic compounds
Ionic compounds contain ionic bonds. Ionic bonds are usually found between metals and
nonmetals, or when one or both of the ions is a polyatomic ion.
In both the names and the molecular formulas of ionic compounds the cation is named first
and the anion is named second. As an example, we know that the substance sodium chloride, NaCl,
is an ionic compound because one of the bonding atoms is a metal (Na) and the other bonding atom
is a nonmetal (Cl). Where do we look to establish whether an element is a metal or a nonmetal? Without
knowing anything else about sodium chloride other than that it is an ionic compound, we know that
sodium is a cation and that chloride is an anion because of the rules of ionic compound
nomenclature.
The following are a few examples of the names of ionic compounds as given by their
molecular formulas.
•
•
•
•
•
KBr: potassium bromide
Ca(NO3)2: calcium nitrate
(NH4)2S: ammonium sulfide
FeSO4: iron (II) sulfate
Zn3(PO4)2: zinc (II) phosphate
69
We must also be able to associate a molecular formula with the corresponding name of the
ionic compound.
•
•
•
•
•
gold (III) cyanide: Au(CN)3
mercury (II) acetate: Hg(C2H3O2)2
titanium (IV) oxide: TiO2
silver (I) perchlorate: AgClO4
sodium phosphate: Na3PO4
Note that parentheses are only used in molecular formulas when there is more than one
polyatomic cation or anion in the compound. For example, KNO3 is correct, while K(NO3) is
incorrect. You're probably wondering how we know how many cations and anions go together in
a particular compound. We'll discuss this below.
The rules of nomenclature are very particular. If, in naming a compound, you forget any of the little things
we have discussed or will be discussing, your work is incorrect and will be marked as such. You must
practice, practice, practice! To help you practice, you can work the Ionic Compound Nomenclature
Self-Quiz found at the SLCC-Science website as well as the problems suggested in Chapter 4 of your text.
The names and molecular formulas of ionic compounds
How do we know how many cations and how many anions go together to form an ionic
compound? Let's look again at the examples of ionic compounds we used earlier:
•
•
•
•
•
•
•
•
•
•
KBr: potassium bromide
Ca(NO3)2: calcium nitrate
(NH4)2S: ammonium sulfide
FeSO4: iron (II) sulfate
Zn3(PO4)2: zinc (II) phosphate
gold (III) cyanide: Au(CN)3
mercury (II) acetate: Hg(C2H3O2)2
titanium (IV) oxide: TiO2
silver (I) perchlorate: AgClO4
sodium phosphate: Na3PO4
Are any of these compounds charged? The answer is no. Charged compounds have a superscript to
the right of the molecular formula indicating whether charge is positive or negative and to what
extent. All ten of the compounds in our examples are not charged.
How is it that ionic compounds can be made up of charged particles and yet remain
electrically neutral?
70
For ionic compounds to be made up of charged particles and yet remain electrically neutral
there must be a balance between the positive charge of the cations and the negative charge of the
anions. In other words, all of the electrons lost by all of the cations in a compound must be offset
by all of the electrons gained by all of the anions in the compound.
If we are given the name of an ionic compound and asked to write its molecular formula,
how can we determine how many cations and how many anions are required to form a neutral
compound? The process can be stated in a series of four steps.
Ask yourself this question: is the compound ionic or covalent?
1.
If ionic, write the symbols for the cation and the anion beneath the name of the compound.
2.
Find the lowest common denominator of the charges and use this to determine how many
cations and anions are needed to form a neutral compound.
3.
Write the formula of the neutral compound.
The ratio of cations to anions in a neutral ionic compound is determined by the ratio of their
charges. Take potassium bromide, which is made up of a potassium ion (K+) and a bromide ion (Br-).
The lowest common denominator (LCD) of the charges is often found by multiplying them together.
(+1) x (-1) = |1|
I consistently use the phrase “lowest common denominator” in these notes but technically speaking it would
be more correct to use the phrase “least common multiple.” By definition, the least common multiple of
two numbers, "a" and "b," is "the smallest positive integer that is a multiple both of a and of b. Since it is
a multiple, it can be divided by a and b without a remainder." (See "Least common multiple." Wikipedia,
The Free Encyclopedia. http://en.wikipedia.org/wiki/Least_common_multiple). This means that multiplying
the charges may not always be the correct way to find the lowest common denominator in the problems
in this chapter, as we will shortly see. The lines on either side of |1| indicate we’re talking about the
absolute value of the number and are not interested in its sign, only the number. In this example,
how many cations are needed to give us a total charge of 1+? How many anions are needed to give
us a total charge of 1-? This means that K+ and Br- go together in a 1:1 ratio.
Let's look at the rest of these examples. The way to do this is to write the symbols for the
cation and the anion in the compound first, calculate the lowest common denominator of the charges,
and then write the molecular formula.
compound
cation
anion
LCD
#cations
needed
#anions
needed
formula
potassium bromide
K+
Br-
1+ x 1- = |1|
1 x 1+ = 1
1 x 1- = 1
KBr
calcium nitrate
Ca2+
NO3-
2+ x 1- = |2|
1 x 2+ = 2
2 x 1- = 2
Ca(NO3)2
ammonium sulfide
NH4+
S2-
1+ x 2- = |2|
2 x 1+ = 2
1 x 2- = 2
(NH4)2S
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iron (II) sulfate
Fe2+
SO42-
2+ x 2- = |2|
1 x 2+ = 2
1 x 2- = 2
FeSO4
zinc (II) phosphate
Zn2+
PO43-
2+ x 3- = |6|
3 x 2+ = 6
2 x 3-= 6
Zn3(PO4)2:
gold (III) cyanide
Au3+
CN-
3+ x 1- = |3|
1 x 3+ = 3
3 x 1- = 3
Au(CN)3
mercury (II) acetate
Hg2+
C2H3O2-
2+ x 1- = |2|
1 x 2+ = 2
2 x 1- = 2
Hg(C2H3O2)2
titanium (IV) oxide
Ti4+
O2-
4+ x 2- = |4|
1 x 4+ = 4
2 x 2- = 4
TiO2
silver (I) perchlorate
Ag+
ClO4-
1+ x 1- = |1|
1 x 1+ = 1
1 x 1- = 1
AgClO4
sodium phosphate
Na+
PO43-
1+ x 3- = |3|
3 x 1+ = 3
1 x 3- = 3
Na3PO4
Let’s be sure we’re clear as to what’s going on in this table.
•
The first three columns are hopefully self-evident although you will need to know the
information discussed previously in this chapter to be able to correctly associate ion names
and symbols.
•
In the LCD (lowest common denominator) column we’re using the cation charge and the
anion charge to determine the lowest common denominator of the charges. We may or may
not actually be multiplying the two charges to find the lowest common denominator. In the
case of iron (II) sulfate, obviously 2 x -2 does not equal 4.
•
In the “#cations needed” column we ask and answer the question: what number multiplied
by the cation charge will equal the lowest common denominator? In the case of zinc (II)
phosphate we see “3 x 2+ = 6" in the column. By working backwards we determine that we
must multiply “2+ ” charge by a factor of “3" to equal “6", the lowest common denominator.
•
In the “#anions needed” column we ask and answer the question: what number multiplied
by the anion charge will equal the lowest common denominator? In the case of zinc (II)
phosphate we see “2 x 3- = 6" in the column. By working backwards we determine that we
must multiply the “3- ” charge by a factor of “2" to equal “6", the lowest common
denominator.
•
In the “formula” column we combine the information from the two previous columns. In the
“#cations needed” column we determined we needed 3 cations for zinc (II) phosphate, while
in the “#anions needed” column we calculated that it would require 2 anions. This tells us
that Zn2+ and PO43- must go together in a 3:2 ratio to give us the correct molecular formula.
At this point you can’t just stare at this stuff and have it make sense. You need to roll up your sleeves, get
out a pencil and some paper, and work through these examples. Do it now! If you don’t work through lots
and lots of examples you will probably *never* figure it out.
You must also be able to state the name of an ionic compound when given its molecular
formula. This too can be determined as a process consisting of four steps.
•
•
•
Ask yourself this question: is the compound ionic or covalent?
If ionic, write the symbols for the cation and the anion beneath the formula of the compound.
Name the cation and anion.
72
•
Write the name of the neutral compound.
In many cases, the charge on the cation will not be immediately apparent since most cations
are capable of two or more different charge states. How do we determine the charge on the cation?
By seeing what the charge on the anion is and how many of them are needed to balance out the
positive charge of the cation(s). Anion charges are fixed. They never vary. This makes it possible
to always determine the charge on the cation in any ionic compound.
Let’s work a few examples and see how we do this. We’ll begin by naming the compound
FePO4.
1. Is this an ionic compound? How do we know? It has a polyatomic ion in its molecular formula. Any
compound in this class with a polyatomic ion in its molecular formula is an ionic compound.
2. Write the symbols for the cation and the anion beneath the formula of the compound.
There are several important things to note. First, as iron (Fe) is a transition metal we can’t guess its
charge from its position in the periodic table. Second, we have a single PO43- group in the molecular
formula of this compound. Remember, the subscripted “4" tells us that this polyatomic anion has
four oxygen atoms in it, not that there are four polyatomic ions. Any time we have more than one
of a given polyatomic ion in the molecular formula of a compound it must be enclosed in
parentheses with a subscript to indicate how many of the polyatomic ion there are in the compound.
As an example, the compound U(PO4)2 contains two PO43- groups.
But, back to FePO4. We have one PO43- group, which has 3 extra electrons. We know this from the
“3-“ charge on the anion. If we multiply the number of anions by the number of extra electrons per
anion, we calculate the total number of extra electrons the anions have brought to the compound.
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Because the compound is electrically neutral (there is no net charge shown in the molecular formula
for the compound), the number of electrons lost by the cations must equal the number of extra
electrons gained by the anions.
So the cations have lost a total of 3 electrons. There is only one cation in the molecular formula of
this compound. By using a bit of simple arithmetic we find
that the charge on the cation must be 3+.
3. Name the cation and anion. The Fe3+ cation is the iron (III) ion. The PO43- anion is the phosphate
ion.
4. To write the name of the neutral compound we combine the name of the cation with the name of
the anion, dropping the word “ion.” So the compound FePO4 is iron(III) phosphate.
Let’s try another one, this time providing a name for UO3.
74
1. Is this an ionic compound? How do we know? As always, take a look at the periodic table. In the
molecular formula of this compound we see that we have metal bonding with a nonmetal. It must be an
ionic compound.
2. Write the symbols for the cation and the anion beneath the formula of the compound.
As “U” is a rare earth metal we can not deduce its charge from its position on the periodic table. But
we can use the anions to help us figure out the charge on the cation. We have three O2- groups, each
of which has 2 extra electrons. We know this from the “2-“ charge on the anion. If we multiply the
number of anions by the number of extra electrons per anion, we calculate the total number of extra
electrons the anions have brought to the compound.
Because the compound is electrically neutral (there is no net charge shown in the molecular formula
for the compound), the number of electrons lost by the cations must equal the number of extra
electrons gained by the anions.
So the cations have lost a total of 6 electrons. There is only one cation in the molecular formula of
this compound. By using a bit of simple arithmetic we find
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that the charge on the cation must be 6+.
3. Name the cation and anion. The U6+ cation is the uranium (VI) ion. The O2- anion is the oxide ion.
4. To write the name of the neutral compound we combine the name of the cation with the name of
the anion, dropping the word “ion.” So the compound UO3 is uranium (VI) oxide.
Let’s work through one more example and come up with the name for Sc2(CO3)3.
1. Is this an ionic compound? How do we know?
2. Write the symbols for the cation and the anion beneath the formula of the compound.
As “Sc” is a transition metal we can not deduce its charge from its position on the periodic table. But
we can use the anions to help us figure out the charge on the cation. We have three CO32- groups,
each of which has 2 extra electrons. We know this from the “2-“ charge on the anion. And
remember: the subscripted “3" immediately next to the O in the molecular formula tells us that there
are 3 oxygen atoms per carbon atom, while the subscripted “3" outside the parentheses is the value
that tells us we have three anions. If we multiply the number of anions by the number of extra
electrons per anion, we calculate the total number of extra electrons the anions have brought to the
compound.
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Because the compound is electrically neutral (there is no net charge shown in the molecular formula
for the compound), the number of electrons lost by the cations must equal the number of extra
electrons gained by the anions.
So the cations have lost a total of 6 electrons. There is only one cation in the molecular formula of
this compound. By using a bit of simple arithmetic we find
that the charge on the cation must be 3+.
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3. Name the cation and anion. The Sc3+ cation is the scandium (III) ion. The CO32- anion is the
carbonate ion.
4. To write the name of the neutral compound we combine the name of the cation with the name of
the anion, dropping the word “ion.” So the compound Sc2(CO3)3 is scandium (III) carbonate.
These examples will hopefully give you a much better idea of why it is so important to know
the names and charges of all of the polyatomic ions and how to find the charges of the monatomic
anions based on their location in the periodic table. But to really learn and retain these concepts you
must practice, practice, practice!
The nomenclature of binary molecular compounds
Binary molecular compounds contain covalent bonds and are made up of two types of atoms.
The order in which elements are named in binary molecular compounds is established by
convention. It starts at the left of the nonmetal elements and works it way from left to right in
periods, and from bottom to top in groups, with the exceptions of oxygen and hydrogen which fall
out of order: B-Si-C-Sb-As-P-N-H-Te-Se-S-I-Br-Cl-O-F. This order is based on the
electronegativity of the elements. You do not need to know this order. It is presented only so that you
understand that there is rhyme and reason to the order in which elements are listed in molecular
compounds. Having said this, the order of the elements in the names and the molecular formulas is
consistent. As you will always be given one or the other, you will always know which order should be
followed.
The compound name states the elements in the same order as the molecular formula and visa
versa. In other words, if the name of a compound is carbon monoxide then the elements are listed
in that same order in the molecular formula, CO, not OC. If the molecular formula of a compound
is NO2 then the name of the compound is nitrogen dioxide, not dioxygen nitride. The name of the
first element is stated as the first element's exact name. The name of the second element ends in the
"ide" suffix. Prefixes are used to indicate the number of each type of atom in the compound. Take
care! Remember that these are molecular compounds. None of the atoms in molecular compounds are
charged, i.e., there are no cations or anions in molecular compounds. The suffix “ide” indicates two
different things, depending on context. In the names of ionic compounds it indicates that the substance is
a monatomic anion (except for hydroxide and cyanide). In the names of covalent compounds it simply
means that element is the second named element, nothing more.
1
mono
2
di
3
tri
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4
tetra
5
penta
6
hexa
7
hepta
8
octa
9
nona
10
deca
There are two exceptions to the use of these prefixes when naming binary molecular compounds:
Exception 1: if the first element is by itself, "mono" is not used e.g. nitrogen dioxide (NO2), not
mononitrogen dioxide
Exception 2: chop "o"/"a" from the end of the prefix if the element name begins with a vowel e.g.
carbon monoxide (CO), not carbon monooxide.
Some illustrations of the names of binary molecular compounds and their corresponding
molecular formulas are as follows.
•
•
•
•
•
XeF6 : xenon hexafluoride
ICl5 : iodine pentachloride
N4S4 : tetranitrogen tetrasulfide
P2O5: diphophorus pentoxide
N2O: dinitrogen monoxide
It surprises many people to learn that noble gas compounds do in fact exist. They are not abundant
and I think that all that are known are synthetic (man-made, not naturally occurring). While their
chemistry is interesting, we will not discuss them in any detail in this class although we will
encounter a few more noble gas compounds in Chapter 5.
A few examples of molecular formulas and their corresponding names:
•
•
•
•
•
iodine heptafluoride: IF7
dinitrogen pentoxide: N2O5
tetraphosphorus decoxide: P4O10
krypton difluoride: KrF2
nitrogen monoxide: NO
79
You've probably noticed that the names of binary molecular compounds are easier than those ionic
compounds. Do not let this keep you from learning the rules and putting them into practice.
For many years the prefixes that are now only correctly used to name molecular compounds
were also used in the naming of ionic compounds. For example, during World War II and the
development of the atomic bomb, a key step in the processing of uranium used in the bombs
involved a compound known then as uranium hexafluoride, UF6. This compound is now called
uranium (VI) fluoride. There are many other examples. Remember: the practice of using prefixes
to indicate numbers of cations and anions in ionic compounds is not acceptable as IUPAC
nomenclature and is therefore also not acceptable in this class.
The nomenclature of acids and bases
There are three common acid-base theories. Each provides a slightly different but
nonetheless distinct definition of what acids and bases are. We will examine these in greater detail
in Chapter 10 but for the time being we will confine ourselves to the Arrhenius theory which defines
acids as substances that can donate a hydrogen ion (H+), and bases as compounds that can donate
a hydroxide ion (OH-). The molecular formulas of acids are usually written with hydrogen listed
first, e.g., HCl, HBr, HC2H3O2 and so on. In this course the names of acids will always include the
word "acid." Some common acids include:
carbonic acid
H2CO3
acetic acid
HC2H3O2
nitric acid
HNO3
nitrous acid
HNO2
sulfuric acid
H2SO4
sulfurous acid
H2SO3
phosphoric acid
H3PO4
hypochlorous acid
HOCl
chlorous acid
HClO2
chloric acid
HClO3
perchloric acid
HClO4
hydrofluoric acid
HF
hydrochloric acid
HCl
80
hydrobromic acid
HBr
hydroiodic acid
HI
The names of acids are common names but are still accepted by IUPAC. Note that the acids named
above are all covalent compounds, not ionic compounds. This is a bit deceptive in that, while they
consist entirely of nonmetal atoms, they must behave to some extent as ionic compounds to act as
acids. Most of them are also oxyacids, i.e., they are acids which contain one or more oxygen atoms.
The common bases are the Group 1A and 2A hydroxides, ammonia, and ammonium
hydroxide.
lithium hydroxide
LiOH
sodium hydroxide
NaOH
potassium hydroxide
KOH
rubidium hydroxide
RbOH
cesium hydroxide
CsOH
beryllium hydroxide
Be(OH)2
magnesium
hydroxide
Mg(OH)2
calcium hydroxide
Ca(OH)2
strontium hydroxide
Sr(OH)2
barium hydroxide
Ba(OH)2
ammonium hydroxide
NH4OH
It is important to note that not every substance that contains hydroxide can behave as a base. There
are many ionic compounds which contain hydroxide ion but which do not behave as bases. Iron (III)
hydroxide, Fe(OH)3, is only one of many dozens of examples.
Ammonia has the molecular formula NH3. Its ability to act as a base is not obvious as there
is no hydroxide ion in its molecular formula. The bases we have listed in the table above are
Arrhenius bases. Ammonia is a Brønsted-Lowry base, not an Arrhenius base. You do not at present
need to understand these distinctions. They will be made more clear in Chapter 10.
Again, you must memorize the names and structures of these compounds if you have any serious
expectations of doing well in this class. I really do try to keep the amount of memorization I require of you
81
to a minimum, but it is not completely unavoidable. And if you refuse to learn these concepts and names
or cannot learn them, your grade will suffer accordingly. This is not a threat. I’m trying to help you
understand the nature of the beast. You *must* master the rules of nomenclature in this chapter.
Post-script: responses to a student questions about nomenclature
Below is the body of a note I sent to a student a while ago who had a question about ionic
nomenclature.
*************************
hi,
the names of transition metal cations should always have a Roman numeral. This is actually true of
*all* metal cations except for the Group 1&2 metals and aluminum.
What you've said as a method of determining cation charge is pretty much on the mark. I've tried to
explain this in the Chapter 4 notes section entitled "The formulas of ionic compounds" but I know
that it still does not always connect for some students. Some tell me working very carefully through
the two tables in this section has been a real help.
We cannot tell transition metal charge from their location on the periodic table because rules besides
the octet rule come into play when we try to explain their behavior. We must use the number of
anions and the charge of the anions to help us figure out the cation charge. Anions are very helpful
because for any given anion, it's charge is always the same and it never changes.
Given your example of AuCl3:
Au is the cation but we do not yet know it's charge because it is not a Group 1 cation, a Group 2
cation, or aluminum; Cl is the anion and we know from its location on the periodic table that it has
a "1-" charge.
There are three Cl- anions. The number of anions (3) multiplied by the anion charge (1-) is equal
to the number of electrons the anions have gained. In the case of AuCl3, the anions have gained (3
x -1 = -3) electrons.
Note that these compounds are electrically neutral, i.e., there is no net charge on them even though
they are made of charged particles (cations and anions). This must mean that the number of electrons
gained by the anions is balanced by the number of electrons lost by the cations. In the case of AuCl3,
the cations have lost three electrons. Since there is only one cation it must have a "3+" charge, hence
Au3+.
The name of the Au3+cation is "gold (III) ion". The name of Cl- anion is "chloride ion. When we put
them together, the name of AuCl3 is gold (III) chloride.
82
As a second example and using the four rules articulated in the chapter 4 notes, take Sb2(SO4)5.
1. Is this an ionic compound? Yes (why?).
2a. The anion is sulfate ion, SO42-. We know this because it is found on the list of polyatomic ions
in the Chapter 4 notes that I have told you you are required to memorize. There are five of them in
the compound. This means that the five anions have gained two electrons each for a total of 10
electrons, i.e., (5 x 2- = 10-)
2b. The cation is Sb (antimony) but we don't know it's charge from its location on the periodic table
because it is not a Group 1 cation, a Group 2 cation, or aluminum. The number of electrons lost by
the two antimony cations must equal the number of electrons gained by the five sulfate anions. In
other words, the two antimony cations have lost 10 electrons.
2c. 2 x ? = 10+.
2d. 2 x 5+ = 10+
2e. This means that each antimony cation must have a "5+" charge, or in other words, the cation in
this compound is Sb5+.
3. The name of the Sb5+cation is "antimony (V) ion. The name of SO42- is sulfate.
4. The name of the compound Sb2(SO4)5 is antimony (V) sulfate.
83
Chapter 5: Molecular Compounds
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as covalent bond, covalent (molecular) compound, double bond,
triple bond, coordinate covalent bond, molecular formula, structural formula, constitutional
isomer, condensed structural formula, backbone or skeletal structural formula, line structural
formula, resonance, VSEPR theory, molecular geometry, axial, equatorial, bonding pairs,
non-bonding (lone) pairs, electronegativity, polar covalent bond, polar molecule, Debye
units, symmetry.
Be able to describe the differences and the similarities between single, double, and triple
covalent bonds.
Be able to describe a coordinate covalent bond and how it differs and is similar to an
ordinary covalent bond.
Be able to differentiate between the molecular formula and the structural formulas of a
compound.
Draw the Lewis structure of a covalent compound given its molecular formula.
Identify which atoms are capable of violating the octet rule and explain why.
Explain what resonance is, why it is important, and identify when it is exists.
Use VSEPR theory to correctly assign molecular geometries to covalent compounds based
simply on their molecular formulas.
Be able to identify whether a covalent bond is polar or nonpolar, given the necessary
electronegativity values.
Be able to identify whether or not a molecule is polar or nonpolar based on its molecular
formula and the necessary electronegativity values.
Be able to tell whether or not a molecule has symmetry.
Covalent bonds and molecular compounds
Covalent bonds are the typical bond formed when both bonding atoms are nonmetals
(NM-NM). The bonding electrons are shared more or less equally. Atoms participating in covalent
bonds gain an octet of electrons through sharing electrons, i.e. through forming chemical bonds,
rather than through ionization and gaining or losing electrons. Covalent bonds result from the
overlap of orbitals. Atoms must be close enough to attract each other and share electrons (through
the overlap of the orbitals in which the bonding electrons are found), but not so close that the
positively-charged nuclei of the bonding atoms repel each other.
There is a strong correlation between the length of a covalent bond (i.e., the distance between
the bonding atoms) and the strength of the bond. Remember Coulomb’s law, the equation that
describes the strength of electrostatic interactions as presented in Chapter 4:
84
F
kq1q2
r2
The closer bonding atoms are to each other (i.e., the shorter the bond length), the smaller the value
of "r" and the stronger the attractive force between them. The further apart bonding atoms are from
each other (i.e., the greater the bond length), the larger the value of "r" and the weaker the attractive
force between them. Not all covalent bonds are equally long, so it stands to reason that not all
covalent bonds are equally strong either. This is more than just theory. This has been experimentally
corroborated using instrumentation and techniques that make it possible to examine lengths and
strengths of bonds in molecules. Why do bond lengths and bond strengths vary? They are
determined by a number of different factors at an atomic level that we will not consider here.
In Chapter 4 we reviewed some of the properties of covalent compounds. They are held
together by covalent bonds. All organic compounds and polymers are covalent compounds. Many
molecular compounds dissolve in water, but most of them that will dissolve do not dissociate.
Remember, there is a difference between dissolution (dissolving) and dissociation (ionizing).
Generally speaking, covalent bonds are not usually as strong as ionic bonds. As a
consequence, the melting points and boiling points of covalent compounds are usually lower than
those of ionic compounds with a similar mass. Why would this be? Think about what has to happen
for any compound, ionic or covalent, to change from a solid to a liquid. The attractive interactions
between the compound's particles must be overcome so that the distance between them can increase.
If the attractive interactions are stronger in one substance than in another, more energy must be
added (i.e., the temperature must be increased to an even greater extent) to convert that substance
from a solid to a liquid. This is the same reason that the boiling points of substances increases as the
strength of the bonds between their particles increases. This means that by comparing the melting
points or the boiling points of two or more compounds, we can obtain an approximate relative idea
as to the strengths of the bonding interactions between the particles in the substances.
Covalent bonds and the Periodic Table
As mentioned in Chapter 4 the number of bonds a nonmetal is likely to form can be predicted
from its position in the periodic table. As a rule the behavior of nonmetals in chemical reactions
tends to be well explained by the octet rule. Atoms are more stable when they are surrounded by an
octet of electrons, whether that octet is obtained through bond formation or through ionization.
Actually, it is more correct to say that atoms are more stable when their outer shell is completely
filled and they are isoelectronic with the nearest Noble Gas, but saying they require an octet of electrons
is very much the same thing in most cases. There are a number of elements with behavior that does not
follow the octet rule that will be discussed later in this chapter.
85
Multiple covalent bonds
It is possible for two bonding nonmetal atoms to share more than one pair of electrons. When
two bonding nonmetal atoms share two pairs of electrons rather than a single pair, a double covalent
bond is formed. When two bonding nonmetal atoms share three pairs of electrons rather than a single
pair, a triple covalent bond is formed. These bonds are shorter and stronger than single bonds.
the strengths of various covalent bonds (kJ/mol)
single bond
double bond
triple bond
C-C
348
612
820
N-N
163
409
945
O-O
146
497
-
C-N
305
613
890
Covalent compounds containing double or triple covalent bonds are common in the real world.
When we draw the structures of these compounds, we indicate single bonds with a dash between the
bonding atoms. Double bonds are represented with two parallel dashes and triple bonds with three
parallel dashes.
Coordinate covalent bonds
Covalent bonds result from the nuclei of two nonmetal atoms sharing a pair of electrons.
Most commonly each of the bonding nuclei donate one electron to the shared pair. An example is
the covalent bond between two hydrogen atoms in molecular hydrogen.
Why is hydrogen satisfied with two electrons, rather than requiring an octet of electrons around it? Each
hydrogen atom donates one electron to the bond and the bonding electrons are shared equally
between the two bonding atoms.
86
There are occasions when one of the bonding atoms donates both of the bonding electrons
to the bond and the other bonding atom does not donate any electrons to the bond. Since the
electrons are shared between the bonding atoms the bond is still a covalent bond. But this type of
covalent bond is called a coordinate covalent bond. Valence electrons always occur in pairs in
molecules in this class. Either the pairs are involved in bond formation and are called bonding pairs,
or they are not involved in bond formation and are called nonbonding pairs, or lone pairs. In the
reaction of ammonia (NH3) and hydrogen ion
the nitrogen atom of ammonia has three bonding pairs of electrons and one nonbonding pair of
electrons. In this reaction nitrogen shares its non-bonding pair of electrons with the hydrogen ion.
A fourth N-H bond is formed and the entire compound, ammonium ion, assumes the positive charge
formerly carried by just the hydrogen ion. This fourth bond, formed in this reaction, is a coordinate
covalent bond because nitrogen donated both of the bonding electrons and hydrogen ion did not
contribute any electrons to the bond.
Coordinate covalent bonds are identical in length and strength to regular covalent bonds.
Their existence can only be predicted from an understanding of how reactant molecules interact with
each other in the formation of a product compound. Still, the formation of coordinate covalent bonds
is often of great importance in many organic and inorganic reactions.
Molecular formulas and structural formulas
Why do we care about the shapes of molecules? Because often, especially in biological
systems, the chemical behavior is determined by molecular shape. The shapes of molecules are often
wildly complex, especially when we examine the shapes of large biological molecules. As you look
at the structure of large biological molecules in your textbooks or on the internet their shape might
not make much sense to you but it makes perfect sense to various enzymes and hormone receptors
in your body. And if the shape of these large biological molecules is changed by even a tiny amount,
they may become completely unrecognizable to those enzymes and hormone receptors. This can
result in serious illness and death. It is very much in the best interests of our good health for our
molecules to keep their proper shapes. In a first semester chemistry class we will not learn about the
shapes of large molecules, but we will take a step in that direction by learning how to predict the
shapes of small molecules.
Molecules can be represented in a variety of ways. Molecular formulas tell us the types of
atoms in a compound as well as the number of each type of atom. The molecular formula of water
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is H2O, which tells us that a single water molecule is made up of two hydrogen atoms and a single
oxygen atom.
There are often two or more compounds that have the same molecular formula, especially
in the study of organic chemistry. We can differentiate between them using various types of
structural formulas, which show the types and numbers of atoms in a compound and also how they
are connected to each other. Structural formulas show all of the bonds between all of the atoms, as
in the following illustration of ethanol and dimethyl ether, both of which have the molecular formula
C2H6O.
Compounds with the same molecular formula but with the atoms connected differently are called
constitutional isomers. Compounds that are constitutional isomers have different physical and
chemical properties, even though they have the same types and numbers of atoms. This is because
the physical and chemical properties of compounds are determined both by the numbers and types
of atoms in the compound as well as by how the atoms are connected, one to another.
Chemists often use an abbreviated type of structural formula, called a condensed structural
formula, in which all of the bonds have been collapsed. A carbon atom bonded to three hydrogen
atoms is called a methyl group and is shown as a CH3 group. A carbon atom bonded to two
hydrogens called a methylene group and is shown as a CH2 group. And the O-H group is shown as
an OH group. The condensed structure of ethanol is CH3CH2OH while that of dimethyl ether is
CH3OCH3.
A more abbreviated representation of a molecule's structure can be achieved using a
backbone or skeletal structure. In backbone structures the bonds between carbon atoms and other
nonmetal atoms are shown, but none of the hydrogen atoms that bond to carbon atoms are shown.
The backbone structure of ethanol is C-C-OH, while that of dimethyl ether is C-O-C.
The most abbreviated representation of molecular structure is achieved with a line structure.
The line structures of ethanol and dimethyl ether are
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in which the end of a line is a methyl (CH3) group and the vertex of a line is a methylene (CH2)
group. The structural formula, condensed structure, backbone structure, and line structure of octane,
a compound with the molecular formula C8H18 appear as follows.
The various types of structural formulas are of greatest use in the study of organic compounds. We seldom
use anything other than the molecular formulas of inorganic compounds to represent them. Constitutional
isomerism is uncommon for inorganic compounds.
The Lewis structures of ionic compounds
We learned in Chapter 4 that the Lewis structures of atoms and ions consist of the elemental
symbol surrounded by one dot for each valence electron of the substance. The Lewis structures of
ions are drawn with the element symbol surrounded by the correct number of valence electrons,
including those it has lost or gained.
Since oppositely charged particles attract each other, cations and anions form ionic bonds as the
resulting ionic compounds are produced. Ionic compounds can be represented using Lewis
structures. The reactions of sodium ion and fluoride, magnesium ion and oxide, and of sodium ion
and oxide, and the resulting ionic compounds, appear as follows:
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Note that in sodium fluoride, fluoride is surrounded by one bonding pair and three nonbonding pairs
of electrons. In magnesium oxide, oxide is also surrounded by one bonding pair and three
nonbonding pairs of electrons. And in sodium oxide, oxide is surrounded by two bonding and two
nonbonding pairs of electrons.
The Lewis structures of covalent compounds
Lewis structures can also show covalent bonds in molecular compounds. As we mentioned
a moment ago, valence electrons always occur as pairs in molecules. This is only true for the molecules
we will discuss in this class. In the real world one will sometimes encounter molecules with unpaired valence
electrons. These electron pairs may be found in bonds, as bonding pairs, or they may occur as
nonbonding pairs (or, lone pairs) that do not participate in bonding. In small covalent compounds
both bonding pairs and nonbonding pairs of electrons affect the shapes of the molecules. We will
be using the Lewis structures of covalent compounds to determine their geometry (shape).
There is a procedure we must follow when drawing the Lewis structures of covalent molecules. It
is:
1.
2.
3.
4.
5.
Find the sum of all of the valence electrons of all of the atoms in the molecule.
Draw the skeleton structure of the molecule by placing the central atom (usually the first
named atom in the molecular formula, except for hydrogen) in the middle and all of the other
atoms (terminal atoms) around it. Draw a line (which represents a single covalent bond)
between the central atom and each of the terminal atoms.
Subtract two electrons for each bond formed in Step 2 from the total valence electrons found
in Step 1. Divide the remainder by two. This tells us the number of nonbonding pairs of
electrons in the molecule.
Assign nonbonding pairs to terminal atoms, until each is surrounded by an octet of electrons.
Assign nonbonding pairs to the central atom so that it also has an octet. In some cases the
central atom may wind up with more than an octet. We'll be discussing what this means in a
moment.
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6.
7.
If there are not enough nonbonding pairs to give the central atom an octet, borrow
nonbonding pairs from terminal atoms and form multiple bonds with the central atom.
Draw resonance structures if appropriate. We'll also be discussing this in one of the examples
below.
Let's see how to use these rules. We'll do a series of examples to illustrate their use.
Note: drawing the Lewis structures of covalent compounds is a very rule-driven exercise. If you follow these
rules exactly every time you draw a Lewis structure you should seldom go wrong.
Example 1: methane has the molecular formula CH4.
While I am providing you with the molecular formulas in these examples I may require you on quizzes and
examinations to be able to write to correct molecular formula simply given a compound's name. Know your
Chapter 4 nomenclature material!
1. Total valence electrons: 4 + (1 x 4) = 8
Carbon has four valence electrons, each of the four hydrogen atoms has one valence electron.
2. Skeletal structure:
Carbon is named first and is the central atom; therefore the hydrogens are terminal atoms.
3. Four covalent bonds were formed. 8 - (4 x 2) = 0.
8 is the number of valence electrons we started with. We formed four bonds and each bond
contains 2 electrons. There are no electrons left to act as nonbonding pairs.
This “0" is the number of valence electrons remaining. Since there are no more electrons to assign,
and since both carbon and all of the hydrogens are satisfied with respect to the octet rule, we're
done! The skeletal structure we drew in Step 2 and the Lewis structure of methane are the same.
In other words we know we're done when (1.) all of the nonbonding pairs have been assigned (i.e., we've
run out of electrons), and (2.) all of the atoms in the molecule are satisfied with respect to the octet rule.
Take note of this. Sometimes students have a hard time knowing when to quit. Again, we’re done when
every atom is satisfied with respect to the octet rule. But methane was rather too easy. Now let's try one
a little more challenging.
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Example 2: phosphorus trichloride has the molecular formula PCl3.
1. Total valence electrons: 5 + (3 x 7) = 26 total valence electrons.
In case you’ve forgotten we can tell how many valence electrons an element has by virtue of its
position in the periodic table. Phosphorus is a group 5A element and has 5 valence electrons.
Chlorine is a group 7A element, so each chlorine atom has 7 valence electrons.
2. Skeletal structure:
Phosphorus is named first in the molecular formula and is therefore the central atom, which leaves
the chlorine atoms as terminal atoms.
3. 26 - (3 x 2) = 20; 20/2 = 10 nonbonding pairs of electrons left to assign.
26 total valence electrons - (3 bonds x 2 electrons per bond) = 20 electrons left;
20 electrons left/2 electrons per nonbonding pair = 10 nonbonding pairs of electrons.
4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. Of the 10 nonbonding pairs, it took 9 of the 10 to give each terminal chlorine atom an octet. The
nonbonding pair is assigned to the central phosphorus atom:
6. Since all of the terminal and central atoms are now surrounded by an octet of electrons, and since
there are no more electrons left, this is the Lewis structure for phosphorus trichloride.
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Example 3: hydrogen cyanide has the molecular formula HCN.
1. Total valence electrons: 1 + 4 + 5 = 10
1 valence electron for hydrogen, 4 valence electrons for carbon, and 5 valence electrons for
nitrogen.
2. Skeletal structure:
While H is listed first it is never a central atom because it can only form one bond. In this case the
next listed atom - carbon - is the central atom.
3. 10 - (2 x 2) = 6; 6/2 = 3 nonbonding pairs remaining.
10 total valence electrons - (2 bonds x 2 electrons per bond) = 6 electrons left;
6 electrons left/2 electrons per nonbonding pair = 3 nonbonding pairs of electrons.
4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. All three nonbonding pairs were assigned to the nitrogen atom. H and N are satisfied with respect
to the octet rule, but carbon is not. Since there are no more nonbonding pairs to assign, one of
nitrogen's nonbonding pairs is converted into a bonding pair and a double bond is formed between
nitrogen and carbon.
6. Hydrogen is still satisfied with respect to the octet rule, as is nitrogen, surrounded by two bonding
pairs and two nonbonding pairs, but carbon is still only surrounded by six electrons (three bonding
pairs of electrons). Another of nitrogen's nonbonding pairs is converted into a bonding pair and a
triple bond is formed between nitrogen and carbon.
7. In this last Lewis structure, all three atoms are satisfied with respect to the octet rule. Therefore,
this last structure is the correct structure for hydrogen cyanide. It is possible to confirm experimentally
that the carbon-nitrogen bond in cyanide is in fact a triple bond. Lewis structures often do a remarkable
job of predicting how atoms are arranged at a molecular level. While they don’t always work well in the
real world, in this class they will prove more than adequate for our needs.
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Example 4: bromate ion has the molecular formula BrO3-.
1. Total valence electrons: 7 + (3 x 6) + 1 = 26 total valence electrons.
7 valence electrons for bromine, 6 for each of the 3 oxygen atoms, and since the compound is a
1- anion, we must add an extra electron to the total to account for the charge of the ion.
2. Skeletal structure:
3. 26 - (3 x 2) = 20; 20/2 = 10 nonbonding pairs of electrons left to assign.
26 total valence electrons - (3 bonds x 2 electrons per bond) = 20 electrons left;
20 electrons left/2 electrons per nonbonding pair = 10 nonbonding pairs of electrons
4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. Of the 10 nonbonding pairs, it took 9 of the 10 to give each terminal oxygen atom an octet. The
nonbonding pair is assigned to the central bromine atom:
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6. In this last Lewis structure, all of the electrons have been used and all four atoms are satisfied
with respect to the octet rule. Therefore, this last structure is the correct structure for bromate ion
except, the Lewis structures of ions must be enclosed in square brackets with the ion charge
indicated as a superscript to the left of the structure.
Note that the square brackets are mandatory for the correct Lewis structures of ions.
Example 5: ammonium ion has the molecular formula NH4+.
1. Total valence electrons: 5 + (4 x 1) - 1 = 8 total valence electrons.
5 valence electrons for nitrogen, 1 for each of the 4 hydrogen atoms, and since the compound is
a 1+ cation, we must subtract an electron from the total to account for the charge of the ion.
2. Skeletal structure:
3. 8 - (4 x 2) = 0.
There are no valence electrons remaining, and so there are no nonbonding pairs to assign. Since
there are no more electrons to assign, and since nitrogen and all of the hydrogens are satisfied with
respect to the octet rule, this is the correct Lewis structure except, since it is an ion, it must be
enclosed in square brackets and show the charge.
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Example 6: sulfur trioxide has the molecular formula SO3.
1. Total valence electrons: 6 + (3 x 6) = 24 total valence electrons
2. Skeletal structure:
3. 24 - (3 x 2) = 18; 18/2 = 9 nonbonding pairs of electrons left to assign
4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. Of the 9 nonbonding pairs, it took all 9 to give each terminal oxygen atom an octet. While they
are satisfied with respect to the octet rule, the central sulfur atom is not. We will borrow a pair of
nonbonding electrons from one of the oxygen atoms and form a double bond to give the sulfur atom
an octet.
6. Wait a minute. Aren't all three oxygen atoms identical to each other? How do we know that sulfur
might not borrow a pair of nonbonding electrons from one of the other oxygen atoms?
The answer is: we don't know. In actual fact there are three equivalent Lewis structures that describe
sulfur trioxide. All are equally probable and equally correct. When a molecule can only correctly
be described by two or more equivalent Lewis structures, we say that it is showing resonance.
There are two very important consequences to resonance in molecules. First, for any
molecule in which resonance occurs, the actual molecule is the average of all of it's resonance
structures. The Lewis structures of molecules exhibiting resonance are also known as resonance
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structures. It does not rapidly switch from one form to the other. It is *always* the average of all of
them. This has some interesting implications that can be proven experimentally. The S-O bond
lengths in sulfur trioxide are shorter than S-O single bond lengths but longer than S-O double bond
lengths. They are, in fact, intermediate between the two. The S-O bond strengths in sulfur trioxide
are also intermediate between those of S-O single and double bond strengths. These bits of
information corroborate our belief that sulfur trioxide cannot be accurately described by any one of
its Lewis structures, or by assuming that it rapidly fluctuates from one structure to the next. It is the
average of its three resonance structures. We represent resonance by drawing all of the correct Lewis
structures, and then by drawing two-headed arrows between them to denote resonance.
Second, molecules are stabilized by resonance. Why this occurs is beyond the scope of this
class, but suffice it to say that resonance is a naturally-occurring event that is energetically favorable
to and which stabilizes those molecules in which it occurs.
Example 7: carbonate ion has the molecular formula CO32-.
1. Total valence electrons: 4 + (3 x 6) + 2 = 24 total valence electrons.
Remember, we add two electrons to the total because this is a 2- anion.
2. Skeletal structure:
3. 24 - (3 x 2) = 18; 18/2 = 9 nonbonding pairs of electrons left to assign.
4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. Of the 9 nonbonding pairs, it took all 9 to give each terminal oxygen atom an octet. While they
are satisfied with respect to the octet rule, the central carbon atom is not. We will borrow a pair of
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nonbonding electrons from one of the oxygen atoms and form a double bond to give the carbon atom
an octet. It is obvious that the double bond required to give carbon an octet can be drawn to any of
the three oxygen atoms, all of which are equivalent to each other.
