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Chapter 2
Atoms, Molecules, and Ions
國防醫學院 生化學科
王明芳老師
2011-9-13
1
Conservation of Mass and the
Law of Definite Proportions
Law of Conservation of Mass: Mass is neither created
nor destroyed in chemical reactions.
Aqueous solutions of mercury(II) nitrate and
potassium iodide will react to form a precipitate of
mercury(II) iodide and aqueous potassium iodide.
3.25 g + 3.32 g = 6.57 g
Hg(NO3)2(aq) + 2KI(aq)
HgI2(s) + 2KNO3(aq)
4.55 g + 2.02 g = 6.57 g
Chapter 2/2
Conservation of Mass and the
Law of Definite Proportions
3.25 g + 3.32 g = 6.57 g
Hg(NO3)2(aq) + 2KI(aq)
HgI2(s) + 2KNO3(aq)
4.55 g + 2.02 g = 6.57 g
Chapter 2/3
Conservation of Mass and the
Law of Definite Proportions
Law of Definite Proportions: Different samples of
a pure chemical substance always contain the
same proportion of elements by mass.
By mass, water is: 88.8 % oxygen
11.2 % hydrogen
Chapter 2/4
The Law of Multiple Proportions
and Dalton’s Atomic Theory
Law of Multiple Proportions: Elements can combine in
different ways to form different substances, whose mass
ratios are small whole-number multiples of each other.
nitrogen monoxide: 7 grams nitrogen per 8 grams oxygen
nitrogen dioxide: 7 grams nitrogen per 16 grams oxygen
Chapter 2/5
The Law of Multiple Proportions
and Dalton’s Atomic Theory
Law of Multiple Proportions: Elements can combine in
different ways to form different substances, whose mass
ratios are small whole-number multiples of each other.
Insert Figure 2.2 p37
Chapter 2/6
The Law of Multiple Proportions
and Dalton’s Atomic Theory
Taking all three laws together --- the law of
mass conservation, the law of definite
proportions, and the law of multiple proportions
--- ultimately led Dalton to propose a new
theory of matter.
•
Elements are made up of tiny particles called
atoms.
•
Each element is characterized by the mass of its
atoms. Atoms of the same element have the same
mass, but atoms of different elements have
Chapter 2/7
different masses.
The Law of Multiple Proportions
and Dalton’s Atomic Theory
•
The chemical combination of elements to make
different chemical compounds occurs when atoms
join in small whole-number ratios.
•
Chemical reactions only rearrange how atoms are
combined in chemical compounds; the atoms
themselves don’t change.
Not everything that Dalton proposed was correct. He though,
for instance, that water had the formula HO rather than H2O
Chapter 2/8
Example 2.1 Using the Law of Multiple Proportions
Methane and propane are both constituents of natural gas. A sample of methane contains 5.70 g of carbon atoms
and 1.90 g of hydrogen atoms combined in a certain way, whereas a sample of propane contains 4.47 g of carbon
atoms and 0.993 g of hydrogen atoms combined in a different way. Show that the two compounds obey the law
of multiple proportions.
Solution
Find the C : H mass ratio in each compound, and then compare the ratios to see whether they are
small whole-number multiples of each other.
Atomic Structure: Electrons
Cathode-Ray Tubes: J. J. Thomson (1856-1940)
proposed that cathode rays must consist of tiny negatively
charged particles. We now call them electrons.
Chapter 2/10
Atomic Structure: Electrons
Deflection of electron depends on three factors:
1. Strength of electric or magnetic field
2. Size of negative charge on electron
3. Mass of the electron
Thomson calculated the electron’s charge to mass
ratio as 1.758820 x 108 Coulombs per gram (C/g).
e/m = 1.758820 x 108 C/g
Chapter 2/11
Atomic Structure: Electrons
Chapter 2/12
Atomic Structure: Electrons
Oil Drop Experiment (Millikan, 1868–1953):
● Applied a voltage to oppose the downward fall of
charged drops and suspend them.
● Voltage on plates place 1.602176 x 10-19 C of charge
on each oil drop.
