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CHEMISTRY
The
Standard
Deviants®
Core
Curriculum
I n s t r u c t o r ’s Guide
Films for the
Humanities & Sciences
®
Introduction
The Core Curriculum Series is designed to be a complete treatment of major topics covered in
an introductory course on a given curricular subject. The series utilizes a unique educational
approach that combines the best of visual and tutorial elements. Through high-end graphics,
animation, and design techniques, academic concepts are broken down, thoroughly explained,
and demonstrated through the creative use of presentational devices and examples.
Each individual subject series consists of up to 10 programs of approximately 20 minutes each.
These programs can be integrated into the lesson plans for an entire course or as stand-alone
supplements for specific topics, as needed.
Each program is accompanied by an instructor’s guide that contains the following elements:
•
Program Outline
•
Key Terms and Formulas
•
Quiz
•
Solutions to Quiz
•
Suggestions for Instructors
The Program Outline lists the main topics covered in each program, and also relates each program to the overall subject series. For example, if you choose to view Program 3, you’ll also
see how Program 3 fits into the overall 10-part series.
Note: Occasionally during the show, you may notice a small heading in the corner of the screen
that makes reference to Part and Section divisions in the program. These headings refer to the
program in its original, extended format and do not correspond to the format of this series.
Copyright © 2000 Films for the Humanities & Sciences® A Films Media Group Company • P.O. Box 2053 • Princeton, NJ 08543-2053
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Program Outline
PROGRAM 1: INTRODUCTION TO MATTER, THE ELEMENTS,
AND UNITS OF MEASURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 6
What is Chemistry?
• Introduction to Matter
• The Elements
Units of Measure
• The Metric System and SI Units
• Uncertainty in Measurement
• Dimensional Analysis
PROGRAM 2: CHEMICAL EQUATIONS AND ATOMIC AND MOLECULAR MASS . . . page 11
Chemical Equations
• Law of Conservation of Mass
• Balancing Equations
Atomic and Molecular Mass
• Mass Number
• Atomic Mass
• Isotopes
• Atomic Mass Units
PROGRAM 3: MOLES, PERCENT COMPOSITION,
AND THE EMPIRICAL FORMULA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 16
Moles
• Molecular Mass
• Molar Mass
Percent Composition
Empirical Formula
Problems Based on Chemical Equations
PROGRAM 4: SOLUTION STOICHIOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 21
Solution Stoichiometry
• Molarity
• Dilutions
• Titrations
• Limiting Reagents
• Yields
PROGRAM 5: THERMOCHEMISTRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 26
Thermochemistry
• Enthalpy
Endothermic Reactions
Exothermic Reactions
Standard Enthalpy
Hess’s Law
3
• Calorimetry
Heat Capacity
Molar Heat Capacity
Specific Heat
PROGRAM 6: ATOMIC STRUCTURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 31
Atomic Structure
• Quantum Mechanics
Electron Orbitals
Quantum Numbers
Types of Orbitals
• Writing Electron Configurations
PROGRAM 7: CHEMICAL BONDING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 35
Chemical Bonding
• Lewis Structures
The Octet Rule
Exceptions to the Octet Rule
• Atomic Bonding
Ionic Bonding
Covalent Bonds
• Bond Energy
PROGRAM 8: MOLECULAR GEOMETRY AND BONDING THEORIES . . . . . . . . . . . page 41
Molecular Geometry and Bonding Theories
• VSEPR Theory
Polarity
• Orbital Overlap
Hybrid Orbitals
Molecular Orbitals
PROGRAM 9: GASES AND STATES OF MATTER . . . . . . . . . . . . . . . . . . . . . . . . . . . page 46
Gases
• Kinetic Molecular Theory
Boyle’s Law
Graham’s Law
• The Ideal Gas Equation
Dalton’s Law of Partial Pressures
States of Matter
• Intermolecular Forces
Ion-dipole
Dipole-dipole
London Dispersion
Hydrogen Bond
• Phase Diagrams
• Vapor Pressure
4
PROGRAM 10: PROPERTIES OF SOLUTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 52
Properties of Solutions
• Solution Formulation
Miscible vs. Immiscible
Solubility
Saturation
Dilution
Concentration
Mass Percentage Composition
Mole Fraction
Molarity
Molality
• Colligative Properties
Freezing-point Depression
Boiling-point Elevation
• Osmosis
5
PROGRAM 1:
INTRODUCTION TO MATTER, THE ELEMENTS, AND UNITS OF MEASURE
Key Terms and Formulas
Chemistry is the study of the interactions between energy and matter.