6. Since we have three equivalent Lewis structures, we know that we have resonance and must
represent all three resonance structures. We also know that, since the structures are of ions, each
must be enclosed in square brackets and show its charge.
Exceptions to the octet rule
There are a number of atoms and molecules that can show behavior that deviates from the
predictions made using the octet rule.
There are numerous molecules with an odd number of valence electrons, such as ClO2, NO,
and NO2. While Lewis structures can be drawn for all of these compounds, we will not consider
them in this class. This is important. If you're drawing a Lewis structure for this class and wind up with
an odd number of valence electrons in Step 1, you've made a mistake. These types of molecules exist and
play important roles in many aspects of the world around us. But, quite simply, you will never be asked to
draw Lewis structures for these compounds in this class. Never.
There are molecules in which an atom might actually be satisfied with less than an octet.
Hydrogen is an example we've already seen. When it has two electrons it is isoelectronic with
helium. Beryllium and boron are two more atoms that are satisfied with less than an octet. Beryllium
forms ionic bonds and is satisfied with a total of four valence electrons around it, as in BeH2. Boron
only forms three covalent bonds and is satisfied with a total of six electrons around it, as in BF3. This
is because beryllium and boron are such small atoms that they do not have room to accommodate
an entire octet around them. Aluminum is also usually satisfied with just six electrons, although it
is large enough to accommodate an octet. Why? I’m really not sure. Perhaps it enjoys being
perverse. I do expect you to remember these four exceptions.
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There are also molecules with an atom with more than an octet of electrons around it. Only
central atoms will ever have more than an octet. Terminal atoms may never have more than an octet
of electrons. And, the central atom may only have more than an octet if it is a nonmetal found in the
third period or higher. This is because elements in the 3rd Period and higher may take advantage of
empty orbitals in their d subshells to accommodate the extra electrons. In other words, if an atom fills
its s and p subshells it can store the extra electrons in empty orbitals in its d subshell. But since d subshells
only occur in the 3rd period and higher, elements in the 1st and 2nd period do not have this option. We
see this most commonly in phosphorus, which may have 10 valence electrons, and in sulfur, which
may have 12 valence electrons. Let's look at two examples. These are only two examples, but before
you're done with this class you will see many other molecules in which the central atom is surrounded by
more than an octet of electrons.
It is important to remember that we never use the formation of double or triple covalent
bonds to violate the octet rule. These violations only occurs when there are left-over non-bonding
pairs and all of the terminal atoms have been satisfied with respect to the octet rule. In this class if
you draw a Lewis structure in which the central atom violates the octet rule due to the
formation of double or triple bonds, you have made a mistake! This is one of the most common
mistakes I see students make with this material. Heed this paragraph well! Terminal atoms *never* violate
the octet rule. And central atoms *only* violate the octet rule when there are an excess of nonbonding
pairs. There, are of course, compounds in the real world for which the preferred Lewis structure is one in
which the central atom violates the octet rule due to the formation of multiple bonds. Sulfate ion is a good
example. But the compounds in which this occurs will not be considered in this class - without exception!
Iodine tetrachloride ion has the molecular formula ICl4-.
1. Total valence electrons: 7 + (4 x 7) + 1 = 36 total valence electrons
2. Skeletal structure:
3. 36 - (4 x 2) = 28; 28/2 = 14 nonbonding pairs of electrons left to assign
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4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. Of the 14 nonbonding pairs, it took 12 to give each terminal chlorine atom an octet. While they
are satisfied with respect to the octet rule, and the central iodine atom is satisfied with respect to the
octet rule, there are still two nonbonding pairs left to assign. They are assigned to (placed around)
the central atom.
6. Since iodine tetrachloride is an ion, it must be placed in square brackets and show its charge.
Phosphorus pentachloride has the molecular formula PCl5.
1. Total valence electrons: 5 + (5 x 7) = 40 total valence electrons
2. Skeletal structure:
3. 40 - (5 x 2) = 30; 30/2 = 15 nonbonding pairs of electrons left to assign
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4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octet rule:
5. Of the 15 nonbonding pairs, it took all 15 to give each terminal chlorine atom an octet. They are
satisfied with respect to the octet rule, as is the central phosphorus atom, which is surrounded by five
bonding pairs of electrons, or, by a total of ten electrons. This is of course a violation of the octet
rule. But, it is possible for phosphorus to do this, and phosphorus is a stable, naturally occurring
element when it behaves this way, unexpected though it may be.
The shapes of molecules and VSEPR - Valence Shell Electron-Pair Repulsion
The shape, or, the geometry of a molecule is determined by the number of bonding and
nonbonding sets of electrons that surround its central atom. Electrons are repelled by other electrons,
since they have the same negative charge. Pairs of electrons in the same orbital are repelled by pairs
of electrons in other orbitals. As a consequence, pairs of electrons will orient themselves as far apart
from each other as is possible in three dimensional space. The theory of molecular geometry based
on this repulsion behavior is called Valence Shell Electron-Pair Repulsion, or VSEPR.
To use VSEPR to determine the shape of a molecule we begin by drawing it's Lewis
structure. The molecule's geometry depends on the total number of sets of electrons that surround
the central atom, by how many of those are bonding sets, and by how many are nonbonding sets.
Two further notes to keep in mind. All of the electrons in multiple bonds count as one set of
electrons. In other words, the presence of multiple covalent bonds does not affect our determination
of the geometry of a molecule. Strangely enough, while double bonds and triple bonds are much
shorter and stronger than single bonds, they seem to play exactly the same role in the determination
of molecular geometry as single bonds.
Nonbonding pairs of electrons take up more space than bonding pairs of electrons. Bonding
electrons are generally confined between the two bonding nuclei. Nonbonding electrons are not
constrained by the presence of a second nucleus, and hence can physically take up more space. As
a result, nonbonding pairs of electrons typically result in the distortion of molecular geometry.
Below is a summary of commonly encountered geometries. You may find it convenient to
refer back to this table as we work our examples. Practice using this table. When I ask you molecular
geometry questions on quizzes or exams you will be permitted to use a copy of it. I will provide you with
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a copy of this table on exams covering this material. Remember that in this table, a "set" may refer either
to a pair of bonding electrons, or to a pair of nonbonding electrons. It may also refer to the two pairs of
bonding electrons in a double bond, or to the three pairs of bonding electrons in a triple bond, each of
which count as only one set.
total sets
bonding sets
nonbonding
sets
geometry
bond angles
examples
2
2
0
linear
180o
BeH2
3
3
0
trigonal
planar
120o
BF3
3
2
1
bent
-
NO2-
4
4
0
tetrahedral
109.5o
CH4
4
3
1
pyramidal
-
NH3
4
2
2
bent
-
H2O
5
5
0
trigonal
bipyramidal
120o, 90o
PCl5
5
4
1
seesaw
-
SCl4
5
3
2
t-shaped
-
ClF3
5
2
3
linear
-
XeF2
6
6
0
octahedral
90o, 90o
SF6
6
5
1
square
pyramidal
-
BrF5
6
4
2
square planar
-
XeF4
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Two Sets of Electrons
When the central atom of a molecule is surrounded by two bonding sets of electrons, as in
beryllium hydride (BeH2, which is also an ionic compound)
the optimal geometry places the bonding sets of electrons at an 180° angle with respect to each
other.
Ask yourself this question: what angle with respect to the central atom places the two terminal atoms as
physically far apart from each other as possible? I suggest that you take a bag of marshmallows or
gumdrops and some toothpicks and use them to make 3-D models of the molecules we discuss. It will help
you visualize the correctness of what I tell you in this section. The marshmallows (gumdrops) represent
atoms. The toothpicks represent bonds between the atoms. Try to optimize the distance between the
terminal marshmallows with respect to the central marshmallow by changing the angles between them.
Which arrangements are best? As a bonus, you may eat your models when your are finished. You won't
get any extra points but it's still a bonus all the same.
This geometry is called linear geometry. There are no nonbonding pairs of electrons around the
central atom, and it is thus an ideal geometry.
“Ideal geometries” are molecular geometries which are not distorted by the presence of nonbonding pairs
of electrons around the central atom. This concept of “ideal geometry” is one you will not find in any text
book. It is, as far as I know, a device of my own creation. But you will find it remarkably useful when we
reach our discussion of molecular polarity at the end of this chapter so pay very close attention to which
geometries are ideal and which are not.
Three Sets of Electrons
When the central atom of a molecule is surrounded by three sets of electrons there are two different
common geometries.
When the central atom of a molecule is surrounded by three sets of electrons and they are all
bonding sets of electrons, as in boron trifluoride,
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the optimal geometry places the bonding sets of electrons at an 120° angle with respect to each
other. This geometry is called trigonal planar geometry. It is an ideal geometry. The three terminal
atoms are found at the corners of a equilateral triangle with the central atom in the middle. All four
atoms are found in the same plane.
When the central atom of a molecule is surrounded by three sets of electrons, two of which are
bonding sets of electrons and one of which is a nonbonding set of electrons, as in nitrite ion,
the optimal geometry ideally places the sets of electrons at an 120° angle with respect to each other.
Note that the double bond only counts as one set. You should also note that nitrite ion has two resonance
structures, which I have omitted, as well as the square brackets and charge information required for the
Lewis structures of ions, which have also been omitted. This is in the interest of trying to keep things simple.
However, the nonbonding pair distorts the trigonal planar geometry and forces the two bonding pairs
closer together than an 120° angle . This distorted trigonal planar geometry is called bent geometry,
or angular geometry. All three atoms are found in the same plane. I do expect you to remember the
bond angles of the ideal geometries, those undistorted by the presence of nonbonding pairs. I do not expect
you to know or remember the bond angles of the distorted geometries. It is sufficient that you know their
names and why they have been distorted.
Four Sets of Electrons
When the central atom of a molecule is surrounded by four sets of electrons there are three different
common geometries.
When the central atom of a molecule is surrounded by four sets of electrons and they are all bonding
sets of electrons, as in methane,
the optimal geometry places the bonding sets of electrons at an 109.5° angle with respect to each
other. This geometry is called tetrahedral geometry. It is an ideal geometry. The four terminal atoms
are found at the corners of a tetrahedron with the central atom in the middle.
Since it is difficult to show three dimensionality with two dimensional figures we use three
different representations of bonds to help achieve the effect. Line bonds, as are the C-H bonds to
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H1a and H1b, are in the plane of the surface on which the drawing is rendered. Wedge bonds, like
the C-H bond to H1c, project out of the surface toward the viewer. And dash bonds, like the C-H
bond to H1d, pass behind the surface away from the viewer.
When the central atom of a molecule is surrounded by four sets of electrons, three of which
are bonding sets of electrons and one of which is a nonbonding set of electrons, as in ammonia
(NH3),
the optimal geometry ideally places the sets of electrons at an 109.5° angle with respect to each
other. However, the nonbonding pair of electrons distorts the tetrahedral geometry and forces the
three bonding pairs closer together than an 109.5° angle. This distorted tetrahedral geometry is
called pyramidal geometry. Note that the bond to H1-a is in the plane of the paper, the bond to H1-b
comes out of the plane toward us, and the bond to H1-c passes behind the plane, away from us.
When the central atom of a molecule is surrounded by four sets of electrons, two of which
are bonding sets of electrons and two of which are nonbonding sets of electrons, as in water,
the optimal geometry ideally places the sets of electrons at an 109.5° angle with respect to each
other. However, the nonbonding pairs distort the tetrahedral geometry and forces the two bonding
pairs closer together than an 109.5° angle . This distorted tetrahedral geometry is called bent
geometry, or angular geometry.
Ok, so there's a certain poverty of imagination in the name of this geometry seeing as how we’ve already
got a bent geometry. Nonetheless, bent geometry may be either due to the effect of one nonbonding pair
on two bonding pairs or due to the effect of two nonbonding pairs on two bonding pairs.
Five Sets of Electrons
When the central atom of a molecule is surrounded by five sets of electrons there are four
different common geometries.
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When the central atom of a molecule is surrounded by five sets of electrons and they are all
bonding sets of electrons, as in phosphorus pentachloride,
the optimal geometry places the bonding sets of electrons at either an 180° angle or an 120° angle
with respect to each other. This geometry is called trigonal bipyramidal geometry. It is an ideal
geometry. Atoms or groups in the 1, 2, and 3 positions are said to be in an equatorial position with
respect to the central atom. Equatorial groups are found at an 120° angle with respect to each other.
The three equatorial groups and the central atom are in the same plane. Atoms or groups in the 4 and
5 positions are said to be in an axial position with respect to the central atom. Axial groups are found
at an 180° angle with respect to each other and are perpendicular to the plane containing the central
atom and the three equatorial atoms. Equatorial and axial, as in the equator and the axis of the earth.
When the central atom of a molecule is surrounded by five sets of electrons, four of which
are bonding sets of electrons and one of which is a nonbonding set of electrons, as in sulfur
tetrachloride,
the nonbonding pair distorts the ideal trigonal bipyramidal geometry and forces the bonding pairs
closer together. When the total number of sets of electrons is five, nonbonding pairs are always
found in the equatorial position. This distorted trigonal bipyramidal geometry is called see-saw
geometry.
Did you notice something curious with sulfur tetrachloride? Given its molecular formula,
SCl4, you might be tempted to leap to the conclusion that it has tetrahedral geometry. But you’d be
wrong. The problem with molecular formulas is that while they do tell us about the numbers and
types of atoms we find in a compound they *don’t* tell us if there are nonbonding pairs of electrons
around the central atom and if so, how many. Don’t take short cuts when determining molecular
geometry. You *must* begin with the correct Lewis structure for the compound. Molecular formulas
alone are not adequate. Notice how often this is the case in the remaining examples we work.
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When the central atom of a molecule is surrounded by five sets of electrons, three of which
are bonding sets of electrons and two of which are nonbonding sets of electrons, as in chlorine
trifluoride,
the nonbonding pairs distort the ideal trigonal bipyramidal geometry and force the bonding pairs
closer together. This distorted trigonal bipyramidal geometry is called t-shaped geometry.
I've never been quite sure how many beers whoever named the see-saw and t-shaped geometries must
have had to drink but I'm assuming it was more than one. As you try to figure out how in the world these
geometries look like a see-saw or the letter "T," remember that when we "look" at molecules with
instruments capable of doing so we do not see the nonbonding pairs of electrons. We only see the atoms
that are present and the effects of nonbonding pairs on the arrangement of the terminal atoms with
respect to the central atom. Nonbonding pairs are invisible. We only see their effects.
When the central atom of a molecule is surrounded by five sets of electrons, three of which
are nonbonding sets of electrons and two of which are bonding sets of electrons, as in xenon
difluoride,
the nonbonding pairs do not distort the ideal trigonal bipyramidal geometry because they offset each
other. The terminal atoms are found in the axial positions, 180° apart from each other. This version
of trigonal bipyramidal geometry is called linear geometry. This geometry, although it has
nonbonding pairs of electrons around the central atom, is an ideal geometry. As all of the
nonbonding pairs are in the equatorial position they cancel each other out and in effect it is as if they
are not present.
Yeah, I know, we've already got a linear geometry. Now we've got another. Don't complain to me, I just
work here. Linear geometry can be found either when a central atom is surrounded by two bonding pairs
of electrons, or when a central atom is surrounded by five pairs of electrons and three of the five are
nonbonding pairs.
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Six Sets of Electrons
When the central atom of a molecule is surrounded by six sets of electrons there are three
different common geometries.
When the central atom of a molecule is surrounded by six sets of electrons and they are all
bonding sets of electrons, as in sulfur hexafluoride,
the optimal geometry places the bonding sets of electrons at either an 180° angle or a 90° angle with
respect to each other. This geometry is called octahedral geometry. It is an ideal geometry. Atoms
or groups in the 1, 2, 3, and 4 positions are said to be in an equatorial position with respect to the
central atom. Equatorial groups are found at a 90° angle with respect to each other. The four
equatorial groups and the central atom are in the same plane. Atoms or groups in the 5 and 6
positions are said to be in an axial position with respect to the central atom. Axial groups are found
at an 180° angle with respect to each other and are perpendicular to the plane containing the central
atom and the four equatorial atoms.
When the central atom of a molecule is surrounded by six sets of electrons, five of which are
bonding sets of electrons and one of which is a nonbonding set of electrons, as in bromine
pentafluoride,
the nonbonding pair distorts the ideal octahedral geometry and forces the bonding pairs closer
together. When the total number of sets of electrons is six, nonbonding pairs are always found in the
axial position. This distorted octahedral geometry is called square pyramidal geometry.
When the central atom of a molecule is surrounded by six sets of electrons, four of which
are bonding sets of electrons and two of which are nonbonding sets of electrons, as in xenon
tetrafluoride,
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the nonbonding pairs do not distort the ideal octahedral geometry because they offset each other.
The terminal atoms are found in the equatorial positions, 90° apart from each other. This version of
octahedral geometry is called square planar geometry. This geometry, although it has nonbonding
pairs of electrons around the central atom, is an ideal geometry. As both of the nonbonding pairs are
in the axial position they cancel each other out and in effect it is as if they are not present.
Let me reiterate the warning I gave you above. You usually cannot look at the molecular formula of a
compound and guess it's geometry. Molecular formulas do not account for the presence of nonbonding
pairs found around the central atom. You have to correctly draw its Lewis structure first and then use the
geometry table. If we did not do this above, we might have thought that bromine pentafluoride has trigonal
bipyramidal geometry, since the central atom forms five bonds. We might have thought that xenon
tetrafluoride has tetrahedral geometry, since the central atom forms four bonds. Take the time to learn
how to do this correctly and practice, practice, practice!
Polar covalent bonds and polar molecules
In covalent bonds each bonding atom donates one electron to the bond and the bonding
electrons are shared. These bonding electrons are shared equally when the bonding atoms are the
same, as for example in H2, N2, O2, and F2. If the bonding atoms are not the same the
electronegativity of the bonding atoms will affect the way in which the bonding electrons are shared.
Electronegativity, as we learned in Chapter 4, is the tendency of an atom to attract electrons to itself.
When a covalent bond is formed between two atoms with different electronegativities the
bonding electrons will still be shared but they will spend a disproportionate amount of time in the
vicinity of the more electronegative atom. Since the more electronegative atom has an extra electron
around it part of the time, the atom develops a partial negative charge because it has one more
electron than it has protons for part of the time. On the other end of the bond the less electronegative
atom is missing one of its electrons part of the time and the atom develops a partial positive charge
because it has one more proton than it has electrons part of the time. When this occurs, we say that
the bond is polarized, or that it is a polar covalent bond. To reiterate: in polar covalent bonds the
bonding electrons are shared but unequally, due to differences in the electronegativities of the
bonding atoms.
Polarity in bonds is indicated using a lowercase Greek letter delta (δ). The partially
negatively charged atom is indicated with a δ-, and the partially positively charged atom is indicated
with a δ+. This is illustrated with hydrochloric acid:
How do we know when the difference in electronegativity is sufficient to cause a bond to be
polar? If you have access to a table of electronegativity values (as, for example, the one found in
Chapter 5 of your text) we can simply compare the values of the bonding atoms. If the difference
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in electronegativity is about 0.7 or greater, the bond is considered to be a polar covalent bond. In
hydrochloric acid, the electronegativity value of hydrogen is 2.1 and of chlorine is 3.0. The
difference is 0.9, which indicates that the bonding electron pair will not be shared equally and that
the bond will be polar. The value 0.7 is not one that is universally agreed upon. Some texts give one value
and others another. Rather, 0.7 lies roughly in the middle of differences in electronegativity from 0.5 to 1.0
that are thought to be the point at which a covalent bond becomes polar.
There are those who believe that if the difference in electronegativities between bonding
atoms is around 1.7 to 2.0 or greater the bonding electrons are shared so unequally that the bond
ceases to be covalent and becomes an ionic bond in nature. There are also those, myself included,
who disagree with this theory. We will disregard it in this class and adhere to what we have said
previously: a bond between a metal and a nonmetal is always an ionic bond, while a bond between
two nonmetals is always a covalent bond regardless of the difference in the electronegativity of the
bonding atoms.
Molecular polarity and symmetry
As a general rule a molecule with one or more polar bonds is itself a polar molecule. In other
words, polar bonds in a molecule can affect the distribution of electrons in the molecule to the extent
that a portion of the molecule develops a partial negative charge while the opposite end of the
molecule develops a partial positive charge. This affects the strength of molecular interactions with
neighboring molecules (which we will discuss in Chapter 8). Molecular polarity is measured in
Debye units (D). Values can range from 0.0 D for nonpolar molecules to values of 2-5 D or more
for very polar molecules.
There are occasions when a molecule with one or more polar bonds is not a polar molecule.
This is the case in symmetric molecules, or, in molecules with a high degree of symmetry.
Molecules have symmetry when they meet two conditions: (1.) the central atom has one of the ideal
geometries - linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, or square planar
and, (2.) all of the terminal atoms are the same. An example of this is phosphorus pentachloride. The
P-Cl bonds are polar, but since there are five chlorine atoms pulling equally on the central
phosphorus atom, they cancel each other out and the molecule as a whole is nonpolar. However, if
we do something to disrupt the symmetry of the molecule, like, for example, substitute a bromine
atom for one of the five chlorine atoms, the molecule will itself become a polar molecule.
Nonbonding pairs of electrons distort the symmetry of the ideal geometries. As a general rule
if a molecule has one or more polar bonds and one or more nonbonding pairs the molecule itself will
be a polar molecule. There are two exceptions to this statement. We saw above that in the linear
form of trigonal bipyramidal geometry and also in square planar geometry, the nonbonding pairs are
aligned in such a manner as to cancel each other out. In this respect these are also ideal geometries.
As such, if the central atom of a molecule with one or more polar bonds and one or more
nonbonding pairs has either of these two geometries, the molecule will be a nonpolar molecule.
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We are treating this concept of molecular polarity as though it is a black-and-white affair when, in reality,
there is a great deal of gray in the real world. As we have done on previous occasions and as we will do
in the future in this class, we are simplifying the situation in an attempt to reduce it to something one can
reasonably understand and learn in the space of a single semester. We are also very much simplifying the
definition of symmetry but we have reduced it to its essence as we need to know it in this course.
Summary of molecular polarity
Is a molecule polar? In this class any molecule with no polar bonds itself is always a
nonpolar molecule. If a molecule has one or more polar bonds then the molecule itself is always
polar unless it has symmetry, i.e., it meets both of the following conditions: (a.) it has an ideal
geometry, and (b.) all of the terminal atoms are the same.
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Chapter 6: Chemical Reactions - Classification and Mass
Relationships
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as molecular formula, empirical formula, molecular weight, amu,
mole, Avogadro's number, macroscopic, microscopic, chemical equation, reactant, product,
coefficient, subscript, mass balance, double displacement reaction, combustion reaction,
stoichiometry, theoretical yield, actual yield, percent yield, strong electrolyte, weak
electrolyte, non-electrolyte, strong acid, weak acid, strong base, weak base, molecular
equation, ionic equation, net ionic equation, spectator ion, spontaneous, "driving force,"
precipitation reaction, solubility rules, Arrhenius theory, acid, base, neutralization reaction,
salt, monoprotic acid, diprotic acid, triprotic acid, electron transfer (redox) reactions,
oxidation, reduction, oxidizing agent, reducing agent, oxidation number.
Given the molecular formula of compound, write its empirical formula.
Calculate the molecular weight of a compound based on its molecular formula.
Calculate the molar mass of a compound based on its molecular formula.
Use Avogadro's number as a conversion factor in the relationship between a certain mass of
a substance and the number of particles of that substance. Also be able to use molecular
formulas as conversion factors.
Express the differences and similarities between the molecular weight and the molar mass
of a substance.
Balance chemical equations.
Predict the outcome of double displacement and combustion reactions and write correctly
balanced equations for these reactions.
Predict the theoretical yield of a reaction using stoichiometry.
Given the actual yield for a reaction and calculate the percent yield for a reaction using
stoichiometry to calculate the theoretical yield.
Be able to identify a substance as a strong electrolyte, weak electrolyte, or non-electrolyte.
Be able to write the molecular equation, ionic equation, and net ionic equation for a reaction.
Be able to predict the likelihood that a chemical reaction is spontaneous based on solubility
rules, formation of water, formation of a gas, or electron transfer.
Predict the outcome of a neutralization reaction.
Be able to assign oxidation numbers, to deduce whether or not a reaction is a redox reaction,
and if so, what is oxidized, reduced, the oxidizing agent, and the reducing agent.
112
Molecular formulas and empirical formulas
Let's review briefly. The molecular formula of a compound tells us what types of atoms are
found in the compound and how many of each type of atom. It is not uncommon for two or more
covalent compounds to have the same molecular formula and in fact this happens frequently. When
this is the case we may use one of the four different types of structural formulas to help differentiate
between the compounds. See chapter 5 for a review of structural formulas. There is usually only one
ionic compound for a given molecular formula.
The empirical formula of a compound is the simplest whole number ratio of atoms in that
compound. If we multiply the empirical formula of a compound by some integer we will obtain its
molecular formula. The empirical formula and the molecular formula of many simple compounds
are the same.
compound
molecular formula
empirical formula
sodium chloride
NaCl
NaCl
ethene
C2H4
CH2
1-octene
C8H16
CH2
glucose
C6H12O6
CH2O
formaldehyde
CH2O
CH2O
imaginary compound
C25H75O15N5S10
C5H15O3NS2
Historically, the empirical formula of a compound was found using a technique called
elemental analysis which showed the proportions of the various elements by mass in a specific
amount of the compound. This knowledge, coupled with the molar mass of the compound (which
is determined using other techniques) revealed the molecular formula of the compound. We will be
discussing molar masses below. This method of determining the molecular formula of compounds is
seldom used any more. Powerful electronic instruments such as mass spectrometers can determine
the molecular formula of most compounds quickly and accurately, often in just a few minutes.
We mention empirical formulas not because they have any great importance to us in class but because
of their historical relevance. You should know what the definition of an empirical formula is and should also
be able to identify the empirical formula of a compound given it's molecular formula. However, you will
never need to reduce the molecular formula of a compound to its empirical formula for anything we do
in this class.
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Molecular formulas, molecular weights, and formula weights
The molecular weight of a compound is the combined mass of all of the atoms in a single
molecule of the compound. The molecular weight of a compound is found by finding the sum of the
masses of all of the atoms in the compound. Molecular weight is abbreviated mw. The units of
molecular weight are amu, atomic mass units. As we said in chapter 3 the mass of individual atoms is
expressed in amu; one amu is equal to 1/12 the mass of single a 12C isotope, or 1.661 x 10-27 kg. The
mass of individual atoms is found in the periodic table. The stated atomic weight for each element
is the average mass of a single atom of the substance in amu. Since the mass of individual atoms is
very small, it is more convenient to express that mass in amu than in grams or kilograms.
compound
molecular
formula
molecular weight
sodium
chloride
NaCl
1 Na atom x 22.989 amu/atom + 1 Cl atom x 35.453
amu/atom = 58.442 amu
ethene
C2H4
2 C atoms x 12.011 amu/atom + 4 H atoms x 1.008
amu/atom = 28.054 amu
1-octene
C8H16
8 C atoms x 12.011 amu/atom + 16 H atoms x 1.008
amu/atom = 112.216 amu
glucose
C6H12O6
6 C atoms x 12.011 amu/atom + 12 H atoms x 1.008
amu/atom + 6 O atoms x 15.999 amu/atom =
180.156 amu
formaldehyd
e
CH2O
imaginary
compound
C25H75O15N5S10
1 C atom x 12.011 amu/atom + 2 H atoms x 1.008
amu/atom + 1 O atom x 15.999 amu/atom =
30.026 amu
25 C atoms x 12.011 amu/atom + 75 H atoms x 1.008
amu/atom + 15 O atoms x 15.999 amu/atom + 5 N
atoms x 14.0067 amu/atom + 10 S atoms x 32.06
amu/atom = 1006.494 amu
The rules for calculating sig figs for addition and subtraction differ from those used to determine the
number of sig figs when multiplying or dividing. However, I will not hold you responsible for correctly
determining sig figs for addition and subtraction. The molecular weight of a compound may be small,
as it is for elemental hydrogen (H2, mw = 2.016 amu). The molecular weight of large molecules,
such as proteins and strands of nucleic acids, may exceed 1 x 106 amu.
The molecular weights of ionic compounds are often called formula weights. Most ionic
compounds do not exist as discrete molecules but rather as interconnected lattices of ions. So,
114
strictly speaking, it is not correct to think of ionic compounds as consisting of individual molecules.
Be this as it may, without going into any further discussion, the formula weight of an ionic
compound is numerically exactly the same as it's molecular weight. In this class we will always refer
to the molecular weight of ionic compounds and will not mention formula weights again.
The mole and Avogadro's number
By looking at the periodic table we can find the mass of an individual atom of any element.
The units of these masses are given in amu (1.00 amu = 1.661 x 10-27 kg). As one amu is roughly
equal to one-billionth of one-billlionth of one-billionth of a kilogram it is obvious that the mass of
an individual atom is extremely small. Since it is difficult to measure the mass of individual atoms,
is there a way to determine the mass of a large number of atoms or even better, to calculate the
number of atoms in a given amount of a substance?
We can count atoms by weighing a substance. Then, using dimensional analysis, we can
determine the number of atoms in the mass of the substance we weighed.
A single carbon atom weighs 12.01 amu. If we take 12.01 amu of carbon, we have one
carbon atom. How many carbon atoms will we have if we take 12.01 grams of carbon rather than
12.01 amu of carbon? Using dimensional analysis, we find
(12.01 g C) x (1 kg C/1000 g C) x (1 amu C/1.6605402 x 10-27 kg C) x (1 C atom/12.01 amu C) =
6.02 x 1023 C atoms
This calculation means that in 12.01 grams of carbon, there are 6.02 x 1023 C atoms, or, there
are about six hundred billion trillion carbon atoms. That means we have a trillion carbon atoms six
hundred billion times That's rather a large number of carbon atoms.
Let's repeat this calculation for another element, this time for gold. From the periodic table
we find the mass of a single gold atom is 196.97 amu. How many gold atoms are there in 196.97
grams of gold?
(196.97 g Au) x (1 kg Au/1000 g Au) x (1 amu Au/1.6605402 x 10-27 kg Au) x (1 Au atom/196.97
amu Au) = 6.02 x 1023 Au atoms
Notice that the number of gold atoms in 196.97 grams of gold is exactly the same as the
number of carbon atoms in 12.01 grams of carbon. How strange! Is this a coincidence?
It is not a coincidence. We find that whenever we take the mass in grams of an element equal
to the mass in amu of one of it's atoms, the number of atoms in that mass is always
6.02 x 1023 atoms.
This is also true of molecules. Whenever we take the mass in grams of a compound equal
to the mass in amu of a single molecule (i.e. the mass in grams equal to the molecular weight of the
115
compound in amu) there are always 6.02 x 1023 molecules present. There are no exceptions to this,
either for atoms or for molecules.
We calculated above that the mass of a single glucose molecule is 180.156 amu (from the
sum of the atomic weights of the elements from the periodic table). How many glucose molecules
are there in 180.156 g of glucose?
(180.156 g glucose) x (1 kg glucose/1000 g glucose) x (1 amu glucose/1.6605402 x 10-27 kg glucose)
x (1 glucose molecule/180.156 amu glucose) = 6.02 x 1023 glucose molecules
The mass in grams of any atom or of any molecule equal to its weight in amu will always
contain 6.02 x 1023 of that atom or molecule. The number 6.02 x 1023 is called Avogadro's number.
Whenever we have Avogadro's number of anything, large or small, we have 6.02 x 1023 of that thing.
We can have Avogadro's number of atoms and we can also have Avogadro's number of stars. And
when we have Avogadro's number of anything, we say that we have 1 mole of that thing.
6.02 x 1023 of (x) = 1.00 mole of (x).
The mole is a unit of quantity. When we have a pair of things - whatever those things might
be - we have two of those things. When we have a dozen objects we have twelve of the object. When
we have a gross of something we have 144 of the thing. And when we have a mole of something we
have 6.02 x 1023 of whatever. Given the small size of atoms, can you see any advantages in having a very
large measure of quantity during the study of chemistry?
The mole is an unimaginably large number. If one does a Google search on the phrase “how
big is a mole” various numbers can be found. While I cannot vouch for the accuracy of these
calculations here are a few examples I found on the Internet that sound more or less correct:
•
•
•
•
one mole of glass marbles would cover the entire surface of the earth (oceans included) to
a depth of several miles.
a mole of beer cans (or Coke cans, if you prefer) would cover the entire surface of the earth
(oceans included) to a depth in excess of 200 miles.
If you count out loud starting with the number "one" at the rate of one count every second
it would take you about 2,000 trillion years to finish.
If you covered the entire state of California with a mole of sand it would bury the state to a
depth of ten feet.
Molecular weight and molar mass
There is an important implication to all of this. The periodic table gives us information at two
levels. The atomic weight on the periodic table is both the mass of a single atom in amu and also
the mass of a mole of the substance in grams. We refer to the mass of one mole of any substance as
it's molar mass. The molar mass of a compound is equal to the sum of the weights of the moles of
atoms in one mole of the compound.
116
compound
molecular
formula
molar mass
sodium
chloride
NaCl
1 mole Na atoms x 22.989 g/mole + 1 mole Cl atoms
x 35.453 g/mole = 58.442 g/mole
ethene
C2H4
2 moles C atoms x 12.011 g/mole + 4 moles H atoms
x 1.008 g/mole = 28.054 g/mole
1-octene
C8H16
8 moles C atoms x 12.011 g/mole + 16 moles H
atoms x 1.008 g/mole = 112.216 g/mole
glucose
C6H12O6
formaldehyd
e
CH2O
1 mole C atoms x 12.011 g/mole + 2 moles H atoms
x 1.008 g/mole + 1 mole O atoms x 15.999 g/mole =
30.026 g/mole
C25H75O15N5S10
25 moles C atoms x 12.011 g/mole + 75 moles H
atoms x 1.008 g/mole + 15 moles O atoms x 15.999
g/mole + 5 moles N atoms x 14.0067 g/mole + 10
moles S atoms x 32.06 g/mole = 1006.494 g/mole
imaginary
compound
6 moles C atoms x 12.011 g/mole + 12 moles H
atoms x 1.008 g/mole + 6 moles O atoms x 15.999
g/mole = 180.156 g/mole
As with the periodic table molecular formulas also give us information at two levels. The
molecular formula of a compound tells us how many of each atom it takes to make a single molecule
of the compound. The molecular formula also tells us how many moles of each type of atom it takes
to make up one mole of molecules of the compound.
Let’s reiterate: the numbers for the molecular weight of a compound and the molar mass of
a compound are exactly the same. Only the units differ. The unit of molecular weight is the amu.
The units of molar mass are grams per mole (g/mol). The abbreviation for mole is mol. Doesn't that
save lots of time and ink?
water, H2O
The mass of a single molecule: 18.02 amu
The mass of a mole of molecules: 18.02 g
carbon dioxide, CO2
The mass of a single molecule: 44.01 amu
The mass of a mole of molecules: 44.01 g
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glucose, C6H12O6
The mass of a single molecule: 180 amu
The mass of a mole of molecules: 180 g
Diazinon is a pesticide with the molecular formula C12H21N2O3PS
The mass of a single molecule: 304.34 amu
The mass of a mole of molecules: 304.34 g
While molar mass and molecular weight are two different things the molar mass of a
compound is frequently referred to as its molecular weight by people who use the terms carelessly.
But as we have seen above, while they are conceptually similar in actual fact they aren’t the same
thing at all.
The linking relationship between the microscopic and the macroscopic is the mole.
Microscopic: at an atomic or molecular level; macroscopic: on a level we can perceive with our unaided
senses. If we know how many moles of a substance we have we also know how many atoms or
molecules of the substance we have. Conversely, if we know how many atoms or molecules of the
substance we have we know how many moles of the substance we have. The following examples
demonstrate how we may convert back and forth using the relationship between Avogadro’s number
and the mole as a conversion factor.
!
!
!
!
!
How many moles of carbon are in one mole of diazinon? how many carbon atoms?
"
there are 12 moles of carbon in every mole of diazinon We know this from diazinon's
molecular formula C12H21N2O3PS
"
(12 moles of carbon) x (6.02 x 1023 C atoms/1 mole carbon) = 7.22 x 1024 carbon
atoms
How many molecules are there in 1.4 moles of ethanol? The condensed structural formula of
ethanol is C2H5OH and it is commonly referred to as EtOH
"
(1.4 moles of EtOH) x (6.02 x 1023 molecules of EtOH/1 mole EtOH) = 8.43 x 1023
molecules EtOH
A sample contains 7.75 x 1015 molecules of ethanol, how many moles is this?
"
(7.75 x 1015 molecules EtOH ) x (1 mole EtOH/6.02 x 1023 molecules EtOH) = 1.29
x 10-8 mole EtOH
A sample of magnesium phosphate weighs 2.50 g, how many molecules is this? To do this
problem you must first begin with the formula of magnesium phosphate (MP).
"
(2.50 g MP) x (1 mole MP/262.87 g MP) x (6.02 x 1023 MP molecules/1 mole MP)
= 5.73 x 1021 MP molecules
If a chemical assay based on the detection of phosphorus can detect 5.65 pg of magnesium
phosphate, how many phosphorus atoms are present? An assay is a chemical test used to detect
the presence of something, usually something present in very small amounts.
"
(5.65 pg MP) x (1 g MP/1012 pg MP) x (1 mole MP/262.87 g MP) x (6.02 x 1023 MP
molecules/1 mole MP ) x (2 P atoms/1 MP molecule) = 2.58 x 1010 P atoms
118
!
Do you notice how we used relationships in the molecular formula of the compound as a
conversion factor in this problem, 2 phosphorus atoms per MP molecule. Is it proper to use
the information in a molecular formula as a conversion factor? Absolutely.
How many mg will 6.1 x 1020 butane molecules weigh? Butane has the molecular formula
C4H10
"
(6.1 x 1020 molecules butane) x (1 mole butane/6.02 x 1023 molecules butane) x
(58.12 g butane/1 mole butane) x (1000 mg butane/1 g butane) = 58.9 mg butane
Chemical equations
Chemical equations are shorthand representations of chemical reactions. For example, consider the
reaction of aqueous silver (I) nitrate and aqueous ammonium chloride which results in the formation
of solid silver (I) chloride and aqueous ammonium nitrate. Describing this reaction verbally is rather
tedious. Alternatively, we can describe it this way:
The substances involved in the reaction are represented by their molecular formulas. If one
of the reactants is an element we use its symbol to represent the element.
The chemicals to the left of the arrow are called reactants. Reactants are the ingredients from
which the products of the chemical reaction are made. Reactants are always found to the left of the
arrow in a chemical equation. Products are the chemicals produced during a reaction and are always
found on the right side of the arrow. Normally we do not label reactants or products as such as it is
understood that anything to the left of the arrow is always a reactant and anything to the right of the
arrow is always a product.
The state of each chemical is indicated with a subscript and enclosed in parentheses, to the
right of it's molecular formula. The four states in which we may find either reactants or products are:
!
!
!
!
the gaseous state, indicated (g)
as a pure liquid, indicated (l)
as a solid, indicated (s)
dissolved in water, or, in aqueous solution, indicated (aq) Note that “aqueous” is not a state
of matter as such. It is simply a reflection of the nature of things we commonly find in chemical
reactions.
119
An important attribute of all correctly written molecular formulas is mass balance.
Remember that mass is *never* created or destroyed in chemical reactions. If we have “x” number
of a given atom on the reactant side of an equation we must *always* have *exactly* “x” of that
same atom on the product side of the equation. This does not preclude them from being found in
different states or in different substances on the two sides of the chemical equation.
In the above reaction we have one silver (Ag) atom on each side of the equation. It does not
matter that it is found in silver (I) nitrate on the reactant side and in silver (I) chloride on the product
side. All that matters is that there is one silver atom on both sides of the equation.
We also have exactly one chlorine atom on each side of the equation, one atom in ammonium
chloride on the reactant side, one atom in silver (I) chloride on the product side.
When it comes to counting the atoms in polyatomic ions we can do one of two things. We
can count the individual atoms in each polyatomic ion which is longer and harder. Or if the
polyatomic ions pass unchanged from the reactant side to the product side we can simply count
polyatomic ions themselves which is quicker and easier.
In the above reaction we have one nitrate ion on each side of the equation and one
ammonium ion on each side of the equation. This means we have mass balance with respect to these
two polyatomic ions.
Let's take a look at the reaction of aqueous barium nitrate and aqueous lithium phosphate:
3 Ba(NO3)2 (aq) + 2 Li3PO4 (aq) => Ba3(PO4)2 (s) + 6 LiNO3 (aq)
Notice that aqueous solutions of barium nitrate and lithium phosphate are the reactants in this
reaction and solid barium phosphate and aqueous lithium nitrate are the products of this reaction.
The numbers in front of the various molecular formulas are called coefficients. A coefficient tells
us how many of each molecule is needed for the reaction to proceed properly. If there is no
coefficient shown in front of a substance it should be inferred that the number "one" is present. In
the above reaction, the equation tells us that three molecules of aqueous barium nitrate react with
two molecules of aqueous lithium phosphate to form a single molecule of solid barium phosphate
and six molecules of aqueous lithium nitrate.
There is a very important difference between the coefficients in chemical equations and the
subscripts in molecular formulas.
The subscripts in molecular formulas tell us how many of each atom are found in a single
molecule of the compound. These numbers are unique for each compound. If we change the subscripts
in a molecular formula we change the compound. Regardless of the type of compound, how the
compound is involved in a reaction, or the type of reaction in which a compound is formed, the
subscripts in a compound's molecular formula are fixed by nature and they *never* vary.
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In ionic compounds the subscripts in the molecular formula of a given compound always
depend exclusively on the charges of the cations and anions in the compound. As we learned in
Chapter 4, cations and anions go together in such ratios as are necessary to form a neutral
compound.
In covalent compounds the subscripts in the molecular formula of a given compound are also
fixed and invariant. We are told how many of each atom occurs in each molecule of compound by
the compound’s name.
As examples, water is always H2O regardless of whether it is a reactant or a product and also
regardless of the reaction in which it is involved. Barium nitrate is always Ba(NO3)2, lithium
phosphate is always Li3PO4, barium phosphate is always Ba3(PO4)2 , and lithium nitrate is always
LiNO3, even if they are involved in reactions other than the one shown above.
Coefficients, on the other hand, always depend on the reaction. A compound may have one
coefficient in one reaction, and a different coefficient in another. As we see in the following reaction
when barium nitrate reacts with lithium carbonate only a single molecule of barium nitrate is
required as a reactant and only two molecules of lithium nitrate are produced. However, the
molecular formulas of the barium nitrate and lithium nitrate are exactly the same in both reactions.