● Millikan calculated the electron’s mass as 9.109382 x
10-28 grams.
e/m = 1.758820 x 108 C/g
m = e/(1.758820 x 108 C/g)
= (1.602176 x 10-19 C )/(1.758820 x 108 C/g)
= 9.109382 x 10-28 g
Chapter 2/13
Atomic Structure: Protons and
Neutrons
Rutherford’s scattering experiment
Chapter 2/14
Atomic Structure: Protons and
Neutrons
Atomic Nucleus: Ernest Rutherford (1871-1937)
bombarded gold foil with alpha particles. Although
most of the alpha particles passed through the foil
undeflected, approximately 1 in every 20,000
particles were deflected. A fraction of those particles
were deflected back at an extreme angle.
Rutherford proposed that the atom must consist
mainly of empty space with the mass concentrated
in a tiny central core—the nucleus.
Chapter 2/15
Atomic Structure: Protons and
Neutrons
a particle:
A type of emission,
naturally occurring
radioactive elements.
Chapter 2/16
Atomic Structure: Protons and
Neutrons
A view of the atom
Chapter 2/17
Atomic Structure: Protons and
Neutrons
Chapter 2/18
Atomic Structure: Protons and
Neutrons
The mass of the atom is
primarily in the nucleus.
Chapter 2/19
Atomic Structure: Protons and
Neutrons
The charge of the proton is
opposite in sign but equal to
that of the electron.
Chapter 2/20
Example 2.2 Calculations Using Atomic Size
Ordinary “lead” pencils actually are made of a form of carbon called graphite. If a pencil line is 0.35 mm wide
and the diameter of a carbon atom is 1.5  1010 m, how many atoms wide is the line?
conversion factors
Atomic Numbers
Atomic Number (Z): Number of protons in an atom’s
nucleus. Equivalent to the number of electrons around
an atom’s nucleus
Mass Number (A): The sum of the number of protons
and the number of neutrons in an atom’s nucleus
Isotope: Atoms with identical atomic numbers but
different mass numbers
Chapter 2/22
Atomic Numbers
99.985%
0.015%
Chapter 2/23
Atomic Numbers
carbon-12
mass number
12
6C
6 protons
6 electrons
6 neutrons
14
6C
6 protons
6 electrons
8 neutrons
atomic number
carbon-14
mass number
atomic number
Chapter 2/24
Example 2.3 Interpreting an Isotope Symbol
The isotope of uranium used to generate nuclear power is
does an atom of
have?
. How many protons, neutrons, and electrons
Solution
An atom of
has 92 protons, 92 electrons, and 235 – 92 = 143 neutrons.
Example 2.4 Writing an Isotope Symbol
Element X is toxic to humans in high concentration but is essential to life in low concentrations. Identify
element X, whose atoms contain 24 protons, and write the symbol for the isotope of X that has 28 neutrons.
Solution
According to the periodic table, the element with atomic number 24 is chromium (Cr). The particular
isotope of chromium in this instance has a mass number of 24 + 28 = 52 and is written
.
Atomic Masses and the Mole
The mass of 1 atom of carbon-12 is defined to be 12 amu.
Atomic Mass (Atomic Weight): The weighted
average of the isotopic masses of the element’s
naturally occurring isotopes
Atomic mass unit (amu), also called a dalton (Da)
1 amu = 1.660539 X 10-24 g
Chapter 2/27
Atomic Masses and the Mole
Why is the atomic mass of the element carbon
12.01 amu?
carbon-12: 98.89 % natural abundance
12 amu
carbon-13: 1.11 % natural abundance
13.0034 amu
mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111)
= 11.87 amu + 0.144 amu
= 12.01 amu
Chapter 2/28
Atomic Masses and the Mole
Avogadro’s Number (NA): One mole of any substance
contains 6.022 x 1023 formula units.
Molar Mass: The mass in
grams of one mole of any
element. It is numerically
equivalent to its atomic
mass.
These samples of helium, sulfur, copper,
and mercury each contain 1 mole.
Chapter 2/29
Example 2.5 Calculating an Atomic Mass
Chlorine has two naturally occurring isotopes:
with a natural abundance of 75.76% and an isotopic mass
of 34.969 amu, and
, with a natural abundance of 24.24% and an isotopic mass of 36.966 amu. What is
the atomic mass of chlorine?
Solution
Atomic mass = (34.969 amu)(0.7576) + (36.966 amu)(0.2424) = 35.45 amu
Example 2.6 Converting from Mass to Numbers of Moles and Atoms
How many moles and how many atoms of silicon are in a sample weighing 10.53 g? The atomic mass of silicon
is 28.0855 amu.