Matter is anything that takes up space and has mass.
Physical properties describe physical attributes or physical changes that we can measure
without changing the identity of the matter. Temperature, height, weight, consistency, odor,
and hardness are all physical properties.
Chemical properties of matter refer to the way the matter acts during chemical reactions,
for instance, water’s ability to decompose into the elements hydrogen and oxygen. You can’t
measure chemical properties without changing the matter.
Homogeneous matter has the same physical and chemical properties throughout it. In other
words, all the stuff that makes up homogeneous matter acts pretty much the same way.
Homogeneous matter is called a “substance.”
Heterogeneous matter is a mixture of substances, like a rock formed from fragments of other
rocks.
A mixture of substances is considered homogeneous if the mixture has the same physical and
chemical properties throughout it. A homogeneous mixture is also called a solution.
An element is a substance that contains only one kind of atom.
A compound is a substance with two or more kinds of atoms combined in fixed proportions.
The molecular formula of a molecule shows the exact number of atoms of each element that
make up a molecule.
Precision measures the reproducibility of a result.
Accuracy refers to how close our measurement is to the theoretical ‘true’ value.
Significant figure s include the numbers on either side of the decimal point that seem reasonably
certain. For addition and subtraction problems, give the result in as many decimal places as the
number in the problem with the fewest decimal
Quantity
Symbol
Name
Common
places. Multiplication and division answers are
Measured
of Unit
of Unit
Metric Prefixes
rounded off to match the number in the problem
10^-6 micro
Length
m
Meter
(one millionth)
with the fewest significant figure s .
Scientific notation is a way of writing really
big or really small numbers. It uses powers of
ten to express a number.
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Mass
kg
Kilogram
10^-3 milli
(one thousandth)
Time
s
Second
10^-2 centi
(one hundredth)
Temperature
K
Kelvin
Volume
L
Liter
Heat
J
Joule
10^3 kilo
(one thousand times)
10^6 mega
(one million times)
Quiz
1. What is chemistry?
2. What is matter?
3. Iron melts at 1535° C. This is a _________________ property.
4. Iron can rust. This is a _________________ property.
5. Explain the difference between homogeneous and heterogeneous mixtures.
6. Classify each item as an element, compound, or mixture:
a. can of soda pop ______________________
b. oxygen ______________________
c. shampoo ______________________
d. rust ______________________
e. helium ______________________
f. water ______________________
7. Determine the number of significant figures in each of the following numbers:
a. 586.3 ______________________
b. 101 ______________________
c. 0.0001 ______________________
d. 20.0 ______________________
e. 0.0150 ______________________
8. Calculate to the proper number of significant figures:
a. 238.41 + 28.2 = ______________________
b. 6.022 x 1023 / 55.85 = ______________________
c. 10.804 – 9.632 = ______________________
d. 0.312 x 28.09 = ______________________
9. Convert 302 miles to kilometers.
10. Convert 0.863 gallons to milliliters.
7
Solutions to Quiz
1. What is chemistry?
Chemistry is the study of matter, its changes, and the energy changes associated with matter.
2. What is matter?
Matter is the material things are made of. Matter has mass and occupies space.
3. Iron melts at 1535° C. This is a physical property.
4. Iron can rust. This is a chemical property.
5. Explain the difference between homogeneous and heterogeneous mixtures.
A homogeneous mixture has all of its components evenly distributed throughout the
mixture. A heterogeneous mixture has an uneven distribution of its components. Salt
water is a homogeneous mixture since all the salt is evenly mixed in the water. Lumpy
gravy is a heterogeneous mixture since pockets of dry flour are found in the gravy.