Chemical equations also provide us with information at two levels. Since molecular formulas
provide both microscopic and macroscopic information, and since chemical equations consist of molecular
formulas, this should come as no great surprise.
A chemical equation tells us how many molecules of each reactant are required to produce
a specific quantity of product molecules.
A chemical equation also indicates how many moles of reactants are required to produce the
indicated number of moles of products.
In the above reaction, the equation shows that a single molecule of barium nitrate may react
with a single molecule of lithium carbonate, and one molecule of barium carbonate and two
molecules of lithium nitrate will be produced. But it also tells us that when one mole of barium
nitrate reacts with one mole of lithium carbonate, one mole of barium carbonate and two mole of
lithium nitrate are produced.
Balancing chemical equations
Let’s repeat some of the important points we just learned about chemical reactions. Mass is
always conserved in chemical reactions. If we have "x" number of a certain type of atoms in a
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reactant compound or compounds there must also be exactly "x" of those atoms in product
molecules as well. Atoms are never lost or gained in chemical reactions. As a consequence chemical
equations must be balanced. There must be equal numbers of all atoms on both sides of the equation.
Let's take a second look at the reaction of barium nitrate and lithium carbonate. The above
equation is balanced. We can tell whether or not an equation is balanced by counting the numbers
of each type of atom on both sides of the equation.
In the balanced equation there is one barium atom and one carbon atom on each side of the equation.
If we prefer to think in terms of moles rather than individual atoms we find one mole of barium
atoms and one mole of carbon atoms on each side of the equation. There are two moles of nitrogen
on each side and two moles of lithium as well. And there are nine moles of oxygen on each side. On
the reactant side, there are three moles of oxygen in every mole of lithium carbonate, and six moles
of oxygen in every mole of barium nitrate. On the product side, there are three moles of oxygen in
every mole of barium carbonate, and three moles of oxygen in every mole of lithium nitrate, of
which we have two moles.
As we said above, when balancing equations with polyatomic ions it is usually easier to
balance the ions if they pass unchanged from one side of the reaction to the other. This will usually
be the case in the reactions we examine in this class.
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A chemical equation must be balanced to be of any use. When writing chemical equations
you must keep this in mind or your work will be incorrect. When balancing chemical equations be
sure to address the following points:
!
!
!
!
Remember mass balance!
Use the Periodic Table and a knowledge of the formulas and charges of the common
polyatomic ions to decide the ratios of ions and how they go together to form ionic
compounds
There is a difference between the subscripts in molecular formulas and the coefficients of
balanced chemical equations.
Coefficients are used to balance an equation.
We're almost to the point where we can begin to balance equations but we first need to
discuss one more topic. There are families of chemical reactions (or classes of chemical reactions)
just as there are families of elements and families of compounds. If we know something about the
general properties of these families of reactions we can easily predict the outcomes for these types
of reactions.
The two most common families of reactions we discuss in this class are double displacement
reactions and combustion reactions.
Double displacement reactions are the common type of reaction that occurs between two
ionic compounds in aqueous solution (i.e., when both ionic compounds are dissolved in water). In
a double displacement reaction the cations swap their anions. Generically, if AB and CD are ionic
compounds in aqueous solution, then in a double displacement reaction between them
AB + CD => AD + CB
AD and CB will be formed as a result. If you look back the three reactions we have just discussed,
you will be able to notice that this is exactly what happened in each case. Keep the constraining
conditions in mind: both reacting compounds must be ionic compounds, and both must be in
aqueous solution (dissolved in water).
Combustion reactions involve the interaction of some material with oxygen and result in the
formation of light and heat energy along with other products. Any time you see something burn
you’re seeing a combustion reaction. All non-nuclear explosions are very fast combustion reactions.
In the combustion of organic materials (materials containing both carbon and hydrogen in their
molecular formulas) two products will always be formed, carbon dioxide gas and water vapor. In
this class these are the *only* two products that will *ever* be formed in a reaction involving the
combustion of an organic material. A generic version of the chemical equation describing the
combustion of carbon-containing compounds is
"organic substance" + O2 (g) => CO2 (g) + H2O(l)
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When balancing a double displacement reaction it is helpful to remember the following steps:
1.
2.
3.
4.
5.
Write the molecular formulas for the reactant compounds.
Write the symbols for the cations and anions in each reactant compound beneath them since
this will help predict the products that are formed.
The charges of the ions determine the ratios in which they combine to form product
molecules; write the formulas of the product compounds.
Place the products on the right side of the arrow.
Use coefficients to balance the equation.
As an example let's consider the double displacement reaction of potassium chloride and lead
(II) nitrate. Hmm, do you think a knowledge of nomenclature might become useful at this point? I mean,
it's not like this is the sort of thing I would put on an exam, is it? Or is it? You've got a 50/50 chance here.
Do you think there is any possibility of seeing problems like these on your exams for this class that one
cannot begin to answer unless one is familiar with nomenclature?
Step 1 - write the molecular formulas for the reactants:
Step 2: Write the symbols for the cations and anions in each reactant compound beneath them since
this will help predict the products that are formed.
Step 3 - The charges of the ions determine the ratios in which they combine to form product
molecules; write the formulas of the product compounds. In this reaction the products will be KNO3
and PbCl2. If you have trouble with this step you need to re-visit Chapter 4.
Step 4 - Place the products on the right side of the arrow.
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Step 5 - Use coefficients to balance the equation.
And that's all there is to it! When balancing the equations of double displacement reactions, it really
doesn't matter where you begin, but you will find it easiest if you start by balancing the highest
charged ions first and ions with the lowest charges last.
When balancing organic combustion reactions you will find it easiest if you begin with
carbon, then balance hydrogen, and balance oxygen last. The number of moles of carbon dioxide
formed is equal to the number of moles of carbon in the organic reactant. The number of moles of
water vapor formed is equal to one-half the number of moles of hydrogen in the organic reactant.
Consider the combustion of pentane, C5H12:
Step 1 - use the generic equation describing the combustion of an organic compound:
C5H12 + O2 => CO2 + H2O
Step 2 - balance C: As there are five carbon atoms in pentane we must have five carbon dioxide
molecules as product:
C5H12 + O2 => 5 CO2 + H2O
Step 3 - balance H: the twelve hydrogen atoms in pentane will form six water molecules, as
(6 water molecules) x (2 hydrogen atoms/water molecule) = 12 hydrogen atoms
C5H12 + O2 => 5 CO2 + 6 H2O
Step 4 - balance O: on the product side we have (5 x 2) + (6 x 1) = 16 oxygen atoms on the product
side. The oxygen gas used in combustion reactions always occurs as a diatomic molecule consisting
of two oxygen atoms. We need 8 oxygen molecules, as
(8 oxygen molecules) x (2 oxygen atoms/oxygen molecule) = 16 oxygen atoms
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so the balanced equation describing the combustion of pentane is
C5H12 + 8 O2 => 5 CO2 + 6 H2O
Some combustion reaction also have a step 5. Let's look at the combustion of ethane, C2H6.
Step 1: use the generic equation describing the combustion of an organic compound:
C2H6 + O2 => CO2 + H2O
Step 2 - balance C: As there are two carbon atoms in ethane we must have two carbon dioxide
molecules as product:
C2H6 + O2 => 2 CO2 + H2O
Step 3 - balance H: the six hydrogen atoms in ethane will form three water molecules, as
(3 water molecules) x (2 hydrogen atoms/water molecule) = 6 hydrogen atoms
C2H6 + O2 => 2 CO2 + 3 H2O
Step 4 - balance O: on the product side we have (2 x 2) + (3 x 1) = 7 oxygen atoms on the product
side. The oxygen gas used in combustion reactions always occurs as a diatomic molecule consisting
of two oxygen atoms. We need 3.5 oxygen molecules, as
(3.5 oxygen molecules) x (2 oxygen atoms/oxygen molecule) = 7 oxygen atoms
so the balanced equation describing the combustion of ethane is
C2H6 + 3½ O2 => 2 CO2 + 3 H2O
Step 5 - the problem with a fractional coefficient is that they are not permitted in balanced equations
except under some very special circumstances which we will not see in this class. We must multiply
all of the coefficients in the equation by two to convert 3½ to an integer value.
2 C2H6 + 7 O2 => 4 CO2 + 6 H2O
Is the equation still balanced? Count the atoms and see. In this class the only time you will encounter
fractional coefficients is while balancing combustion reactions equations and then, only
occasionally. You will never balance the equation of a double displacement reaction in this class and
wind up with a fractional coefficient. Unless you’ve made a mistake.
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Ok, let’s get some practice. Write a balanced chemical equation for each of the following reactions.
No peeking until you've solved the problem yourself first! At this point you have not yet been
introduced to the rules that permit you to predict the state of a compound, so don't worry about them yet. Patience, my children, patience. This will be covered a little later in this chapter.
1.
2.
3.
4.
5.
6.
the double displacement reaction of sodium bromide and iron (III) nitrate
the double displacement reaction of aluminum chloride and barium sulfate
the double displacement reaction of ammonium phosphate and calcium sulfite
the double displacement reaction of sodium cyanide and gold (III) nitrate
the combustion of methane gas (CH4)
the combustion of propanol (C3H7OH)
Answers:
1.
3 NaBr + Fe(NO3)3 => FeBr3 + 3 NaNO3
2.
2 AlCl3 + 3 BaSO4 => Al2(SO4)3 + 3 BaCl2
3.
2 (NH4)3PO4 + 3 CaSO3 => Ca3(PO4)2 + 3 (NH4)2SO3
4.
3 NaCN + Au(NO3)3 => Au(CN)3 + 3 NaNO3
5.
CH4 + 2 O2 => CO2 + 2 H2O
6.
2 C3H7OH+ 9 O2 => 6 CO2 + 8 H2O
How did you do? Do you need more practice?
Stoichiometry
Let’s examine the combustion of liquid ethanol, C2H5OH(l):
C2H5OH(l) + 3 O2 (g) => 2 CO2 (g) + 3 H2O(g)
The balanced equation conveys information at two levels. It tells us that, on a microscopic level, one
ethanol molecule will react with three oxygen molecules to form two carbon dioxide molecules and
three water molecules. On a macroscopic level the balanced equation tells us that if one mole of
ethanol reacts with three moles of oxygen, two moles of carbon dioxide and three moles of water
vapor will form. But the balanced equation does not convey information about mass amounts. It
would be utterly incorrect to assume from the balanced equation that if we take one gram of ethanol
and react it with three grams of oxygen that two grams of carbon dioxide and three grams of water
would form. No chemical equation, even a balanced one, ever provides that sort of information
directly.
And yet mass to mass relationships are of a great deal of interest to chemists in the real
world. A common question one might ask is how much product will be formed during the course
of a reaction based on the starting amount of reactants. Or, how much reactant must one begin with
to guarantee the production of a certain amount of a compound. Using the above example of the
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combustion of ethanol, we might ask if 100.0 grams of ethanol are burned how many grams of
carbon dioxide will be produced? Or, in order to produce 100.0 grams of water vapor, how many
grams of ethanol must be used?
The answers to questions of this sort rely on a technique known as stoichiometry, which has
been defined as the "calculation of the quantities of reactants and products involved in a chemical
reaction." I prefer to say that stoichiometry is the calculation of the quantities of reactants and
products based on their mole-to-mole relationships in a balanced chemical equation. We will see that
a balanced chemical equation is essential to stoichiometry, as is a knowledge of molar masses of the
compounds of interest.
The calculations in stoichiometry problems are based on the mole-to-mole relationships
derived from a balanced chemical equation. There is an equivalence between all of the reactants and
all of the products in every balanced chemical equation. These equivalences are used as conversion
factors to set up and solve stoichiometric questions.
From this balanced chemical equation above we obtain the following sets of equivalences.
We use the equal sign (=) to indicate equivalence. It does not mean that the two compounds are the
same.
•
•
•
•
•
•
1 mol C2H5OH(l) = 3 mol O2 (g)
3 mol O2 (g) = 2 mol CO2 (g)
2 mol CO2 (g) = 3 mol H2O(g)
1 mol C2H5OH(l) = 2 mol CO2 (g)
3 mol O2 (g) = 3 mol H2O(g)
1 mol C2H5OH(l) = 3 mol H2O(g)
These equivalences depend on the particular reaction and generally vary from one reaction
to the next. While there is a 2:3 relationship between carbon dioxide gas and water vapor in this
reaction, we found in the combustion of pentane (which is discussed above) that they existed in a
5:6 relationship to each other.
As I said, we can use these relationships as conversion factors. They are useful when
calculating amounts of reactants needed and/or amounts of product formed. Pay close attention to
the following examples. All are similar and yet each a little different from the next. Note that when
mass to mass conversions must be made that they always go through mole to mole conversions first.
They absolutely cannot ever be made directly. In other words, before we can do anything with a
mass given to us in grams (or any other unit of mass) we must first convert it to moles using the
molar mass of the substance.
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•
•
How many moles of ethanol (EtOH) must be burned to produce 16.7 moles of carbon
dioxide?
(16.7 mol CO2) x (1 mol EtOH/2 mol CO2) = 8.35 mol EtOH
The combustion of 2.78 moles of ethanol will produce how many moles of water vapor?
(2.78 mol EtOH) x (3 mol H2O/1 mol EtOH) = 8.34 mol H2O
•
How many moles of oxygen are required for the complete combustion of 33.6 moles of
ethanol?
(33.6 mol EtOH) x (3 mol O2/1 mol EtOH) = 100.8 mol O2
•
The combustion of ethanol produces 16.62 moles of carbon dioxide. How many moles of
water vapor are also produced?
(16.62 mol CO2) x (3 mol H2O/2 mol CO2) = 24.93 mol H2O
•
How many grams of ethanol must be burned to produce 125 grams of carbon dioxide?
(125 g CO2) x (1 mol CO2/44.01 g CO2) x (1 mol EtOH/2 mol CO2) x (46.07 g EtOH/1 mol
EtOH) = 65.4 g EtOH
•
How many grams of oxygen are required for the complete combustion of 1500 grams of
ethanol?
(1500 g EtOH) x (1 mol EtOH/46.07 g EtOH) x (3 mol O2/1 mol EtOH) x (32.0 g O2/1 mol O2) =
3125.7 grams of oxygen
•
The combustion of a certain amount of ethanol produces 6.5 milligrams of carbon dioxide.
How many milligrams of water vapor are also formed?
(6.5 mg CO2) x (1 g CO2/1000 mg CO2) x (1 mol CO2/44.01 g CO2) x (3 mol H2O/2 mol CO2) x
(18.02 g H2O/1 mol H2O) x (1000 mg H2O/1 g H2O) = 3.99 mg H2O
Theoretical yield and percent yield
The masses calculated in stoichiometry problems represent theoretical possibilities. In the
real world the actual amount of a substance produced might differ significantly from what is
predicted through calculations. There are a number of reasons for this ranging from unanticipated
chemical effects to equipment problems to physical errors made during the actual experiment. If,
for example, during the course of a reaction you accidentally spilled some of your product on the
lab bench, do you think your final results would exactly match the amount you predicted you would
have? What if the substance clings to the interior wall of the reaction vessel so that you cannot
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transfer all of it? What if not all of the reactants react? What if there are impurities in the reactants?
What if a product changes into another chemical before you have the opportunity to assess the
results of your experiment? And so on.
The amount of a chemical calculated using stoichiometry is called the theoretical yield. The
amount of chemical that is actually measured in the lab is called the actual yield. The ratio of the
actual yield to the theoretical yield, multiplied by 100, is called the percent yield.
(actual yield/theoretical yield) x 100 = percent yield
In the real world, chemists hope for high percent yields from their reactions. Of course a 100%
percent yield is ideal and cannot be surpassed unless mistakes have been made. There are a number
of ways to end up with a percent yield greater than 100% when doing an experiment, but none of them
is correct. Remember that mass is always conserved in chemical reactions and is never created or
destroyed. Actual yields are often much lower. Percent yields of 70-90% or better are common for
some types of reactions. For others, the percent yields might be much less. For some reactions,
especially those that take place in a series of steps, the overall percent yield might be only 10-15%
or less.
•
We calculated above that 65.4 g of EtOH must be burned to produce 125 g of CO2. If only
88.6 g of CO2 are produced, what is the percent yield of the reaction?
(88.6 g CO2/125 g CO2) x 100 = 70.9% percent yield
•
We calculated above that 3125.7 g of O2 are required for the complete combustion of 1500
g of EtOH. If the percent yield of CO2 in this reaction is only 65.0%, how many grams of
CO2 are produced?
since (actual yield/theoretical yield) x 100 = percent yield,
then actual yield = (percent yield x theoretical yield)/100
since we are not given the theoretical yield of carbon dioxide it must be calculated
(1500 g EtOH) x (1 mol EtOH/46.07 g EtOH) x (2 mol CO2/1 mol EtOH) x (44.01 g CO2/1 mol
CO2) = 2865.86 g CO2
(65.0 x 2865.86)/100 = 1862.8 g CO2
Strong electrolytes, weak electrolytes, and non-electrolytes
All compounds are classified as strong electrolytes, weak electrolytes, or non-electrolytes.
This is true of all ionic compounds and also of all covalent compounds.
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Electrolytes are substances that form aqueous solutions capable of conducting electricity. A
substance does this by dissociating (ionizing is a synonym) when in aqueous solution, which means
it breaks apart into ions when it dissolves in water. The ions serve as charge carriers. Pure water is
non-conductive unless it has ions in it. Of course pure water with ions in it is no longer pure water, is
it?
Strong electrolytes are substances that dissociate completely in aqueous solution. Nearly
100% of all of the molecules of a sample of strong electrolyte will dissociate in aqueous solution
if they dissolve. All ionic compounds are strong electrolytes, as are all strong acids and bases.
Weak electrolytes are substances that dissociate incompletely in aqueous solution. Typically
only1-10% of a sample of weak electrolytes will dissociate when they dissolve in water (although
weak electrolytes may also dissociate more than 10% or far less than 1%, depending on the
substance and on conditions). The most common weak electrolytes are the weak acids and bases,
which comprise a far more extensive list than that of strong acids and bases. Organic acids (also
known as carboxylic acids) are weak acids, and there are thousands of them. Ammonia and a class
of compounds known as amines are the most common weak bases.
Non-electrolytes do not ionize at all in aqueous solution, even though they may dissolve.
Nearly all organic compounds are non-electrolytes, with the exceptions noted in the previous
paragraph.
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strong electrolytes
general
acids
bases
weak electrolytes
non-electrolytes
nearly all organic
compounds except
organic acids and
bases
all ionic compounds
strong acids: HCl,
HBr, HI, HNO3,
H2SO4, HClO4
weak acids: HF,
H3PO4, H2CO3,
H2SO3, H2S, all
organic acids, usually
any acid not named
as a strong acid
strong bases: all
Group I and Group II
hydroxides
weak bases:
ammonia (NH3),
NH4OH, amines,
organic bases
(usually contain C,
H, N in molecular
formula), usually any
base not named as a
strong base,
This table is not absolutely correct but it does provide us with a very good set of generally accurate
guidelines that are more than adequate for a class at this level. You would be very well advised to commit
the contents of this table to memory. In fact, I’ll tell you right now: when you take exams that cover this
material, plan on including the information in this table on your 3" x 5" card. You’ll need it. And - I know
this would never happen, but just in case you may have forgotten about acids and bases: we talked about
them near the end of Chapter 4. Do yourself a favor and brush up.
Note that the distinction between strong and weak electrolytes lies in the extent to which they
dissociate. There is absolutely no correlation between an acid or base being strong or weak and how
dangerous it might be. Many of the most dangerous chemicals you may encounter in a chemistry
laboratory are weak acids and bases. I am confident that the number of annual injuries and deaths
caused by weak acids and bases far exceed those caused by strong acids and bases. In over 30 years
of chemistry the most serious injury I have received was caused by the mishandling of a weak acid. This
is also true for most of my friends who have been injured by acids or bases. The weak acids and bases can
be remarkably corrosive to flesh. Handle them with a great deal of care!
We should also note that dissolution, or, the process of dissolving, and dissociation are two
different things. Before a substance can dissociate it must first be able to dissolve. We will explain
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this in a later chapter. If a substance can not dissolve it cannot dissociate. Having said this, there are
many ionic compounds that do not dissolve in aqueous solution but which are still classed as strong
electrolytes because if they could dissolve they would also dissociate. But there are many
non-electrolytes that do dissolve in water but which still do not dissociate. Table sugar is a good
example. Molecules of sugar dissolve readily in water, but because of their chemical nature they do
not ionize at all. Compounds that dissolve in water are said to be soluble and are indicated with the
(aq) subscript in chemical equations. Compounds that do not dissolve in water are said to be
insoluble and are indicated with the (s) subscript in chemical equations. This information will be
valuable to you very shortly.
Molecular, ionic, and net ionic equations
We learned above that chemical equations are a shorthand way of describing a chemical
reaction. You should know that there are different types of chemical equations.
Molecular equations are equations in which all reactants and products are written as
complete molecules even though they may exist as ions in solution. In other words, in a molecular
equation all compounds are represented with their molecular formulas. To this point in this course
nearly all of the chemical equations you have seen in this class have been molecular equations.
Complete ionic equations, or, ionic equations as they are sometimes called, differ from
molecular equations in that strong electrolytes are written as ions if they are in aqueous solution.
Ionic equations reflect the real state of things in aqueous solution.
Net ionic equations are ionic equations in which the spectator ions are canceled out. The
actual reaction that takes place is what remains. A net ionic equation often represents what we would
see happening if we watched the reaction happen. Spectator ions are those ions which are needed
to balance the mass and charge of a reaction but which do not actually take place in the reaction.
Spectator ions are ions which are exactly the same in number and form on both the reactant side and
the product side of the equation.
Let's do some examples, beginning with the double displacement reaction of sodium chloride
and silver (I) nitrate.
•
molecular equation:
NaCl(aq) + AgNO3 (aq) => AgCl(s) + NaNO3 (aq)
Note that all four compounds are written in their molecular form. Further, note that we always begin
with a correctly balanced molecular equation, although this one was pretty easy to balance as all of
the coefficients are “one.”
•
ionic equation:
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Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) => AgCl(s) + Na+(aq) + NO3-(aq)
When writing ionic equations, in deciding whether to write compounds in their molecular form or
as ions we must answer yes to two questions: (1.) is the compound a strong electrolyte? and (2.) is
it soluble? If we answer yes to both questions the compound is written as ions. If we answer no to
one or both questions, the compound is represented with its molecular formula. As NaCl, AgNO3,
and NaNO3 are all strong electrolytes and are all also soluble (as represented by the (aq) subscript),
they are written in ionic form since, as strong electrolytes, this is the form in which they really exist
in aqueous solution. Soluble strong electrolytes don’t float around as intact molecules. They move
about as ions. While AgCl is also a strong electrolyte, it is insoluble and as such it is left in its
molecular form. How do we know that AgCl is insoluble? That's what the (s) subscript means. (s) does
not stand for soluble. It stands for solid, and compounds classed as solids are insoluble in aqueous solution.
But again, how do we know? This is what we’ll learn in the next section of this chapter.
•
net ionic equation:
Cl-(aq) + Ag+(aq) => AgCl(s)
In the ionic equation, Cl- and Ag+ combine to form solid AgCl while Na+ and NO3- do nothing in the
actual reaction. They are exactly the same in number and form on both the reactant side and the
product side of the equation. This makes them spectator ions, and as such they are canceled from
the ionic equation to leave the net ionic equation.
Does it matter which order the ions are written in? Would it also be correct to write
Ag+(aq) + Cl-(aq) => AgCl(s)
The answer is yes, this is also correct. The order of reactants and products in a chemical equation
does not matter as long as reactants are found on the reactant side and products are found on the
product side.
Now let's write the molecular, complete ionic, and net ionic equations for the
double-displacement reaction of barium nitrate and ammonium sulfide.
•
molecular equation:
Ba(NO3)2 (aq) + (NH4)2S(aq) => BaS(s) + 2 NH4NO3 (aq)
We start with a properly balanced equation.
•
ionic equation:
134
Ba2+(aq) + 2 NO3-(aq) + 2 NH4+(aq) + S2-(aq) => BaS(s) + 2 NH4+(aq) + 2 NO3-(aq)
Are Ba(NO3)2, (NH4)2S, and NH4NO3 strong electrolytes? Yes, they are ionic compounds. Are they soluble?
Yes, their (aq) subscript indicates that they are soluble. They are therefore represented as ions in an ionic
equation. While its true, BaS is a strong electrolyte, it is insoluble and so we leave it in its molecular form.
If it can’t dissolve neither can it ionize.
•
net ionic equation:
Ba2+(aq) + S2-(aq) => BaS(s)
In the ionic equation we notice that NO3-(aq) and NH4+(aq) occur in exactly the same number and form on
both the reactant side and the product side of the equation. They are therefore spectator ions and are
canceled, leaving us with our net ionic equation.
As a final example, let's write the molecular, complete ionic, and net ionic equations for the
double-displacement reaction of lithium sulfate and lead (II) acetate. See if you can do this one by
yourself and then check your answer below.
•
molecular equation:
Li2SO4 (aq) + Pb(C2H3O2)2 (aq) => PbSO4 (s) + 2 LiC2H3O2 (aq)
•
ionic equation:
2 Li+(aq) + SO42-(aq) + Pb2+(aq) + 2 C2H3O2-(aq) => PbSO4 (s) + 2 Li+(aq) + 2 C2H3O2-(aq)
•
net ionic equation:
SO42-(aq) + Pb2+(aq) => PbSO4 (s)
So, how did you do? Do you “get it,” or do you need more practice?
Predicting the likelihood of chemical reactions
We now have gained the ability to predict the outcome of two types of reactions, double
displacement reactions and the combustion of organic compounds. But, how do we know whether
or not what we predict will really happen?
Why do reactions occur when two or more chemicals are mixed together? For reactions to
take place there has to be more than simply a mixing of chemicals. There has to be a "driving force"
that makes the reaction happen. While the real explanation for why reactions do or do not occur lies
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in the study of thermodynamics and changes in the energy of the system, we can still look for clues
that serve as indicators that a reaction will occur.
There are four events that indicate that a reaction will take place as written. These are called
"driving forces" by the authors of older chemistry texts, but that is a bit over-enthusiastic. In fact, it’s
so over-enthusiastic that the phrase isn’t used much any more. But we’ll go ahead and use it anyway,
cautiously, understanding all the while its limitations. These things we’re about to discuss don’t actually
make the reaction happen, as such. It is more accurate to think of them as diagnostic clues or as
indicators that a reaction will occur. These "driving forces" are:
•
•
•
•
the formation of a solid (also known as a precipitate) in double displacement reactions
the formation of a pure liquid (always water in this class) in double displacement reactions
the formation of a gas in double displacement reactions
electron transfer in reactions that are not double displacement reactions
Electron transfer will *never* happen in a double displacement reaction you see in this class. At
the same time, *every* combustion reaction you see in this class will *always* be an electron
transfer reaction and will have none of the other three driving forces as they are *only* applicable
to double displacement reactions. Something to remember for the Chapter 6 quiz and MT-2.
We look for these “driving forces” on the product side of the chemical equation. If we see
that a reaction results in the formation of a solid, water, or a gas, or some combination of these three,
it is usually an accurate indication that the reaction is spontaneous. It is important to note that the
formation of precipitates, pure liquids, and gases are only valid "driving forces" i.e., accurate
indicators that a reaction is spontaneous when it is a double displacement reaction between two ionic
compounds in aqueous solution. If a reaction is not a reaction between two ionic compounds in
aqueous solution, even if we find a solid, liquid, or gas on the product side of a chemical equation,
we cannot use these to indicate that the reaction is spontaneous. Electron transfer reactions include
a great many common and important reactions in the real world and are not limited by the
constraints just enumerated. We’ll talk about how to tell if a reaction involves electron transfer
below. But first things first.
Precipitation reactions and solubility rules
The mixing of two aqueous solutions of ionic compounds may result in the formation of a
new insoluble ionic compound called a precipitate. Remember: insoluble means it will not dissolve in
water. This type of reaction is known generally as a precipitation reaction. The "driving force" in a
precipitation reaction is the formation of a precipitate, i.e., we form an insoluble solid on the product
side of the chemical equation.
We can predict the solubility of ionic compounds using a set of solubility rules. While these
rules are limited in that they only describe the solubility of certain ionic compounds in water, they
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still provide us with a surprisingly broad and accurate tool. Note: different text books state these rules
in slightly different ways, even though they’re mostly similar. Regardless of what your text says, I hold you
accountable for these rules in these notes only.
Rule 1: ionic compounds with Group I cations and ammonium ion are always soluble. This rule takes
precedence over all other rules. Regardless of anion, if an ionic compound has one of these cations, it will
*always* be soluble.
Rule 2: ionic compounds with acetate, nitrate, and perchlorate as anions are always soluble.
Rule 3: ionic compounds with halogens as anions are always soluble unless the cation is Ag+, Hg22+,
Hg2+, or Pb2+ Don’t forget these exceptions! And be sure you know and remember the differences between
mercury (I) ion and mercury (II) ion.
Rule 4: ionic compounds with sulfate as an anions are always soluble unless the cation is Ag+,
Hg22+, Hg2+, Pb2+, Ca2+, Sr2+, or Ba2+ These are the same exceptions as those for the halogen anions in
Rule 3 plus the addition of the larger Group 2 cations.
Rule 5: ionic compounds with carbonate, phosphate, sulfide, and hydroxide as anions are always
insoluble unless the cation is a Group 1 cation or ammonium ion or unless the compound is a strong
base - and as I’m sure you recall, earlier in this chapter we told you that the strong bases are the Group
1 and Group 2 hydroxides.
While I really do try to keep those things you should memorize to a minimum, you must be familiar with
and memorize these rules, and also those used to assign oxidation numbers, which we will discuss below.
I will of course provide you with all of this information on exams, but you really, really should know it.
Will a reaction occur if the aqueous solutions of the following ionic compounds are mixed?
Assume that if a reaction occurs it is a double-displacement reaction. Remember, the typical reactions
of two ionic compounds in aqueous solution is a double displacement reaction. And in this chapter, the only
other type of reaction we talk about is combustion reactions. So don’t make things harder for yourself than
is necessary. When I ask you about a reaction in this chapter, you’ve got a 50-50 chance of guessing what
type of reaction it will be, either double displacement or combustion. If you really haven’t picked up on the
differences between double displacement and combustion reactions, pull a coin out of your pocket and flip
it. You’ve going to be right 50% of the time simply by guessing and with a bit of luck. We begin by writing
the balanced molecular equation for the reaction. Then, using the solubility rules, we assign the state
of each reactant and product. If one or both of our products is a solid (insoluble), the reaction will
occur as written and it's "driving force" is the formation of a precipitate. A very important caveat:
if one or both of the reactants is insoluble, the reaction will not occur as written because the ions are
137
not free to mix and recombine as is required in a double displacement reaction. This, however, does
not prevent us from predicting the outcome of the reaction *if* it could occur.
!
Ammonium phosphate and iron (III) nitrate
"
write the balanced equation: (NH4)3PO4 + Fe(NO3)3 => FePO4 + 3 NH4NO3
"
is it soluble or solid? (NH4)3PO4 (aq) + Fe(NO3)3 (aq) => FePO4(s) + 3 NH4NO3 (aq)
"
the reaction does occur as written; the "driving force" is the formation of solid iron
(III) phosphate
!
Sodium iodide and mercury (II) acetate
"
write the balanced equation: 2 NaI + Hg(C2H3O2)2 => HgI2 + 2 NaC2H3O2
"
is it soluble or solid? 2 NaI(aq) + Hg(C2H3O2)2 (aq) => HgI2 (s) + 2 NaC2H3O2 (aq)
"
the reaction does occur as written; the "driving force" is the formation of solid
mercury (II) iodide
!
Ammonium iodide and potassium hydroxide
"
NH4I + KOH => KI + NH4OH
"
NH4I(aq) + KOH(aq) => KI(aq) + NH4OH(aq)
"
the reaction does not occur as written; there is no "driving force" as both of the
products are (aq) and not (s), (l), or (g)
!
Silver (I) acetate and cesium sulfate
"
2 AgC2H3O2 + Cs2SO4 => Ag2SO4 + 2 CsC2H3O2
"
2 AgC2H3O2 (aq) + Cs2SO4 (aq) => Ag2SO4 (s) + 2 CsC2H3O2 (aq)
"
the reaction does occur as written; the "driving force" is the formation of solid silver
(I) sulfate
!
Rubidium carbonate and chromium (III) chloride
"
3 Rb2CO3 + 2 CrCl3 => Cr2(CO3)3 + 6 RbCl
"
3 Rb2CO3 (aq) + 2 CrCl3 (aq) => Cr2(CO3)3 (s) + 6 RbCl(aq)
"
the reaction does occur as written; the "driving force" is the formation of solid
chromium (III) carbonate
!
Gold (III) carbonate and nickel (IV) sulfide
"
2 Au2(CO3)3 + 3 NiS2 => 2 Au2S3 + 3 Ni(CO3)2
"
2 Au2(CO3)3 (s) + 3 NiS2 (s) => 2 Au2S3 (s) + 3 Ni(CO3)2 (s)
"
the reaction does not occur as written; both reactants are solid and therefore cannot
react, even though we can (and do) predict what would form if the two reactants
weren’t both insoluble.
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Water formation in double-displacement reactions
Water can be formed in two common types of double displacement reactions, the reaction
of an acid with a base, or in the reaction of an acid with an ionic hydroxide compound. While most
of the acids we discuss in this class are covalent compounds, in aqueous solution they behave like
ionic compounds in that they do dissociate to some extent depending on whether they are strong
acids or weak acids.
We can define what acids and bases are in a variety of ways and have mentioned this
previously. You’ll find this in Chapter 4 as well as earlier in this chapter. We will discuss three different
acid-base theories in detail in Chapter 10, but for the time being we will content ourselves with the
definitions provided by the Arrhenius theory. In aqueous solution, an acid is a substance that donates
a hydrogen ion (H+), and a base is a substance that donates a hydroxide ion (OH-). All of the strong
and weak acids and all of the strong bases described in the electrolyte table we gave you earlier in
the chapter are Arrhenius acids and bases. Ammonia and amines are not Arrhenius bases, but we
won't worry about that until Chapter 10.
When Arrhenius acids and bases in aqueous solution are mixed, the resulting reaction is
called a neutralization reaction. It is a double-displacement reaction in which water is formed as a
pure liquid. The "driving force" in neutralization reactions is the formation of a pure liquid, water.
An ionic compound, known as a salt, is also formed.
Salts are a very large family of chemical compounds. By definition salts are ionic compounds
which *may* be formed as the result of neutralization reactions. All salts are ionic compounds;
however, not all ionic compounds are salts. Table salt, sodium chloride, is of course a member of
the salt family, but so are thousands of other ionic compounds. That a compound is classed as a salt
does not necessarily mean that it *was* formed by a neutralization reaction. It means that it *can*
be formed by a neutralization reaction.
The generic representation of a neutralization reaction looks something like this:
acid + base => H2O(l) + salt
An example of a neutralization reaction is the reaction of hydrochloric acid and sodium hydroxide:
•
•
•
molecular equation: HCl(aq) + NaOH(aq) => H2O(l) + NaCl(aq)
ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) => H2O(l) + Na+(aq) + Cl-(aq)
net ionic equation: H+(aq) + OH-(aq) => H2O(l)
Note in the molecular equation that the salt formed in this reaction is our old friend sodium chloride,
without whom french fries or popcorn would never taste the same. However, not all salts are this tasty.
Some are actually quite foul. There are even those that, when ingested, cause vomiting or more serious
illnesses, like - death.
139
Hydrochloric acid can donate a single proton to a neutralization reaction. We categorize all
acids, strong or weak, that can donate one hydrogen ion (i.e., one proton) to a reaction as monoprotic
acids.
Acids with two hydrogen ions available to participate in reactions are categorized as diprotic
acids. Sulfuric acid (H2SO4) is an example of a diprotic acid.
Acids with three available hydrogen ions are categorized as triprotic acids. Phosphoric acid
(H3PO4) is an example of a triprotic acid.
Let's look at examples of the neutralization reactions of diprotic and triprotic acids. When
sulfuric acid is mixed with potassium hydroxide, we observe the following:
•
•
•
molecular equation: H2SO4 (aq) + 2 KOH(aq) => 2 H2O(l) + K2SO4(aq)
ionic equation: 2 H+(aq) + SO42-(aq) + 2 K+(aq) + 2 OH-(aq) => 2 H2O(l) + 2 K+(aq) + SO42-(aq)
net ionic equation: 2 H+(aq) + 2 OH-(aq) => 2 H2O(l) *or* H+(aq) + OH-(aq) => H2O(l)
For net ionic equations it is common to reduce the coefficients of the equation to the lowest
common denominator. This is the *only* time is it permissible to do this. You *never* adjust the
coefficients in molecular equations in this class.
And when phosphoric acid reacts with calcium hydroxide:
•
•
•
molecular equation: 2 H3PO4 (aq) + 3 Ca(OH)2 (aq) => 6 H2O(l) + Ca3(PO4)2 (s)
ionic equation: 2 H3PO4 (aq) + 3 Ca2+(aq) + 6 OH-(aq) => 6 H2O(l) + Ca3(PO4)2 (s)
net ionic equation: 2 H3PO4 (aq) + 3 Ca2+(aq) + 6 OH-(aq) => 6 H2O(l) + Ca3(PO4)2 (s) there are no
spectator ions, so the ionic equation and the net ionic equation are the same.
You may be wondering why phosphoric acid is left in its molecular form in the ionic and net
ionic equations. Remember, we only write a compound as ions if it passes the two tests: (1.) is it a
strong electrolyte? and (2.) is it soluble? Phosphoric acid is a weak acid, which means it is also a
weak electrolyte. Since, in aqueous solution, far more phosphoric acid remains in its molecular form
than dissociated as ions, it is more correct to represent it in its molecular form. This is always true
when representing weak electrolytes in ionic and net ionic equations: leave them in their molecular
form.
A couple of things to notice with the above reaction. There are two driving forces in this
particular neutralization reaction, the formation of a pure liquid and the formation of a precipitate.
It is possible for one reaction to have multiple "driving forces." You should also pay attention to the
behavior of Group II bases, which can donate two hydroxide ions in a reaction.
Finally, let's consider the reaction of acetic acid, HC2H3O2, a weak acid, and copper (II)
hydroxide: How do we know phosphoric and acetic acids are weak acids? Very simply, because they're
140
not on the list of strong acids I gave you earlier. And in the case of acetic acid, it’s an organic acid; organic
acids are always weak acids.
•
•
•
molecular equation: HC2H3O2 (aq) + Cu(OH)2 (s) => H2O(l) + Cu(C2H3O2)2 (aq)
ionic equation: HC2H3O2 (aq) + Cu(OH)2 (s) => H2O(l) + Cu2+(aq) + 2 C2H3O2-(aq)
net ionic equation: HC2H3O2 (aq) + Cu(OH)2 (s) => H2O(l) + Cu2+(aq) + 2 C2H3O2-(aq)
Again there are no spectator ions, and the ionic and net ionic equations are the same.
It is important to note that in these double displacement reactions between an acid and a base
or an ionic hydroxide, it is possible for one of the reactants to be a solid and for a reaction to still
occur. The previous reaction is an example. Here’s another. Let's go to the kitchen and try an
experiment to prove this to ourselves. Pour a small amount of baking soda in a dish. Pour some
vinegar onto the baking soda. Does a reaction take place? Vinegar is a dilute aqueous solution of
acetic acid. Baking soda is solid sodium hydrogen carbonate (NaHCO3), a base. What you see when
you perform this experiment is proof that a reaction can occur between a solid base and a liquid acid.
These reactions can also occur liquid or aqueous bases and solid acids.
Gas formation in double-displacement reactions
Double displacement reactions in aqueous solution which result in gas formation only take
place under a very limited set of circumstances. There are many reactions other than those that we
will discuss in this section which may result in the formation of a gas, but nearly all of them are
electron transfer reactions and will be considered after this section. For gas formation to be used as
a “driving force” three conditions *must* be met:
•
•
•
double-displacement reactions of ionic compounds in aqueous solution
one of the reactants must be an acid
the anion of the other reactant must be either carbonate, sulfite, sulfide, or cyanide.
As I have said, the "driving force" in these reactions is the formation of a gas, although other
“driving forces” may also occur. Let's look at the behavior of each of these four anions in turn.
Consider the reaction of hydrobromic acid and sodium carbonate. It doesn't really matter which
acid we select. All acids will react in essentially the same way with carbonate ion, although there might be
slight differences in the resulting net ionic equations. At first glance it might seem that the molecular
equation should appear as
2 HBr(aq) + Na2CO3 (aq) => H2CO3 (aq) + 2 NaBr(aq)
141
but carbonic acid, H2CO3, is unstable when formed by double displacement reactions. It
spontaneously decomposes to form liquid water and carbon dioxide gas
H2CO3 (aq) => H2O(l) + CO2 (g)
The resulting equations appears as:
•
•
•
molecular equation: 2 HBr(aq) + Na2CO3 (aq) => H2O(l) + CO2 (g) + 2 NaBr(aq)
ionic equation: 2 H+(aq) + 2 Br-(aq) + 2 Na+(aq) + CO32-(aq) => H2O(l) + CO2 (g) + 2 Na+(aq) + 2 Br-(aq)
net ionic equation: 2 H+(aq) + CO32-(aq) => H2O(l) + CO2 (g)
Sulfite ion behaves similarly to carbonate ion when it reacts with an acid. The result is the
formation of sulfurous acid, H2SO3, which is also unstable when formed in aqueous solution as the
product of a double displacement reaction. Sulfurous acid generated in a double displacement
reaction will spontaneously decompose to form water and sulfur dioxide gas
H2SO3 (aq) => H2O(l) + SO2 (g)
Sulfur dioxide is a very sharp-smelling irritating gas used to preserve fruits and vegetables. It is produced
by the combustion of sulfur and is an important contributor to the production of acid rain, as when rain
water combines with atmospheric sulfur dioxide, sulfurous acid is formed. And unfortunately, when
sulfurous acid is formed this way it is stable. So it falls to the ground and attacks and corrodes stone and
kills living things such as trees. To illustrate the behavior of sulfite ion in the presence of an acid, let's
examine the reaction of perchloric acid and ammonium sulfite.