The fact that the atomic mass of silicon is 28.0855 amu means that 1 mol of silicon has a mass of
28.0855 g. Use this molar mass to convert between mass and number of moles, and then use
Avogadro’s number to find the number of atoms.
Solution
.
Nuclear Chemistry: The Change
of One Element Into Another
Nuclear Equation: A reaction that changes an atomic
nucleus
14
C
6
carbon-14 nucleus
14
N
7
+
0
e
-1
nitrogen-14 nucleus
In a nuclear equation, the element symbols represent only the nuclei of atoms
Rather than the entire neutral atoms, so the subscript represents only the
number of nuclear charges (protons). An emitted electron is written as 0-1e
, where the superscript o indicates that the mass of an electron is essentially
zero when compared to that of a proton or neutron, and the subscript
indicates that the charge is -1.
Chapter 2/32
Nuclear Chemistry: The Change
of One Element Into Another
Comparisons Between Nuclear and Chemical Reactions
•
A nuclear reaction changes an atom’s nucleus usually
producing a different element. A chemical reaction, by
contrast, involves only a change in the way that different
atoms are combined.
•
Different isotopes of an elements have essentially the
same behavior in chemical reactions but often have
completely different behavior in nuclear reactions.
•
The energy change accompanying a nuclear reaction is
far greater than that accompanying a chemical reaction.
Chapter 2/33
Radioactivity
The effect of an electric field on a, b, and g radiation
Chapter 2/34
Radioactivity
Alpha (a ) Radiation
An alpha particle is a helium-4 nucleus (2 protons
and 2 neutrons).
Alpha particle, a
Chapter 2/35
Radioactivity
Beta (b ) Radiation
A beta particle is an electron.
Beta particle, b Chapter 2/36
Radioactivity
Gamma (g ) Radiation
A gamma particle is a high-energy photon
Gamma radiation almost always accompanies a and
b emission as a mechanism for the release of
energy, but it is often not shown when writing
nuclear equations because it changes neither the
mass number nor the atomic number of the product
nucleus.
Chapter 2/37
Radioactivity
Positron Emission
A positron has the same mass as an electron but
the opposite charge.
Positron, b +
Chapter 2/38
Radioactivity
Electron Capture
A process in which the nucleus captures an innershell electron, thereby converting a proton to a
neutron.
Chapter 2/39
Radioactivity
Chapter 2/40
Example 2.7 Balancing Nuclear Equations
Write a balanced nuclear equation for each of the following processes:
(a) Alpha emission from curium-242:
(b) Beta emission from magnesium-28:
(c) Positron emission from xenon-118:
Solution
(a) In a emission, the mass number decreases by 4 and the atomic number decreases by 2,
giving plutonium-238:
(b) In b emission, the mass number is unchanged and the atomic number increases by 1,
giving aluminum-28:
(c) In positron emission, the mass number is unchanged and the atomic number decreases by 1,
giving iodine-118:
Nuclear Stability
The band indicates
various
neutron/proton
combinations that
give rise to nuclei
that are either
nonradioactive or
that are radioactive
but decay slowly
enough to exist for
a measurable time.
The band of nuclear stability
Chapter 2/42
Nuclear Stability
•
Every element in the periodic table has at least one
radioactive isotope.
•
Hydrogen is the only element whose most abundant
stable isotope, hydrogen-1, contains more protons (1)
than neutrons (0).
Chapter 2/43
Nuclear Stability
•
The ratio of neutrons to protons gradually increases,
giving a curved appearance to the band of stability.
•
All isotopes heavier than bismuth-209 are radioactive,
even though they may decay slowly and be stable
enough to occur naturally.
Chapter 2/44
Nuclear Stability
This look at the region
from Z = 66
(dysprosium) through
Z = 79 (gold) shows
the types of
radioactive processes
that various
radioisotopes undergo.
A close-up look at the band of nuclear stability
Chapter 2/45
Mixtures and Chemical
Compounds
A mixture is simply a blend of two or more substances added together in some
arbitrary proportion without chemically changing the individual substances
themselves. (H2 and O2; sugar+ salt)
Most substances are chemical compounds, formed when atoms of two or more
elements combine a chemical reaction.
Chapter 2/46
Molecules and Covalent Bonds
Covalent Bond: Results when two atoms share
several (usually two) electrons. Typically a
nonmetal bonded to a nonmetal
A covalent bond between atoms is analogous to tug-of-war
Chapter 2/47
Molecules and Covalent Bonds
Covalent Bond: Results when two atoms share
several (usually two) electrons. Typically a
nonmetal bonded to a nonmetal
Molecule: The unit of matter that results when two
or more atoms are joined by covalent bonds.