6. Classify each item as an element, compound, or mixture:
a. can of soda pop: mixture
b. oxygen: element
c. shampoo: mixture
d. rust: compound
e. helium: element
f. water: compound
7. Determine the number of significant figures in each of the following numbers:
a. 586.3: 4 (All nonzero numbers)
b. 101: 3 (The zero is a “trapped” zero [between two nonzero digits], so it’s significant.)
c. 0.0001: 1 (The leading zeros are not significant.)
d. 20.0: 3 (The last zero is significant because it is to the right of a nonzero digit and it
is to the right of the decimal point.)
e. 0.0150: 3 (The last zero is significant because it is to the right of a nonzero digit and
it is to the right of the decimal point. The zeros before the “1” are not significant
because they are leading zeros.)
8. Calculate to the proper number of significant figures:
a. 238.41 + 28.2 = 266.6
Because this is addition/subtraction, you need to look at the number that has the least
number of decimal places. Here, the answer can only go to the tenths place.
8
b. 6.022 x 1023 / 55.85 = 1.078 x 1022
Because this is multiplication/division, you need to look at the number that has the least
number of significant digits. The answer can have no more significant digits than that.
c. 10.804 – 9.632 = 1.172
d. 0.312 x 28.09 = 8.76
9. Convert 302 miles to kilometers.
Make the conversion using dimensional analysis. 1 mile = 1.609 kilometers
302 miles x 1.609 kilometers = 486 kilometers
1 mile
10. Convert 0.863 gallons to milliliters.
1 gallon = 3.7854 liters, 1 liter = 1000 milliliters
0.863 gallons x
9
liters x 1000 milliliters = 3270 milliliters
(3.784
)
1 gallon
1 liter
Suggestions for Instructors
1. Begin by getting the students to organize their problems. Have them write down what they
know, what they need to know, and any formulas that may help. Sometimes a problem may
have numbers or other information that is not important for a calculation. Help the students
discern what is and what is not relevant to solve a problem.
2. A math review would probably be advisable. Review basic algebra, scientific notation, log
functions, and how to use a calculator. Many students know how to do the math, but do not
know the algebraic order of operations, so they tend to use their calculator incorrectly.
3. An understanding of metric prefixes and their meanings is crucial for conversions between
units. I use dimensional analysis to do metric-metric conversions, stressing that many different types of calculations can be done using dimensional analysis. I will also intentionally set
up a problem incorrectly and show the students that the units do not cancel, demonstrating
the importance of checking one’s work to see if the answer makes sense.
4. Students typically have problems distinguishing between chemical and physical changes.
Frying an egg is an example that confuses students—they think it’s a physical change because
of the change from liquid to solid. When I ask them how they usually make a liquid turn to
a solid, their answer is to cool it. Then, when I ask how they made the egg solid, and they
reply that they applied heat, they realize the reaction isn’t as simple as it looks. I have the
students observe the egg as it fries. Common observations are that an odor is produced and
that the clear yellow liquid becomes a white solid. This observation of changes of physical
p roperties can indicate a chemical change.
5. To demonstrate the difference between homogeneous and heterogeneous mixtures, I use a
lab that separates a sand/salt mixture. This gives the student some experience in separation
techniques and percent calculations. After I collect the reports from the students, I show
them that while no one had the same results, the average of all the data gives the ratio of
sand/salt I mixed. The sand/salt mixture is a heterogeneous mixture.
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PROGRAM 2:
CHEMICAL EQUATIONS AND ATOMIC AND MOLECULAR MASS
Key Terms and Formulas
Stoichiometry deals with the quantities of stuff in chemical reactions. We use stoichiometry to
answer the question “how much?”
A chemical equation is a short-hand way to describe a reaction, giving the formulas for all of
the reactants and the products.
Chemical equations have to be consistent with the Law of Conservation of Mass, which says
that atoms are not created or destroyed in chemical reactions. To avoid breaking the Law of
Conservation of Mass, you have to learn how to balance equations.