•
•
molecular equation: 2 HClO4 (aq) + (NH4)2SO3(aq) => H2O(l) + SO2 (g) + 2 NH4ClO4 (aq)
ionic equation: 2 H+(aq) + 2 ClO4-(aq) + 2 NH4+(aq) + SO32-(aq) => H2O(l) + SO2 (g) + 2 NH4+(aq)
+ 2 ClO4-(aq)
•
net ionic equation: 2 H+(aq) + SO32-(aq) => H2O(l) + SO2 (g)
Sulfide ion, S2-, reacts directly with hydrogen ion in double displacement reactions to form
hydrogen sulfide gas, H2S(g). I realize that, as a covalent compound, the correct name of H2S should be
dihydrogen sulfide, but it's historic name, while systematically incorrect, is the accepted name of the
compound. Hydrogen sulfide is also known as rotten egg gas. You often smell it when passing oil refineries
and around geothermal features such as geysers and hot springs. Our noses are extremely sensitive to it
and can detect it in concentrations of around 1 ppb (part per billion!) or less. At levels of 10-50 ppm (parts
per million), it can cause serious injury and death. This is a serious problem on oil rigs when pockets of
hydrogen sulfide gas are hit while drilling for petroleum or natural gas. Hydrogen sulfide gas is naturally
produced by the decay of dead plant and animal life, as are oil and natural gas. Fortunately, the stench
of hydrogen sulfide is so intolerable at toxic concentrations that most people don't wait around to be injured
by it. One very nasty attribute of the gas is its ability to induce nasal tetany, paralysis of the olfactory
nerves. Sometimes people exposed to it get one good sniff, lose their sense of smell, think that the gas has
142
disappeared, and continue to work in a dangerous environment until they collapse and die. You are highly
unlikely to ever be in this position unless you work on an oil rig.
When, for example, sulfuric acid reacts with cesium sulfide, we observe the following:
•
•
•
molecular equation: H2SO4 (aq) + Cs2S(aq) => H2S(g) + Cs2SO4 (aq)
ionic equation: 2 H+(aq) + 2 SO42-(aq) + 2 Cs+(aq) + S2-(aq) => H2S(g) + 2 Cs+(aq) + SO42-(aq)
net ionic equation: 2 H+(aq) + S2-(aq) => H2S(g)
Cyanide ion behaves like sulfide ion in that it reacts directly with hydrogen ion in double
displacement reactions to form hydrogen cyanide gas, HCN(g). This is another very nasty and
dangerous gas. It inhibits respiration at a cellular level. This is, in fact, the gas that was used in the gas
chamber to execute people. There is an interesting article on gas chambers at Wikipedia, if you’re
interested. But it is not required reading. Hydrogen cyanide gas is also naturally produced by the
combustion of organic compounds that also contain nitrogen - and many do. Hydrogen cyanide occurs in
very small concentrations in tobacco smoke, and also in camp fire smoke. But as long as you don’t inhale
too much you don’t have anything to worry about.
When, for example, hydrochloric acid reacts with potassium cyanide, we observe the following:
•
•
•
molecular equation: HCl (aq) + KCN(s) => HCN(g) + KCl(aq)
ionic equation: H+(aq) + Cl-(aq) + K+(aq) + CN-(aq) => HCN(g) + K+(aq) + Cl-(aq)
net ionic equation: H+(aq)+ CN-(aq) => HCN(g)
As is the case with neutralization reactions, it is possible for one of the reactants to be a solid
and for a gas-forming reaction to occur. We see this in the previous example. And remember our
example of the reaction of baking soda and vinegar? The bubbles formed are carbon dioxide gas.
In this case the acetic acid reacts with sodium hydrogen carbonate according to the equation
HC2H3O2 (aq) + NaHCO3 (aq) => H2O(l) + CO2 (g) + NaC2H3O2 (aq)
Electron transfer (redox) reactions
Electron transfer reactions are chemical reactions in which one or more electrons is
transferred from one atom to another. Electron transfer reactions are also known as
oxidation-reduction reactions, or as redox reactions. pronounced REE-dox The transfer of electrons
serves as the "driving force" for these types of reactions.
In the real world, redox reactions are far more common than double displacement reactions
which result in precipitation, neutralization, or gas formation. Electron transfer may result in the
formation of a solid, a pure liquid, or a gas, but these are not the "driving forces" for the reaction.
It is the transfer of electrons from one atom to another. Redox reactions are not constrained to occur
143
in aqueous solution, nor do the reactants have to be ionic compounds. Electron transfer lies at the
heart of many important processes, from respiration and photosynthesis and the reactions in batteries
to corrosion and combustion and making things pretty and shiny by applying a thin layer of metal
coating to them.
Let's begin with some definitions.
•
•
•
•
Oxidation (oxidized): the loss of one or more electrons
Reduction (reduced): the gain of one or more electrons oxidation and reduction are
inextricably tied to each other. One never occurs without the other also taking place in the
same reaction. Absolutely never!
Oxidizing agent: a chemical that oxidizes something else and reduces itself
Reducing agent: a chemical that reduces something else and oxidizes itself
A commonly used mnemonic device for remembering oxidation and reduction is "OIL RIG" which
stands for “oxidation is loss, reduction is gain.” Of course, we’re talking about oxidation being the
loss of electrons, and reduction is the gain of electrons.
How can we tell when electron transfer occurs? By using a concept known as oxidation
numbers. Oxidation numbers are a theoretical book-keeping method devised as a way of keeping
track of electrons in reactions. With a specific set of rules we assign an oxidation number to every
atom in every substance in the reaction. If the oxidation number of the same atom is different on the
product side than it was on the reactant side, then it has either been oxidized, if the atom has lost
electrons, or reduced, if the atom has gained electrons.
Rules for determining oxidation numbers:
Rule 1: the oxidation number of atoms in their elemental state is zero.
Rule 2: the oxidation number of a monatomic ion is equal to its charge.
Rule 3: the oxidation number of oxygen is always equal to -2 unless in it's molecular form (see
Rule 1) or in a peroxide (we will not discuss peroxides in this class, but O in peroxides has an
oxidation number of -1).
Rule 4: the oxidation number of hydrogen is always +1 unless in it's molecular form (see Rule 1)
or in a hydride (we will not discuss hydrides in this class, but H in hydrides has an oxidation number
of -1).
Rule 5: Fluorine always has an oxidation number of -1. The other halogens always have an oxidation
of -1 as anions in ionic compounds or as the second named atom in binary molecular compounds.
Halogens listed as the first member of a binary molecular compound or involved in oxyanions have
positive oxidation numbers and must be determined using Rule 6 below.
144
Rule 6: for either a neutral compound or for any polyatomic ion, the sum of the oxidation numbers
of the atoms in the molecule is equal to the net charge of the compound or ion.
You must be familiar with and able to use these rules. As with the solubility rules, I will provide them for
you on exams.
Let's practice assigning oxidation numbers to a few compounds before we use it in equations.
!
NO3"
"
"
"
"
"
!
!
NO2"
"
"
"
"
"
H2SO4
"
"
"
"
"
"
"
!
oxygen is always -2, and there are three of them
the net charge on the molecule is -1
we do not have a rule specifically for nitrogen, but we can figure out it's oxidation
number using Rule 6
assume the oxidation number of nitrogen is X
using Rule 6: X + (3)(-2) = -1
X = +5, meaning the oxidation number of the nitrogen atom in nitrate ion is +5
oxygen is always -2, and there are two of them
the net charge on the molecule is -1
we do not have a rule specifically for nitrogen, but we can figure out it's oxidation
number using Rule 6. That the nitrogen atom in nitrate ion has an oxidation number of
+5 doesn’t mean a thing here. We have to figure it out. Perhaps it will be same. Perhaps
not. Let’s see.
assume the oxidation number of nitrogen is X
using Rule 6: X + (2)(-2) = -1
X = +3, meaning the oxidation number of the nitrogen atom in nitrite ion is +3
oxygen is always -2, and there are four of them
hydrogen is always +1, and there are two of them
the molecule is neutral, i.e., there is no net charge
we do not have a rule specifically for sulfur, but we can figure out it's oxidation
number using Rule 6
assume the oxidation number of sulfur is X
using Rule 6: (2)(+1) + X + (4)(-2) = 0
X = +6, meaning the oxidation number of the sulfur atom is +6
Fe(OH)2
"
oxygen is always -2, and there are two of them
"
hydrogen is always +1, and there are two of them
"
the molecule is neutral, i.e., there is no net charge
145
"
"
"
"
we do not have a rule specifically for iron, but we can figure out it's oxidation
number using Rule 6
assume the oxidation number of iron is X
using Rule 6: X + (2)(+1) + (2)(-2) = 0
X = +2, meaning the oxidation number of the iron atom is +2
Alternatively, remember that this is an ionic compound. When the compound dissociates it forms
Fe2+ ion and two OH- ions, but we already know how to figure out the oxidation numbers of
hydrogen and oxygen According to Rule 2, the oxidation number of the iron cation is equal to it's
charge, +2. This works for all ionic compounds and in fact it is often simpler to break an ionic
compound down into its component ions before assigning oxidation numbers to its atoms.
!
Li3PO4
"
oxygen is always -2, and there are four of them
"
the molecule is neutral, i.e., there is no net charge
"
we do not have rules specifically for lithium or phosphorus, but we can figure out
their oxidation numbers using Rule 6 and by observing that this is an ionic
compound consisting of lithium ion and phosphate ion
"
the cation is Li+, so the oxidation number of lithium is +1
"
assume the oxidation number of phosphorus is X
"
the net charge on phosphate is -3
"
using Rule 6: X + (4)(-2) = -3
"
X = +5, meaning the oxidation number of the phosphorus atom is +5
!
HClO3
"
"
"
"
"
"
"
!
oxygen is always -2, and there are three of them
hydrogen is always +1, and there is one of them
the molecule is neutral, i.e., there is no net charge
we do not have a rule specifically for chlorine in a molecule made up of three atoms,
but we can figure out it's oxidation number using Rule 6
assume the oxidation number of chlorine is X
using Rule 6: (1)(+1) + X + (3)(-2) = 0
X = +5, meaning the oxidation number of the chlorine atom is +5
W2(SO3)3
"
the molecule is neutral, i.e., there is no net charge
"
we do not have rules specifically for tungsten or sulfur, but we can figure out their
oxidation numbers using Rule 6 and by observing that this is an ionic compound
consisting of a tungsten ion and sulfite ion
"
sulfite ion has a 2- charge, and since there are three of them the tungsten ion must be
W3+, which tells us that its oxidation number is also 3+
"
the net charge on sulfite is -2
"
oxygen is always -2, and there are three of them in a single sulfite ion
"
assume the oxidation number of sulfur is X
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"
"
using Rule 6: X + (3)(-2) = -2
X = +4, meaning the oxidation number of the sulfur atom is +4
It is important to remember the distinction between charge and oxidation number. Charge
is real and is caused by the loss or gain of electrons. Oxidation number is imaginary and is a device
used to help track the movements of electrons from one atom to another in redox reactions. This is
how it is possible to find positive and negative oxidation numbers for the atoms in covalent
compounds. We know that atoms are electrically neutral in covalent compounds, but they can still
participate in electron transfer. Sometimes the charge and the oxidation number of an atom are
numerically the same, but do not forget that charge and oxidation number are conceptually two
different things.
There are many different electron transfer reactions in the world around us, but there are four
common classes of reactions that are often - not always, but often - "driven" by electron transfer.
Prove to yourself that each of the example reactions is in fact an electron transfer reaction by
assigning oxidation numbers to all of the atoms in all of the compounds and observing which atoms
are oxidized and which atoms are reduced.
Combination reactions are those in which two smaller substances combine to form a third
larger substance. An example of a combination reaction "driven" by electron transfer is the reaction
of sodium and chlorine to form sodium chloride:
2 Na(s) + Cl2 (g) => 2 NaCl(s)
The oxidation numbers of sodium metal and chlorine gas are 0, since both are in their elemental
state. The oxidation number of sodium ion is +1 and the oxidation number of chloride ion is -1. In
this reaction sodium is oxidized and chlorine is reduced. Sodium is the reducing agent and chlorine
gas is the oxidizing agent. Note that oxidizing agents and reducing agents are only ever found on the
reactant side of the equation.
In decomposition reactions a single (relatively) larger compound reacts to form two or more
(relatively) smaller substances. The formation of hydrogen and oxygen from water is an example
of a decomposition reaction "driven" by electron transfer:
2 H2O(l) => 2 H2 (g) + O2 (g)
The oxidation numbers of the hydrogen and oxygen in water are +1 and -2 respectively. The
oxidation numbers of hydrogen gas and oxygen gas are both 0, since each is in it's elemental state.
In this reaction hydrogen is reduced and oxygen is oxidized, and water is both the oxidizing agent
and the reducing agent.
Displacement reactions, or single displacement reactions, are reactions in which an element
reacts with a compound and replaces an element in the compound. They have the general form
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AB + C => A + BC
The reaction of chlorine and hydrogen sulfide is a single displacement reaction in which oxidation
and reduction occur:
8 H2S(g) + 8 Cl2 (g) => 16 HCl(g) + S8 (s)
In hydrogen sulfide the oxidation number of hydrogen is +1 and of sulfur is -2, while the oxidation
number of chlorine gas is 0 since it is in it's elemental state. The oxidation number of hydrogen in
hydrogen chloride gas is +1 while that of chloride ion in -1. The oxidation number of sulfur in S8
is 0, since this is the elemental state of sulfur (yes, we did mention this, but it was a long time ago.
Check your Chapter 1 notes). Sulfur is oxidized and chlorine is reduced. Chlorine gas is the
oxidizing agent and hydrogen sulfide is the reducing agent.
We've already learned a little bit about combustion reactions, in which a substance reacts
with oxygen, resulting in a rapid release of heat and light energy. While many inorganic elements
and compounds can be oxidized, we have learned specifically about organic oxidation reactions, in
which carbon dioxide gas and water vapor are formed. Oxygen is reduced in organic combustion
reactions while carbon is oxidized. Take, for example, the combustion of methane (CH4):
CH4 (g) + 2 O2 (g) => CO2 (g) + 2 H2O(g)
The carbon atom in methane has an oxidation number of -4 while the hydrogen in methane has an
oxidation number of +1. The oxidation number of oxygen gas is 0 since it is in it's elemental state.
The oxidation number of carbon in carbon dioxide is +4 while that of oxygen is -2. The oxidation
numbers of the hydrogen and oxygen in water are +1 and -2 respectively. Carbon is oxidized in this
reaction and oxygen is reduced. Oxygen gas is the oxidizing agent and methane is the reducing
agent.
Post-script: responses to student questions about Chapter 6 concepts
Below are five notes I have sent to students over the years in response to material in this
chapter. While the questions were asked quite some time ago, there is useful information in these
notes for every student taking this class, no matter how well they may think they know this material.
*************************
Letter 1: this first note addresses a question about balancing equations.
hi,
I would follow the sequence of steps suggested in the chapter 6 notes.
When balancing a double displacement reaction:
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Step 1: write the molecular formulas for the reactant compounds
Step 2: write the cations and anions in each reactant compound beneath them, since this will help
predict the products that are formed
Step 3: the charges of the ions determine the ratios in which they combine to form product
molecules; write the formulas of the product compounds
Step 4: place the products on the right side of the arrow
Step 5: use coefficients to balance the equation
There is one example of how to employ these steps in the lecture notes. Here's another.
Provide a balanced molecular formula for the double displacement reaction of barium hydroxide and
phosphoric acid.
Step 1: Ba(OH)2 and H3PO4
Step 2: Ba(OH)2 consists of Ba2+ and OH-; H3PO4 consists of H+ and PO43Step 3: Ba2+ and PO43- combine to form Ba3(PO4)2; H+ and OH- combine to form HOH, or H2O as
it is correctly written
Step 4: Ba(OH)2 + H3PO4 -> Ba3(PO4)2 + H2O
Step 5: 3 Ba(OH)2 + 2 H3PO4 -> Ba3(PO4)2 + 6 H2O
*************************
Letter 2: this second note is in response to a question about driving forces.
hi,
Although this is only a first semester chemistry class, with a modest amount of effort you can learn
to predict (a.) the outcome of chemical reactions, and (b.) whether or not the outcome will actually
happen.
The outcome of a reaction can be predicted theoretically if you know it is a double displacement
reaction or a combustion reaction. These are discussed in the Chapter 6 notes under the heading
"Balancing chemical equations." There are other types of reactions besides double displacement and
combustion reactions, but we will not discuss them in this class and you are not responsible for
them.
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Once you have learned how to predict the outcome of a reaction, you can also predict whether or
not the reaction will actually take place as written, i.e., whether or not the reaction will occur
"spontaneously." This is described in the Chapter 6 lecture notes beneath the heading "Predicting
the likelihood of chemical reactions." In the reaction of two ionic compounds in aqueous solution,
if a precipitate, pure liquid, or a gas is formed, the reaction will take place spontaneously as written.
These three events are sometimes referred to as "driving forces." It is more accurate to think of them
as symptoms or indicators of spontaneity. Electron transfer is also a "driving force," although it is
never a driving force in the double displacement reactions we see in this class.
On the chapter 6 quiz you are asked trios of questions about various reactions. The wording for the
second question in each trio is:
The driving force(s) (i.e., the indication(s) that the reaction is spontaneous) for the reaction in
Problem 1 is:
I. Formation of a precipitate
ii. Formation of a pure liquid
iii. Formation of a gas
iv. Electron transfer
Unfortunately while the information is there these options have not formatted correctly when
transferred to Vista. I apologize for any confusion that may have resulted.
*************************
Letter 3: in this third note I discuss how to use solubility rules.
hi,
regarding the reaction between lithium sulfate and lead (II) acetate on page 135 of the lecture notes.
You’ve asked how we can tell whether or not each of the reactant and product compounds is soluble.
The solubilities are determined by the solubility rules. Although these are general rules, they're still
quite useful. We need to look at each of the four compounds in the reaction and compare it to what
the solubility rules tell us. Note that these rules depend on the particular compound and are
independent of the coefficient in the balanced equation or even the reaction. In other words,
according to the rules, lead (II) sulfate will *always* be a solid in aqueous solution, regardless of
the reaction, regardless of whether it's a reactant or a product, regardless of its coefficient in the
balanced equation.
Ok, according to the rules:
Li2SO4 - according to Rule 1, compounds with a group 1 cation are always soluble in water. If we
check the periodic table, we see that Li+ is a group 1 cation. So this compound receives an (aq)
subscript.
Pb(C2H3O2)2 - according to Rule 2, compounds with acetate as an anion are always soluble in
water, regardless of the cation. This compound also receives an (aq) subscript.
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PbSO4 - according to Rule 4, compounds with sulfate as an anion are always soluble in water
*except* when the cation is Ag^+, Pb^2+, Hg2^2+, Hg^2+, Ca^2+, Sr^2+, or Ba^2+. Sulfate
compounds with one of these seven cations are insoluble in water and receive an (s) subscript. Hence
PbSO4 is labeled (s).
LiC2H3O2 - this compound has a rule 1 cation and is therefore (aq). It also has a rule 2 anion, which
also means it's (aq).
Remember, you *don't* need to memorize the rules. I'll give you a copy on the exam. You just need
to know how and when to use them.
*************************
Letter 4: in this fourth note I attempt to address a question about the essential concepts in Chapter
6.
hi,
It is important to master this material. You will see several sets of questions exactly like these Quiz
6 questions on the midterm and on the final as well. If you're going to make mistakes - and nearly
everyone does on this material - it's better to make them on the quiz where it doesn't count against
your grade as much as it does when an exam question is missed.
Since you have asked general questions, I can only respond in a general manner. But it's a starting
point.
First, there are a number of concepts addressed in Chapter 6. This is part of what makes it such a
challenging chapter. Overall, the two most important concepts are those of stoichiometry and
predicting the outcome of chemical reactions. But there are other concepts such as the mole concept,
strong and weak electrolytes, different types of chemical equations, and etc. that are also important.
Despite how it may seem, I have only asked you about two types of reactions, combustion reactions
and double displacement reactions. Don't make things harder for yourself than it already is. If a
reaction you are asked about is *not* a combustion reaction, it *must* be a double displacement
reaction. Yes, it is as simple as that.
Double displacement reactions take place between two ionic compounds in aqueous solution. The
acid-base reactions described in the notes are all double displacement reactions. The bases are ionic
compounds. The acids are most commonly covalent compounds *but* acids *always* behave as
though they are ionic compounds in that they ionize to a greater or lesser extent in aqueous solution.
You absolutely *must* know the Chapter 4 nomenclature stuff - as has been pointed out by others
- or you haven't got a prayer at doing these problems. That is in part why I choose to emphasize this
material, as well as for its relevance in its own right.
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I would strongly suggest that anyone having trouble with the Chapter 6 quiz to go back, as a starting
point, to my lecture notes and *very* *carefully* re-read the sections entitled "Strong electrolytes,
weak electrolytes, and non-electrolytes," "Molecular, ionic, and net ionic equations," "Predicting
the likelihood of chemical reactions," " Precipitation reactions and solubility rules," "Acids, bases,
and neutralization reactions," and "Gas formation." There is no fluff here. You *must* know
*everything* in these sections to do these problems. If you skip anything, even a seemingly small
detail, it will probably come back to cause trouble. As a part of your study of these sections, you
really *must* work the examples I provide. Don't just read them. That will not help. You must sit
down with a paper and pencil and work through each and every example with a great deal of care.
I don't know if this is helpful or not. This is exactly the same information I provide my lecture
classes, and these are exactly the same instructions I give my classroom students. This material is
extremely demanding and the only way to master it is to roll up your sleeves and get dirty with it.
I know it's hard but don't get discouraged. Every semester students who persevere eventually figure
it out. Remember that the tutors in the Student Resource Center (SI 359) may not answer quiz
questions but they are permitted to help you master the concepts. Please continue to post notes here
on the discussion board when you get stuck. And many, many thanks to those of you who read these
notes and who respond with helpful hints of your own.
*************************
Letter 5: in this fifth note I discuss redox reactions.
hi,
to determine whether or not a reaction is an electron transfer reaction you must assign an oxidation
number to every atom in every substance in the reaction and see if they change as an atom passes
from the reactant side to the product side of the equation. If none of the atoms have oxidation
numbers that change, it is not a redox reaction. If any of the atoms have oxidation numbers that
change, it is a redox reaction.
Oxidation is the loss of electrons. If an oxidation number becomes more positive, oxidation has
taken place. If an oxidation number becomes less negative, oxidation has taken place. As an
example, if in a reaction hydrogen has an oxidation number of 0 on the reactant side and an
oxidation number of +1 on the product side, this means that hydrogen has been oxidized, i.e., it has
lost electrons.
Reduction is the gain of electrons. If an oxidation number becomes more negative, reduction has
taken place. If an oxidation number becomes less positive, reduction has taken place. As an example,
if in a reaction chlorine has an oxidation number of 0 on the reactant side and an oxidation number
of -1 on the product side, this means that chlorine has been reduced, i.e., it has gained electrons.
Oxidation and reduction are a paired event. You *never* have one without the other.
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The word "agent" conveys a sense of what one compound does to another. An oxidizing agent
oxidizes something else and in the process reduces itself. A reducing agent reduces something else
and in the process oxidizes itself. The oxidizing agent is the entire compound that contains the atom
that is reduced in the reaction. The reducing agent is the entire compound that contains the atom that
is oxidizes in the reaction. Oxidizing and reducing agents are always on the reactant side of the
equation. Products are *never* oxidizing or reducing agents - at least, not in this class.
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Chapter 7: Chemical Reactions - Energy, Rates, and
Equilibrium
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as spontaneous, non-spontaneous, thermodynamics, kinetics,
energy, work, heat, Joule, calorie, potential energy, kinetic energy, chemical potential
energy, conservation of energy, system, surroundings, internal energy, enthalpy, heat of
reaction, exothermic, endothermic, intensive properties, extensive properties,
thermochemical equation, entropy, Gibbs free energy, reaction rate, rate law, activation
energy, activated complex, transition state, catalyst, adsorp, enzymes, "lock and key"
analogy, reversibility of chemical reactions, chemical equilibrium, equilibrium constant,
equilibrium constant expression, Kc, Kp, product-favored reactions, reactant-favored
reactions, LeChatelier's principle.
Be able to describe the relationship between enthalpy, entropy, temperature, Gibbs free
energy, and the spontaneity or non-spontaneity of any reaction.
Describe the factors that affect the rate of a chemical reaction, particularly the relationship
between temperature and reaction rate and why temperature has the effect it does.
Describe the effect of a catalyst on the rates of both forward and reverse reactions.
Given the necessary information, be able to write the equilibrium constant expression for a
reaction and calculate the equilibrium constant for the reaction.
Explain the relationship between the equilibrium constant for a reaction and whether it is
product-favored or reactant-favored.
Use LeChatelier's principle to explain how a system at equilibrium will shift as a result of
changes in concentration or temperature.
An introduction to thermodynamics and kinetics
In Chapter 6 we discussed "driving forces," behaviors that help us predict whether or not a
chemical reaction will spontaneously occur as it is written in a chemical equation. As we mentioned
at the time, these "driving forces" may be valuable as indicators but are often ultimately of limited
usefulness. Do you remember what limitations we face when relying on the Chapter 6 "driving forces" to
predict whether or not a reaction will occur? If not, you need to review. In reality, it is the changes in
energy and molecular orderliness that occur during a chemical reaction that determine whether or
not the reaction occurs naturally.
Naturally occurring reactions are said to be spontaneous. Reactions that do not occur
naturally are said to be non-spontaneous. The double displacement reactions of ionic compounds
in aqueous solution in which there is the formation of a precipitate, a pure liquid, or a gas are usually
spontaneous reactions. Many of the common reactions in which electron transfer takes place are also
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spontaneous reactions. How can we reliably and consistently predict whether or not a reaction is
spontaneous without relying on "driving forces"?
Thermodynamics is the branch of chemistry in which we study the energetic feasibility of
chemical reactions. Thermodynamics is defined as the study of the transformations of energy,
especially the transformation of heat into work and work into heat. By observing whether a chemical
reaction emits or absorbs energy, and whether it results in the formation of substances that are more
orderly at a molecular level (such as solids) or less orderly (such as gases) we can predict with
perfect correctness whether the reaction will be spontaneous or non-spontaneous. We also turn to
thermodynamics to help us predict how much energy will be released by a spontaneous reaction, and
how much energy we will need to put into a nonspontaneous reaction to make it happen.
In addition to predicting whether or not reactions are spontaneous or non-spontaneous it is
often important to have some idea of the time frame of a reaction. Comparatively speaking, different
reactions occur at different rates. Some reactions are fast, and some are slow. Whether a reaction is
spontaneous or non-spontaneous does not have any apparent correlation with how fast or slow it
might be. Some reactions, such as those that generate electricity in batteries or that take place during
combustion, are both spontaneous and fast. Some of the chemical reactions that occur during an
explosion might take place on a picosecond or even a femtosecond time scale. Remember that a
picosecond is a trillionth of a second and a femtosecond is a thousand-trillionth of a second. Other
reactions, such as the oxidation of iron or the weathering of rocks and other geological processes
are spontaneous reactions that might take place over days, weeks, months, or even years. Kinetics
is the branch of chemistry dedicated to the study of the rates at which reactions occur and those
factors that affect the rates of reactions.
Energy, work, and other thermodynamic definitions
One of the quirky aspects of thermodynamics is that words we use one way in our everyday
communications sometimes have a somewhat different and very precise meaning in
thermodynamics. The study of thermodynamics began during the early 19th century as scientists
attempted to make steam engines perform more efficiently. Surprisingly, their work on steam
engines is both relevant and accurate in describing events that take place in a collection of atoms or
molecules. It sometimes helps to keep this history in mind as we encounter the definitions of
thermodynamics terms.
In thermodynamics we divide the universe into two regions, the system and the surroundings.
The system is the part of the world in which we have an interest. The system is what we measure
or observe. We make our measurements or observations from the surroundings. In chemistry the
system might be an event taking place in a test tube or a beaker. A system can be as small as a cell
or bacterium (or even smaller), or as large as a planet or star or galaxy (or even larger). The system
in chemistry is where ever the reaction of interest is taking place, however big or small that reaction
might be. The surroundings are everything else. In essence, the surroundings are the rest of the
universe.
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We also need to say a word or two on sign conventions in thermodynamics. If a quantity such
as heat or work is transferred from the system to its surroundings, the quantity is given a negative
(-) sign. If, on the other hand, a thermodynamic property such as work or heat is transferred from
the surroundings to the system, it is given a positive (+) sign.
from system to surrounds: (-)
from surroundings to system: (+)
As a part of our discussion of thermodynamics we must spend a few moments talking about
energy, work, and heat.
Energy (commonly represented with the symbol “E”) is defined as the capacity to do work.
Work (represented with the symbol “w”) is defined as a transfer of energy that can be used
to the change the height of a weight somewhere in the surroundings. Remember that steam engines
were not only used to propel trains and boats. They were also used to lift coal out of mines in northern
England and Wales and to help drain water in mines that would otherwise have flooded. While we typically
could care less about the height of weights, scientists trying to improve the efficiencies of steam engines
found this a compelling topic.
Heat (represented with the symbol “q”) is defined as a transfer of energy as a result of a
temperature difference between the system and the surroundings.
Energy is measured in Joules (J) in the SI system and in calories (cal) in the British system.
The relationship between Joules and calories is 4.184 J = 1 cal. Dietary calories are actually equal to
1,000 calories as we discuss them in this chapter. I'm not entirely sure as to why things are done this way,
but can you imagine the shock and panic you might experience when you discover that the Snickers bar
you just ate contained 280,000 calories? Perhaps there is some merit to doing things this way after all.
Energy governs everything that happens in chemistry. The lower the energy of a system, the
greater its stability. The higher the energy of a system, the greater its instability. If events result in
the system having a lower energy at the end than it did in the beginning, the process will be
spontaneous. Events that result in the system having more energy in the end than it started with are
nonspontaneous. This is true both of chemical reactions and also of physical changes.
There are two types of energy, potential energy and kinetic energy.
Potential energy is stored energy based on position. Most commonly we think of potential
energy in terms of gravitational potential energy. If we pick an object up and hold it above the
surface upon which it rested, we have given it gravitational potential energy. If we release the object,
gravity will pull it back to the surface as the object's potential energy is converted to kinetic energy.
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This occurs spontaneously not only because gravity is acting on the object, but also because the
object returns to a lower energy state as it falls. While gravity plays an essential role in the behavior
of large objects, it has no effect on the outcome of chemical reactions. This is true as far as we know.
However, recent experiments on the Space Shuttle and in the International Space Station suggest that
gravity may in fact play a small role in the chemical behavior of atoms and molecules that we do not yet
understand. Stay tuned over the next few decades.
In chemistry we are interested in chemical potential energy. This is the energy stored in
chemical bonds, which is based on the strength of the attractive interactions between the bonding
atoms, which in turn is related to their charges and to their positions with respect to each other.
Remember the equation that describes bonding forces given in the "Chemical bonds and chemical
compounds" section of Chapter 4 to understand this statement. According to this equation the closer two
oppositely charged particles are to each other the stronger the attractive interactions between them. The
further two oppositely charged particles are from each other, the weaker the attractive interactions they
experience.
Kinetic energy is energy associated with motion and movement. We are most familiar with
translational motion, which is the energy of any object moving in a straight line from one point in
space to another. Translational motion is possible in the gaseous and liquid states but not in the solid
state, as particles are more or less locked into place.
There are also three other important types of motion that may impart kinetic energy to atoms
and molecules.
Bonded atoms vibrate with respect to each other in molecules. The behavior of two bonded
atoms is similar to that of two masses at opposite ends of a spring. Vibrations are a source of kinetic
energy to molecules in all three states of matter.
Molecules in the liquid and gas phases are free to rotate, or tumble, as they move. There is
kinetic energy associated with rotational motion.
And, within molecules, atoms and their nuclei and electrons are free to spin independently
of the motion of the molecule in which they are found. This rotation is also a source of kinetic
energy to particles in the solid, liquid and gaseous states.
The total kinetic energy of a molecule is the sum of all of its translational, vibrational,
rotational, and spin motions. All molecules always have some kinetic energy, no matter how low
the temperature. And the higher the temperature of a molecule, the greater its total kinetic energy.
It is important to remember that energy is neither created nor destroyed during chemical
reactions. This principle is called the conservation of energy. While energy is not created or
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destroyed, it can be transferred (as heat, light, or electrical energy) or transformed (from heat to
light, etc.) during chemical reactions.
As stated above, work (w) is a transfer of energy that can be used to the change the height
of a weight somewhere in the surroundings. In chemical reactions it is most common for work to
be done by a gas which expands against an external force, such as steam driving a piston against
atmospheric pressure. This sort of work is called expansion work, or PV work. In situations in which
a system does work on its surroundings through the expansion of a gas created in the system
w = - PΔV
where P is the applied pressure and ΔV is the change in volume that results from the expansion of
the gas. While chemical reactions are capable of performing other types of work, such as electrical
work, these are not considered during fundamental discussions of thermodynamics.
Ok, I think that's it. Now that we're equipped with the language of thermodynamics, we can begin to
discuss it.
Internal energy
There are four important fundamental quantities used in thermodynamics to predict the
spontaneity of reactions. These are internal energy (U), enthalpy (H), entropy (S), and Gibb's free
energy (G).
Internal energy is the total energy of a system, the sum of all of the potential and kinetic
energies of all of the reactant particles and product particles in the system. During the course of a
chemical reaction the internal energy always changes.
ΔU = Ufinal - Uinitial
Remember that the Greek letter ) is used to indicate a change in the value that it precedes. As any
chemical reaction proceeds from reactants to products, bonds are broken in the reactant particles and
new bonds are formed in the product particles. As bonds are storehouses of chemical potential
energy, it should not surprise you to learn that the amount of chemical potential energy stored in the
bonds of the reactants is always different from the amount of chemical potential energy stored in the
bonds of the products. If the temperature of the system changes as the reaction progresses because
the reaction emits or absorbs heat, the kinetic energy of the reactant and product particles changes
as well.
In other words, as a reaction proceeds from its initial state to its final state (i.e., from
reactants to products), the change in free energy in the system is equal to the difference between the
free energy of the system in its initial state and in its final state. This change in energy occurs as heat
and (or) work are transferred from the system to its surroundings, or from the surroundings to the
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system. Work and heat are equivalent ways of changing the internal energy of the system. If the only
interactions of a system with its surroundings consist of transfers of heat to the system and the
performance of work on the system, then
ΔU = q + w
Enthalpy
In a reaction in a system at constant pressure, the heat-associated change in internal energy
is called enthalpy. Enthalpy comes from the Greek word "enthalpein" which means "to warm." Since
many important reactions, including a substantial fraction of all chemical reactions that occur in
living systems, occur at atmospheric pressure (which is more or less constant) enthalpy is an
important property of substances.
H = U + PV
We cannot directly measure the enthalpy of a substance, but we can measure changes in
enthalpy so we are interested in ΔH, which is equal to
ΔH = Hfinal - Hinitial
or, alternatively
ΔH = ΔU + PΔV
Let me interject two notes at this point. First, we will not prove this equation or demonstrate its origin, but
you can find a proof of it in any thermodynamics textbook if you're really interested. Ok, I didn't think so.
Second, we’re not going to use these equations in this class. But, as mathematics is the language of
chemistry, we actually find ourselves handcuffed if we attempt to discuss some of these principles without
the use of a few equations. So take a deep breath and stop hyperventilating! The change in enthalpy
during a chemical reaction is equal to the difference between the enthalpy of the final state and the
enthalpy of the initial state. We can also say that the change in enthalpy of a system during a
reaction is equal to its change in internal energy and any PV work done to or by the system.
Different types of enthalpy changes can be encountered during chemical reactions. This is
to say, there are different events that, at constant pressure, lead to changes in internal energy due to
releases or absorbing of heat. Some of the more common processes which result in changes in
enthalpy are:
•
•
the enthalpy of reaction, or ΔHrxn which is also known as the "heat of reaction"
the standard enthalpy of formation, associated with the enthalpy changes of forming
compounds from the elements in their standard elemental state, ΔH°f
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•
•
there are enthalpies associated with physical changes of state (phase change), such as the
enthalpy of fusion and melting, and of vaporization and condensation
the processes of gaining or losing electrons also affect the enthalpy of compounds
In this class we are most interested in the enthalpy of reaction, ΔHrxn. The units of ΔHrxn are
usually expressed in kJ/mol. The enthalpy of reaction is the net change in heat-related energy that
results from the breaking and making of chemical bonds that occurs during chemical reactions.
We categorize all chemical reactions based on whether they emit or absorb heat.
In exothermic reactions, heat is given off by the reaction and it flows from the system to its
surroundings. The products of exothermic reactions have less chemical potential energy than the
reactants. Heat is essentially a product of the reaction. The sign of ΔHrxn for exothermic reactions
is negative, -ΔHrxn
In endothermic reactions, heat is absorbed as it flows from the surroundings to the system.
The products of endothermic reactions have more chemical potential energy than the reactants, so
heat is essentially a reactant. The sign of ΔHrxn is positive for endothermic reactions, +ΔHrxn.
If a chemical reaction is exothermic it is usually spontaneous. If a reaction is endothermic
it is usually non-spontaneous. A knowledge of the heat of reaction for a specific chemical reaction
is far more useful in predicting whether or not it is spontaneous than the use of solubility rules and
looking for the “driving forces” described in Chapter 6.
In chemistry, properties can be classified as intensive or extensive. Intensive properties are
independent of amount, in other words, intensive properties are the same whether we have a lot or
only a little bit of a substance. A few common intensive properties are color, odor, state of matter,
density, temperature, melting point, and boiling point. Most of the properties we discuss in this class
are intensive properties. Extensive properties depend on the amount of substance present. Extensive
properties include mass and volume.
Enthalpy is an extensive property. In other words, the more matter we have in a reaction, the
greater the enthalpy change. Take, for example, the reaction of hydrogen gas and oxygen gas as they
form liquid water:
2 H2 (g) + O2 (g) => 2 H2O(l)
ΔHrxn = -571.6 kJ
A balanced chemical equation that also includes thermodynamic information is called a
thermochemical equation. The above thermochemical equation tells us that when two moles of
hydrogen gas and one mole of oxygen gas react to form two moles of liquid water vapor, 571.6
kilojoules of energy is released. So, is this an exothermic or endothermic reaction? If we cut the amount
of reactants by one-half, we find that the amount of energy released during the reaction diminishes
proportionally as well.
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H2 (g) + ½O2 (g) => H2O(l)
ΔHrxn = -285.8 kJ
Note that while we never use fractional coefficients in chemical equations, it is both correct and
necessary on occasion to use them in thermochemical equations. This is the only time in this class
that the use of fractional coefficients is acceptable, as we study and discuss thermodynamics. This
is true because enthalpy is extensive. If we multiply through by two to eliminate the fractional coefficient
of oxygen, we must also multiply the enthalpy of reaction by a factor of two as well. And if we increase
the amount of reactants and products by a factor of ten, the enthalpy of reaction also increases by
a factor of ten:
20 H2 (g) + 10 O2 (g) => 20 H2O(l)
ΔHrxn = -5716 kJ
But you already knew the enthalpy was extensive although you may not have realized it. Think about it.
If enthalpy was not extensive, which would give off more heat, a very large fire, a very small fire, or would
they in fact give off equal amounts of heat? (Answer: if enthalpy was intensive, not extensive, a big fire and
a small fire would give off equal amounts of heat) If enthalpy was not extensive, what would happen when
you stepped on the gas pedal in your car? (Answer: nothing) Giving the engine more fuel would not produce
any more energy than giving it a little bit of fuel. And so on. Be grateful that enthalpy is extensive!
Enthalpies of reaction depend on the physical states of the reactants and products. If we
examine the following reaction in which water vapor is formed rather than liquid water:
2 H2 (g) + O2 (g) => 2 H2O(g)
ΔHrxn = -483.6 kJ
The reaction of hydrogen and oxygen to form water vapor is an exothermic reaction. If we
want to decompose water vapor to form hydrogen gas and oxygen gas, will it be an exothermic or
endothermic process? And, will the process be spontaneous or non-spontaneous? To write the
thermochemical equation for this reverse reaction, we simply place the products on the reactant side
of the equation, and the reactants on the products side:
2 H2O(g) => 2 H2 (g) + O2 (g)
ΔHrxn = +483.6 kJ
This reversal of the reaction involves the same amount of energy as the forward reaction. Note that
the value of the heat of reaction has remained the same. However, the sign has changed. To break
two moles of water vapor down into two moles of hydrogen gas and one mole of oxygen gas will
require us to put 483.6 kilojoules of energy into the system.
This illustrates an important point. In theory, nearly all chemical reactions are reversible,
although there may be practical considerations that make it unlikely. Given any exothermic reaction,
it's reverse reaction is always endothermic. The reverse reaction for any endothermic reaction is
always exothermic. The magnitude of the enthalpy change is constant. Only the sign changes.
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We can use the relationships in thermochemical equations as conversion factors in
stoichiometry problems. As an example, we can ask how much energy will be released by the
reaction of 100.0 g of hydrogen gas to form water vapor.
(100.0 g H2) x (1 mole H2/2.016 g H2) x (-483.6 kJ/2 moles H2) = -11,994 kJ
We can also predict how much energy will be required to break 100.0 grams of water vapor down
into hydrogen and oxygen gases:
(100.0 g H2O) x (1 mole H2O/18.02 g H2O) x (+483.6 kJ/2 moles H2O) = +1,342 kJ
I probably will not require you to do this sort of problem. Probably. I’d still advise you to be sure you
understand conceptually how do to these sorts of problems, just in case.
Entropy
We said above that if a chemical reaction is exothermic it is usually spontaneous, and that
if a reaction is endothermic it is usually non-spontaneous. This is a good general rule but there are
some exceptions. This is because changes in the order of the system also affect the spontaneity of
chemical reactions.
One of the fundamental laws of thermodynamics is that the order of a system decreases
during spontaneous reactions. Although, as we will see in a moment, there are a few exceptions to this.
In the world around us there are many spontaneous processes in which order decreases. Melting is
one example. Vaporization is another. Dissolving table salt or sugar, either one, in water is yet
another example. And this is also true of many chemical reactions.
So what do we mean by order and randomness in thermodynamics? Order and randomness
are measures of the structure in a system at an atomic level. An ordered system is a highly structured
system, much like a very large orchard. If you stand in the middle of a well laid out orchard the view
looks the same in every direction, no matter which way you look. If you could shrink yourself down
to an atomic level and then stand in the middle of an ordered substance the view would look the
same in every direction, no matter which way you look, including overhead and beneath your feet.
In other words, in ordered substances there is a high degree of repetition of particles and spacing
between the particles in all three dimensions, for many hundreds (or thousands or even millions) of
particle diameters.
We can make distinctions between the three common states of matter based on the
differences in their internal order.
Solids are highly ordered (structured) substances. They have both short range and long range
order. Solids are, at an atomic level, like a very large three dimensional orchard. No matter where
you look in a solid, the view is exactly the same in all directions for many hundreds or thousands
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(or millions, billions, or more) of atoms or molecules. This is true of crystalline solids. All ionic
compounds form crystalline solids, as do many other substances in the solid state. Amorphous solids, such
as glass, are structurally more like liquids. The atoms and molecules in different types of glass are not
nearly as well-ordered and highly structured as the ions in a salt (sodium chloride) crystal, although this is
not apparent to the naked eye. But you don’t need to worry about it. In this class, in this chapter, we’ll
assume all solids are crystalline solids.