Chapter 2/48
Molecules and Covalent Bonds
Insert Figure 2.12 p56
Structural formula: Uses lines between atoms to
indicate the covalent bonds
49
Example 2.8 Drawing a Structural Formula
Propane, C3H8, has a structure in which the three carbon atoms are bonded in a row, each end carbon is bonded
to three hydrogens, and the middle carbon is bonded to two hydrogens. Draw the structural formula, using lines
between atoms to represent covalent bonds.
Solution
Example 2.9 Visual Representations of Mixtures and Compounds
Which of the following drawings represents a mixture, which a pure compound, and which an element?
Solution
Drawing (a) represents a mixture of two diatomic elements, one composed of two red atoms and
one composed of two blue atoms.
Drawing (b) represents molecules of a pure diatomic element because all atoms are identical.
Drawing (c) represents molecules of a pure compound composed of one red and one blue atom.
Ions and Ionic Bonds
Ionic Bond: A transfer of one or more electrons from
one atom to another. A strong electrical attraction
between charged particles. Typically a metal bonded
to a nonmetal.
Ion: A charged particle
Cation: A positively charged particle. Metals tend to
form cations.
Anion: A negatively charged particle. Nonmetals tend
to form anions.
Chapter 2/52
Ions and Ionic Bonds
In the formation of sodium chloride, one electron is
transferred from the sodium atom to the chlorine
atom.
1
Na + 2 Cl2
Na+ + Cl-
Ionic solid
Chapter 2/53
Example 2.10 Identifying Ionic and Molecular Compounds
Which of the following compounds would you expect to be ionic and which molecular (covalent)?
(a) BaF2
(b) SF4
(c) PH3
(d) CH3OH
Remember that covalent bonds generally form between nonmetal atoms, while ionic bonds form between
metal and nonmetal atoms.
Solution
Compound (a) is composed of a metal (barium) and a nonmetal (fluorine) and is likely to be ionic.
Compounds (b)–(d) are composed entirely of nonmetals and therefore are probably molecular.
Naming Chemical Compounds
Cation Charges for Typical Main-Group Ions
1+ 2+
3+
Chapter 2/55
Naming Chemical Compounds
Anion Charges for Typical Main-Group Ions
3- 2- 1-
A cation bears the same name as the element it is
derived Form; An anion name has an –ide ending
Chapter 2/56
Naming Chemical Compounds
Ionic Compound: A neutral compound in which
the total number of positive charges must equal
the total number of negative charges.
Binary Ionic Compounds
Na+
Cl-
NaCl
magnesium oxide:
Mg2+
O2-
MgO
aluminum sulfide:
Al3+
S2-
Al2S3
sodium chloride:
Chapter 2/57
Naming Chemical Compounds
Some transition metals form more than one cation
Only ions that exist in aqueous solution are shown.
Chapter 2/58
Naming Chemical Compounds
Use Roman numerals in parentheses to indicate
the charge on metals that form more than one
kind of cation.
Binary Ionic Compounds
iron(III) oxide:
Fe3+
O2-
Fe2O3
tin(II) chloride:
Sn2+
Cl-
SnCl2
lead(II) fluoride:
Pb2+
F-
PbF2
Chapter 2/59
Naming Chemical Compounds
Fe2+ Iron(II) ion; Ferrous ion
Fe3+ Iron(III) ion; Ferric ion
Sn2+ Tin(II) ion; Stannous ion
Sn4+ Tin(IV) ion; Stannic ion
FeCl2
Iron(II) chloride; Ferrous chloride
FeCl3
Iron(III) chloride; Ferric chloride
Chapter 2/60
Example 2.11 Naming Binary Ionic Compounds
Give systematic names for the following compounds:
(a) BaCl2
(b) CrCl3
(c) PbS
(d) Fe2O3
Try to figure out the number of positive charges on each cation by counting the number of negative
charges on the associated anion(s)
Solution
(a) Barium chloride
No Roman numeral is necessary because barium, a group 2A element,
forms only Ba2+.
(b) Chromium(III) chloride The Roman numeral III is necessary to specify the +3 charge on
chromium (a transition metal).
(c) Lead(II) sulfide
The sulfide anion (S2) has a double negative charge, so the lead cation
must be doubly positive.