Atoms are made up of protons, which have a positive charge, neutrons, which have no
charge, and electrons, which have a negative charge.
The nucleus is the central core of the atom, and is made up of the atom’s protons and
neutrons. The nucleus contains 99.9% of an atom’s mass.
Elements are arranged on the Periodic Table of Elements in order of increasing atomic number. They are organized into 18 vertical columns called groups, and 7 main horizontal rows
called periods.
The number of protons in an atom is called the atomic number.
The total number of protons and neutrons in an atom’s nucleus is called the mass number.
The mass number is described in terms of Atomic Mass Units, or amu’s.
An isotope is an atom that has a different number of neutrons than you’re led to expect by
the periodic table.
11
Quiz
1. Balance the following equations:
a. H2 + O2 <=> H2O
b. Fe + O2 <=> Fe2O3
c. H2SO4 + NaOH <=> Na2SO4 + H2O
d. C3H8 + O2 <=> CO2 + H2O
2. Give the general notation for the isotope that has 7 neutrons and 6 protons.
3. Give the general notation for the isotope that has 33 neutrons and 27 protons.
4. Calculate the average atomic mass of chlorine from the following data.
ISOTOPE
35
Cl
37
Cl
ISOTOPIC MASS
34.96885 amu
36.96590 amu
RELATIVE ABUNDANCE
75.771%
24.229%
5. Determine the atomic masses for each of the following elements:
a. H
b. O
c. Fe
d. Ag
e. Cu
6. Calculate the molecular/formula weights for the following substances:
a. CO2
b. N2O4
c. SF6
d. BaCl2
e. H2SO4
f. Na2S2O3
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Solutions to Quiz
1. Balance the following equations:
a. 2 H2 + O2 <=> 2 H2O
b. 4 Fe + 3 O2 <=> 2 Fe2O3
c. H2SO4 + 2 NaOH <=> Na2SO4 + 2 H2O
d. C3H8 + 5 O2 <=> 3 CO2 + 4 H2O
This is material balance. Remember that each side of the equation needs to have the same
number of the atoms from every element represented.
2. Give the general notation for the isotope that has 7 neutrons and 6 protons.
13
6
C
3. Give the general notation for the isotope that has 33 neutrons and 27 protons.
60
27
Co
4. Calculate the average atomic mass of chlorine from the following data.
ISOTOPE
35
Cl
37
Cl
ISOTOPIC MASS
34.96885 amu
36.96590 amu
RELATIVE ABUNDANCE
75.771%
24.229%
The contribution of 35Cl is: 34.96885 amu x 0.75771 = 26.496 amu
The contribution of 37Cl is: 36.96590 amu x 0.24229 = 8.9565 amu
The total average mass of a chlorine atom is the sum of the contributions of the individual
isotopes: 26.496 amu + 8.9565 amu = 35.453 amu
This number should be the same as listed on the periodic table. Remember that the percent
must be converted to its decimal equivalent before using it in a calculation.
5. Determine the atomic masses for each of the following elements:
a. H: 1.008 amu
b. O: 16.00 amu
c. Fe: 55.85 amu
d. Ag: 107.9 amu
e. Cu: 63.55 amu
These are straight off the periodic table. The mass is in amu and not grams. This is the mass
of one atom.
13
6. Calculate the molecular/formula weights for the following substances:
a. CO2: 44.01 amu
C: 1 x 12.01 amu = 12.01 amu
O: 2 x 16.00 amu = 32.00 amu
44.01 amu
b. N2O4: 92.02 amu
c. SF6: 146.07 amu
d. BaCl2: 208.2 amu
e. H2SO4: 98.09 amu
f. Na2S2O3: 158.12 amu
These masses are for one unit of the substance in amu. Remember, the mass of the whole is
the sum of the masses of the pieces.
14
Suggestions for Instructors
1. To illustrate the Law of Conservation of Mass (matter), use ball-and-stick models for a simple
reaction to show that the atoms are simply rearranged in a chemical reaction. Unlike trying
to fix something at home, we should not have any spare parts when our reaction is done.