Liquids are much less ordered (structured) at the atomic level than solids. Liquids have short
range order but do not have long range order. This is similar to standing in an orchard where no
matter which way you look the view is the same for three or four trees, but then the lines of trees
become crooked and perhaps the spacing between the trees changes and you find them closer
together or further apart than the trees in your immediate vicinity. Because liquids have less order
than solids, we say that they are more disordered, or more random, or more chaotic. We use disorder,
randomness, and chaos more or less synonymously in thermodynamics. The more disordered (or random
or chaotic) something is, the less structure (order) it has.
Gases have absolutely no order (structure) at the atomic level. Gases have neither short range
order or long range order. At an atomic level gases are like an orchard that was laid out by some one
walking randomly around the grounds and throwing a handful of seeds here and there every now and
then. It is like an orchard that has been hit by a bomb. No matter where you stand, nothing looks the
same in any direction.
Entropy is a measure of the randomness (or disorder or chaos) of a substance. A relationship
exists between free energy and entropy, and we will discuss it in a moment.
As is the case with enthalpy, we cannot directly measure entropy. We can only measure the
changes in entropy that occur within a system as the system changes. We are particularly interested
in the changes in entropy that result from chemical reactions. The change in entropy during a
chemical reaction is equal to the difference between the entropy of the final state and the entropy
of the initial state of the system
ΔSrxn = Sfinal - Sinitial
Any process which results in an increase in the entropy of the system has a value of +ΔS, while any
process the results in a decrease in the entropy of the system has a value of -ΔS. These sign
conventions are typically something of a challenge to reconcile with what we have said above about signs
in thermodynamics. So I’m going to let you take it on faith that when an entropy change results in greater
chaos (less order or structure) the sign of )S is positive, and when an entropy change results in less chaos
(more order or structure) the sign of )S is negative.
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Entropy increases when there are changes from more ordered states to less ordered states,
such as the formation of a gas from solid reactants. Entropy also increases when the number of
moles of products is greater than the number of moles of reactants. Entropy decreases when, as a
result of a chemical reaction, products are formed that are more ordered than the reactants or when
the number of moles of products formed are less than the moles of reactants initially present. The
units of ΔSrxn are usually in J/molAK. We could discuss entropy in a far more quantitative manner, i.e.,
with math and equations and all that fun stuff, but it will be sufficient for our purposes if you understand
entropy in a more general, qualitative sense.
Gibbs free energy and the spontaneity of reactions
The ultimate indicator of the spontaneity or non-spontaneity of any (and all) chemical
reactions is Gibb's free energy. The Gibbs free energy of a reaction, indicated with the symbol G,
is the energy available to do useful work in a chemical reaction. As is true with enthalpy, we cannot
measure Gibbs free energy directly, but we can observe how it changes during the course of a
chemical reaction.
ΔGrxn = Gfinal - Ginitial
The sign of ΔGrxn for spontaneous reactions is always negative, -ΔGrxn. The sign is positive for
non-spontaneous reactions, +ΔGrxn . The units of ΔGrxn are generally in kJ/mol. In other words, if
we know the sign of ΔGrxn for any reaction, we also know with absolute certainty whether the
reaction is spontaneous or nonspontaneous as written. The magnitude of ΔGrxn tells us how much
energy released by a spontaneous reaction is available to do useful work, or how much energy must
be put into a nonspontaneous reaction to make it happen.
It is important to note that nonspontaneous reactions can and do occur. It’s just that they
don’t happen to any great extent. Reactions with negative values of Gibbs free energy will typically
have a great deal of product and relatively little reactant remaining. Reactions with positive values
of Gibbs free energy will form very little (but not necessarily zero) product and have a great deal
of reactant that remains.
The Gibbs free energy of reaction ΔGrxn is always dependant on the enthalpy of the reaction
ΔHrxn, the entropy of the reaction ΔSrxn, and the temperature at which the reaction is performed. This
relationship is described by the equation
ΔGrxn = ΔHrxn - TΔSrxn
Let's examine the implications of this equation.
If a reaction is exothermic and if the entropy of the system increases during the reaction, the
reaction will always be spontaneous. Think about why this is true. If entropy increases as a
consequence of the reaction, ΔSrxn is positive. Temperature is always a positive number. We're
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working with the Kelvin scale during our study of thermodynamics. The lowest possible temperature on the
Kelvin scale is 0 K. It is not possible to have a negative Kelvin temperature. A positive number multiplied
by a positive number is always a positive number, i.e., TΔSrxn is a positive number. If we multiply
a positive number by (-1), as we do when using this equation, the quantity TΔSrxn becomes a
negative number. If we add this negative number to the enthalpy of reaction, which for an
exothermic reaction is always a negative number (-ΔHrxn), ΔGrxn cannot be other than a negative
value, which means the reaction is spontaneous. Remember, the sum of two negative numbers is always
itself a negative number: a negative plus a negative equals a negative.
On the other hand, if a reaction is endothermic and if the entropy of the system decreases
during the reaction, the reaction will always be non-spontaneous. If entropy decreases as a
consequence of the reaction, ΔSrxn is negative. Temperature is, as we said above, always a positive
number . A positive number multiplied by a negative number is always a negative number, i.e.,
TΔSrxn is a negative number. If we multiply TΔSrxn by (-1), the quantity TΔSrxn becomes a positive
number. If we add this positive number to the enthalpy of reaction, which for an endothermic
reaction is always a positive number, ΔGrxn will always have a positive value, and the reaction is
non-spontaneous. Remember, the sum of two positive numbers is always itself a positive number: a
positive plus a positive equals a positive.
This leaves us to discuss those reactions in which the enthalpy and entropy of the reaction
are both either positive or negative. Let's consider an exothermic reaction in which the entropy
decreases. This means that ΔHrxn is negative, ΔSrxn is negative, and -TΔSrxn is positive. Whether
ΔGrxn is negative or positive depends on the temperature of the reaction. If the reaction is carried out
at a high temperature, the -TΔSrxn term will be large enough to override the negative enthalpy term
and ΔGrxn will be positive. On the other hand, if the reaction is carried out at a low temperature, the
-TΔSrxn term will be too small to override the negative enthalpy term and ΔGrxn will be negative.
Finally, let's examine an endothermic reaction in which the entropy increases. This means
that ΔHrxn is positive, ΔSrxn is positive, and -TΔSrxn is negative. Again, whether ΔGrxn is negative or
positive depends on the temperature of the reaction. If the reaction is carried out at a high
temperature, the -TΔSrxn term will be large enough to override the positive enthalpy term and ΔGrxn
will be negative. If the reaction is carried out at a low temperature, the -TΔSrxn term will be too small
to override the positive enthalpy term and ΔGrxn will be positive. The following table summarizes
the relationship between the signs of ΔHrxn, ΔSrxn, temperature, and ΔGrxn.
ΔHrxn
ΔSrxn
T
ΔGrxn
-
+
any
always -
+
-
any
always +
-
-
low/high
-/+
+
+
low/high
+/-
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Do you need to memorize this table? No. But you should be able to use it to help you answer questions.
You need to know what this stuff means and how it is used. An example of how it may be used is found
on the Chapter 7 quiz..
So how do we use this information? In the conversion of ozone gas to oxygen gas
2 O3 (g) => 3 O2 (g)
the sign of ΔHrxn is negative and the sign of ΔSrxn is positive. This means, according to the table, that
the sign of ΔGrxn is negative at all temperatures (i.e., at any temperature) and that the reaction is
therefore spontaneous at all temperatures.
As another example, the process of dissolving solid sodium chloride in water has an enthalpy
of reaction of +3.6 kJ/mol and an entropy of reaction of +43.2 J/mol.K. given that the signs of both
of these are positive and based on the above table, we can surmise that the sign of for ΔGrxn this
process is positive at low temperatures and negative at high temperatures. In other words, sodium
chloride will spontaneously dissolve in water at high temperatures but not at low temperatures.
"High" temperature and "low" temperature are very much relative terms in this situation. Exactly how high
or low a temperature must be for a reaction of this type to be spontaneous or non-spontaneous depends
on the magnitudes of the enthalpy and entropy terms which depend, in turn , on the particular reaction
of interest. In some cases “high” temperature might be many thousands of degrees. The combustion of
methane is an example of reaction that is exothermic and which also results in a decrease in entropy. For
this particular combustion reaction it will remain spontaneous as long as the reaction occurs at a
temperature less than 160,000 K. This is good news unless you plan on using methane to heat your home
in some place very warm, like, say for example, the surface of the sun. In other reactions “the “high”
temperature may be only a few degrees above absolute zero.
The rates of chemical reactions
The rates of chemical reactions can be expressed several different ways. It is common to
express the rate of a reaction in terms of how fast a reactant disappears as a function of time, or in
terms of how quickly a product appears as a function of time. Mathematically, this can be expressed
as
rate = Δ[reactant]/Δt
or
rate = Δ[product]/Δt
where the square brackets [] represent the molar concentration of the substance contained within the
brackets and "Δ" represents "the change in." Molar concentration is the number of moles of a substance
per liter of solution. It is discussed in Chapter 9 of your text. As an example, if we are discussing the molar
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concentration of sodium chloride in a solution, we would it express it as [NaCl]. This means that the
mathematical statement "Δ[reactant]/Δt " means “the change in reactant concentration with the
change in time.” The rate of any chemical reaction depends on these changes in concentration as
time progresses.
Let's examine the data from a fairly simple reaction. Hydrogen peroxide decomposes to form
water and oxygen according to the equation
2 H2O2 (aq) => 2 H2O(l) + O2 (g)
Out of curiosity, what's the "driving force" in this reaction? If you answered electron transfer, give yourself
a big pat on the back!
If we begin with roughly 0.9 moles of hydrogen peroxide in 1.00 L of water, we can measure the
decrease in it's concentration with the passage of time. The concentration of hydrogen peroxide
decreases because it is being consumed as the reaction progresses. If we plot this data graphically,
we see the following:
There are several important things to note in this plot. You can see that the trend line described by
the data is not linear, but gently curving. This behavior is not unique to this particular reaction but
is in fact characteristic of many reactions. An important implication of this non-linear behavior is
that the concentration of hydrogen peroxide will never reach zero, i.e., all of the hydrogen peroxide
will never be completely consumed. We will discuss why this is so a little later.
At any given time, the average rate of the reaction is equal to the slope of the line described
by the data. We can see in the graph that the rate of the reaction changes as the reaction proceeds.
We can calculate the slope between any two of the adjacent data points, so let's choose the first two
data points. Remember your algebra: given two points, (x1, y1,) and (x2, y2) the slope m between the points
is equal to m = (y2-y1)/(x2-x1). And a further word about slopes: this equation is the way we calculate the
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average slope. This is fine for our purposes in this class. In truly serious studies of kinetics we are usually
more interested in the instantaneous slope, the slope of the curve at any given instant, but since it requires
a bit of calculus to determine it, we simply will not worry about it.
m = (0.70 - 0.88)/(60 - 0) = -0.00300
If we do this for all of the pairs of data we will find that the reaction has slowed to about 15% of it's
initial rate by the time 10 minutes (600 seconds) has passed. This is consistent with the slope of the
line decreasing as the reaction progresses.
Reaction rates and molecular collisions
Why does the rate of the decomposition of hydrogen peroxide slow as it is consumed? Why
doesn't the reaction continue on at the same rate until all of the peroxide is consumed?
In most reactions particles must be able to collide if they are to react. Remember, we use the
word “particle” to generically represent atoms, molecules, and ions. Various factors affect reaction rates.
Nearly all are factors that pertain to collisions between particles and include:
•
•
•
•
•
reactant concentration: how many particles there are in a given volume of space. The more
particles in a given volume, the more frequently they will collide and the faster the reaction.
reactant pressure (if a reactant is in the gas phase): as pressure increases in the gas phase, the
number of reacting particles per unit volume of space will also increase, and therefore so will
the frequency of collisions and the rate of the reaction.
spatial orientation: how the particles are aligned with respect to one another when they
collide. You can think of the interactions between particles in many reactions as being
similar to the interaction between a key and a lock. While there are numerous ways for a key
and a lock to interact with each other, only when the key interacts with the lock in precisely
the right place - the key hole - will something happen. If the key strikes the lock anywhere
but the keyhole, nothing will happen.
system temperature affects both the speed of the particles, and therefore how often they
collide, but also the kinetic energy involved in these collisions. As a general rule, for most
common reactions the rate increases as temperature increases, and reaction rate slows as
temperature decreases.
the presence of catalysts, which make reactions proceed more rapidly. For reasons we will
soon explain.
Chemical reactions are, at the atomic level, events in which chemical bonds are broken and
made. Bonds must be broken for the reactant molecules to react. New bonds are formed as the
“broken” reactant molecules recombine and product molecules are created. The energy used to break
chemical bonds is often the kinetic energy of the colliding molecules. This means that during most
chemical reactions, the reacting molecules must collide with enough energy to result in the breaking
of bonds.
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The reacting molecules must also be oriented properly, positioned in such a way that the
collisions between the molecules happen with the parts of the molecules that have the ability to
react. Why this is so may not be apparent but we will not elaborate further on this topic in this class.
Suffice it to say that, tiny as they are, not all parts of most molecules are equally reactive. As I have
described above you can think of reacting particles as very tiny locks and keys.
Concentration has a significant effect on the number of effective collisions. As a reactant is
consumed the number of reactant molecules decreases so there are fewer available to collide and
react.
Temperature also has a profound affect on the rate of chemical reactions. As temperature
increases, the kinetic energy of the molecules also increases and they collide with more energy,
making the breaking of chemical bonds easier. As a general rule, reaction rates double with every
10°C increase in temperature, because the fraction of molecules in the system with sufficient energy
to break bonds when they collide with other molecules increases with the increase in temperature.
Rate Laws
Rates of reaction can also be expressed using rate laws. Rate laws are mathematical
expressions of the relationship between reaction rate, temperature, and reactant concentration. For
the reaction wA + xB <=> yC + zD, where w, x, y, and z are the stoichiometric coefficients of the
reactants A and B and the products C and D, the reaction’s rate law is written
rate = k[A]m[B]n
Rate laws are independent of time. They can be used to calculate the rate of a reaction at any point
during the reaction as long as reactant concentrations are known. The value k is a rate constant that
depends on the particular reaction and on the temperature at which the reaction is performed. The
exponents m and n are called the orders of the reactants A and B respectively. They describe exactly
how the concentrations of A and B affect the reaction rate and they must be determined
experimentally. Which we will not do in this class but if you feel cheated, plan on taking Chem 1220 from
me some time and I'll show you how this is done. I should also tell you that the lack of a connection
between rate and time in rate laws can be something of a nuisance. But fear not. Using calculus we can
integrate rate laws and establish a relationship between reaction rate, reactant concentrations, and time,
something wonderfully useful in chemistry but sadly, something we will not discuss in this class.
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Activation energy and activated complexes
The process of breaking old bonds and making new bonds is integral to all chemical
reactions. The process is generally rapid, but it is not instantaneous.
During a reaction a reacting substance will typically find itself very briefly in an abnormally
high energy state. We're talking milliseconds to femtoseconds here and in some cases even faster,
although in many cases it may also be substantially slower. There are very good reasons for this
temporary increase in energy. Some of the reactant's atoms may briefly have more or fewer bonds
than are required for them to obey the octet rule. Or, one or more of the reactant atoms may gain or
lose an electron (or, electrons) temporarily, again causing these reactant atoms to violate the octet
rule. Whatever the reason, when atoms violate the octet rule their energy increases. This in turn
raises the energy of the entire reactant molecule in which the atoms are found. There are of course
other ways for reactant molecule energy to increase besides violations of the octet rule but we will not
consider them here. This temporary high energy state is always an unstable one, as it is higher in
chemical potential energy than either the starting potential energy state of the reactant molecules or
the final potential energy state of the product molecules. Remember that molecules have chemical
potential energy, the energy stored in the bonds between the atoms. And also remember the correlation
between energy and stability: in relative terms, high energy equals unstable and low energy equals stable.
This temporary high energy state is called the transition state. The unstable substance in the
transition state is called an activated complex. The increased potential energy of the activated
complex is called activation energy and is represented with the symbol Ea. By definition the
activation energy of a reaction is the difference in energy between the energy of the reactants and
the energy of the activated complex in the transition state.
Activation energy can be thought of as an energy barrier between the reactant and product
states. Spontaneous reactions often require a bit of a "nudge" to get them started, usually in the form
of a small amount of energy. For example, combustion reactions are spontaneous reactions, yet they
seldom begin without a small initial input of energy to get them started. You will never enjoy a
campfire without initially striking a match. And yet, this is still classified as a spontaneous reaction.
Spontaneous reactions are reactions that result in a lower overall system energy in the product state than
in the reactant state. The small amount of energy a spontaneous reaction may require to get it started is
generally insignificant compared to the overall changes in energy in the system as the reactants become
products.
A car parked at the very edge of a steep hill will spontaneously roll down the hill if it is not
in gear and the emergency brake is not engaged. Some of you have probably learned this the hard way,
through experience. All it takes is the slightest of pushes to start it rolling. If there is a rock in front
of one of the tires one must push a little harder to make the spontaneous rolling occur. The larger
the rock, the harder one must push to make something that is ostensibly spontaneous happen. This
rock is like the activation energy of a reaction. And if one wishes to make the reverse reaction occur,
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one must not only supply enough energy to push the car back up the hill, but also enough to get the
tires over the rock at the top of the hill.
Let's look at the change in the energy of a system during an exothermic reaction. In an
exothermic reaction (a reaction that emits heat) the chemical potential energy of the reactants is
greater than the chemical potential energy of the products. The difference in potential energy
between the two is emitted as heat as reactants are converted to products. This emitted heat is the
enthalpy of reaction, or, the heat of reaction. Even though exothermic reactions are generally
spontaneous (this also depends on entropy changes in the reaction), reactant molecules must collide
with sufficient energy to be able to break bonds and form the activated complex that exists in the
transition state. Here we have a plot of the changes in the energy of a system as an exothermic
reaction progresses from reactant to products. The “progress of reaction” axis is, really, just a measure
of time as it increases.
During an endothermic reaction (a reaction that absorbs heat) the starting chemical potential
energy of the reactants is less than the final chemical potential energy of the products. The difference
in potential energy between the two is absorbed, usually as heat, as reactants are converted to
products. This absorbed heat is the enthalpy of reaction. Endothermic reactions are generally
non-spontaneous, because reactant molecules must collide with sufficient energy to be able to break
bonds and form the activated complex that exists in the transition state.
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The reversibility of chemical reactions
At this point we need to learn a rather unexpected but important truth: nearly all chemical
reactions are reversible. This is true no matter how improbable the likelihood of the reserve reaction
occurring may seem. And there is excellent experimental evidence to confirm this statement.
For any exothermic reaction, it's reverse reaction will always be endothermic. The magnitude
of the enthalpy change is constant. Only the sign changes. In other words, if a reaction has an
enthalpy of reaction of “-X” then the enthalpy of reaction for the reverse reaction will be exactly
“+X.” Remember that exothermic reactions have negative enthalpies of reaction and endothermic
reactions have positive enthalpies of reaction. The reason that this observation is true is due to the
extensive nature of enthalpy. Do you remember what extensive properties are and why they’re important?
For the reverse reaction of an exothermic process to occur, enough energy must be added to the
products to (a.) restore them to the higher potential energy state of the reactants, and (b.) overcome
the activation energy of the transition state.
Endothermic reactions are also reversible. For any endothermic reaction, it's reverse reaction
is always exothermic. The magnitude of the enthalpy change is constant. Only the sign changes. For
the reverse reaction of an endothermic reaction to occur, enough energy must be added to the
products to overcome the activation energy of the transition state. Energy will be emitted as the
higher energy product molecules are converted to lower energy reactant molecules.
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It is extremely important to realize that in saying that, in theory, all reactions are reversible,
we are *not* saying that reverse reactions are equally as probable - or, if you’d rather, as
energetically favorable - as the forward reaction. Far from it.
As we established above, all spontaneous reactions have a negative Gibbs free energy
associated with them, i.e., a -ΔG value. This means that the Gibbs free energy change associated
with the reverse reaction will be equal in magnitude but opposite in sign. This is, as with enthalpy,
due to the extensive nature of Gibbs free energy. In other words, if a reaction has an ΔG value of
“-X” then the value of ΔG value for the reverse reaction will be exactly “+X.” In other words, the
reverse reaction of a spontaneous reaction is nonspontaneous, meaning, it will typically only happen
to a slight extent.
Also, as we learned above, all nonspontaneous reactions have a positive Gibbs free energy
associated with them, i.e., a +ΔG value. The value of the Gibbs free energy change associated with
the reverse reaction will be equal in magnitude but opposite in sign. In other words, the reverse
reaction of a nonspontaneous reaction is itself spontaneous, meaning, it will typically happen to a
far greater extent than the forward reaction.
Catalysts
A catalyst is a substance that increases the rate of a reaction without any net change in its
concentration. In other words, catalysts make chemical reactions happen faster and are usually not
consumed during the course of the reaction. Or if a catalyst is consumed at one point in a reaction,
it is immediately regenerated during another portion of the same reaction so that there is no net
change in its concentration.
Catalysts make reactions rates increase by lowering the activation energy of the reaction.
This can be accomplished in one of two ways. Some catalysts work by providing an alternative
reaction pathway with a lower energy transition state. Perhaps more common, catalysts provide a
surface on which the reaction takes place. Molecules adsorp to the surface of the catalyst. "Adsorp"
means to chemically bond to the surface; this is different from "absorb", which is what occurs when one
material incorporates another material into itself, as when one absorbs spilled water with a paper towel.
Features of the catalyst's surface result in the stretching of chemical bonds in the adsorped
molecules. This results in weakening of the bonds which makes them easier to break when other
reactant molecules collide with them.
173
In looking at this image it is hopefully apparent that enzymes catalyze both the forward and
the reverse reactions equally. If a catalyst lowers the activation energy for a given reaction by “X”
kilojoules, it makes both the forward and the reverse reactions that much easier.
Enzymes are biological catalysts. They play an essential role in making biological reactions
happen that would otherwise not occur, or that would happen so slowly as to cause serious health
problems. Like, for example, death. Enzymes are very specific as to the substances upon which they
act. In some cases, simply changing one or two of the atoms of a large molecule can change its shape
to the extent that an enzyme can no longer act upon it. This specificity of enzymes for their target
molecules often results in the use of a "lock and key" model when discussing them. Enzymes are
likened unto a lock, and the molecules that they act on, called substrates, are analogous to a key. If
you change the shape of a key ever so slightly, it will often no longer unlock the lock for which it
was intended. Conversely, if you change the configuration of the tumblers in the lock (in other
words, if you do something to change the shape of the enzyme), the key will no longer work
properly. Although the “lock and key” model has been replaced with the “induced fit” model, I do not
think the lock and key model is so far from correct that it will hurt you to think of things this way. See the
wikipedia entry for enzymes at http://en.wikipedia.org/wiki/Enzyme for more information, if you’re
interested.
174
Chemical equilibrium
Let's return to our discussion of the decomposition of hydrogen peroxide. We saw that the
rate of the reaction slows as hydrogen peroxide is consumed. This is indicated by a leveling of the
trend line that connects the data points. If we plot the change in the oxygen concentration of the
system as a function of time, we eventually see similar leveling behavior in the trend line.
At first the rate of the production of oxygen is relatively fast and the slope of the line is relatively
steep. But the rate of production of oxygen slows as the reaction progresses, and the slope of the line
begins to level off. The graph shows that rate of oxygen production begins to level off at about the
same time in the experiment that the rate of the decomposition of hydrogen peroxide begins to slow.
Why is it that the rate of hydrogen peroxide decomposition and the rate of oxygen production
both slow as the reaction proceeds? When we discussed this behavior above, we said that as the
concentration of hydrogen peroxide decreases the number of molecules available to participate in
collisions decreases and that the reaction slows as a consequence. This is true, but it is only part of
the story. It also does not explain why the hydrogen peroxide is never entirely consumed.
The complete answer to this question lies in the reversibility of chemical reactions. As
hydrogen peroxide decomposes, liquid water and oxygen gas are produced. We know this from the
balanced chemical equation for the reaction. As molecules of water and oxygen are produced,
collisions between them will sometimes result in the reverse reaction taking place, i.e., in the
production of hydrogen peroxide. As the concentrations of water and oxygen increase, the number
of collisions between water and oxygen molecules also increases, and the rate of the reverse reaction
increases as well.
In every reaction a point is reached at which the rate of the forward reaction is equal to the
rate of the reverse reaction. Remember that the slopes of the trend lines in the plot of the data are
related to the rates of the reactions. When the slopes of the trend lines are parallel, the rates of the
two reactions are equal. In the above plot we do not have quite enough data to show this leveling of the
175
trend lines. However, if we extrapolate from the data that we have, we can guess that both trend lines will
level out and become parallel between 600 and 1200 seconds. When we reach this point we say we
have reached equilibrium in the reaction. By definition, in any reaction a state of dynamic
equilibrium exists when the rate of a forward reaction is equal to the rate of its reverse reaction. A
reaction at equilibrium is indicated with a two-headed arrow between the reactants and the products.
All chemical reactions always proceed toward a state of equilibrium. For example, once the
decomposition of hydrogen peroxide is at equilibrium, the chemical equation can be written
2 H2O2 (aq)
W2 H O
2
(l)
+ O2 (g)
Equilibrium is as a state of dynamic equilibrium. A reaction does not simply "shut off" when
equilibrium is reached. At equilibrium both the forward and the reverse reaction continue to occur,
but at the same rate. As a consequence, the concentrations of reactants and products become
constant, because as quickly as reactant molecules are consumed in the forward reaction to form
products, product molecules are consumed in the reverse reaction in the production of the original
reactants.
How long does it take a reaction to reach equilibrium? It varies, depending on the reaction
and on the temperature at which the reaction takes place. Some reactions proceed to equilibrium
rapidly, reaching it in seconds or even quicker. Other reactions might require a period of hours, days,
or longer before equilibrium is attained.
How does a catalyst affect equilibrium? Catalysts speed up the rate of both the forward and
the reverse reactions. A catalyst will help the system reach equilibrium more rapidly but they do not
cause a shift in the ratio of the concentrations of the products to the concentrations of the reactants.
A catalyst will help a slow reaction reach equilibrium more quickly, and a fast reaction reach
equilibrium faster still.
The equilibrium constant
Equilibrium can be expressed mathematically using rate laws. Let's assume we have the
following reaction
wA + xB
W yC + zD
and that the rate laws for the forward and reverse reactions are
ratef = kf[A]w[B]x
and
rater = kr[C]y[D]z
176
where ratef and rater are the rates of the forward and reverse reactions, respectively, kf and kr are the
rate constants of the forward and reverse reactions, respectively, and [A]w, [B]x, [C]y and [D]z are
the concentrations of the reactants and products raised to the powers of their respective coefficients.
At equilibrium the rates of the forward and reverse reactions are equal to each other. Therefore, at
equilibrium,
ratef = rater
which can be re-written as
kf[A]w[B]x = kr[C]y[D]z
and rearranged as
kf
kr

[ C ] y [ D] z
[ A] w [ B] x
There is, actually, an advantage to expressing things this way. This last equation tells us that
for any chemical reaction at equilibrium, the ratio of the concentrations of the products and the
reactants is always a constant value. We call this constant the equilibrium constant, Kc. For any
reaction at equilibrium we can write an equilibrium expression or, equilibrium constant expression
that describes the relationship between the equilibrium constant and the balanced chemical equation
for the reaction. As an example, for the reaction wA + xB W yC + zD, the equilibrium constant
expression is
Kc 
kf
kr

[ C ] y [ D] z
[ A] w [ B] x
Why is this useful? Because for any given reaction Kc will only change if we change the temperature
at which the reaction is performed or if we change to a completely different reaction. This means
that regardless of how much reactant and product with which we start a reaction, when equilibrium
is reached, the ratio of the concentrations of the products and reactants will always equal the
equilibrium constant.
To demonstrate this, let's look at the results of the reaction of dinitrogen tetroxide and
nitrogen dioxide. The balanced chemical equation for this reaction is
N2O4 (g)
W 2 NO
2 (g)
The equilibrium constant expression for this reaction is
[ NO2 ]2
Kp 
[ N 2 O4 ]
177
Note that we are calculating Kp in these experiments rather than Kc. There are a number of different
type of equilibrium constants. Kp is used when we are studying equilibrium in gaseous systems. Kc
is used when we are studying equilibrium in aqueous solution. The "p" subscript means that the
equilibrium constant was calculated using partial pressures. The "c" subscript means that the
equilibrium constant was calculated using molar concentrations. We’ll explain these concepts of partial
pressure and molar concentration in Chapters 8 and 9 respectively. I should also tell you that in addition
to equilibrium constants calculated using molar concentrations, Kc, and equilibrium constants calculated
using partial pressures, Kp, that there are equilibrium constants that describe the dissociation behavior of
acids in water, Ka, bases in water, Kb, and ionic compounds in water, Ksp, and etc. The point is that there
are a variety of ways in which we find it useful to apply equilibrium theory in chemistry. But we don’t have
a chance to discuss them in this class, and that’s truly a shame.
Let’s look at the data from three sets of experiments. In the first set of experiments we’ll start
with different amounts of dinitrogen tetroxide and with no nitrogen dioxide. In the second set of
experiments we’ll start with different amounts of nitrogen dioxide and no dinitrogen tetroxide. And
in the third set of experiments we’ll start with varying amounts of both compounds.
Here we have the data for five experiments. In each of these we have an initial amount of
dinitrogen tetroxide (in units of atmospheres, N2O4 I, the “subscript “I” stands for “initial.”) and no
nitrogen dioxide is initially present (again in units of atmospheres, NO2 I).
Once the reaction begins we let it progress until it reaches equilibrium. When we measure the
amount of each compound at equilibrium we find that the concentration of dinitrogen tetroxide has
decreased, as we would expect of a reactant (still in units of atmospheres, N2O4 eqb in which the “eqb”
subscript stands for “equilibrium”). We also find that at equilibrium the concentration of nitrogen
dioxide has increased, which is consistent behavior for a product (also in units of atmospheres,
NO2 eqb). To calculate the numerical value of the equilibrium constant we take the equilibrium
concentrations of the reactant and the product from each of the five experiments and plug them into
the equilibrium constant expression for this reaction. Here we use the data from the first experiment:
Kp 
[ NO2 ]2 (140
. )2

 6.45
[ N 2 O4 ] 0.30
The amazing thing is that in each of the five experiments, regardless of the amount of dinitrogen
tetroxide with which we began, the ratio of products to reactants at equilibrium is always the same
178
value, 6.45. Note that there are no units associated with this equilibrium constant, as by convention we
always express equilibrium constants as unitless numbers.
In our next set of five experiments we begin with no dinitrogen tetroxide and with differing
amounts of nitrogen dioxide.
In these experiments it is not initially possible for the forward reaction to occur, as there is no
dinitrogen tetroxide. We would expect the reverse reaction to be the prominent reaction, and this is
what we see in the results. In each experiment the concentration of nitrogen dioxide decreases as we
progress from our initial conditions to equilibrium, and the concentration of dinitrogen tetroxide
increases as we move from initial to equilibrium conditions. And, as with the first sets of
experiments, again we see that if we take the equilibrium amounts of product and reactant and plug
them into the equilibrium constant expression, we wind up with exactly the same ratio of products
to reactants in each case, independent of starting amounts.
In our last five experiments we begin with equal amounts each of reactant and product.
It’s interesting to note that for experiments 11, 12, and 13 dinitrogen tetroxide is consumed and
nitrogen dioxide is produced as the experiment proceeds from initial to equilibrium conditions. In
experiments 14 and 15 we see the reverse behavior: nitrogen dioxide is consumed and dinitrogen
tetroxide is produced as the experiments move from initial to equilibrium conditions. But in all five
experiments the ratio of products to reactants as calculated using the equilibrium constant expression
remains the same as in our first ten experiments.
So what’s the point of all of these experiments? To underline what we said above:
“regardless of how much reactant and product with which we start a reaction, when equilibrium is
reached, the ratio of the concentrations of the products and reactants will always equal the
equilibrium constant.” The only two things that will change the equilibrium constant for a reaction
is a change in temperature or a change to a completely different reaction.
179
Writing equilibrium constant expressions
I expect you to be able to write equilibrium constant expressions for reactions and to
calculate the value of equilibrium constants. Let’s look at a few more examples.
Let’s begin with a reaction called the Haber process in which nitrogen gas reacts with
hydrogen gas to form ammonia gas according to the following chemical equation.
W 2 NH
N2 (g) + 3 H2 (g)
3 (g)
Based on this equation the equilibrium constant expression is
Kc 
 NH 32
[ N 2][ H 2]3
If we’re told that the equilibrium concentrations of nitrogen, hydrogen, and ammonia are 1M, 3M,
and 5M respectively, then the equilibrium constant is equal to
Kc 
52
[1][3]3
 0.96
The unit “M” indicates that we are using molar concentration (molarity) for these values, again, a topic we
will discuss in Chapter 9. I should point out that this is not the real value of the equilibrium constant for
this reaction. I have arbitrarily selected some numbers to illustrate how they are used in these calculations.
But even though these are not actual values, the way in which they are used is precisely the same as
though they are in fact, real and correct values.
Another example would be the reaction of hydrogen gas and iodine vapor to form hydrogen
iodide gas.
H2 (g) + I2 (g)
W 2 HI
(g)
Based on this chemical equation the equilibrium constant expression would be
Kc 
 HI 2
[ H 2][ I 2]
If, at equilibrium, the molar concentrations of hydrogen, iodine, and hydrogen iodide are 0.1 M, 0.25
M, and 3.7 M respectively, then the value of the equilibrium constant would be
180
Kc 
3.72
[01
. ][0.25]
 548
Again, I have arbitrarily selected numbers at random. This is not the actual value of the equilibrium
constant for this reaction.
It is important to note that when writing equilibrium constant expressions we only include
terms for those substances in the gaseous or aqueous state. We do not include terms for substances
in the liquid or the solid state as the concentrations of materials in these states are constant and, as
such, unchanging during the course of a chemical reaction. This undoubtedly seems an odd statement
to you but it is one I will not justify in these notes. If you have questions, please contact me privately. I am,
quite simply, going to ask you to trust me on this one. Let’s look at a few examples of how this affects
the way in which we write equilibrium constant expressions.
Given the reaction
CO2 (g) + H2 (g) WCO(g) + H2O(l)
the equilibrium constant expression is
Kc 
[CO]
[CO2 ][ H2 ]
For the reaction
SnO2 (s) + 2 CO(g) W Sn(s) + 2 CO2 (g)
the equilibrium constant expression is
Kc 
[CO2 ]2
[CO]2
For the reaction
3 Fe(s) + 4 H2O(g) W Fe3O4 (s) + 4 H2 (g)
the equilibrium constant expression is
Kc 
[ H 2 ]4
[ H 2 O]4
181
And finally, for the reaction
2 Al(s) + Fe2O3 (s) W 2 Fe(s) + Al2O3 (s)
there is no equilibrium constant expression, as all of the substances in the reaction are in the solid
state.
Equilibrium constants and Gibbs free energy
Why is it that an equilibrium constant represents a seemingly “magic number” for a reaction?
Because the total energy of a system is at a minimum at equilibrium, and energy governs everything
that happens in chemistry. All physical changes and chemical reactions move spontaneously from
states of higher energy to states of lower energy. And if the energy of a system is at its lowest
possible point at equilibrium, then that system will spontaneously progress to equilibrium.
Equilibrium constants provide insight into the favorability of chemical reactions. This is
because for any chemical reaction (or physical change) there is an inverse relationship between the
Gibbs free energy of a reaction and the equilibrium constant for that reaction. Any reaction with a
negative Gibbs free energy of reaction will have an equilibrium constant greater than 1. And any
reaction with a positive Gibbs free energy of reaction will have an equilibrium constant less than 1.
If the equilibrium constant for a reaction is much greater than 1 (Kc > 1), this means that
more products are formed in the forward reaction than reactants are formed in the reverse reaction.
For a fraction to be greater than 1 the value of the numerator must be greater than the value of the
denominator, as is the case in 2/1. The equilibrium mixture will contain a greater amount of products
than reactants. Reactions in which there are more products formed than reactants are said to be
product-favored reactions. The forward reaction is the energetically favorable reaction in
product-favored reactions.
The larger the value of Kc the more favored the forward reaction must be. If we compare two
reactions, one with an equilibrium constant of 1 x 105 and another with an equilibrium constant
of 1 x 1010, we know that the reaction with the larger equilibrium constant is the more productfavored reaction.
If the equilibrium constant for a reaction is much less than 1 (Kc < 1), this means that more
reactants are formed in the reverse reaction than products are formed in the forward reaction. For a
fraction to be less than 1, the value of the numerator must be less than the value of the denominator, as
is the case in ½. The equilibrium mixture will contain a greater amount of reactants than products.
Reactions in which there are more reactants formed than products are said to be reactant-favored
reactions. The reverse reaction is the energetically favorable reaction in reactant-favored reactions.
The smaller the value of Kc the more favored the reverse reaction must be. If we compare
two reactions, one with an equilibrium constant of 1 x 10-5 and another with an equilibrium constant
182
of 1 x 10-10, we know that the reaction with the smaller equilibrium constant is the more
reactant-favored reaction.
If the equilibrium constant for a reaction is around 1 (Kc ~ 1), this means that there are
roughly equal amounts of reactants and products in the equilibrium mixture. Values of Kc and Kp are
never negative. Why is that? (Hint: is it possible to have a negative concentration or a negative partial
pressure? Answer: no!)
LeChatelier's principle
Why do all reactions always proceed toward equilibrium? As I said above, it is because
systems at equilibrium are in a relatively low energy state, compared to the energy of the system
when it is in any state other than a state of equilibrium. In other words, a system at equilibrium is
more stable than the same system when it is not at equilibrium.
Knowing this, it should not surprise us to learn that if a system is at equilibrium and is
disturbed, i.e., the equilibrium is disrupted or, some books refer to this as "stressing" the system, the
system will spontaneously react so as to restore its state of equilibrium. This is called LeChatelier's
le SHAT-lee-ay principle. LeChatelier's principle states that a system at equilibrium, if disturbed, will
shift so as to minimize the effects of the disturbance. Or in other words, the system will do
everything it can to return to a state of equilibrium.
We can disturb a system at equilibrium by adding or removing reactants or products, by
changing the temperature of the system, or by changing the pressure of the system. To understand
the effect of changing the concentration of reactants or products in an equilibrium, remember that
at equilibrium the ratio of the products and reactants is a constant number (i.e., the equilibrium
constant).
Let's say we have a system at equilibrium and we add more reactants to it. The ratio of
products to reactants is no longer equal to the equilibrium constant. To restore things to a state in
which the ratio of products to reactants is equal to the equilibrium constant, the forward reaction
must consume some of the added reactant and form more products. This will continue until the ratio
of products and reactants is equal to the equilibrium constant and equilibrium has been restored.
Assume we have a system at equilibrium and we add more products to it. The ratio of
products to reactants is no longer equal to the equilibrium constant. To restore things to a state in
which the ratio of products to reactants is equal to the equilibrium constant, the reverse reaction
must consume some of the added product and form more reactants. This will continue until the ratio
of products and reactants is equal to the equilibrium constant and equilibrium has been restored.
What will happen if we somehow remove reactant from a system at equilibrium? Again, the
ratio of products to reactants is no longer equal to the equilibrium constant. To restore things to a
state in which the ratio of products to reactants is equal to the equilibrium constant, the reverse
183
reaction must consume some of the product and form more reactants. This will continue until the
ratio of products and reactants is equal to the equilibrium constant and equilibrium has been
restored.
Finally, how will the equilibrium system respond if we remove some of the product? As in
the three cases above, the ratio of products to reactants is no longer equal to the equilibrium
constant. To restore things to a state in which the ratio of products to reactants is equal to the
equilibrium constant, the forward reaction must consume some of the reactant and form more
products. This will continue until the ratio of products and reactants is equal to the equilibrium
constant and equilibrium has been restored.
Consider the reaction
A+B
WC+D
with the equilibrium constant expression
Kc 
[C][ D]
[ A][ B]
The effects of various disturbances to this system at equilibrium are summarized in the following
table:
if we
the system will shift by
and form more
through the
add A and/or B
consuming A and B
C and D
forward reaction
add C and/or D
consuming C and D
A and B
reverse reaction
remove A and/or B
consuming C and D
A and B
reverse reaction
remove C and/or D
consuming A and B
C and D
forward reaction
How will changes in temperature affect a system at equilibrium? Of course, a change in
temperature will result in a change in the equilibrium constant for a reaction in nearly all cases, but
let's ignore that effect during this discussion. In fact, let's just forget about it all together. We can
actually ignore it safely in this case. If you really want to learn more about this, you should take a proper
thermodynamics class. We learned above that exothermic reactions give off heat and endothermic
reactions absorb heat. To understand the effect of temperature change on a system at equilibrium,
let's consider heat as a product in exothermic reactions and as a reactant in endothermic reactions.
In other words, we can write a chemical equation for an exothermic reaction
A+B
W C + D + heat
184
and an endothermic reaction as
heat + A + B
WC+D
It is easy to appreciate that we can add heat to a system by doing something to increase it's
temperature. We remove heat from a system by doing something to lower it's temperature, i.e., by
cooling it. The effects of temperature changes to this system at equilibrium are summarized in the
following table:
if we
to/from an ________ reaction
the system will shift
toward
through the
add heat
exothermic
reactants
reverse reaction
remove heat
exothermic
products
forward reaction
add heat
endothermic
products
forward reaction
remove heat
endothermic
reactants
reverse reaction
How will changes in pressure affect a system at equilibrium? If the system is in aqueous
solution and an inert (unreactive) gas is added to the system, there will be no effect. This is also true
for gaseous systems: the addition of an inert gas will have no effect on the equilibrium. However,
if a gas that participates in the reaction is added to or removed from the system, the system will
behave exactly the same as a system in aqueous solution. We will ignore changes in pressure that
result from changing the volume of the system in this class.
How will addition of a catalyst affect a system at equilibrium? Quite simply, it will not affect
it at all. A catalyst speed up the rates of both the forward and the reverse reactions of a system, so
adding a catalyst could help a system reach equilibrium more quickly. But, once at equilibrium the
catalyst will no longer have any effect.