(d) Iron(III) oxide
The three oxide anions (O2) have a total negative charge of –6, so the
two iron cations must have a total charge of +6. Thus, each is Fe(III).
Example 2.12 Converting Names into Formulas
Write formulas for the following compounds:
(a) Magnesium fluoride
(b) Tin(IV) oxide
(c) Iron(III) sulfide
Solution
For transition metal compounds, the charge on the cation is indicated by the Roman numeral in the
name. Knowing the number of positive charges, you can then figure out the number of necessary
negative charges for the associated anions.
(a) MgF2
Magnesium (group 2A) forms only a 2+ cation, so there must be two fluoride ions (F)
to balance the charge.
(b) SnO2
Tin(IV) has a +4 charge, so there must be two oxide ions (O2) to balance the charge.
(c) Fe2S3
Iron(III) has a +3 charge and sulfide ion a 2 charge (S2), so there must be two irons and
three sulfurs.
Naming Chemical Compounds
Binary Molecular Compounds
Because nonmetals often
combine with one another in
different proportions to form
different compounds, numerical
prefixes are usually included in
the names of binary molecular
compounds.
Chapter 2/63
Naming Chemical Compounds
N2F4
The prefix is added to the front of each to
indicate the number of each atom.
dinitrogen tetrafluoride
Chapter 2/64
Naming Chemical Compounds
Binary Molecular Compounds
Whenever the prefix ends in ―a‖ or ―o‖ and
the element name begins with a vowel,
drop the ―a‖ or ―o‖ in the prefix.
N2O4
dinitrogen tetroxide
Whenever the prefix for the first element
is ―mono,‖ drop it.
CO2
carbon dioxide
CO
carbon monoxide
Chapter 2/65
Example 2.13 Naming Binary Molecular Compounds
Give systematic names for the following compounds:
(a) PCl3
(b) N2O3
(c) P4O7
(d) BrF3
Solution
Look at a periodic table to see which element in each compound is more cationlike (farther to the
left or lower) and which is more anionlike (farther to the right or higher). Then name the
compound using the appropriate numerical prefix.
(a) Phosphorus trichloride
(c) Tetraphosphorus heptoxide
(b) Dinitrogen trioxide
(d) Bromine trifluoride
Naming Chemical Compounds
1.
2.
Cation is
identified,
then the
anion
Oxoanions
fewer O, -ite
more O, -ate
less O, hypomore O, per-
“H”
hydrogen
 bi-
3. Presence/absence
Chapter 2/67
Naming Chemical Compounds
Polyatomic Ionic Compounds
Na+
OH-
NaOH
Mg2+
CO32-
MgCO3
sodium carbonate:
Na+
CO32-
Na2CO3
iron(II) hydroxide:
Fe2+
OH-
Fe(OH)2
sodium hydroxide:
magnesium carbonate:
Chapter 2/68
Example 2.14 Naming Compounds with Polyatomic Ions
Give systematic names for the following compounds:
(a) LiNO3
(b) KHSO4
(c) CuCO3
(d) Fe(ClO4)3
Solution
Unfortunately, there is no alternative: The names and charges of the common
polyatomic ions must be memorized.
(a) Lithium nitrate Lithium (group 1A) forms only the Li+ ion and does not need a Roman numeral.
(b) Potassium hydrogen sulfate (potassium bisulfate) Potassium (group 1A) forms only the K+ ion.
(c) Copper(II) carbonate The carbonate ion has a 2 charge, so copper must be +2. A Roman
numeral is needed because copper, a transition metal, can form more than one ion.
(d) Iron(III) perchlorate There are three perchlorate ions, each with a 1 charge, so the iron must have
a +3 charge.
Example 2.15 Writing Formulas of Compounds with Polyatomic Ions
Write formulas for the following compounds:
(a) Potassium hypochlorite
(b) Silver(I) chromate
(c) Iron(III) carbonate
Solution
(a) KClO
Potassium forms only the K+ ion, so only one ClO is needed.
(b) Ag2CrO4
The polyatomic chromate ion has a 2 charge, so two Ag+ ions are needed.
(c) Fe2(CO3)3
Iron(III) has a +3 charge and the polyatomic carbonate ion has a 2 charge,
so there must be two iron ions and three carbonate ions. The polyatomic
carbonate ion is set off in parentheses to indicate that there are three of
them.