2. Isotope notation can be odd to the student; it’s usually the terminology that gets them. Mass
number (A) has to do with mass. It is just the number of protons (Z) plus neutrons.
A
Z
X
Remind students that the bigger number goes on top.
3. Students sometimes think that the atomic mass given on the periodic table is the mass a
single atom can have. In truth, the mass on the periodic table is an average, so no atom
can have this mass (unless there is only one isotope).
4. Though not mentioned in the video, properly naming compounds is crucial. A student needs
to be able to read a name of a compound and write the formula. In the balanced equation,
the student must remember not to change a formula to get the reaction to balance. Once the
correct formula is written in the equation, it is written in stone.
15
PROGRAM 3:
MOLES, PERCENT COMPOSITION, AND THE EMPIRICAL FORMULA
Key Terms and Formulas
A mole is a unit of measurement that represents 6.022 times 10 to the 23rd atoms, molecules,
or other small things like that. This number is known as Avogadro’s Constant.
The atomic mass of an element is the mass in grams of 1 mole of that element’s atoms. You
can look up the atomic masses of atoms in the periodic table, which gives the atomic mass in
atomic mass units, or AMU’s.
Molecular weight is the combined mass of the group of atoms forming a single molecule. To
find the molecular weight of a substance, simply add the atomic masses of the atoms that form
the molecule.
Formula weight finds the weight of atomic structures which are not molecules, like crystals.
The percent composition of a compound gives the ratio of elements that make up the compound. In other words, it specifies what percentage of the total compound each component
element represents.
Once you know the percent composition, you can write a molecular formula, which tells
you the exact number of each type of element in one molecule of the compound.
Empirical formulas show atom ratios in a molecule rather than exact quantities.
16
Quiz
1. How many atoms are in 6.38 grams of helium?
2. What is the molar mass of 0.250 moles of a substance that has a mass of 5.75 grams?
3. Calculate the percent composition by mass of each element in the following compounds:
a. Fe2O3
b. C2H5OH
c. BaSO4
d. NH4NO3
4. What is the empirical formula of a compound containing 76.57% carbon, 6.43% hydrogen,
and 17.00% oxygen by mass?
5. Assuming gasoline is C8H18 and during the combustion process gasoline is combined with
atmospheric oxygen and converted to carbon dioxide and water, how many grams of carbon
dioxide and water are pumped into the atmosphere from a 21.0 gallon gas tank? The density
of gasoline is 0.690 g/ml.
17
Solutions to Quiz
1. How many atoms are in 6.38 grams of helium?
Each mole of atoms contains 6.022 x 1023 atoms, so you need to find the number of moles
and then the number of atoms.
6.38g x
mole
(14.003
g)
10 atoms
( 6.0221xmole
)
23
x
= 9.60 x 1023 atoms
2. What is the molar mass of 0.250 moles of a substance that has a mass of 5.75 grams?
Molar mass is the mass in grams divided by number of moles with the units, g/mol.
5.75g
g
=
23.0
0.250 moles
mol
3. Calculate the percent composition by mass of each element in the following compounds:
a. Fe2O3
Fe: 2 x 55.85 g/mol = 111.7 g
%Fe = (111.7g/159.7g) x 100 = 69.94% Fe
O : 3 x 16.00 g/mol = 48.00 g
% O = (48.00g/159.7g) x 100 = 30.06% O
159.7 g
b. C2H5OH
c. BaSO4
d. NH4NO3
52.14% C
58.83% Ba
35.00% N
13.13% H
13.74% S
5.037% H
34.73% O
27.43% O
59.96% O
Problems b, c, and d are calculated in the same manner as problem a. Remember that all
the percentages should add up to 100 percent.
4. What is the empirical formula of a compound containing 76.57% carbon, 6.43% hydrogen,
and 17.00% oxygen by mass?
Since no mass is given, assume the sample is 100 grams. It does not matter how much
sample is present, the composition must still be the same (Law of Constant Composition).
This assumption gives the following:
C: 76.57 g
H: 6.43 g
O: 17.00 g
First find the number of moles of each element in the compound.