185
Chapter 8: Gases, Liquids, and Solids
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as fluid, vapor, temperature, pressure, psi, bar, torr, mm Hg, Pascal,
ideal gas, ideal behavior, ideal gas law, standard temperature and pressure (STP), combined
gas law, Dalton's law of partial pressures, total pressure, partial pressure, mole fraction,
states of matter, phase transitions, enthalpies of phase transitions, Coulomb's law,
intramolecular force, intermolecular force, London (or, dispersion or induced dipole-induced
dipole) forces, dipole-dipole interactions, hydrogen bonds, polarizable, dipole moment,
solution, solvent, solute, solvation, "like dissolves like."
Explain how gases, liquids, and solids differ at a molecular level.
Explain the relationship between temperature and pressure.
Be able to interconvert between the various temperature units.
Be able to interconvert between the various pressure units.
Explain ideal behavior and the assumptions made relating to ideal behavior.
Given the appropriate information, use the ideal gas law to calculate the pressure, volume,
number of moles, temperature, or molar mass of a gas.
Use the combined gas law in calculations involving changing conditions.
Use Dalton's law of partial pressures to calculate the total pressure of a system or the partial
pressures of gases in the system.
Calculate the mole fraction of a system and use that information, along with the total system
pressure, to calculate partial pressures.
Describe the three common intermolecular forces discussed in this chapter and explain how
they are similar and how they differ.
Given a molecular formula of a simple compound, predict the types of intermolecular forces
of which it is capable.
Explain what "like dissolves like" means and how it can be used to predict whether or not
one substance will dissolve in another, and if not, why.
Caution: let me warn you from the beginning of this chapter. The Chapter 8 quiz focuses on gases and
there are no questions about intermolecular forces. These do not show up until the Chapter 9 quiz. I don’t
know that there is a good reason for this, but it is what it is. Students sometimes seem to feel they are only
responsible for the gas portion of the chapter and that they can ignore the very useful and important
information treating intermolecular forces. Rest assured, you are responsible for *all* of the material in this
chapter and you should plan to see questions pertaining to *all* of the various topics discussed in this
chapter, both in the quizzes and also in the final exam.
186
General gas properties
There are three common states of matter, solids, liquids, and gases. At a molecular level the relative
differences between these three states of matter generally depend on the distance between the
particles and on the strengths of the attractive interactions between the particles (remember, we use
the word "particle" in a generic sense to describe a substance that at an atomic level may consist of
individual atoms, ions, or molecules):
•
•
•
in gases the particles are far apart and there are no attractive interactions between the gas
particles
in liquids the particles are close together and there are relatively weak attractive interactions
between the liquid particles
in solids the particles are very close together and there are relatively strong attractive
interactions between the gas particles
Substances in a gas or liquid state are often referred to as fluids, substances with no fixed
volume or shape but which take the shape and volume of their container. The words vapor and gas
are used synonymously in general vocabulary but there is a distinction in science, even though they
have reference to the same physical state. Substances that are usually in a liquid or solid state at
room temperature and pressure but which have been converted to a gaseous state are called vapors.
Substances which are found in the gaseous state at room temperature and pressure are called gases.
This is why we talk about hydrogen gas and nitrogen gas but water vapor.
To better understand the states of matter at a molecular level, particularly gases, we must
discuss temperature and pressure. Temperature, T, can be defined as a measure of how hot or cold
something is. There is a direct correlation between the temperature of a substance and the kinetic
energy of its particles. The kinetic energy (K.E.) of atoms and molecules can be described
mathematically as
K.E. = 3/2 kT
where k is Boltzmann's constant and is equal to 1.38 x 10-23 JAK-1 We won’t be solving this equation,
so you don't need to remember this constant in this class. Kinetic energy can also be described as
K.E. = ½ mv2
where m is the mass of the particle and v is its velocity. Since both of these equations are equal to
the same thing, kinetic energy, they are also equal to each other
½ mv2 = 3/2 kT
187
If we multiply both sides by 2 we obtain
mv2 = 3 kT
and if we rearrange the equations in terms of velocity we see that
v = (3 kT/m)½
This equation tells us that the velocity of atoms and molecules is directly dependent on their mass
and on the system temperature. As temperature increases the velocity of particles will also increase.
And as temperature deceases the velocity of the particles will also decrease. Given the relationship
between velocity and kinetic energy, this means that as particles move faster they have greater
kinetic energy. Therefore, as the temperature of particles increases, they have greater kinetic energy.
Conversely, as the temperature of particles decreases, their kinetic energy also declines. We will not
be using these equations quantitatively in this class, but it will help some of you to more clearly understand
the relationship between temperature and kinetic energy.
In our study of gases, temperature is always expressed in Kelvin. While we commonly work
with the Celsius scale in chemistry, especially when making measurements, you will never be able
to correctly make calculations when studying gases without first converting Celsius temperatures
to the Kelvin scale. This is a common source of error for students in this chapter. Always think Kelvin!
Pressure, P, is defined as the ratio of a force applied to a surface divided by the area over
which the force is applied, i.e.,
Pressure = Force/Area over which the force is applied
At a molecular level, when molecules collide with a surface they exert a force upon it. The force
exerted by molecular collisions depends on the kinetic energy of the collisions. It also depends on
the frequency of the collisions. The more frequent the collisions, the greater the pressure. We can
re-write our equation describing pressure as
Pressure = (K.E. of collisions) x (# of collisions per unit time)/Area over which collisions occur
The role of temperature in the behavior of gases becomes more apparent here. We proved
above the relationship between temperature, velocity, and kinetic energy. If pressure is dependent
on both the kinetic energy of collisions and on the frequency of collisions, then it should be apparent
that as temperature increases, pressure will also increase. This is because as temperature increases,
the particles move more rapidly and can collide more frequently. It is also because as the particles
move more rapidly, they have greater kinetic energy. Knowing these things, can you explain why
pressure decreases as the temperature decreases?
188
The units in which pressure may be expressed are a bit more varied than those of
temperature. Pressure is commonly expressed in units called atmospheres (atm), but there are other
units also used.
Pressure may be expressed in pounds per square inch (psi, or lbs/in2 ).
1 atm = 14.7 psi
This is because the air contained in a column with a surface area of one inch by one inch that
stretches from sea level to the upper reaches of the atmosphere will weigh exactly 14.7 pounds.
Pressure may be expressed in units of bars:
1 bar = 14.5 psi
This unit is not typically used in elementary chemistry and will not be used in this class. However,
it is commonly used in meteorology. If you are watching the weather on television, you will
commonly hear barometric pressure expressed in units of millibars (thousandths of bars).
Pressure may be expressed in units of torr, T, or in millimeters of mercury, mm Hg. Since
temperature is also indicated with the symbol T, this is sometimes confusing. You must establish whether
temperature or the pressure of a gas in units of torr is being discussed by paying close attention to the
context of the discussion or problem in which the symbol is used. The unit torr takes its name from the
Italian physicist Evangelista Torricelli who invented the mercury barometer. A simple barometer
consists of a sealed glass tube filled with mercury. The tube is inverted in a dish of liquid mercury
with the end of the tube submerged beneath the surface of the mercury in the dish. Surprisingly,
when the tube is inverted, not all of the mercury runs out of the tube. The pressure of air on the
surface of the mercury in the dish is sufficient to force much of the mercury back up into the tube.
In other words, the pressure exerted by air on the surface of the mercury in the dish is sufficient to
offset the force of gravity pulling the mercury down the tube. Whenever air pressure changes the
height of the column of mercury in the tube changes as well. If atmospheric pressure increases, the
height of the column of mercury in the tube also increases as air pressure forces the column of
mercury slightly further up the tube. If atmospheric pressure decreases, the height of the column of
mercury in the tube also decreases. The relationship between atmospheres, torr, and mm Hg is
1 atm = 760 T = 760 mm Hg
Note that 1 torr is exactly equal to 1 mm Hg. You can see photographs and other images and learn
more about barometers at the Wikipedia address http://en.wikipedia.org/wiki/Barometer.
The SI unit of pressure is the Pascal (Pa). We will not use this unit in this class but you should
still know what it is, as you may bump into it at some future point outside of class.
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1 atm - 101,325 Pa
In summary, you should be familiar with the following relationships between the various units of
pressure:
1 atm = 14.7 psi = 760 T = 760 mm Hg = 101,325 Pa
Finally. When studying gases, experiments are commonly performed at standard temperature
and pressure (STP). For gases, these conditions are 1.00 atm and 273 K.
Ideal gases and the Ideal Gas Law (Universal Gas Law)
It will be easier to study gases if we make two assumptions about them. We will assume that
gas particles do not occupy any volume. Which is, of course, completely untrue. Gases are mostly empty
space. But while in any sample of gas the volume occupied by the actual molecules of gas is extremely
small, it is still a finite number. However, this assumption does make it possible to make some remarkable
calculations that are generally quite accurate as we study most gases. And we will assume that gas
particles do not interact with each other. Again, the absolute correctness of this assumption is rather poor
in most cases, but the short-term benefits we derive from making it make it well worth our while to do so.
And surprisingly, in most cases, the error introduced is small. When we make these two assumptions
about a gas, about any gas, we state that the gas exhibits ideal behavior. In the real world nearly all
gases show ideal behavior under appropriate conditions, usually at low pressures and at high
temperatures. The implication of this is that, regardless of composition, all gases behave roughly the
same under the same set of conditions. This is a truly remarkable implication, if you think about it!
Historically, the study of gases lies at the heart of the science of modern chemistry. These
studies began in the 17th century and continued to shape the development of chemistry during the
18th and 19th centuries. It is common to acknowledge the contributions of Robert Boyle, Jacques
Charles, Joseph Gay-Lussac, and Amedeo Avogadro by using the laws and relationships they
discovered to explain the behavior of gases. But, in actual fact, it is not necessary to know or
remember their laws individually. They can be synthesized into one relationship, the ideal gas law.
This law explains the relationship between pressure, volume, the amount (number of moles) of gas,
and temperature as follows:
PV = nRT
•
•
•
•
P is the pressure of the gas, units = atmospheres (atm)
V is the volume the gas occupies, units = liters (L)
n is the number of moles of gas in the sample under study, units = moles (mol)
T is the temperature of the gas, units = Kelvin (K)
190
•
R is a constant, called the gas constant and is equal to 0.0821 LAatm/molAK. This is rather an
odd set of units, but it derives from the origin of R which is calculated using the equation
R = PV/nT.
How do we use the ideal gas law in the study of chemistry? This is an equation with four
variables and one constant. In problems in which we study gases using the ideal gas law, we will be
given three of the four variables. Using algebra, we will re-arrange the equation and solve for the
variable of interest. Some examples of ideal gas law problems are as follows.
•
If 1.00 mole of a gas occupies a volume of 1.00 L at 298 K, what is the pressure inside the
container?
Start by using a bit of algebra to rearrange the ideal gas law:
P = nRT/V
P = (1.00 mol) x (.0821 LAatm/molAK) x (298 K)/1.00 L = 24.46 atm
Be sure to check your units carefully when you set up and work ideal gas law problems! These are, really,
just dimensional analysis problems. And smile - that’s supposed to make you feel better about things.
•
A 5.00 L container holds a gas at 810 T and 62oC. How many moles of gas are present?
Start by using a bit of algebra to rearrange the ideal gas law
n = PV/RT
convert 810 T to atm: 810 T/760 T/atm = 1.066 atm
convert 62oC to K: 62oC + 273 = 335 K
n = [(1.066 atm) x (5.00 L)]/[(.0821 LAatm/molAK) x (335 K)] = 0.194 moles
•
If you blow exactly 2.00 moles of air into a balloon at a pressure of 780 mm Hg and at a
temperature of 35oC, what volume must the balloon hold to keep from bursting? Note: an 11"
balloon holds a volume of 11.4 L, and an 18" balloon holds a volume of 50.0 L.
Start by using a bit of algebra to rearrange the ideal gas law
V = nRT/P
convert 780 T to atm: 780 mm Hg/760 mm Hg/atm = 1.026 atm
convert 35oC to K: 35oC + 273 = 308 K
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V = (2.00 mol) x (.0821 LAatm/molAK) x (308 K)/(1.026 atm) = 49.3 L
You're probably thinking to yourself, why in the world didn't we just use the 18" balloon and save ourselves
the trouble of these calculations? It's true, we could have done things that way, but just think of the fun you
would have cheated yourself out of!
•
6.75 moles of an unknown gas occupy a volume of 13.3 L at a pressure of 17.6 atm. What
is the temperature of the gas?
Start by using a bit of algebra to rearrange the ideal gas law
T = PV/nR
T = [(17.6 atm) x (13.3 L)]/[(6.75 mol) x (.0821 LAatm/molAK)] = 422.4 K
If you're having problems using algebra to rearrange the equation in these examples, I can't help you.
You're going to need to review your basic algebra on your own.
Let me point something out that you may not have noticed. We just worked four examples
using the ideal gas law. In any one of the problems, did we ever give any consideration to what the
gas in the problem was? Looking back, you can see that in three of the four problems, the gas isn't
even identified. Why can we do this? Because the ideal gas law operates on the assumption that all
gases behave ideally. This means that they all behave the same, regardless of their chemical
composition, under the same physical conditions.
The ideal gas law and molar mass calculations
The Ideal Gas Law can be used to find the molar mass of substances. This is based on the
variable n in the ideal gas law. If we have a certain amount of a substance, we can calculate the
number of moles of the substance if we know how much of it (i.e. it's mass in grams) we have and
if we know it's molar mass. In equation form
n = number of moles = mass of substance (grams)/molar mass (g/mol)
For example, if we have 36 grams of water, and given that the molar mass of water is 18 g/mol, then
we have 2 moles of water.
Given that
n = g/Mm
where g is the grams of substance and Mm is it molar mass, we can substitute this into the ideal gas
law
192
PV = gRT/Mm
and by re-arranging the ideal gas law and solving for molar mass, we have
Mm = gRT/PV
Here are two examples of problems that use this form of the ideal gas law:
•
44.9 g of an unknown gas exert a pressure of 2.55 atm within a 10.2 L container at 325 K.
Is the unknown gas nitrogen dioxide or sulfur dioxide?
To answer this particular question we need to know the molar masses of the two gases of
interest. The molar masses are 46.01 g/mol for NO2 and 64.06 g/mol for SO2.
Mm = gRT/PV
Mm = [(44.9 g) x (.0821 LAatm/molAK) x (325 K)]/[(2.55 atm) x (10.2 L)] = 46.06 g/mol
Based on the similarities of their molar masses, the unknown gas must be nitrogen dioxide.
•
A damaged unmarked cylinder of compressed gas is found in the ruins of a burned building.
You suspect that the gas is either argon, nitrogen, helium, or carbon dioxide. A 5.00 L mylar
bag is filled with the gas to a pressure of 1900 T at 298 K. When weighed, the mass of the
gas is 20.42 g. What is it?
Again, to answer this particular question you'll need to know the molar masses of the four
named gases that the unknown gas may possibly be.
Mm = gRT/PV
convert 1900 T to atm: 1900 T/760 T/atm = 2.50 atm
Mm = [(20.42 g) x (.0821 LAatm/molAK) x (298 K)]/[(2.50 atm) x (5.00 L)] = 39.97 g/mol
The unknown gas is argon.
The ideal gas law and stoichiometry
We can also use the ideal gas law to help in the solution of stoichiometry problems. Two
examples follow.
•
An 18" balloon has a volume of 50.0 L. How many grams of water must be decomposed via
electrolysis to fill the balloon with hydrogen to a pressure of 1.05 atm at 298 K?
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By rearranging the ideal gas law and using the information given find n
n = PV/RT = [(1.05 atm) x (50.0 L)]/[(.0821 LAatm/molAK) x (298 K)] = 2.146 moles hydrogen
Now we need a balanced chemical equation for the reaction
2 H2O(l) => 2 H2 (g) + O2 (g)
Use the balanced chemical equation and the information used from solving the ideal gas law
to answer the question
(2.146 moles H2) x (2 moles H2O/2 moles H2) x (18.02 g H2O/1 mol H2O) = 38.7 grams of water
•
The reaction of 75.0 g of iron (III) sulfide with excess hydrochloric acid will produce what
volume of gas at 755 T and 293 K?
In this problem we need to begin with a balanced chemical equation but then, we always
need a balanced equation to do stoichiometry problems:
Fe2S3 (s) + 6 HCl(aq) => 2 FeCl3 (aq) + 3 H2S(g)
Use the balanced chemical equation to determine the number of moles of gas that will be
produced.
(75.0 g Fe2S3) x (1 mol Fe2S3/207.87 g Fe2S3) x (3 mol H2/1 mol Fe2S3) = 1.082 moles H2S
By rearranging the ideal gas law and using the information given, find V:
convert 755 T to atm: 755 T/760 T/atm = 0.993 atm
V = nRT/P = [(1.082 mol H2S) x (.0821 LAatm/molAK) x (293 K)]/(0.993 atm) = 26.2 L
The combined gas law
We can use the ideal gas law to help us solve problems in which conditions are constant. But
how do we solve problems in which one or more conditions change during the course of the
problem?
Let's say that we have two sets of conditions, our initial conditions and our final conditions.
We can describe both the initial and final conditions using the ideal gas law:
initial: PiVi = niRTi
and
final: PfVf = nfRTf
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If we express these equations in terms of the gas constant R
R = (PiVi)/(niTi)
and
R = (PfVf)/(nfTf)
Since R is a constant and therefore the same in both cases, we can say that
(PiVi)/(niTi) = (PfVf)/(nfTf)
This form of the ideal gas law is called the combined gas law. I’ve also heard students call it
other things, but none of them should be repeated in polite company. This is the equation we use when
we solve gas problems in which one or more conditions change during the course of the problem.
This provides for Boyle's Law, Charles's Law, Gay-Lussac's Law, and Avogadro's Law all in the
same equation. If you remember this relationship you don't need to remember all of the others.
Below are five examples of problems solved using the combined gas law.
•
7.5 moles of a gas at 2.25 atm of pressure occupy a volume of 2.00 L. If the gas is
compressed to a volume of 1.00 L what is the new pressure of the gas?
To solve problems using the combined gas law, I suggest setting up a table in which you list
the information given to you in the problem. This will often help clarify exactly which
variable we need to solve for
Pi
2.25 atm
Pf
?
Vi
2.00 L
Vf
1.00 L
ni
7.5 mol
nf
Ti
Tf
Note that temperature is not mentioned in the problem. We infer from this that temperature
does not change and therefore that the initial and final temperatures are the same. Also note
that the initial number of moles is mentioned, but it is not mentioned again. We infer from
this that this value also does not change during the course of the problem. We begin with
(PiVi)/(niTi) = (PfVf)/(nfTf)
Since the temperatures and numbers of moles are unchanging and are therefore constant,
they can be cancelled and
PiVi = PfVf
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We're interested in finding Pf so we rearrange the equation and arrive at
Pf = PiVi/Vf
[(2.25 atm) x ( 2.00 L)]/(1.00 L ) = 4.50 atm
Does it make sense, physically, for the pressure to increase when the volume decreases? Why? Think in
terms of collisions if you have difficulty answering this question.
This is actually a Boyle's Law problem. Boyle's law is the equation PiVi = PfVf, and yet we were able to solve
this problem without even realizing we needed Boyle's law, simply because we know how to use the
combined gas law. The next few problems we work will also be problems that could be solved using the
appropriate law - Charles's law, Avogadro's relationship, Gay-Lussac's law - if we knew them. But we don't
have to know them if we know how to use the combined gas law. We don't even have to worry about
them.
•
A sample of gas at 373 K exerts a pressure of 1500 T. If the temperature drops to 273 K,
what will the new pressure be?
Let's begin with our table of values, remembering to convert our pressures from torr to atm:
Pi
1.974 atm
Pf
Vi
Vf
ni
nf
Ti
373 K
Tf
?
273 K
(PiVi)/(niTi) = (PfVf)/(nfTf)
Since the volumes and numbers of moles are unchanging and are therefore constant, they can
be cancelled and:
Pi/Ti = Pf/Tf
Once again we're interested in solving for Pf so we rearrange the equation and arrive at
Pf = PiTf/Ti
[(1.974 atm) x ( 273 K)]/(373 K) = 1.44 atm
Is the answer physically consistent with what we know about the relationship between temperature and
pressure to see the pressure decrease with a decrease in temperature?
196
•
10.0 moles of helium at 298 K occupy a volume of 20.5 L at a pressure of 11.93 atm. If the
temperature of the gas increases to 328 K, how must the container volume change for the
pressure to remain the same?
Pi
11.93 atm
Pf
11.93 atm
Vi
20.5 L
Vf
?
ni
10.0 mol
nf
Ti
298 K
Tf
328 K
(PiVi)/(niTi) = (PfVf)/(nfTf)
If the pressure is to stay the same, this means that the initial and final pressure must be the
same. The number of moles is constant. Our equation becomes
Vi/Ti = Vf/Tf
We want to solve for Vf and so the equation becomes
Vf = ViTf/Ti
[(20.5 L) x ( 328 K)]/(298 K) = 22.56 L
We know that as temperature increases, pressure will also increase. The only way to prevent the pressure
from increasing with an increase in temperature is to increase the size of the container holding the gas,
which will prevent an increase in the number of collisions by spreading the gas particles further apart.
•
A 3.6 mole sample of gas in a sealed vessel has a pressure of 4.25 atm. If the number of
moles of gas is increased to 4.9 moles, what will the new pressure be?
Pi
4.25 atm
Pf
Vf
Vi
ni
?
3.6 mol
nf
Ti
4.9 mol
Tf
(PiVi)/(niTi) = (PfVf)/(nfTf)
The volume and temperature are constant in this problem. Our equation becomes:
Pi/ni = Pf/nf
197
To solve for Pf the equation becomes
Pf = Pinf/ni
[(4.25 atm) x ( 4.9 mol)]/(3.6 mol) = 5.78 atm
Why should pressure increase if we add more particles at the same temperature to a gas constrained by
a constant volume?
•
10.0 moles of a gas at 26.0 atm and 303 K occupy a volume of 9.57 L. If the number of
moles of gas is suddenly changed to 15.0 moles while the temperature increases to 373 K
and the volume decreases to 5.00 L, what will the new pressure in the container be?
Pi
26.0 atm
Pf
?
Vi
9.57 L
Vf
5.00 L
ni
10.0 mol
nf
15.0 mol
Ti
303 K
Tf
373 K
(PiVi)/(niTi) = (PfVf)/(nfTf)
Nearly everything changes in this problem. Can we still solve it? Of course we can. The
combined gas law has eight variables. As long as we can account for seven of the eight, we
can solve the problem. To solve for Pf the equation becomes
Pf = PiVinfTf/VfniTi
[(26.0 atm) x ( 9.57 L) x (15.0 mol) x (373 K)]/[(5.00 L) x (10.0 mol) x (303 K)] = 91.9 atm
Increasing the temperature increased both collision frequency and the kinetic energy of the collisions.
Decreasing the volume increased the collision frequency, as did increasing the number of moles of gas. On
the basis of these things we would expect the pressure to increase, and it did.
Dalton's Law of Partial Pressures and mole fractions
Given a system that contains two or more gases, the total pressure of the system is equal to
the sum of pressures of all of the gases present. This is described by the equation
PT = P1 + P2 + P3 + . . . .
where PT is the total pressure of the system and P1, P2, and P3 are the pressures of the gases present
in the system. In other words, the pressures of multiple gases in a system are additive. They do not
198
cancel each other out or affect each other in any other way. This equation is true regardless of
whether a system contains only two gases or one hundred or more gases. A fundamental assumption
is being made here, that none of the gases react with each other. If they do, the things we are about to
discuss do not work.
The pressure that each of the individual gases contributes to the total pressure is called its
partial pressure. The partial pressure of a gas is in no way incomplete for that gas, but rather, it is
simply a part of the total pressure of the system.
In other words, any time you hear someone refer to the pressure of a gas as a partial pressure, they’re
implying that it is not the only gas in the system. As an example, when you study physiology you read of
the partial pressure of oxygen in the blood. This means, quite simply, that there are other gases dissolved
in the blood besides oxygen.
The partial pressures of the gases in a system can be determined using the ideal gas law. As
examples:
•
A sealed flask contains nitrogen at a pressure of 500 T and oxygen at a pressure of 250 T.
What is the total pressure in the flask?
PT = 500 T + 250 T = 750 T
You may have noticed that in this case we did not convert from torr to atmospheres. In problems in which
the ideal gas law is not used, we can get away with it. If you don't believe me, convert the two values from
torr to atm, add them together, and then convert them back to torr. And to think you doubted me.
•
A 10.0 L flask contains 1.0 mole of carbon dioxide, 2.0 moles of methane, and 4.0 moles of
water vapor. If the temperature inside the flask is 303 K, what is the total pressure? In this
problem we're going to need to use the ideal gas law to find the pressure of each of the three
gases.
PCO2 = nRT/V = [(1.0 mol) x (.0821 LAatm/molAK) x (303 K)]/(10.0 L) = 2.49 atm
Pmethane = nRT/V = [(2.0 mol) x (.0821 LAatm/molAK) x (303 K)]/(10.0 L) = 4.98 atm
PH2O = nRT/V = [(4.0 mol) x (.0821 LAatm/molAK) x (303 K)]/(10.0 L) = 9.95 atm
PT = PCO2 + Pmethane + PH2O = 2.49 atm + 4.98 atm + 9.95 atm = 17.4 atm
There's one final concept we need to mention while we're discussing partial pressures. In a
system containing two or more gases, the mole fraction of a gas can be described as either the ratio
of that gas's partial pressure divided by the total pressure of the system, or as the ratio of the number
199
of moles of that gas to the total number of moles of gas in the system. In a system of two or more
gases, the mole fraction X1 of gas 1 is equal to
X1 = (P1/PT) = (n1/nT)
where P1 is the partial pressure of gas 1, PT is the total pressure in the system, n1 is the number of
moles of gas 1, and nT is the total number of moles of gas in the system. The same would hold true
for the second gas in the system, and the third, and so on. For gas 2,
X2 = (P2/PT) = (n2/nT)
In the example we just worked there is 1.0 mole of carbon dioxide gas, 2.0 moles of methane gas,
and 4.0 moles of water vapor. There are therefore 7.0 (1.0 + 2.0 + 4.0) total moles of gas in the
system. The mole fraction of carbon dioxide XCO2 is therefore 0.143 (1.0/7.0), the mole fraction of
methane Xmethane is 0.286 (2.0/7.0), and the mole fraction of water vapor XH2O is 0.571 (4.0/7.0).
The sum of the mole fractions of a system will always be equal to exactly 1. In a system with
“n” different gases,
X1 + X2 + X3 + ... Xn = 1
In our example, 1/7 + 2/7 + 4/7 = 1. Mole fractions are always unitless values.
The concept of mole fractions becomes useful in the calculation of partial pressures. In a system
with two or more gases, the partial pressure of any gas is equal to the product of its mole fraction
and the total system pressure. In other words, for gas 1,
and for gas 2
P1 = X1PT
P2 = X2PT
Now let's apply this concept to the problem we just worked. Let's suppose we're told that we
have a system containing 1.0 mole of carbon dioxide, 2.0 moles of methane, and 4.0 moles of water
vapor and that the total pressure of the system is 17.4 atm. Having already calculated the mole
fractions of each of these three compounds, then
PCO2 = XCO2 PT = (0.143) x (17.4 atm) = 2.49 atm
Pmethane = Xmethane PT = (0.286) x (17.4 atm) = 4.98 atm
PH2O = XH2OPT = (0.571) x (17.4 atm) = 9.95 atm
200
This concept is often a useful and easy way to calculate partial pressures, given the proper
information.
States of matter, phase transitions, and enthalpies of phase transitions
We have discussed the three common states of matter in Chapter 1, in Chapter 7, and again
at the beginning of this chapter. Let's summarize the things we’ve discussed throughout this course
thus far:
Gases
•
are fluids
•
have no fixed volume or shape - a gas will fill and take the shape of its container
•
have no intermolecular potential energy: there are no attractive or repulsive interactions
between gas particles
•
gas particles are very far apart compared to their size
•
the total energy of a perfect gas is the sum of its translational and internal energy (vibrations,
rotations, spins, etc.)
•
there is no short-range order or long-range in gases
Liquids
•
are fluids
•
have a fixed volume but no fixed shape - a liquid will take the shape of its container
•
intermolecular potential energy: there are attractive or repulsive interactions between liquid
particles
•
liquid molecules are close together compared to their size, which is why liquids are largely
incompressible
•
the translational energy of liquid particles is much lower than in gas (i.e., it's harder for them
to move around than in the gas phase)
•
in liquids there is short-range order but no long-range order
Solids
•
fixed volume and shape
•
intermolecular potential energy: there are attractive or repulsive interactions between solid
particles
•
solid particles are, on average, closer together than in particles in the liquid phase, which is
why solids are incompressible
•
the translational energy of solids is much lower than in liquids (i.e., it's harder for them to
move around than in the liquid phase)
•
in solids there is both short-range order and long-range order (especially in crystalline solids)
201
We also learned in Chapter 1 that there are specific names for the transition of one state of
matter to another.
In Chapter 7 we learned that there are enthalpies associated with physical changes of state
(phase changes, or phase transitions), such as the enthalpy of fusion, melting, vaporization, and
condensation. Since enthalpy is associated with processes that result in a system gaining or losing
energy, it should not surprise us to learn that we can also think of phase transitions in terms of
enthalpy. Below is a plot of the changes in the state of water with the addition of heat.
The plot begins with solid water (ice) at a temperature less than -50°C.
202
As a side question, is it possible to cool water to a temperature less than it’s freezing point? More generally
speaking, is it possible to cool any substance to a temperature below its freezing point? The answer is yes.
We can cool any thing and all things to a temperature slightly above 0 K, absolute zero, regardless of their
freezing points. So water at a temperature less than its freezing point should not come as much of a
surprise to you.
As heat is added to the ice its temperature increases rapidly. However, once we reach 0°C we see
something curious. The temperature of the ice remains constant, even though heat is added to the
system. This is represented by the first plateau, or leveling, on the left side of the line. Why does this
occur? Because before the solid can be converted to a liquid we must add enough energy to the
system to help the water molecules overcome the strong attractive interactions that hold them
together in the solid state. The energy that must be put into the system before the solid can transform
to a liquid is called the enthalpy of melting. Since we must put energy into the system to convert a
solid into a liquid, we say that melting is an endothermic process.
Once the solid has been converted to a liquid, the temperature of the liquid increases as heat
is added to the system. The slope of this line is different from the slope of the solid line because ice and
liquid water have different heat capacities. That's the explanation, but you needn't worry about it in this
class. The increase in temperature continues until the liquid water reaches a temperature of 100°C.
At this point we reach a second leveling of the line. This plateau is much longer than the first. It
represents the addition of heat to the system but without an increase in the temperature of the
system. This occurs because energy must be added to the system to break the strong interactions that
take place between the water molecules in the liquid state. These interactions must be broken before
the liquid can transform to a vapor (gas). The energy that must be put into the system before the
liquid can transform to a vapor is called the enthalpy of vaporization. Since we must put energy into
the system to convert a liquid into a vapor, we say that vaporization is also an endothermic process.
Once the liquid has been converted to a vapor, the temperature of the vapor continues to
increase as heat is added to the system. The slope of this line is different from the slope of the liquid line
because liquid water and water vapor have different heat capacities. Again, you don't need to worry about
it in this class.
What about the reverse processes, condensation and freezing?
Ok, here’s another question. Is it possible to heat water above its boiling point at 100oC? And again, more
generally speaking, is it possible to heat any substance to a temperature in excess of its boiling point?
Again, the answer is yes. Exactly how hot can water be heated? Any compound, regardless of whether it
is ionic, covalent, or metallic, will eventually decompose into the elements of which it is made if you heat
it to a temperature hot enough to break the bonds holding the atoms together. At a temperature in excess
of 2000oC water becomes so hot that the covalent bonds holding the hydrogen atoms and oxygen atoms
together break and water decomposes. This decomposition temperature will differ for each compound.
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Assume we begin a second experiment with water vapor heated to a temperature in excess
of 150°C. You can follow this experiment by following the line on the plot from right to left, rather than
from left to right as we did just a moment ago. As we remove heat from the system (i.e cool the
system) the temperature of the system decreases. This continues until our vapor reaches a
temperature of 100°C. At this point, we reach a leveling of the line. We continue to remove heat
from the system but the system temperature does not decrease. This is because the relatively high
energy vapor molecules must release some of their kinetic energy before they are able to settle into
the (relatively speaking, much more sedate, stable, lower energy) liquid phase. The energy the
system must lose before the vapor can condense into a liquid is called the enthalpy of condensation.
Thus we see that condensation is an exothermic process. The enthalpies of vaporization and
condensation are related. They are equal in magnitude but opposite in sign.
Once the vapor has been converted to a liquid, the temperature of the liquid decreases as heat
is removed from the system. The decrease in temperature continues until the liquid water reaches
a temperature of 0°C. At this point we reach a second leveling of the line. It represents the removal
of heat from the system but without a decrease in the temperature of the system. This occurs because
energy must be removed from the system before it is sufficiently stable for the strong interactions
that take place between water molecules in the solid phase to form. The energy the system must lose
before the liquid can freeze into a solid is called the enthalpy of fusion. Fusion (freezing) is also an
exothermic process. The enthalpies of fusion and melting are also related. They too are equal in
magnitude but opposite in sign. Note that fusion can mean one of two things. In thermodynamics fusion
means freezing. But outside of thermodynamics, which fortunately is most of the time, fusion refers to
nuclear reactions as described in Chapter 11, and not to chemical reactions or physical changes
Intermolecular forces: attractive interactions between molecules
To this point in this class, when we've discussed chemical bonds we have been talking about
intramolecular bonds (or, intramolecular forces), although you did not know it. Intramolecular bonds
are the bonding interactions that occur between atoms within a molecule. Intramolecular forces
include metallic, ionic, and covalent bonds.
There are attractive interactions besides intramolecular forces. These are the interactions that
take place between molecules, rather than within molecules. Intermolecular forces are the bonding
interactions between molecules. They are found primarily between particles in the liquid and solid
phases. If there were no intermolecular forces, all substances would behave as ideal gases. Liquids
and solids are held together by intermolecular forces.
Intramolecular forces and intermolecular forces are similar in that they are both based on the
attraction of oppositely charged particles. In Chapter 4 we first saw the equation, which is called
Coulomb's law, that describes the force with which oppositely charged particles attract each other:
F
kq1q2
r2
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Intramolecular forces result from the attraction of ions for one another (ionic bonds) or for the
attractions between electrons and protons (covalent bonds, metallic bonds). These attractive
interactions are based on full integer charges. Intermolecular forces result from molecular polarity.
We discussed this in Chapter 5. If you don’t remember now would be an extremely good time to do a bit
of review, as you’re going to be terribly lost if you don’t recall the things you hopefully learned back then.
Molecular polarity result from a molecule having one or more polar bonds. The charges of polar
molecules are only partial charges. As a result, intermolecular bonds are much weaker than
intramolecular bonds. The strongest intermolecular bonds are only about 10-25% as strong as a
typical covalent bond.
We will discuss four types of intermolecular forces in this chapter, London (or, dispersion)
forces, dipole-dipole interactions, hydrogen bonds, and ion-dipole interactions, although there are
other types of intermolecular forces besides these. These first three of these intermolecular
interactions are known collectively as van der Waal's forces (interactions).
London (dispersion) forces
London forces, or dispersion forces, are also known as induced dipole-induced dipole
interactions. They take their name from Fritz London, who identified their role in chemical bonding
in 1930. London forces are generally the weakest type of intermolecular interaction, although they
can be significant, especially in large, nonpolar molecules. London forces are the only
intermolecular bonding force that occur between nonpolar molecules. However, all substances are
capable of London forces. Even if other stronger forces predominate the interactions between
molecules, London forces may still be a major contributor to the interactions between polar
molecules. They are the most common intermolecular forces. London forces are one of the principle
intermolecular forces in living systems.
London forces are weak because (a.) partial charges are involved and (b.) because they are
temporary. They arise from the random movement of electrons in atoms and molecules. Let's
consider two nonpolar molecules. This we see in the below image.
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Note: nonpolar molecules are not necessarily round. I have simply represented them this way for ease of
illustration. While electrons are constrained by the laws of quantum mechanics to specific shells,
subshells, and orbitals, depending on their energy, they do have a certain amount of freedom to
move about within atoms and molecules. This movement is extremely rapid and, in nonpolar
molecules, it is also more or less random. In general, we can describe the electron distribution in a
neutral, nonpolar molecule as a " 'cloud' of negative charge" surrounding the nucleus, which is
positively charged. ("Physical Chemistry;" Walter J. Moore; Prentice-Hall, Inc. 4th Ed., p. 914.)
While this electron cloud is, on average, distributed symmetrically with respect to the nucleus,
distortions of this cloud may arise and a nonpolar molecule may temporarily have a slight excess
of electrons in one portion of the molecule. We see this in the image below.
This means that, for the instant in which this charge imbalance exists, the molecule becomes polar,
with a partial negative charge at the end of the molecule with the excess of electrons and with a
partial positive charge at the end of the molecule that is temporarily deficient in electrons. The
polarity of this molecule, temporary though it is, has an effect on those molecules that are its
neighbors. This is shown in this image:
Electrons in the neighboring molecules that are adjacent to the negative end of the temporarily polar
molecule are slightly repelled, causing the neighboring molecule to also become temporarily polar.
Electrons in molecules that are adjacent to the positive end of the temporarily polar molecule are
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slightly attracted, which also causes these neighboring molecules to become temporarily polar.
During the brief interval that these temporarily polar molecules exist, they are attracted to each
other. This is a London force, as represented by the dashed lines between the spheres in the above
image.
Then, as the electrons in the initial molecule return to a random distribution
the polarity disappears, the bond is broken, and the electron distribution in the neighboring molecule
returns to the random state in which its electrons were initially found.
As we said, London forces are weak because the charges are only partial charges and also
because these partial charges are only temporary. The duration of the temporary polarity is generally
described as "instantaneous" in most books, although in actuality this instantaneous lifetime must
be finite albeit very brief. Let's assume for the sake of illustration that these temporarily polar
molecules only last for a few microseconds. In actual fact they may last for a longer or shorter period
of time, but humor me for just a moment, if you please. If this is the case, given the very high velocity
of electrons in atoms, every nonpolar molecule in a system may be capable of becoming temporarily
polar and forming these short-lived interactions with their neighbors many thousands of times every
second. In a collection of a mole of molecules, this would result in an unimaginably large number
of these interactions, even in a short period of time. By themselves London forces are weak, but
when you consider the enormous number of interactions that can take place between a large number
of molecules in a few seconds, you can see that collectively these interactions are not
inconsequential.
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The magnitude of London forces increases with molecular size and surface area
As molecules (and individual atoms as well) become larger, they have more electrons. We
know from previous discussions that the further we find electrons from the nucleus the more loosely
they are held. This is also true of molecules. We say that large molecules are more polarizable than
smaller molecules. Polarizability is a measure of the ease of distortion of the electron cloud of atoms
and molecules. Molecules that are more easily polarized can develop larger partial charges during
the brief interval in which they have asymmetric electron distributions.
Surface area also plays a role in the magnitude of London forces. We will only describe this
qualitatively. Imagine two sets of refrigerator magnets. Imagine trying to stick two of the refrigerator
magnets together by their ends. Now imagine sticking two of the magnets together face to face.
Which will stick together more forcefully? This is analogous to the effect of surface area on
intermolecular bond strength.
Dipole-dipole interactions
With very few exceptions, molecules that have one or more polar bonds are polar molecules.
Again, if you need to review this information you'll find it in Chapter 5. One important property of polar
molecules is called the dipole moment, which is a measure of the polarity of polar molecules. The
dipole moment of a polar molecule gives information about the magnitude of the partial positive and
negative charges within it. The larger the dipole moment of a polar molecule, the more polar the
molecule. The units of dipole moment are Debye units (D). Some examples of substances and their
dipole moments follows:
compound
dipole moment (D)
HCN
2.98
H2O
1.84
NH3
1.47
HCl
1.08
SF4
0.63
CO2
0
SF6
0
Ne
0
Note that the last three named compounds are nonpolar compounds, either due to a lack of polar
covalent bonds or due to symmetry. Symmetry, as explained at the end of Chapter 5 in these notes.
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Dipole-dipole interactions occur when a molecule with a permanent dipole moment forms a bond
with another molecule with a permanent dipole moment.
Dipole-dipole interactions only occur between permanently polar molecules. Compared with London
forces, dipole-dipole interactions are relatively strong because while polar molecules only bear
partial charges, they are permanently polar, as compared to the temporary polarity found in London
forces. Note: polar molecules are not necessarily oval shaped. I have simply represented them this way
for ease of illustration.
Hydrogen bonds - a special case of dipole-dipole interactions
Within a molecule, when a hydrogen atom is bonded directly to either a nitrogen, oxygen,
or fluorine atom, the resulting covalent bond is so polar that a very strong dipole-dipole bond is
formed when these molecules interact with other similar molecules. This type of dipole-dipole
interaction between molecules is called a hydrogen bond. Let me point something out. Based on the
electronegativity values in your text, a C-N covalent bond does not qualify as a polar bond. But in the real
world bond polarity actually depends on a number of factors. The difference in electronegativity in bonding
atoms is important, but it’s not the only thing that needs to be considered. Quite simply, we’re not going
to discuss these other factors in this class. You’re going to have to trust me on this one.
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Let's use an alcohol as an example. Alcohols are a very large family of organic compounds
with hundreds of family members. All alcohols are characterized by the presence of at least one
hydroxyl group. A hydroxyl group is an O-H group and differs from a hydroxide ion, OH-, in that
hydroxyl groups are uncharged. In the illustration below the "R" group bonded to the hydroxyl
groups can stand for any organic group and is unimportant in our consideration of the bonding that
takes place. The O-H bonds in hydroxyl groups are extremely polar due to the differences in
electronegativity in oxygen and hydrogen.
When two alcohol molecules approach each other with the correct orientation, the hydrogen
atom of one alcohol, which has a relatively large partial positive charge, begins to share one of the
two non-bonding pairs of electrons on the oxygen atom (which has a large partial negative charge)
of the hydroxyl group of the other alcohol molecule. This results in the formation of a partial
coordinate covalent bond and a very strong intermolecular interaction between the two alcohol
molecules. The dashed lines in the image below represent hydrogen bonds between the pairs of
alcohol molecules.
In case you don’t recall, we discussed coordinate covalent bonds in Chapter 5. Still, while hydrogen
bonds are a very strong intermolecular force, they are on average only about 10-25% as strong as
an average covalent bond. This is because the charges involved in hydrogen bonds are only partial
charges.
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Remember that hydrogen bonds are intermolecular bonds. There are exceptions to this,
especially in very large biological molecules such as proteins, but we will not consider those
exceptions in this class. The bond between the hydrogen atom and the nitrogen, oxygen, or fluorine
atom within the molecule is simply a polar covalent bond.