18
Moles C: 76.57 g x
mole
= 6.376 mol C
(112.01
g)
Moles H: 6.43 g x
mole
= 6.38 mol H
(11.008
g)
Moles O: 17.00 g x
mole
= 1.063 mol O
(116.00
g)
Now we have a mole ratio: C6.376H6.38O1.063
Divide each number by the smallest number in the formula. In this case, divide by 1.063 to
give C6H6O. This is the empirical formula.
5. Assuming gasoline is C8H18 and during the combustion process gasoline is combined with
atmospheric oxygen and converted to carbon dioxide and water, how many grams of carbon
dioxide and water are pumped into the atmosphere from a 21.0 gallon gas tank? The density
of gasoline is 0.690 g/ml.
The first thing to do is to write the equation and balance it.
2 C8H18 + 25 O2 <=> 16 CO2 + 18 H2O
The next thing is to find the mass of gasoline. You need to convert the volume to milliliters
and use density to find the mass.
21.0 gal x
liters
ml
0.690 g
x (
= 54800 g
( 3.785
) x (1000
1 gal
1 liter )
1 ml )
Now you need to find the molar mass and number of moles of gasoline.
C: 8 x 12.01 g/mol =
96.08 g
H: 18 x 1.008 g/mol = 18.144 g
114.22 g/mol
54800 g x
1 mole
= 480 mol C8H18
(114.22
g)
Now calculate the number of moles produced of water and carbon dioxide.
480 mol C8H18
( 16 mol CO2 ) = 3840 mol CO2
2 mol C8H18
(
480 mol C8H18 18 mol H2O
2 mol C8H1
) = 4320 mol H2O
Now convert each to grams by multiplying by molar mass.
19
3840 mol CO2
g = 169000 g
)
( 44.01
1 mol
4320 mol H2O
( 18.02 g ) = 77850 g
1 mol
CO2
H 2O
Suggestions for Instructors
1. Much confusion has arisen out of the term “molecular weight,” or “molecular mass.” The
trend is to use the term “molar mass.” Things like NaCl and other ionic compounds do not
occur in molecules. They have formula units, not molecules. We use the mass of one mole
of the formula or formula mass. The term molar mass can be used with ionic or molecular
compounds.
2. Why do we use moles? Moles are a unit like dozen. It is the amount of something. We can
go to the candy store and buy candy by the pound. If we know the mass of each piece of
candy (like a jelly bean), we can take the total mass of candy and divide by the mass of
each piece to calculate the total number of pieces. You can take a bag of candy, find the
average mass using a few pieces, and divide the mass of all the candy by the average mass
of a piece to find the total number of pieces of candy. Count the number of pieces and see
how close you are. You won’t be exact, but you should be close. We use molar mass and
mass to count pieces. The balanced equation is a recipe based on number of pieces.
3. When you have your students calculate molar mass for a problem, instruct them to carry
the significant figures of the molar mass to have one more digit than the number in the
problem. This will insure that your molar mass calculation does not limit the significant figure s .
4. Many students insist on doing calculations on paper and do not use the units. Insist that
they use the units. It is easier for the students (and the teacher) to find a problem if the
answer doesn’t seem right. Sometimes the wrong method gives the right answer.
5. Work many problems for the students in class. Do what you can to get the students to do
their homework.
6. Chemistry is a subject that relies on moles. Make sure the students can convert to and from
moles.
20
PROGRAM 4:
SOLUTION STOICHIOMETRY
Key Terms and Formulas
Molarity, or M, is the number of moles of solute in one liter of solution. Molarity tells you
about the concentration of a solution.
Changing the concentration of a solution by adding more solvent, giving you a higher percentage of solvent and a lower percentage of solute, is called a dilution.
Acids have lots of positive hydrogen ions, and bases have lots of negative hydroxide ions.
Put acids and bases together, and they neutralize each other by creating a salt and H20.
Titration is a method for determining the concentration of a solution. Titration involves adding
an unknown concentration to a known concentration until the mixture reaches the equivalence
point.