Ion-dipole interactions
An ion-dipole interaction is the intermolecular bonding force that occurs between a
permanently polar molecule and an ion. As one might expect, these are on average stronger
intermolecular forces than hydrogen bonds, as they involve interactions between a polar molecule,
which has its partial charges, with an ion, which has a full charge.
Predicting the intermolecular forces in compounds
Intermolecular forces play an important role in many important physical properties. We will
expand on this topic below. To predict the role of intermolecular forces in the physical properties
of a substance, we need to be able to accurately state the intermolecular forces of which the
substance is capable. Remember the general order of intermolecular bond strengths: on average,
ion-dipole bonds are the strongest, followed by hydrogen bonds, and then by dipole-dipole
interactions. London forces are generally the weakest intermolecular force, but all atoms and
molecules are capable of London forces.
To correctly identify the intermolecular forces of which a substance is capable we must be
able to correctly identify the substance as being polar or nonpolar. To do this we need to be able to
draw its Lewis structure and to identify whether or not any of its bonds are polar bonds, and also
whether or not the compound has symmetry.
Let's look at two examples of problems in which we need to be able to identify the
intermolecular forces of which compounds are capable.
•
Rank the following in order of the strength of their intermolecular forces: CH4 (methane),
CH2F2 (difluoromethane), H2 (hydrogen), and CH3OH (methanol).
Methane.
Methane is a nonpolar molecule. It therefore has London forces, because all
substances are capable of London forces. As a nonpolar molecule it cannot have
dipole-dipole interactions or hydrogen bonds. It is not capable of forming ion-dipole
interactions. Note: while we can and will find London forces, dipole-dipole interactions,
and hydrogen bonds in pure substances, in this class we will only find ion-dipole
interactions occurring in aqueous solutions of dissolved ionic compounds and never - in this
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class - in samples of pure substances, such as pure water, or pure methane, or pure - well,
I hope you get the idea.
Difluoromethane.
Note that the nonbonding pairs of electrons do not appear on the fluorine atoms in this
Lewis structure - sorry! I need a new software package that permits me to do better Lewis
structures. Difluoromethane is capable of London forces because all compounds,
polar and nonpolar alike, are capable of London forces. The C-F bonds in the
compound are polar bonds. This molecule has two polar bonds and does not have
symmetry and is therefore a polar molecule. It is also capable of dipole-dipole
interactions. It is not capable of forming hydrogen bonds.
Hydrogen.
Hydrogen is capable of London forces, as all substances are capable of London
forces. It is a nonpolar molecule so it is not capable of dipole-dipole interactions,
neither is it capable of forming hydrogen bonds.
Methanol.
Methanol is capable of London forces, as all substances are capable of London
forces. The C-O and O-H bonds are polar, and the molecule does not have symmetry.
It is therefore a polar molecule and is capable of forming dipole-dipole interactions.
Further, as there is a covalent O-H bond within the molecule, this is a compound that
can form hydrogen bonds with other similar molecules.
Ok, we’re now ready to rank these four compounds in order of the strength of their
intermolecular forces.
Methanol is capable of hydrogen bonding, which are the strongest intermolecular forces we
see in this collection of compounds, so it sits at the top of the list for these four compounds.
Difluoromethane cannot form hydrogen bonds but it can form dipole-dipole interactions,
which are stronger than London forces, so we rank it second in this group of four compounds.
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Methane and hydrogen both form London forces and nothing else, but we would expect the
London forces to be stronger in methane than in hydrogen because methane is larger molecule. So
we’d rank methane third and hydrogen last (or weakest) in ranking the strengths of the
intermolecular forces for these four compounds. No, I do not expect you to remember the names or
molecule formulas of the compounds used in this example, or in the next example either.
•
Rank the following in order of the strength of their intermolecular forces:
CH2OHCHOHCH2OH (glycerin), Ar, N2, CH3CH2F (fluoroethane), and CH3NH2
(methylamine).
We must again begin with the Lewis structures for the compounds in which we are
interested and use these to determine whether the compounds are polar or nonpolar.
Glycerin.
Again, I need to remind you that these Lewis structures do not show the nonbonding pairs
of electrons on the oxygen atoms in glycerine, the fluorine atom in fluoroethane, or the
nitrogen atom in methylamine. And again, I apologize. Also: yes, glycerine does have a
Lewis structure a bit more complicated than those we have seen in this class. Don’t panic.
I’m not going to ask you to do a Lewis structure this complicated on your own in this class.
We’re using it for the purposes of this example only. Glycerine is capable of London
forces, as all substances are capable of London forces. The C-O and O-H bonds are
polar, and the molecule does not have symmetry. It is therefore a polar molecule and
capable of forming dipole-dipole interactions. Further, as there is a covalent O-H
bond within the molecule, this is a compound that can form hydrogen bonds with
other similar molecules. Moreover, as there are three O-H groups on this one
molecule, we would expect this molecule to be able to form as many as three
hydrogen bonds per molecule.
Argon. The Lewis structure for this noble gas atom consists of a single atom (Ar)
surrounded by four nonbonding pairs of electrons. Argon is capable of London
forces, as all substances are capable of London forces. It is nonpolar (how could an
individual atom be otherwise than nonpolar?) so it is not capable of dipole-dipole
interactions, neither is it capable of forming hydrogen bonds.
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Neon. The Lewis structure for this noble gas atom consists of a single atom (Ne)
surrounded by four nonbonding pairs of electrons. Neon is capable of London forces,
as all substances are capable of London forces. It, as argon, is nonpolar so it is not
capable of dipole-dipole interactions, neither is it capable of forming hydrogen
bonds.
Fluoroethane.
Fluoroethane is capable of London forces, as all substances are capable of London
forces. The C-F bond is polar and the molecule does not have symmetry. It is
therefore a polar molecule and is capable of forming dipole-dipole interactions. But
at this point take care. Even though this a molecule which contains both fluorine and
hydrogen, as there is no hydrogen covalently bonded to the fluorine atom, it is *not*
capable of forming hydrogen bonds with other molecules.
Methylamine.
Methylamine is capable of London forces, as all substances are capable of London
forces. The C-N and N-H bonds are polar (remember what I said earlier about the
C-N bond), and the molecule does not have symmetry. It is therefore a polar
molecule and is capable of forming dipole-dipole interactions. Further, as there is a
covalent N-H bond within the molecule, this is a compound that can form hydrogen
bonds with other similar molecules.
We can now rank these five compounds in order of the strength of their intermolecular
forces.
Both glycerine and methylamine are both capable of hydrogen bonding, the strongest of the
intermolecular forces we see in pure substances. But as glycerine can form up to three hydrogen
bonds per molecule while methylamine can only form one, we would predict that the intermolecular
forces are stronger in glycerine than in methylamine.
Next we come to the dipole-dipole interactions in fluoroethane.
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Argon and neon both form London forces and nothing else. We expect the London forces
to be stronger in argon because it is larger molecule. So we’d say that glycerine has the strongest
intermolecular forces in this collection of substances, followed in order by methylamine,
fluoroethane, argon, and neon.
The relative strengths of intermolecular forces can be estimated by comparing the boiling
points of compounds. For boiling to occur in any substance enough energy must be added to the
system to overcome the intermolecular forces between the liquid phase molecules. The stronger the
intermolecular forces, the more energy must be added in order to overcome those forces. This is
reflected as a higher boiling point.
As an example, three substances have boiling points of 175°C, -30°C, and 50°C respectively.
Rank these three in order from strongest to weakest intermolecular forces:
(strongest) 175°C > 50°C > -30°C (weakest)
Did you notice that it doesn't even matter if we know what the compounds are. As long as we know
their boiling points we can make correct assumptions about the strengths of the intermolecular forces
in the substances. We cannot, however, come to any correct conclusions about the types of
intermolecular forces in these three substances. For this we need molecular formulas and Lewis
structures. A comparison of melting points and other physical properties can also provide insight
into the relative strengths of the intermolecular forces between compounds. We will discuss this
below.
Intermolecular forces and solutions
We first learned of solutions in Chapter 1 and we will study them in greater detail in
Chapter 9. Intermolecular forces are an essential feature of all solutions, but let's become familiar
with the vocabulary of solutions before we begin discussing them. You'll be seeing this first bit again
in Chapter 9, so save yourself some time and pay attention now.
•
•
•
•
solutions are homogeneous mixtures that are equally dispersed at the molecular level and
that are uniform throughout in their physical and chemical properties
solvent: does the dissolving in a solution and is generally the substance present in greatest
abundance. There is only ever one solvent per solution.
solute: gets dissolved and is the substance (or substances) that is less abundant. A solution
may have more than one solute.
solvation is the molecular process by which solutes are dissolved by solvents
A knowledge of the intermolecular forces of which a substance is capable can help us predict
the substance's solubility behavior. The phrase "like attracts like," or "like dissolves like" is a very
good general rule that can help us predict whether or not one substance will dissolve in another.
"Like dissolves like" means that substances with similar intermolecular forces will mix and form
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solutions, while substances with different intermolecular forces will not form solutions well, if at
all. Let's work a few examples of this.
Will the following pairs of substances form solutions?
•
•
•
•
•
•
Table salt and water. Table salt, sodium chloride, is an ionic compound. Water is a polar
liquid. Since the hydrogen bonds that hold water together in the liquid phase are similar to
the ionic bonds that hold sodium chloride together in the solid state, we predict that sodium
chloride and water should dissolve in each other.
Water and alcohol. Water is a polar liquid capable of forming hydrogen bonds. This is also
true of alcohols. Alcohols and water dissolve in each other.
Water and oil. Water is a polar liquid. Oil is a homogeneous mixture of various nonpolar
compounds. Since the intermolecular forces are dissimilar, oil and water do not dissolve in
each other.
Oil and vinegar. Oil is a homogeneous mixture of various nonpolar compounds. Vinegar is
an aqueous solution of acetic acid, a polar compound. Since the intermolecular forces are
dissimilar, oil and vinegar should not dissolve in each other.
Oil and hexane. Oil is a homogeneous mixture of various nonpolar compounds. Hexane is
a nonpolar organic compound. Since the intermolecular forces are similar, oil and hexane
should dissolve in each other.
Oil and carbon tetrachloride. Oil is a homogeneous mixture of various nonpolar compounds.
Carbon tetrachloride is a nonpolar organic compound. Since the intermolecular forces are
similar, oil and carbon tetrachloride should dissolve in each other.
Intermolecular forces and physical properties
Many of the physical properties of substances are determined by the intermolecular forces
between the substance’s particles. Freezing point, boiling point, vapor pressure and the ease with
which a substance evaporates, density, and the solubility - or lack of solubility - of one compound
in another are all examples of properties directly determined by intermolecular forces. Let’s take a
look at the boiling points and aqueous solubilities of some organic compounds.
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alkanes (hydrocarbons)
alcohols
substance
BP (oC)
solubility
substance
BP (oC)
solubility
CH4
-162
no
CH3OH
65
yes
C2H6
-89
no
C2H5OH
78
yes
C3H8
-42
no
C3H7OH
97
yes
C4H10
-1
no
C4H9OH
117
moderately
C5H12
36
no
C5H11OH
138
slightly
C6H14
69
no
C6H13OH
158
no
Although it may not be immediately apparent to you, there is a wealth of information contained in
this table. Relax! You don’t have to memorize this table or the names and molecular formulas of the
compounds we discuss in this table. But you do need to understand what we’re looking at in this table and
I’m going to explain it to you.
Let’s begin with the alkanes, methane (CH4), ethane (C2H6), propane (C3H8), butane (C4H10),
pentane (C5H12), and hexane (C6H14). Alkanes are all nonpolar compounds. The only types of
bonding forces that can occur between hydrocarbon molecules are London forces. Their structures
may be seen in the following image.
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Note that the boiling points of these hydrocarbon compounds increase as the size of the molecules
increases. This is consistent with the behavior of London forces: the larger the molecules, the
stronger the London forces between the molecules. It is important to remember that boiling point
of a substance, which is a physical property, is determined by the strength of the intermolecular
forces between the particles of the substance (as we mentioned earlier). And as we said, in a general
sense, if we compare the boiling points of any two substances, the one with the higher boiling point
will have the stronger intermolecular forces between its particles, while the one with the lower
boiling point will have the weaker intermolecular forces between its particles.
If we examine the solubilities of the six hydrocarbons we see that none of them is soluble
in water. But as water is polar and the hydrocarbons are nonpolar, this should not surprise us.
Now let’s look at the six alcohols, methanol (CH3OH), ethanol (C2H5OH), propanol
(C3H7OH), butanol (C4H9OH), pentanol (C5H11OH), and hexanol (C6H13OH). All alcohols are polar
compounds which are also capable of forming hydrogen bonds. Their structures may be seen in the
following image.
If we compare the boiling points of the alkane and the alcohol with the same number of
carbons i.e., methane and methanol, ethane and ethanol, propane and propanol, etc., we see that the
alcohols have much higher boiling points than the corresponding alkanes. The alcohols are slightly
larger, it’s true, but the principal reason for the difference in boiling points is a reflection of the
difference in strength of the stronger hydrogen bonds of the alcohols and the weaker London forces
of the alkanes.
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Notice that the boiling points of the alcohols increase as the size of the molecules increases.
This is similar to the behavior demonstrated by the alkanes, and for exactly the same reason: the
larger the molecules, the stronger the London forces between the molecules. It may seem odd to you
to think that London forces play a role in the behavior demonstrated by alcohols but remember that
all substances experience London forces. And, the larger the molecules the stronger the London
forces between them.
We need to remember this to explain the aqueous solubility of the alcohols. The first three
alcohols are soluble in water and this should come as no great surprise, as both water and alcohols
are polar and also capable of forming hydrogen bonds. But it is curious to note that butanol is less
soluble than the three smaller alcohols, pentanol even less so, and hexanol is insoluble water. How
can it be that these three polar hydrogen-bond forming compounds have difficulty dissolving in
water? Look at the structures of the alcohols. As they become larger, they begin to resemble the
alkanes as the carbon backbone of the molecules becomes longer. And remember: the chains of
carbon atoms bonded to hydrogen atoms are not polar. Although the alcohols retain their polar
hydroxy group and the capability to form hydrogen bonds, the longer nonpolar hydrocarbon chains
become more strongly attracted to each other than the polar hydroxy groups are to each other or to
water. In other words, at around 5-6 carbon atoms in alcohols, we reach the point at which polar
compounds are more stable behaving as though they were nonpolar.
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Chapter 9: Solutions
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as solution, homogeneous mixture, solution, solvent, solute,
solvation, hydration, dissolution, enthalpy of solvation, hydration, bulk solute, clustering,
soluble, insoluble, solubility, molar solubility, saturated solution, unsaturated solution,
supersaturated solution, miscible, percent composition, parts per million, parts per billion,
molality, molarity, dilution, surface tension, capillary action, vaporization, vapor pressure,
normal boiling point, diffusion, Brownian motion, osmosis, semi-permeable membrane,
reverse osmosis.
Describe the molecular processes of solvation and clustering.
Be able to calculate to molar concentration of a solution.
Given the molar concentrations of several solutions, be able to calculate the relative numbers
of particles in each.
Be able to calculate the new concentration of a diluted solution.
Explain how properties such as vapor pressure and boiling point are related to intermolecular
forces.
General solution properties and definitions
Let's begin our study of solutions with some definitions.
Solutions are homogeneous mixtures that are evenly dispersed at the molecular level and that
are uniform throughout in their physical and chemical properties. They consist of a single phase.
Most of us are familiar with solutions made by dissolving a solid in a liquid, but a solution may be
made by dissolving:
•
•
•
•
•
•
•
a solid in another solid (all alloys)
a solid in a liquid (all aqueous solutions)
a liquid in a solid (mercury in gold)
a liquid in another liquid (rubbing alcohol, for example)
a gas in a solid (hydrogen in palladium is a good example although most of us in this class
don't have any real world experience with this mixture)
a gas in a liquid (carbon dioxide in soda, beer, and champagne)
a gas in another gas (oxygen and nitrogen in air)
There are always at least two components in every solution. The solvent is the material
present in greater abundance in the solution. The solute is present in lesser abundance. If solvation
occurs, the solvent does the dissolving and the solute is the substance that is dissolved. A solution
can only have one solvent, but it may have more than one solute. If you doubt this, read the label
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for a can of your favorite soda, which, regardless of the product you select, is an aqueous solution
of carbon dioxide, sugar, and various other substances. Unless it's diet soda - then it has an artificial
sweetener in place of the sugar.
Dissolution (the processing of dissolving ) is often a spontaneous process. At a molecular
level, the dispersion of solute molecules throughout the solvent results in an increase in the entropy
of the solution system. For many solutes dissolution in aqueous solution is an endothermic process,
although it is not uncommon for dissolution to be an exothermic process. The enthalpy of solvation
is dependent both on the solvent and on the solute and thus varies from one solution system to
another.
Solvation is a molecular process by which a solute may become dispersed throughout a
solvent. This is an especially important process when discussing the dissolution of solids in liquids.
For solvation to occur, solvent particles (atoms, ions, or molecules) must form attractive
intermolecular interactions with solute particles at the surface of the bulk solute. The bulk solute is
the big mass of solute particles as it is placed in the solvent, before the individual particles are dispersed
throughout the solvent. These interactions take place most easily at the edges and corners of the bulk
solute. As each particle leaves the bulk solute, other solvent particles crowd around it, forming a
number of solvent-solute intermolecular bonds. This phenomenon is called clustering. The number
of solvent particles that cluster around solute particles depends on various attributes of the solution,
such as the particular solvent and solute, solute concentration, and temperature. For an aqueous 1M
sodium chloride solution at room temperature and pressure, it is common for four to seven water
molecules to cluster around each of the ions as they are pulled away from the solute surface. As
solvated solute particles are randomly dispersed through the solvent they are accompanied by their
cluster of solvent particles at all times. Bare ions almost never exist in nature and especially in
solution. They are nearly always in the company of solvent clusters. This dispersion can occur due to
collisions with other molecules, or through physical mixing of the solution, or both.
For solvation to occur, the solvent particles must successfully attract individual solute
particles away from the bulk solute by forming intermolecular interactions with them. If the
interactions between individual particles in the bulk solute are too strong, or if the solute particles
are too small, it will not be possible for a sufficient number of solvent particles to form
intermolecular bonds with individual solute particles to induce them away from the bulk solute
surface. When this occurs, a solute will not dissolve in a solvent. A solute that will not dissolve in
a solvent is said to be insoluble.
Hydration is the specific name for the process of solvation when water is the solvent. An
illustration of a hydrated (solvated) ion may be found at http://en.wikipedia.org/wiki/Solvation. In
this example of the hydration of sodium ion, water molecules form ion-dipole interactions with the
sodium ion. The water molecules orient themselves such that their oxygen atoms, which have a
partially negative charge due to the polarity of the water molecules, are attracted to a positively
charged sodium cation.
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As we stated in Chapter 8, "like attracts like" or the equivalent "like dissolves like" is a very
good general rule that can help us predict whether or not one substance will dissolve in another. It
qualitatively describes the probability of the formation of the favorable intermolecular interactions
between solvent-solute particles required for dissolution to occur. "Like dissolves like" means that
generally speaking, substances with similar intermolecular forces will mix and form solutions, while
substances with different intermolecular forces will not form solutions well, if at all.
The solubility of a substance is the maximum amount of that substance that will dissolve in
a stated amount of a specific solvent to give a thermodynamically stable solution. Solubility is
usually expressed in grams of solute for a given amount of solvent, at specific temperature and
pressure conditions. As examples:
•
•
•
the solubility of NaCl is 35.7 g per 100 mL of water at 0°C and 1 atm of pressure
the solubility of CO2 is 171.3 g per 100 mL of water at 0°C and 1 atm of pressure
the solubility of barium sulfate is 0.000222 g per 100 mL of water at 18°C and 1 atm of
pressure. This small number tells us that barium sulfate does not dissolve very well in water at this
temperature.
Solubility may also be expressed as the number of moles, rather than grams, that will dissolve in a
given amount of solvent. This is called the molar solubility of a substance.
A saturated solution is a solution in which as much solute as is physically possible has been
dissolved in the solvent, with some solute remaining, undissolved in the container holding the
solution. The point at which a solution becomes saturated is dependent on the particular
solvent-solute system. It also depends on the temperature of the solution. An unsaturated solution
is one in which more solute may be dissolved. Occasionally, supersaturated solutions may be
formed. These are solutions in which more solute has been dissolved than is thermodynamically
stable. This does not necessarily mean that they are dangerous, but it does mean that they can be a
challenge to prepare and store. Portable chemical hand warmers and liquid infant heel warmers
contain a supersaturated solution of sodium acetate dissolved in water.
Substances that are miscible can form solutions in all proportions. In other words, if two
substance are miscible, they are infinitely soluble in each other.
A colloid consists of small aggregates of particles (clusters of particles) of one material
dispersed throughout another. Small aggregates are defined as those that are about 500 nm or smaller
in size. ("Physical Chemistry," Peter W. Atkins; W. H. Freeman and Co. 5th Ed., p. 971) These
collections of particles often behave differently than they do as individual particles dispersed
throughout a solvent. There are different types of colloids, depending on the types of materials
involved. A dispersion of solid particles in a solid, or of solid particles in liquid, is called a sol. A
dispersions of solid particles in a gas, or of liquid particles in gas, is called an aerosol. A dispersion
of liquid particles in a liquid is called an emulsion. Colloids are often referred to as suspensions,
which contain molecular-level aggregations of particles physically suspended in liquid that are often
so small that gravity has no effect on the aggregates.
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Calculating concentration
One of the most important attributes of a solution is its concentration. Concentration is a
measure of the amount of solute dissolved in either a specific amount of solvent or solution.
Concentration can be calculated in a variety of ways. Concentration can be stated in terms
of percent composition which is, in general, the ratio of amount of solute to some amount of solution
multiplied by 100. There are three types of percent composition:
•
•
•
Weight/weight (w/w) = (grams solute/grams solution) x 100
Weight/volume (w/v) = (grams solute/volume solution) x 100
Volume/volume (v/v) = (volume solute/volume solution) x 100
Some examples of the calculation of percent composition are as follow.
•
What is the (w/w) concentration of 100. g of solution containing 1.22 g of sodium chloride?
(w/w) = (1.22 g NaCl/100 g solution) x 100 = 1.22% w/w
•
What is the (w/v) concentration of 50.0 mL of solution containing 1.22 g of sodium
chloride?
(w/v) = (1.22 g NaCl/50.0 mL solution) x 100 = 2.44% w/v
•
What is the (v/v) concentration of a solution containing 100 mL of isopropyl alcohol that is
diluted with water to 150 mL?
(v/v) = (100 mL isopropyl alcohol/150.0 mL solution) x 100 = 66.7% v/v
The concentrations of dilute solutions can be expressed in terms of mg%, parts per million
(ppm), and parts per billion (ppb).
•
•
•
mg % = (mg solute/100 mL solution) x 100
ppm: = (mg solute/L solution)
ppb = (ug solute/L solution)
When concentration is expressed in parts per million it is equivalent to one part solute for
every one million parts of solution. Another way of looking at this is to think of it in terms of
populations of people. If the state of Utah has a population of three million people, then 1 ppm is
equal to three individuals out of the state's population. Parts per billion can also be thought of along
these lines. If the country of China has a population of one billion people, then 1 ppb is equivalent
to 1 person out of the total population of the country.
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While ppm and ppb may seem like very small concentrations, they still are often significant.
For example, many substances are toxic at a ppm level, some lethally so, and there are numerous
chemicals that can induce physiological effects at a ppb level. As another example, our noses are
extremely sensitive to many chemicals. Many of the chemicals we smell in the world around us are
only present in ppm concentrations. We can smell the isopropyl alcohol in rubbing alcohol
beginning at concentrations of about 22 ppm and the acetone in nail polish remover at concentrations
of about 5 ppm and higher. Of course we're talking about the airborne concentrations of these
compounds as they evaporate; this ability to smell these compounds in not directly linked to the
concentrations of the compounds in the products which I have mentioned. We have a particularly acute
sense of smell when it comes to a class of compounds called sulfides, all of which contain a sulfur
atom. The smallest member of the sulfide family is hydrogen sulfide, H2S. Note: this is the common
name of compound, not the IUPAC name. But this particular compound is known almost exclusively by its
common name and so we’ll use it here. This compound is also known as rotten egg gas, and it is what
one smells when one drives by oil refineries or visits geysers, hot springs, and mud pots as are found
in Yellowstone National Park. The average person can smell H2S gas beginning at a concentration
of about 0.5 ppb, or 500 parts per trillion! The levels at which we can begin to smell chemicals are called
odor thresholds. There are a some good references available online, if you find this topic interesting. I'd
suggest you begin with the most current edition of 3M's "Respirator Selection Guide," which can be found
at their corporate website and at no charge.
Molality is also known as molal concentration. It is the ratio of moles of solute per kg of
solvent. Molal concentration is indicated with a lower case "m".
•
What is the molal concentration of 125 grams of glucose that is dissolved in 750 grams of
water?
m = mol solute/kg solvent
convert 125 grams of glucose to moles of glucose
convert the mass of water to kg of water
(125 g glucose) x (1 mole glucose/180.2 g glucose) x (1/750 g water) x
(1000 g water/1 kg water) = 0.925 m
Molarity is also known as molar concentration. It is the ratio of moles of solute per liter of
solution. Molar concentration is indicated with an upper case "M".
•
What is the molar concentration of 125 grams of glucose that is dissolved in 750 mLs of
water?
M = mol solute/L solution
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convert 125 grams of glucose to moles of glucose
convert the volume of water to liters of water
(125 g glucose) x (mole glucose/180.2 g glucose) x (1/0.750 L solution) = 0.925 M
It is not uncommon for the molar and the molal concentrations of a solutions to be the same,
but it does not always work out that way. Molality is not as commonly used as it once was. Molarity
is much more commonly encountered. Some examples of typical molarity calculations follow.
•
If 0.100 moles of H2SO4 is dissolved in 450 mL of water, what is the molarity of the
resulting solution?
convert 450 mL of water to liters
0.100 mol/0.450 L = 0.222 M
•
If 37.6 grams of copper (II) nitrate is dissolved in 500 mL of water, what is the molarity of
the resulting concentration?
convert 37.6 g of copper (II) nitrate to moles; the Mm = 187.56 g/mol
convert 500 mL of water to L
•
(37.6 g copper (II) nitrate) x (1 mol copper (II) nitrate/187.56 g copper (II) nitrate) x
(1/0.500 L) = 0.401 M
How many moles of copper (II) nitrate are contained in 25.0 mL of 2.5 M solution?
if M = mol/L then mol = M x L
convert volume to L
2.5 mol/L x 0.025 L = 0.0625 mol copper (II) nitrate
•
How many grams of ammonium hydroxide are contained in 75 mL of concentrated
(15M) solution?
To do this problem we need to calculate the number of moles of ammonium hydroxide
found in 75 mL of concentrated solution and then convert it to a mass in grams using the
molar mass of the compound (Mm = 35.05 g/mol)
again, if M = mol/L then mol = M x L
convert volume to L
15 mol/L x 0.075 L x (35.05 g/mol) = 39.4 g ammonium hydroxide
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Dilutions
It is often necessary to take concentrated solutions and to lessen or lower the concentrations.
This is typically done by adding the concentrated acid or base to water. In doing so, the number of
moles of acid or base remains constant but as the volume of the solution increases, the concentration
of the solution decreases.
compound
solution strength
acetic acid
17.5 M
hydrochloric acid
12 M
sulfuric acid
18 M
nitric acid
16 M
ammonium hydroxide
15 M
No, you don't need to remember these concentrations, they are simply provided to help illustrate a point.
Too, we should point out explicitly what is being implied: all of the substances listed in the table are actually
aqueous solutions of the various compounds for which they are named.
Assume we begin with exactly 1.00 L of 12 M hydrochloric acid. This means that we have
exactly 12.0 moles of hydrochloric acid in each liter of solution.
12 M HCl = 12 mol HCl/1.00 L solution
Now let’s assume we begin to add water to the solution. The number of moles of HCl is unaffected
by the addition of water but the solution volume begins to increase. In this solution, water is the solvent
and HCl is the solute. We’re making an assumption here: that the solvent does not react with the solute
and therefore does not consume it as more solvent is added. In terms of concentration, this means that
the numerator (the number of moles of acid) remains constant while the denominator (the solution
volume) becomes larger. As a consequence, the overall solution concentration decreases.
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moles of HCl
solution volume
solution concentration
12.0
1.00 L
12.0 M
12.0
2.00 L
6.0 M
12.0
3.00 L
4.0 M
12.0
4.00 L
3.0 M
12.0
6.00 L
2.0 M
12.0
12.00 L
1.0 M
We can dilute any solution by adding more solvent to the solution. This does not change the
total number of moles of solute in the solution, again, operating under the constraint that the solvent
and solute do not react with each other but it does lessen the ratio of the amount of solute per amount
of solvent. In other words, when a solution is diluted, the solute concentration becomes less.
A useful equation when studying dilutions is
M1V1 = M2V2
in which M1 is the initial concentration of the solution, V1 is the initial volume of the concentrated
solution, and M2 andV2 are the final concentration and final volume of the diluted solution. If we
think of this equation in terms of the units involved
(mol1/L) x (L) = (mol2/L) x (L)
remembering that the units of molarity are mol/L, it can be simplified to
mol1 = mol2
since the units of volume cancel. This corroborates what was stated above: during the course of a
dilution the total number of moles of solute does not change. It is constant.
Here are a few examples of how this dilution equation can be used.
•
How many milliliters of concentrated NH4OH are required to form 100. mL of 1.0 M
solution?
make a table of the given values to help clarify what you have and what you are solving for
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M1
15 M
V1
?
M2
1.0 M
V2
100 mL
if M1V1 = M2V2
thenV1 = (M2V2)/M1
(1.0 mol/L x 100. mL)/15 mol/L = 6.67 mL of concentrated ammonium hydroxide
This means that if 6.67 mL of 15 M ammonium hydroxide are diluted to 100 mL, the
concentration of the diluted solution will be 1.0 M
Note that it was not necessary to convert the volume from milliliters to liters. As long as one
is consistent, either unit may be safely used.
•
If 10.0 mL of 12 M HCl is diluted to 600 mL, what is the new concentration of the acid?
make a table of the given values to help clarify what you have and what you are solving for
M1
12 M
V1
10.0 mL
M2
?
V2
600 mL
if M1V1 = M2V2
then (M1V1)/V2 = M2
(12.0 mol/L x 10. mL)/600 mL = 0.200 M HCl
In plain English, by taking 10 mL of 12 M HCl and adding another 590 mL of water to the
solution (so that the total solution volume is 590 + 10 = 600. mL), we change the concentration of
HCl to 12 M to 0.200 M.
•
The dilution of 50.0 mL of a solution of phosphoric acid results in 1.00 L of a 0.75 M
solution. What was the initial concentration of the acid?
make a table of the given values to help clarify what you have and what you are solving for
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M1
?
V1
50.0 mL
M2
0.75 M
V2
1.00 L
if M1V1 = M2V2
then M1 = (M2V2)/V1
(0.75 mol/L x 1.00 L)/0.050 L = 15 M phosphoric acid
Note that we have to be consistent in our choice of units of volume. We chose to solve this
problem using L, but we could also have correctly solved it using mL.
Some properties of liquids and solutions: surface tension and capillary action
For any substance, in the vapor phase (gaseous state) its particles are, relatively speaking,
far apart, possess high kinetic energy, and have few if any interaction with each other. For this same
substance in the liquid phase, the particles will be lower in kinetic energy than in the vapor phase,
closer together than in the vapor phase, and there will be attractive interactions between the
particles.
The particles at the boundary of a liquid and it's vapor behave differently than particles far
away (5-10 particle diameters) from the phase boundary. In the vapor state, away from the interface
with the liquid, the spacing between the particles is so great that, in relative terms, they seldom have
the opportunity to interact. In the bulk liquid, away from the interface with the gas, liquid particles
attract each other equally and, in turn, are equally attracted in all directions. But at the phase
boundary the attraction in the liquid is one-sided. The liquid particles are attracted inwardly, toward
other molecules in the liquid phase. The liquid particles at the surface can engage in intermolecular
interactions with other particles in the liquid phase but they can form few or no such interactions
with particles in the gas phase, due to the low density of the particles in the vapor phase and also
because of the relatively high kinetic energy of the vapor phase particles. This one-sided attraction
makes the liquid resist expansion. It also drives the bulk liquid to attempt to minimize its surface
area. The particles at the interface pack together more closely than in the bulk liquid. The molecules
closest to the interface wind up acting like a kind of "skin" on the surface of the liquid. This explains
the behavior of water droplets on the waxed surface of a car. Like attracts like. The molecules at the
surface of a water droplet are attracted to the hydrogen-bonding interior of the droplet and not to the
waxy surface of the car, which can only engage in dispersion forces. As a consequence, the droplets
bead up. This is also the reason that rain drops falling through the air assume a more or less spherical
shape. And this attractive tendency is the reason that liquids in containers form a meniscus at their
surface.
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Surface tension is related to this attribute of liquids attempting to make their surface area as
small as possible. Surface tension is the energy required to overcome the tendency of liquids to
minimize their surface area.
There are exceptions to this behavior. If the intermolecular forces in the substances on both
sides of phase interface have similar bonding interactions, the meniscus may shrink to the point of
being barely observable. As an example, the meniscus in a container of a nonpolar liquid is often
less pronounced than that seen with polar liquids. And as nonpolar liquids have similar
intermolecular forces as those observed in air, this is not surprising.
There may also be attractive intermolecular interactions between particles in the liquid phase
and those on the surface of the container holding the liquid, if they are similar. The surface of glass
vessels is polar. If the diameter of the glass container is very small, as is the case with a thin glass
tube, attractive interactions between a polar liquid - such as water - and the tube's interior surface
can cause the liquid to creep up the interior walls of the tube. This phenomenon is called capillary
action.
Some properties of liquids and solutions: vaporization and vapor pressure
Let's discuss the behavior of a substance in the liquid state in a sealed container. The
container will hold not only the substance in its liquid form, but it will also hold some of the
substance in its vapor (gas) phase.
An important difference between molecules in the liquid and vapor phases is the kinetic
energy of the particles. Particles in the vapor phase of a substance generally have higher kinetic
energies than particles in the liquid phase of the same substance.
If we examine the kinetic energy of all of the particles in any phase of any substance, be it
solid, liquid or gas, we find that the particles do not all have a single kinetic energy. Rather, there
is a (more or less) Gaussian distribution of energies amongst the particles. A Gaussian distribution is
a bell-shaped curve. In other words, in a collection of the particles in any phase, some of the particles
have high energies, some have low energies, and most have a kinetic energy intermediate between
high energy and low energy. This kinetic energy is affected by a number of factors but most notably
by temperature. We talk about this relationship in the “General gas properties” section of Chapter 8.
Particles in the liquid phase of a substance that have (a.) a high kinetic energy and (b.) that
are near the liquid-gas phase boundary can actually have enough kinetic energy to "escape" from
the liquid phase and join the vapor phase. This is why heating a liquid facilitates the transition of
the liquid to the vapor phase. As the liquid is heated, the kinetic energy of the particles increases and
the fraction of particles with sufficient kinetic energy to escape the liquid phase also increases.
Molecules in the vapor phase travel randomly. They collide with the walls of their container
and with each other. They also collide with the liquid surface during their random travels. If these
colliding vapor phase particles have high kinetic energy they generally bounce off liquid surface
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when they bump into it. But vapor phase particles with low kinetic energy may "stick" and become
part of liquid phase when they collide with it. By cooling a vapor we lower the average kinetic
energy of the molecules in the vapor phase and make it easier for them to " stick" to the liquid
surface if they collide with it.
At any given temperature, an equilibrium exists between the rates of particles leaving the
liquid phase and joining the vapor phase and the reverse process, particles leaving the vapor phase
and rejoining the liquid phase. As per our discussion of equilibrium in the Chapter 7 notes under the
heading “Chemical equilibrium.” The pressure exerted by the vapor when this equilibrium exists is
called the vapor pressure.
A liquid will boil when its vapor pressure is equal to atmospheric pressure. We call the
temperature at which this occurs the normal boiling point of the liquid. In other words, the normal
boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to
exactly 1 atm of pressure. That water boils at 100°C tells us that at this temperature, the vapor
pressure of water is equal to exactly 1 atm. The boiling point of isopropyl alcohol, a compound
found in rubbing alcohol, is 82.4°C. This means that in pure isopropyl alcohol at this temperature,
the vapor pressure of the liquid will equal exactly 1 atm of pressure.
There is an interesting implication to this. As atmospheric pressure decreases, any liquid will
boil at an increasingly lower temperature. Water boils at 100oC when atmospheric pressure is exactly
1.00 atm. If one tries to boil water in the mountains, where atmospheric pressure is significantly less
than 1.00 atm, the water will boil at a temperature less than 100oC.This is why it takes longer to boil
an egg or to cook pasta when one is camping in the mountains. Conversely, if one were vacationing
in a place where atmospheric pressure was greater than 1.00 atm, water would then boil at a
temperature greater than 100oC and your egg or pasta would cook more quickly. There is a online
video clip at YouTube that demonstrates the boiling of water at room temperature. The demonstrator
lowers the pressure by using a vacuum pump. Take a look if you’re interested at the link found at
http://www.youtube.com/watch?v=LxtAeGtL9SE.
In a comparative respect vapor pressure is rather like boiling point. It gives a general sense
of the relative strengths of intermolecular forces. If we compare the vapor pressure of two
substances at the same temperature, the substance with the highest vapor pressure is the substance
with the weakest intermolecular forces. These weaker intermolecular forces make it easier for
particles to escape the liquid phase and enter the vapor phase. The vapor pressure of such a
substance will be higher than that of a substance with stronger intermolecular forces, which make
it more difficult for particles to escape the liquid phase.
Some properties of liquids and solutions: diffusion
In the solid state the particles of a substance are more or less locked in place, but in pure
liquids and gases the particles of a substance tend to move randomly. In solutions this random
movement results in solute particles moving from areas of high concentration to areas of low
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concentration until they are uniformly distributed throughout the solvent particles in the solution.
The solvent particles in like manner tend to move from areas of high concentration to areas of low
concentration until they too are uniformly distributed throughout the solute particles. It is statistically
possible for the solvent and solute molecules in a solution to behave in some other manner, but
mathematically it is extremely unlikely. This behavior is called diffusion. This is how a solute achieves
uniform distribution throughout a solvent in the process of forming a solution, whether the solution
is a gas/gas solution, a gas/liquid solution, a liquid/solid solution, or etc.
If we add a single crystal of sodium chloride to a container of water, the salt will dissolve
and ionize. The ions will eventually become uniformly distributed throughout the container of water.
At a molecular level, upon addition of the crystal, the concentration of sodium chloride within the
crystal is much higher than the concentration of sodium chloride in the pure water. Diffusion
disperses the sodium ions and chloride ions throughout the water until they are evenly dispersed
throughout the water. Conversely, the concentration of pure water outside the crystal is much greater
than the concentration of water within the crystal. Diffusion disperses the water molecules
throughout the sodium chloride until they are evenly dispersed throughout the dissolving solid.
Diffusion makes the system in which it occurs more random, i.e., the entropy of the system
increases as a consequence of diffusion. This means that diffusion should usually be a spontaneous
process. It is, in fact, one of the fundamental laws of thermodynamics that the overall entropy of a system
will increase during a spontaneous process. This randomizing effect will occur independent of any
external forces (e.g. mixing) and is due to random collisions between molecules. These random
collisions are called Brownian motion (see Einstein's Explanation of Brownian Motion at
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/
brownian/brownian.html). The rate of diffusion depends on those factors that affect the rates at
which these random collisions may occur and include temperature, viscosity, pressure, the molecular
weight of the solute, and solute concentration.
Some properties of liquids and solutions: osmosis
Let's begin with a few definitions.
•
•
•
A membrane is a sheet-like structure, often porous, that can regulate the passage of
substances from one side of the membrane to the other based on molecular size or charge.
Membranes that allow some substances to pass but not others are called semi-permeable.
Osmosis is the flow of water through a semi-permeable membrane from areas of low to high
solute concentration.
I would strongly advise you to view the short video entitled “Osmosis Experiment” at
http://www.youtube.com/watch?v=7WX8zz_RlnE. It will take you less than a minute. In the video
a column of blue liquid moves up a tube over a period of several hours. Why does this occur?
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The blue liquid is probably a concentrated solution of table sugar, dissolved in water, with
a few drops of blue food coloring added to make it easier to see the behavior of the solution. The
sugar solution sits in a beaker of distilled water. A semi-permeable membrane separates the sugar
solution from the distilled water. It is the natural tendency of water to pass from areas of high
concentration to areas of low concentration. As water is at a higher concentration in the beaker, and
at a lower concentration within the sugar solution, water molecules pass through the semi-permeable
membrane into the solution. As the semi-permeable membrane does not stretch, the addition of water
to the solution increases its volume, which forces it slowly but surely up the tube.
Diffusion is a two-way street. Just as it is the natural tendency of water to diffuse from an
area of high concentration (pure water) to an area of low concentration (sugar solution), it is also
the natural tendency of the sugar in the solution to diffuse from areas of high concentration (sugar
solution) to areas of low concentration (pure water). But in this example it cannot occur. Water
molecules are small enough to pass freely through the pores of the semi-permeable membrane. But
the sugar molecules are too large. Hence the sugar remains trapped in the solution, while water can
flow freely into the solution.
Osmotic pressure is a force that is generated by osmosis. It is the pressure that would be
required to stop the net flow of water (i.e., the force required to stop osmosis) from one side of the
semi-permeable membrane to the other.
Reverse osmosis is a process which forces water across a semi-permeable membrane from
areas of high to low solute concentration. Reverse osmosis is used to purify sea water. Sea water,
which is salty, this comes as a real revelation to you, doesn't it? is placed on one side of a
semi-permeable membrane and then forced through a membrane with small enough pores to prevent
ions from passing through it. The natural tendency of the purified water is to diffuse back through
the membrane, to areas of high salt concentration, but means are devised to prevent this from
occurring. While reverse osmosis is a very effective way to purify water, it takes a great deal of
energy. This can make the cost of purifying water using reverse osmosis very expensive. And yet
in areas like the Persian gulf and Saudi Arabia, there are entire cities that obtain their drinking and
household water from large reverse osmosis plants. Reverse osmosis is also used to purify water on
a limited scale in many American cities, like Los Angeles.
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Chapter 10: Acids and Bases
Chapter Objectives: After completing this chapter you should at a minimum be able to do the
following. This information can be found in my lecture notes for this and other chapters and also in
your text.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Correctly answer all of the questions in the quiz for this chapter.
Define basic terms such as Arrhenius acid, Arrhenius base, Brønsted-Lowry acid,
Brønsted-Lowry base, Lewis acid, Lewis base, strong acid, weak acid, strong base, weak
base, hydronium ion, conjugate acid, conjugate base, monprotic acid, diprotic acid, triprotic
acid, polyprotic acid, amphoteric, amphiprotic, ion product constant for water, pH, pOH,
neutralization, hydrolysis, buffer.
Describe which acids are Arrhenius acids, Brønsted-Lowry acids, and Lewis acids.
Describe which bases are Arrhenius acids, Brønsted-Lowry bases, and Lewis bases.
Given a list of acids and bases, identify which are strong acids and bases and which are weak
acids and bases.
Describe how water behaves as both an acid and a base. To what extent does this occur in
a neutral solution of water at room temperature and pressure? Is it a product-favored or
reactant-favored reaction?
Calculate the concentrations of strong acids and bases in aqueous solution.
Explain how the pH scale is based on water.
Calculate the pH of a solution of a strong acid or base.
Explain what neutralization and hydrolysis reactions are.
Explain what a buffer is and how it works.
General acid-base information: a review
We have already spent a bit of time discussing acids and bases in Chapter 4. Let’s summarize
what was said.
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There are three common acid-base theories.
The Arrhenius theory defines acids as substances that can donate a hydrogen ion (H+) and
bases as compounds that can donate a hydroxide ion (OH-).
The molecular formulas of acids are usually written with hydrogen listed first, e.g., HCl,
HBr, HC2H3O2 and so on.
The names of acids are common names but are still accepted by IUPAC.
The names of Arrhenius acids always include the word “acid.”
The common bases are the Group 1A and 2A hydroxides, ammonia, and ammonium
hydroxide.
Not every substance that contains hydroxide can behave as a base.
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The Arrhenius acid-base theory
There are a number of different acid-base theories. We will concern ourselves with three that
are more commonly used, the Arrhenius theory, the Brønsted-Lowry theory, and the Lewis theory.
The Arrhenius theory is named for the Swedish chemist Svante Arrhenius, who not only
contributed to our knowledge of acids and bases but also to our understanding of strong and weak
electrolytes in aqueous solution. He received a Nobel Prize for his work in Chemistry in 1903, the third
awarded in this branch of science. A brief biography and description of his work may be found at a web
site maintained by the Nobel Foundation which awards the annual prizes. If you’ve never visited this site,
even if you don’t like chemistry, it’s rather interesting all the same. Arrhenius’s page can be found at
http://nobelprize.org/nobel_prizes/chemistry/laureates/1903/arrhenius-bio.html. As we have said,
according to the Arrhenius theory an acid is a substance that can donate a hydrogen ion in aqueous
solution and a base is a substance that can donate a hydroxide ion in aqueous solution. This is an
extremely useful working definition and in fact, the one most commonly used in elementary and
general chemistry. But as it only describes things as acids and bases in aqueous solution, it is often
rather limited in its ability to be useful when studying subjects outside the purview of elementary
chemistry.
The Brønsted-Lowry acid-base theory
In the 1920s, several decades after Arrhenius’s work was published, Johannes Brønsted and
Martin Lowry, working independent of each other and hundreds of miles apart, arrived at the same
conclusion at more or less the same time. According to their theory an acid is a substance which can
donate a proton in chemical reactions and a base is a substance which can accept a proton in
chemical reactions. Proton donation is accomplished through the donation of a hydrogen ion and not
through the donation of protons from the nuclei of other atoms, which would be nuclear (nonchemical) events requiring huge amounts of energy.
Let’s be sure we’re clear on this. A hydrogen atom consists of its nucleus, in which we find
a single proton, and a single electron which is found in an orbital in the 1s shell and subshell outside
the nucleus. If we remove the electron from a hydrogen atom all that remains is its nucleus - in
which we find one proton. Knowing this single fact can make a particular aspect of physiology a bit more
clear. An important class of medicinal compounds are the proton pump inhibitors, which include such
medicines as Nexium and Prilosec. What does it mean to say that they inhibit the production of protons?
It’s not that they interfere with nuclear reactions taking places in human stomachs. Rather, they simply
mediate the production of hydrochloric acid which is produced in cells lining the stomach and which is
essential to digestion.
The principal advantage of the Brønsted-Lowry theory is that it is a broader, more
encompassing theory than that of Arrhenius. The Brønsted-Lowry theory can be used to explain
acid-base behavior in aqueous and also in non-aqueous solution, i.e., solutions in which the solvent
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is not water. Everything that is an Arrhenius acid is also a Brønsted-Lowry acid. Everything that is
an Arrhenius base is also a Brønsted-Lowry base. While it is theoretically possible for a substance
to be a Brønsted-Lowry acid and not be an Arrhenius acid, I am not aware of any exceptions. But
there are many thousands of substances that can accept a proton in chemical reactions besides
hydroxide ion and in doing so, behave as Brønsted-Lowry bases.
In this image we see that ammonia and sulfide ion can accept a proton (hydrogen ion), just as
hydroxide ion. In doing so they behave as Brønsted-Lowry bases. We also note that hydrogen sulfide
ion, HS-, also possesses the ability to accept a proton and to form hydrogen sulfide as a result. All
polyatomic anions can behave as Brønsted-Lowry bases. There are also many neutral molecules
which can accept protons, such as ammonia and water. We’re going to be talking a little more about
the acid-base behavior of water later in this chapter.
The Lewis acid-base theory
In 1923, the same year that Brønsted and Lowry proposed their acid-base theory, Gilbert
Lewis proposed his own acid-base theory in which he described acids as electron-pair acceptors and
bases as electron pair donors. This is the broadest and most useful of the acid-base definitions.
Although it is one we will little use in this course (and in fact, it is not even mentioned in your text
book), we want to spend a few minutes trying to better understand it.
As is the case with Brønsted-Lowry acid-base reactions, Lewis acid-base reactions may take
place either in aqueous or non-aqueous solution.
As the broadest acid-base definition, we can say that anything that is an Arrhenius acid is
also a Brønsted-Lowry acid and therefore also a Lewis acid. But, there are many substances that
can act as Lewis acids but which do not donate protons (hydrogen ions) and which are therefore not
Brønsted-Lowry or Arrhenius acids. As an example, all transition metal cations have the capacity
to act as Lewis acids but as they do not have hydrogen ions (or protons) to donate, they cannot act
as Brønsted-Lowry or Arrhenius acids.
And, as the broadest definition, we can also say that anything that is a Arrhenius base and
a Brønsted-Lowry base is also a Lewis base. But, there are substances which can act as Lewis bases
but which do not accept protons (hydrogen ions) in the process and which are therefore not
Brønsted-Lowry bases.
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The reaction of silver (I) ion and ammonia (NH3) is an example of a Lewis acid-base
reaction. In this reaction
Ag+(aq) + 2 NH3 (aq) => Ag(NH3)2+(aq)
This is a rather odd and completely unexpected reaction, to us in this class at least. And yet it
happens easily and spontaneously. In this reaction silver (I) ion is acting as a Lewis acid and
ammonia is acting as a Lewis base. The resulting silver product is a called a complex ion. Complex
ion chemistry is a topic on which we could easily spend an entire semester, if we had the time and
if you as students had the interest. There are hundreds of these complex ions that occur naturally in
the world around us. But it will suffice to point out that the formation of each of these many different
complex ions is the result of a Lewis acid-base reaction.
The typical attribute of a Lewis acid is one or more empty orbitals. In order to accept an
electron pair there must be a place for the accepting atom (molecule) to place it and empty orbitals
are typically used. The typical attribute of a Lewis base is the presence of one or more atoms with
at least one nonbonding pair of electrons. The pi bond electrons found in double and triple covalent
bonds make it possible for substances with these types of bonds to act as Lewis bases as well. But as we
have not discussed pi electrons, now that I’ve mentioned it, you can promptly forget about it - unless you
study organic chemistry outside of this class. The sharing of a nonbonding pair by the Lewis base with
the empty orbital of the Lewis acid results in the formation of a coordinate covalent bond, a covalent
bond in which one atom donates both of the bonding electrons. Sound familiar? This is a topic we
discussed in Chapter 5. And as we have seen in the reaction of silver (I) ion and ammonia, it is
possible for a single Lewis acid to form coordinate covalent bonds with more than one Lewis base,
depending on the number of empty orbitals in the Lewis acid.
If what we’ve said is correct, that all Arrhenius acid-base reactions are Lewis acid-base
reactions, and that all Brønsted-Lowry acid-base reactions are also Lewis acid-base reactions, then
it would be well worth our time to take a look at a few reactions to prove that this concept is correct.
We need to use Lewis structures to make this possible.
Let’s begin with the reaction of hydrogen ion and hydroxide ion, an Arrhenius acid-base
reaction.
In this reaction we see that the oxygen atom of the hydroxide ion possesses two nonbonding pairs
of electrons. The 1s orbital of the hydrogen ion is empty. Remember: the full electron configuration of
hydrogen is 1s1. If it loses its electron in becoming an ion, it now has a configuration of 1s0. One of
oxygen’s two nonbonding pairs is used to form a coordinate covalent bond and water is formed as
a product. In this illustration the curved arrow running from the oxygen atom in hydroxide ion to the
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hydrogen ion indicates that it is the atom which is sharing a nonbonding pair with the hydrogen ion. And
as we said in Chapter 5, the coordinate covalent bond formed in this reaction is identical in length and
strength to regular covalent bond that existed in the hydroxide ion. In this reaction hydrogen ion is acting
as a Lewis acid and hydroxide ion is acting as a Lewis base. To be clear: hydrogen ion is acting as an
Arrhenius acid as well as a Brønsted-Lowry acid and a Lewis acid, and hydroxide ion is acting as an
Arrhenius base, as a Brønsted-Lowry base, and also as a Lewis base.
In the reaction of hydrogen ion and ammonia
the nitrogen atom of the ammonia molecule possesses one nonbonding pair of electrons. As in the
previous example, the 1s orbital of the hydrogen ion is empty. Nitrogen’s nonbonding pair is used
to form a coordinate covalent bond with hydrogen ion. Again, the curved arrow running from the
nitrogen atom in ammonia to the hydrogen ion indicates that it is the atom sharing a nonbonding pair with
the hydrogen ion. As a result, ammonium ion is formed. In this reaction hydrogen ion is acting as a
Lewis acid and ammonia is acting as a Lewis base. Hydrogen ion is acting as an Arrhenius acid as well
as a Brønsted-Lowry acid and a Lewis acid. Ammonia is acting as a Brønsted-Lowry base and also as a
Lewis base.
Finally, as an example, let’s look at the reaction between silver (I) ion and ammonia.
In this reaction the nitrogen atom in each ammonia molecule is using its nonbonding pair of
electrons to form a coordinate covalent bond with one silver (I) ion’s empty orbitals. As a result, the
silver diammine complex ion is formed. In this reaction silver (I) ion is acting as a Lewis acid and
ammonia is acting as a Lewis base. Silver (I) ion is acting only as a Lewis acid. Ammonia is acting only
as a Lewis base and not also as a Brønsted-Lowry base in this particular reaction. Why? Because it’s not
accepting a proton in this reaction.
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Strong and weak acids and bases
As we covered this material back in Chapter 6, we will review a few things you have
hopefully learned and remember which are relevant for the remainder of this chapter. If you don’t
remember this stuff, turn back to Chapter 6 and spend a few minutes reviewing it. You’re going to need
it.
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All compounds are classified as strong electrolytes, weak electrolytes, or non-electrolytes.
This is true of all ionic compounds and also of all covalent compounds.
Electrolytes are substances that form aqueous solutions capable of conducting electricity. A
substance does this by dissociating when in aqueous solution. The ions serve as charge
carriers.
Substances must dissolve before they can dissociate. Dissolution, or, the process of
dissolving, and dissociation are two different things. A substance can dissolve without
dissociating if it is a non-electrolyte. But no substance can dissociate unless it first dissolves.
Strong electrolytes are substances that dissociate completely in aqueous solution. This means
that roughly100% of all of the molecules in a sample of strong electrolyte will dissociate in
aqueous solution if they dissolve.
All ionic compounds are strong electrolytes, as are all strong acids and bases.
Not all ionic compounds are soluble in aqueous solution. But we still classify them as strong
electrolytes based on their potential for complete ionization if they did dissolve.
Weak electrolytes are substances that dissociate incompletely in aqueous solution. This
means that far less than 100% of all of the molecules in a sample of weak electrolyte will
dissociate in aqueous solution if they dissolve. Typical values usually range from 1-10%.
This means that in a sample of weak electrolyte, only about 1-10% of the particles will
ionize if they dissolve in water.
The most common weak electrolytes are the weak acids and bases.
Non-electrolytes do not ionize at all in aqueous solution, even though they may dissolve.
Nearly all organic compounds are non-electrolytes, except for the organic acids and organic
bases.
In chemistry the words “strong” and “weak” are used almost exclusively to indicate the
extent to which a compound ionizes in water. There is no apparent correlation between
strong and weak and how corrosive or dangerous a compound may be.
Hopefully this is all familiar and well understood to you. And hopefully you retain the ability
to categorize substances as strong acids, weak acids, strong bases, and weak bases, based on the list
you were given in Chapter 6.
The behavior of acids and conjugate bases in water
The generic behavior of any Arrhenius acid in aqueous solution, whether strong or weak, is
described by the following equation:
HA(aq) + H2O(l) => H3O+(aq) + A-(aq)
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In this equation HA represents all Arrhenius acids. The anion of the acid HA is A- and is called the
conjugate base of the acid HA. We will discuss conjugate bases further in just a moment. The compound
H3O+is called hydronium ion. The reaction by which hydronium ion is formed is as follows:
The oxygen atom in water shares one of its two nonbonding pairs of electrons with hydrogen ion,
forming a coordinate covalent bond. It should be noted that bare ions of any sort are extremely
unstable and almost never occur in aqueous solution without being surrounded by a cluster of
solvent (water) molecules. As was the case for sodium ion in Chapter 9, it is also true for hydrogen
ion in that it seems to move about in aqueous solution surrounded by four to seven water molecules.
Studying clustering in solutions is one of the more difficult problems chemists attempt. Hence the lack of
precision in relating the number of water molecules per cluster. And it does not help when one learns that
cluster size, i.e., the number of solvent molecules clustering around the solute molecule, depends not only
on the given solvent and the solute but also on temperature, solute concentration, and other factors. So
hydronium ion probably does not exist precisely as such. But as far as we’re concerned in this class,
it’s far easier to use the molecular formula of hydronium when balancing equations than the
molecular formula for a small, four water molecule cluster, which would be expressed as H9O4+.
The anion of any acid is called its conjugate base, as mentioned above. This is not a random
choice of words. The anion of any acid, by itself, will behave in the following way in aqueous
solution:
A-(aq) + H2O(l) => HA(aq) + OH-(aq)
As we know, any substance that forms hydroxide ion in aqueous solution can be correctly classified
as a base. By labeling the anions of acids as conjugate bases we are simply acknowledging their
capacity to do just that: to behave as bases.
This generic equation showing the behavior of any and all conjugate bases in water is a
special type of reaction called a hydrolysis reaction. Hydrolysis reactions are water-splitting
reactions.
In this reaction the conjugate base shares one of its nonbonding pairs of electrons with one of the
hydrogen atoms of a water molecule and forms a coordinate covalent bond with it. The hydrogen
atom in water loses its electron. The O-H covalent bond in water is broken. Both electrons in the
broken bond are transferred to the oxygen atom which now belongs to a hydroxide ion. There are
numerous reactions in which hydrolysis reactions occur besides the reactions of conjugate bases. In
physiology hydrolysis plays an important role in many common and important reactions. For example, when
protein is ingested one of the first steps in the process of digestion is hydrolysis of the bonds between the
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various amino acids in the protein. There are many other examples. You should know, by definition, what
a hydrolysis reaction is.
You need to be able to identify the conjugate base of any acid. This is far easier than you
might believe. We simply remove the acidic “H” at the beginning of the molecular formula for an
acid. The remainder will have a net negative charge as the acidic hydrogen atom leaves its electron
behind when the acid dissociates. Here are a few examples:
acid
conjugate base
hydrofluoric acid, HF
fluoride ion, F-
hydrochloric acid, HCl
chloride ion, Cl-
hydrobromic acid, HBr
bromide ion, Br-
hydroiodic acid, HI
iodide ion, I-
nitric acid, HNO3
nitrate ion, NO3-
perchloric acid, HClO4
perchlorate ion, ClO4-
acetic acid, HC2H3O2
acetate ion, C2H3O2-
sulfuric acid, H2SO4
hydrogen sulfate ion, HSO4-
hydrogen sulfate ion, HSO4-
sulfate ion, SO42-
phosphoric acid, H3PO4
dihydrogen phosphate ion, H2PO4-
dihydrogen phosphate ion, H2PO4-
hydrogen phosphate ion, HPO42-
hydrogen phosphate ion, HPO42-
phosphate ion, PO43-
There are several important things to note in this table.
The relationship between an acid and it conjugate base is the same for all acids, regardless
of whether they are strong acids or weak acids. Simply remove the acidic “H” from the molecular
formula and add a negative charge to the acid’s anion to determine its conjugate base.
Note that the molecular formula for sulfuric acid is H2SO4, something you’ve known for
some time now. The number “2" following the initial “H” in the molecular formula tells us that a
single molecule of sulfuric acid can donate two protons in a chemical reaction. In a broader sense,
this is true of all acids. The subscript following the initial “H” in the molecular formula of any acid
tells us how many protons one molecule of that acid can donate in a reaction. So the first seven acids
in the above table (hydrofluoric acid, hydrochloric acid, hydrobromic acid, hydroiodic acid, nitric
acid, perchloric acid, and acetic acid) can donate one hydrogen ion per molecule of acid. The “1" is
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not written expressly but is implied. This means that phosphoric acid, H3PO4, can donate three
hydrogen ions per molecule. And an imaginary acid, H4A, could donate how many hydrogen ions
per molecule of acid? (4)
Acids that can donate one hydrogen ion per molecule of acid are called monoprotic acids.
Or if you’d rather, we can also think in terms of moles and say that a monoprotic acid is an acid that can
donate one mole of hydrogen ion per mole of acid molecules. Acids that donate two hydrogen ions per
molecule of acid are classified as diprotic acids, while those that can donate three hydrogen ions per
acid molecule are referred to as triprotic acids. Most of the common acids in the world around us
are monoprotic or diprotic acids. There are a few common triprotic acids, such as phosphoric acid
and citric acid. And there are one or two tetraprotic acids known, which can donate four hydrogen
ions per acid molecule. In general, acids that are not monoprotic may be referred to as polyprotic
acids, acids that can donate more than one hydrogen ion per molecule.
The seven monoprotic acids in the above table behave exactly as described when predicting
their conjugate bases. But when we reach sulfuric acid, we find that it’s conjugate base is the
hydrogen sulfate ion and not simply sulfate as we might have predicted. This is because a diprotic
acid does not lose both of its acidic protons at once. Instead, they dissociate in a series of steps. The
number of dissociative steps is equal to the number of protons a molecule of acid can donate in a
reaction. Sulfuric acid can donate two protons in a chemical reaction. It will therefore lose these
protons in a series of two steps.
In the first step sulfuric acid reacts with water to form hydronium ion and sulfuric acid’s conjugate
base, hydrogen sulfate ion. In the second step hydrogen sulfate ion, which is itself an acid, reacts
with water to form hydronium ion and sulfate ion, the conjugate base of hydrogen sulfate ion. How
do we know hydrogen sulfate ion is an acid? It’s molecular formula begins with “H.” If we add these two
steps together we obtain the observed reaction, which tells us that one molecule of sulfuric acid will
react with two molecules of water to form two molecules of hydronium ion and one molecule of
sulfate ion. The hydrogen sulfate ion does not cancel out in this case because it is a spectator ion. It is
simply a function of the arithmetic and nothing more.
We have observed that hydrogen sulfate ion can behave both as an acid and also as a base.
Substances with this ability are classed as amphoteric, possessing the capacity to act either as an acid
or as a base. We can also classify hydrogen sulfate are amphiprotic, possessing the capacity to either
donate or accept a proton. Amphoteric and amphiprotic do not mean exactly the same thing, but the
two definitions are sufficiently close that we, for our purposes in this class, will consider them to
be the same. Amphiprotic substances are actually quite common in the world around us.
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We have observed that phosphoric acid is a triprotic acid. We might therefore reasonably
deduce that it will not lose all three acid protons at once but rather, in a series of three dissociative
steps which are as follows:
In the first step phosphoric acid and water react to form hydronium ion and dihydrogen phosphate
ion, the conjugate base of phosphoric acid. In the second step dihydrogen phosphate ion, which is
amphoteric, reacts with water to form hydronium ion and hydrogen phosphate ion, the conjugate
base of dihydrogen phosphate ion. And in the third step hydrogen phosphate ion, which is also
amphoteric, reacts with water to form hydronium ion and phosphate ion, the conjugate base of
hydrogen phosphate ion. In the overall reaction one molecule of phosphoric acid reacts with three
water molecules to form three hydronium molecules and a molecule of phosphate ion. As was the
case above, hydrogen phosphate ion and dihydrogen phosphate ion do not cancel because they are
spectator ions, but rather, simply as a function of the arithmetic.
The behavior of bases and conjugate acids in water
The Arrhenius bases, which consist of the Group 1 and 2 hydroxides and ammonium
hydroxide, behave as ionic compounds in aqueous solution. They dissolve and they also dissociate.
The Group 1 and group 2 hydroxides are strong bases, and as such they ionize completely.
Ammonium hydroxide is a weak base, and while it does dissolve readily, only a fraction of its
solvated molecules also ionize. Using sodium hydroxide to serve as a model for all the Arrhenius
bases we see that they behave as follows:
NaOH(aq) => Na+(aq) + OH-(aq)
The generic behavior of any Brønsted-Lowry base in aqueous solution (except for the
Arrehenius bases), whether strong or weak, is described by the following generic equation:
B(aq) + H2O(l) => HB+(aq) + OH-(aq)
I hope you noticed that this is a hydrolysis reaction! In this equation B represents all Brønsted-Lowry
bases (except for the Arrhenius bases). The protonated form of the base HB+ is called the conjugate
acid of the base B. The conjugate acid of any base, by itself, will behave in aqueous solution in a
manner described by this generic equation:
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HB+(aq) + H2O(l) => B(aq) + H3O+(aq)
You need to be able to identify the conjugate acid of any base. We simply add a hydrogen
atom to the molecular formula of the base and change its charge by the amount of “+1". Here are
a few examples:
base
conjugate acid
ammonia, NH3
ammonium ion, NH4+
hydroxide ion, OH-
water, H2O
chloride ion, Cl-
hydrochloric acid, HCl
bromide ion, Br-
hydrobromic acid, HBr
nitrate ion, NO3-
nitric acid, HNO3
acetate ion, C2H3O2-
acetic acid, HC2H3O2
sulfate ion, SO42-
hydrogen sulfate ion, HSO4-
hydrogen sulfate ion, HSO4-
sulfuric acid, H2SO4
phosphate ion, PO43-
hydrogen phosphate ion, HPO42-
hydrogen phosphate ion, HPO42-
dihydrogen phosphate ion, H2PO4-
dihydrogen phosphate ion, H2PO4-
phosphoric acid, H3PO4
As with the table of acids and conjugate bases, there are several things to take note of in this
table as well.
Many of these conjugate acids are amphiprotic.
Conjugate acids do not always have a net positive charge. Regardless of the base’s charge,
the charge on its conjugate acid is always “1+" more than that of the base. Ammonia is a base with
no net charge. It’s conjugate acid ammonium ion has a net charge of 1+. Chloride ion has a net
charge of 1-. It’s conjugate acid hydrochloric acid has no net charge as it is neutral. Phosphate ion
has a net charge of 3-. It’s conjugate acid, hydrogen phosphate, has a net charge of 2-. Hydrogen
phosphate ion has a net charge of 2- while its conjugate acid, dihydrogen phosphate, has a net charge
of 1-. And so on.
You have been given a total of four generic equations in this section and the previous section.
These are the equations showing the behavior of all acids in water, the behavior of conjugate bases
in water, the behavior of Brønsted-Lowry bases in water, and the behavior of the conjugate acids
of Brønsted-Lowry bases in water. I do not expect you to remember the specific equations
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describing the behavior of sulfuric acid and phosphoric acid in aqueous solution, but you should
remember and be able to apply all four of these generic equations to any acid or base.
The acid-base behavior of water
Water is amphiprotic. It can behave as an acid or as a base, depending on circumstances. We
know that water is an acid as its molecular formula, H2O, begins with “H.” As we see in the table
in the previous section water is the conjugate acid of hydroxide ion. Water behaves as a base in the
formation of hydronium ion, which we discussed above, and also in many other reactions we have
not discussed.
The equation describing the equilibrium behavior of pure water is
2 H2O(l) W H3O+(aq) + OH-(aq)
This equation describes behavior referred to as the auto-ionization of water (it is also called the
auto-dissociation of water, the self ionization of water, and etc.).
The equilibrium constant expression for this reaction is
K w  [ H 3O  ][OH  ]
in which the equilibrium constant Kw is called the auto-ionization constant of water (or the autodissociation constant of water, the self-ionization constant of water, etc.). The numerical value of
this auto-ionization constant has been measured and is known to be 1.0 x 10-14 at roughly 0 oC.
Knowing this number permits us to make several very important observations. That Kw is
less than 1 tells us that this is a reactant-favored reaction. The forward reaction in which hydronium
ion and hydroxide ion are formed only happens to a very slight extent. The reverse reaction, in
which water remains in its undissociated state, is energetically favored and therefore spontaneous.
We discussed the meaning of equilibrium constants in the Chapter 7 section entitled “Equilibrium constants
and Gibbs free energy.”
Knowing the value of the equilibrium constant permits us to calculate the concentrations of
hydronium ion and hydroxide ion. In pure water
[H3O+] = [OH-]
as there is a one to one relationship between the two compounds in the balanced equation that
describes the auto-ionization of water. And by using the equilibrium constant expression and a bit
of arithmetic, we can calculate that
[H3O+] = [OH-] = 1.0 x 10-7
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The pH scale
In chemistry we often find it useful to employ logarithms when we work with numbers that
are either very large or very small. This is because in the days before calculators and computers, slide
rules and logarithm tables were commonly used to facilitate work with very large and very small numbers.
If we take the log (which is short for “logarithm”) of the hydrogen ion concentration in pure water
and change the sign of the resulting calculation
-log (1.0 x 10-7) = 7.00
we have calculated the pH of pure water. Note: throughout the remainder of this chapter when we speak
of taking the log of something, and from this point on we’ll mention it frequently, we mean the base 10
log. To do this on your calculator you use the “log” button, not the “ln” button which is a different type of
logarithmic function. This value serves as a reference point for determining whether things in the
world around us are acidic or basic. In an acidic solution the concentration of hydronium ion is
always greater than the concentration of hydroxide ion
[H3O+] > [OH-]
and the pH of an acidic solution is always less than 7. In a basic solution the concentration of
hydroxide ion is always greater than the concentration of hydronium ion
[H3O+] < [OH-]
and the pH of basic solutions is always greater than 7. In a neutral solution, the hydronium ion
concentration and hydroxide ion concentrations are equal
[H3O+] = [OH-]
and the pH is exactly equal to 7, as is the case in pure water.
Most of the things in the world around us are either acidic or basic, not neutral. There are
various places you can turn to get a better idea of which things are acidic and which things are basic.
Your text has a number of examples, and the Wikipedia entry for pH has several illustrations worth
examination. (http://en.wikipedia.org/wiki/PH note -should you look at this reference I would encourage
you to confine yourself to looking at the illustrations, as the accompanying text is something that will be
rather confusing and frustrating to most of you)
One important thing to note is that the pH scale is a logarithmic scale. This means that little
differences in pH correspond to big differences in acidity or basicity. One can calculate the relative
difference in the acidities of two solutions using the following equation
difference = 10ΔpH
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where ΔpH is the difference between the two pH values being compared.
If we want to know how much more acidic a solution with a pH of 6 is than a solution with
a pH of 7,
difference = 10(7-6) = 101
This means that the solution with a pH of 6 is ten times more acidic than the solution with a pH of 7.
If we compare two solutions, one with a pH of 7 and the other with a pH of 5, the difference in pH
is
difference = 10(7-5) = 102
or, the solution with the pH of 5 is 100 (102) times more acidic than the solution with a pH of 7. If
we compare the pH of pure water, with it a value of 7, with the pH of 1.0 M hydrochloric acid, with
a pH of 0, the difference in acidities is
difference = 10(7-0) = 107
or, the solution of hydrochloric acid is 10,000,000 (107) times more acidic than the pure water.
Calculating [H3O+], [OH-], and pH
Before we begin, you need to have a calculator that can do logarithms. A TI-30, a Casio
FX-260, or something equivalent is necessary. And you need to know how to use it. This is
something I can’t help you with. You need to sit down and take a look at the manual or to have
someone help you. You should also be sure you know how to use the exponential notation feature
of your calculator. For many calculators there is an “EE” button or an “EXP” button which should
be used to enter a number using scientific notation. Students are often surprised to learn that if they
simply type, for example, “1 x 10-14" into their calculator that the instrument understands the input
differently than if they used “1 EE ± 14" which is something the calculator will correctly interpret.
We will only discuss how to calculate the pH for strong acids and bases. It is not difficult
to calculate the pH of weak acids and bases but it requires the use of equilibrium theory in a way we
do not discuss in this class.
To calculate the pH of a strong acid solution we must know the equilibrium hydronium ion
concentration of that solution. Fortunately there is always a direct correlation between the starting
concentration of the acid and the equilibrium hydronium ion concentration which we can use to
make these calculations.
Let’s take hydrochloric acid as an example. The equation describing its behavior in water
is
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HCl(aq) + H2O(l) => H3O+(aq) + Cl-(aq)
This balanced equation shows us that there is a 1:1 correlation between the starting acid
concentration and the equilibrium hydronium ion concentration. So if we begin with 0.25 M HCl,
the concentration of hydronium ion at equilibrium is also 0.25 M. If we are given 1.50 M HCl, the
concentration of hydronium ion at equilibrium is also 1.50 M. And so on.
More generally speaking, what we have just seen for hydrochloric acid is also true for any
strong monoprotic acid. There is always a 1:1 relationship between the starting acid concentration
and the equilibrium hydronium ion concentration.
Diprotic acids are a bit different in that every mole of strong diprotic acid will produce two
moles of hydronium ion at equilibrium. Using sulfuric acid as an example the balanced equation
which shows its overall behavior in water is
H2SO4 (aq) + 2 H2O(l) => 2 H3O+(aq) + SO42-(aq)
We can see that there is a 1:2 correlation between the starting acid concentration and the equilibrium
hydronium ion concentration. So if we begin with 0.25 M H2SO4 , the concentration of hydronium
ion at equilibrium will be 0.50 M (2 x 0.25 M). If we are given 1.50 M H2SO4 , the concentration of
hydronium ion at equilibrium will be 3.00 M. And so on.
There are no strong triprotic or tetraprotic acids. But if there were, the relationship between
starting acid concentration and equilibrium hydronium ion concentration would be 1:3 and 1:4
respectively. In other words, if we had a 0.25 M concentration of a strong triprotic acid, the
equilibrium hydronium ion concentration would be 0.75 M (3 x 0.25). And if we had a 0.25 M
solution of a strong tetraprotic acid solution, the equilibrium hydronium ion concentration would
be 1.00 M (4 x 0.25).
Remember, this is only true for strong acids. As weak acids do not dissociate completely we
never see the direct relationships between starting weak acid concentration and equilibrium
hydronium ion concentration that we see with strong acids. If we write the equation that describes
the behavior of acetic acid in water
HC2H3O2 (aq) + H2O(l) => H3O+(aq) + C2H3O2-(aq)
we see the 1:1 relationship between the starting acid concentration and the equilibrium hydronium
ion concentration. But as a weak acid, acetic acid does not dissociate completely. Typically only
about 1% of the acetic acid molecules ionize. This means that if we start with 1.00 M of acetic acid,
our equilibrium hydronium ion concentration will only be around 0.01 mol. And we see similar
behavior with all of the other weak acids as well.
We take a similar approach to that we used with strong acids when calculating the hydroxide
ion concentration of solutions of strong bases. We can establish the relationship between the starting
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base concentration and the equilibrium hydroxide ion concentration by examining the equation that
describes the base’s behavior in water. If we consider sodium hydroxide as an example we see that
NaOH(aq) => Na+(aq) + OH-(aq)
This balanced equation shows us that there is a 1:1 correlation between the starting base
concentration and the equilibrium hydroxide ion concentration. So if we begin with 0.25 M NaOH,
the equilibrium hydroxide ion concentration is also 0.25 M. If we are given 1.50 M NaOH, the
equilibrium hydroxide ion concentration is also 1.50 M. And so on.
The behavior shown by sodium hydroxide is the same for any other Group I hydroxide.
There is a 1:1 relationship between the starting base concentration and the equilibrium hydroxide
ion concentration. This is correct because these compounds behave as strong bases.
The Group II hydroxides differ from those in Group I in that one mole of base will produce
two moles of hydroxide ion at equilibrium. Using calcium hydroxide as an example, the balanced
equation which shows its behavior in water is
Ca(OH)2 (aq)=> Ca2+(aq) + 2 OH-(aq)
We can see that there is a 1:2 correlation between the starting base concentration and the equilibrium
hydroxide ion concentration. So if we begin with 0.25 M Ca(OH)2 , the concentration of hydroxide
ion at equilibrium will be 0.50 M (2 x 0.25 M). If we are given 1.50 M Ca(OH)2 , the equilibrium
concentration of hydroxide ion will be 3.00 M. And so on.
As was the case for acids, what you have just read is only true for strong bases. Weak bases
do not dissociate completely. We therefore never see the direct relationships between starting weak
base concentration and equilibrium hydroxide ion concentration that we see with strong bases. If we
are given a 1.00 M solution of ammonium hydroxide, we cannot say that the equilibrium hydroxide
ion concentration will also be 1.00 M. We only know that, as a weak base, the equilibrium hydroxide
ion concentration will be considerably less than 1.00 M.
When we calculate pH we need a hydronium ion concentration. Is it possible to determine
the hydronium ion concentration for a strong base solution? The answer is yes. We have two
different techniques, both of which will produce exactly the same results when used correctly.
We saw above that
K w  [ H3O  ][OH  ]
As Kw is a constant and always equal to 1 x 10-14 (at least, this is always the value we will assume
for our calculations in this class), this means that any time we know the hydroxide ion concentration
of a solution we can rearrange this equation to find the hydronium ion concentration
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Kw

[OH ]
 [ H 3O  ]
Let’s work a few examples. We mentioned above that the equilibrium hydroxide ion concentration
of a 0.25 M solution of sodium hydroxide will also be 0.25 M. What is the hydronium ion
concentration of this solution? We can simply plug numbers into this equation to find the answer.
1x10 14
 [ H 3O ] 
 4.00 x10 14

0.25
[OH ]
Kw

If we begin with 0.25 M Ca(OH)2 , the concentration of hydroxide ion at equilibrium will be 0.50 M,
as we said above. What is the hydronium ion concentration of this solution? Again, we can simply
plug numbers into the equation used in the previous example to calculate our answer.
1x10 14
 [ H 3O ] 
 2.00 x10 14

0.50
[OH ]
Kw

Once we know the hydronium ion concentration of a solution we can calculate the pH of that
solution. Again, let’s work through a few examples. At this point you need to get out your calculator
and actually try to get the same answers as I do in the following table.
solution
[H3O+]
pH = -log [H3O+]
0.25 M HCl
0.25 M
0.60
1.50 M HCl
1.50 M
-0.18
0.25 M H2SO4
0.50 M
0.30
1.50 M H2SO4
3.00 M
-0.48
0.25 M NaOH
4.00 x 10-14 M
13.4
0.25 M Ca(OH)2
2.00 x 10-14 M
13.7
How did you do with your calculations? A point of interest may be seen in this table. You observe that
the 1.50 M solutions of hydrochloric acid and of sulfuric acid both have negative pH values. Is that
possible? Yes, it is. The pH scale is often depicted as having endpoints at pH values of 0 and 14 but
this is not correct. Any acid solution with a hydronium ion concentration of 1.0 M or greater will
have a pH less than zero, i.e. a negative pH. Any base solution with a hydroxide ion concentration
of 1.0 M or greater will have a ph above 14. Concentrated strong acid solutions may have pH values
ranging as low as -2 to -3. Concentrated strong base solutions may have pH values as high as 15 to
16.
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Buffer solutions
If you’ve studied human physiology you know that our bodies operate at a very slightly basic
pH of around 7.4. If our physiological pH climbs above 7.45 we enter respiratory alkalosis. If our
physiological pH drops below 7.35 it causes respiratory acidosis. And if our pH changes much
beyond either of these values our bodies stop working properly and we die. The difference in
hydronium ion content between these values is very small, so in theory it should not take too much
acid or base to cause a dramatic and very unhealthy change in pH for the average person.
pH
[H3O+]
7.35
4.46 x 10-8 M
7.40
3.98 x 10-8 M
7.45
3.54 x 10-8 M
The pH of pure water is exactly 7.0. If we take 1.00 L of pure water and add just 1.00 mL
of 1.0 M hydrochloric acid to it, the pH of the resulting solution will drop from 7.0 to 3.0. As the
pH scale is a logarithmic scale this means that adding this small amount of a rather dilute acid
solution makes the resulting solution 10,000 times more acidic than pure water.
We see something similar occur if we add just 1.00 mL of 1.0 M sodium hydroxide to 1.00 L
of pure water. The solution pH will jump from 7.0 to 11.0, 10,000 times more basic than pure water.
Think about the implications of this. Every day most of what we eat and drink is either acidic
or basic. Take a look at the following web page kept by the U.S. Food and Drug Administration
which lists the pH values of some of the foods we commonly eat (it can be found at
http://www.cfsan.fda.gov/~comm/lacf-phs.html). We need to maintain a pH of about 7.4 but most
of what we eat has a pH far different from this. If we could only eat foods with a pH of about 7.4,
according to this web page our diets would consist mostly of Camembert cheese, cooked frozen
corn, graham crackers, chocolate cake, cooked lobster, and tofu. We would risk death were we to
eat those things that most of us prefer in our diets. We couldn’t even drink pure water, let alone tap
water or bottled water which are usually much more acidic than pure water. And no soft drinks,
milk, coffee, beer, wine, or hard liquors either. Life would get rather thirsty for most of us very
quickly.
So how is it that we can eat and drink these things we love without running the risk of
imminent death? Buffers (or, buffer solutions) are solutions with the ability to resist changes in pH
when acids or bases are added to them. And our body contains a buffer that makes it possible to live
in a world where a neutral pH is seldom ever found.
A buffer solution usually consists of a weak acid and its conjugate base. Although it is also
possible to have a buffer solution made with a weak base and its conjugate acid. The buffer solution we
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find in our bodies consists of carbonic acid, H2CO3, and its conjugate base bicarbonate ion, HCO3-.
Remember that bicarbonate ion is also known as hydrogen carbonate ion which is one of the polyatomic
ions you hopefully learned and remember from Chapter 4. The carbonic acid in our blood is formed
from the combination reaction of the carbon dioxide produced as a waste product in cellular
metabolism and the water in our plasma. Bicarbonate ion is formed by the ionization of carbonic
acid. While this is certainly the most interesting and important buffer system to us, there are many
other buffer systems that play a role in other living systems, biochemistry, and other studies in
chemistry.
How does a buffer resist a change in pH when an acid or base is added? When an acid is
added to a buffer solution it reacts with the conjugate base of the buffer. And when a base is added
to a buffer solution it reacts with the acid of the buffer.
Let’s take a closer look at this. We can imagine the way in which an acid is neutralized by
a buffer solution as occurring in three steps. Assume we add some generic acid HX to a carbonic
acid/bicarbonate ion buffer solution. In water the acid will dissociate to some extent, depending on
whether it is a strong or a weak acid
HX(aq) + H2O(l) W H3O+(aq) + X-(aq)
The hydronium ion formed in this step will then react with bicarbonate ion found in the buffer
solution and will be completely consumed, regardless of whether the added acid is a strong acid or
a weak acid
H3O+(aq) + HCO3-(aq) W H2CO3(aq) + H2O(l)
The carbonic acid formed in this reaction will then behave as all acids behave in water
H2CO3(aq) + H2O(l) W H3O+(aq) + HCO3-(aq)
but as carbonic acid is a weak acid it will only ionize slightly, which means the solution hydronium
ion concentration will increase only slightly, and the solution pH will also only change to a slight
extent.
We can also imagine the way in which a carbonic acid-bicarbonate ion buffer solution acts
on a base as taking place in a series of three steps. In the first step the base “B” either forms
hydroxide ion directly or forms hydroxide ion through the hydrolysis of water
B(aq) + H2O(l) W HB+(aq) + OH-(aq)
In the second step carbonic acid reacts with the hydroxide ion formed in the first step and consumes
it completely
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H2CO3(aq) + OH-(aq) W H2O(l) + HCO3-(aq)
In the third step the bicarbonate ion hydrolyzes water and forms carbonic acid and hydroxide ion
as products
HCO3-(aq) + H2O(l) W H2CO3(aq) + OH-(aq)
but as bicarbonate ion is a weak base, the increase in the hydroxide ion concentration of the solution
will be small, which means the resulting change in pH will also be small.
A common buffer solution in chemistry is made with acetic acid and acetate ion. If we have
1.00 L of a buffer solution that is 0.70 M in acetic acid and 0.60 M in acetate ion, the pH of this
buffer will be 4.68. We saw above that if we added 1.00 mL of 1.0 M HCl to a liter of water that the
pH changed from 7.0 to 3.0. But if we add the same volume of this same hydrochloric acid to our
liter of acetic acid-acetate buffer solution, there is no measurable change in the buffer solution pH.
I’m not going to show you how we calculate this as it is beyond the scope of this course. But if you’re
interested you can see how we do this at my Chem 1220 web site. Ask me and I’ll tell you where to find
it.
Every buffer solution has its limits. This limit is called the buffer capacity and it depends on
the amounts of weak acid and conjugate base found in the buffer solution. In the acetic acid-acetate
buffer solution we just discussed, if we have exactly 1.00 liter of this solution then we have 0.70
moles of acetic acid and 0.60 moles of acetate ion. When we add an acid to this solution it reacts
with the acetate ion on a 1:1 basis. If, then, we add any amount of acid beyond 0.600 moles, say for
example, 0.601 moles of acid to a liter of this solution, we will consume all of the acetate ion in the
solution and we now will see a large change in pH as a consequence. The same is true for the acetic
acid. If we add any amount of base beyond 0.70 moles, say, for example, 0.701 moles of a base to
a liter of this solution, the acetic acid - which reacts with the base on a 1:1 ratio, will all be
consumed and a large change in pH will result.
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