The equivalence point is the point in an acid-base titration when equal amounts of hydrogen
ions and hydroxide ions have been mixed.
In a chemical reaction, the reactant that gets used up first is the limiting reagent. Most chemical reactions will have a limiting reagent, since the reactants are not added in the perfect ratio
that is indicated by the balanced equation. The limiting reagent limits the amount of product
you can get from your chemical reaction, or yield.
There are three different kinds of yields:
The theoretical yield is the amount of product you predict you would get from a reaction
based on its balanced equation.
The actual yield is the amount of the product that you actually get from an experiment.
The percent yield is the actual yield divided by the theoretical yield, times one hundred,
giving you an answer that’s a percent.
21
Quiz
1. What is the molarity of a 253 ml solution containing 0.0562 moles of NaCl?
2. How many milliliters of 0.25 M HCl will give 0.163 moles of HCl?
3. How many moles are in 125 ml of 0.150 M KBr?
4. How many milliliters of 2.0 M HCl are required to make 250.0 ml of 0.36 M solution?
5. How many milliliters of 0.55 M HCl are required to neutralize 125 ml of 0.33 M Ca(OH)2?
6. Calculate the amount, in grams, of ZnS(s) made from 1.000 g Zn(s) and 1.000 g S8(s).
What is the limiting reactant?
7. If the theoretical yield is 10.5 g and your actual yield is 9.80 g, what is your percent yield?
22
Solutions to Quiz
1. What is the molarity of a 253 ml solution containing 0.0562 moles of NaCl?
First, the volume must be converted to liters. Then you divide the moles by the volume.
253 ml x
1L
(1000
ml)
= 0.253 L
0.0562 mol
= 0.222 M
0.253 L
2. How many milliliters of 0.25 M HCl will give 0.163 moles of HCl?
You can use the molarity as a conversion.
1L
1000 ml
0.163 mol x
= .65 L
.65 L x
0.25 mol
1L
(
)
(
)
= 650 ml
3. How many moles are in 125 ml of 0.150 M KBr?
First, the volume must be converted to liters.
125 ml x
1L
(1000
ml)
= 0.125 L
0.125 L x
mol
( 0.150
) = 0.0188 moles
1L
4. How many milliliters of 2.0 M HCl are required to make 250.0 ml of 0.36 M solution?
Use the formula M1V1 = M2V2 . You do not need to convert to liters, because the units work
themselves out. This is because the ratio of ml to ml is the same as L to L. First define what is
M1, V1, M2, and V2. Here, let’s define the following:
M1 = 2.0 M
V1 = ? (unknown)
(2.0 M)V1 = (0.36 M) (250.0 ml)
M2 = 0.36 M
Solve for V1.
V2 = 250.0 ml
V1 = 45 ml
To make the dilution, take 45 ml of 2.0 M HCl and add enough water to make 250.0 ml
of solution.
5. How many milliliters of 0.55 M HCl are required to neutralize 125 ml of 0.33 M Ca(OH)2?
First you must write a balanced equation:
2 HCl(aq) + Ca(OH)2(aq) <=> CaCl2(aq) + 2 H2O(l)
23
Next, find the number of moles of what you know, Ca(OH)2. You will need to convert the
volume to liters.
0.125 L x
( 0.331 Lmol) = 0.041 mol Ca(OH)2
Now use the balanced equation to find the number of moles of HCl.
2 mol HCl
0.041 mol Ca(OH)2 x
= 0.082 mol HCl
1 mol Ca(OH)2
(
)
We have 0.082 mol HCl neutralized in the reaction. Now use the molarity to find the volume,
then convert the liters to milliliters.
1L
1000 ml
0.082 mol x
= 0.15 L HCl
0.15 L x
= 150 ml
0.55 mol
1L
(
)
(
)
6. Calculate the amount, in grams, of ZnS(s) made from 1.000 g Zn(s) and 1.000 g S8(s).
What is the limiting reactant?
First, write the balanced equation: 8 Zn(s) + S8(s) <=> 8 ZnS(s)
Next, find the number of moles of each reactant.
1 mol
1.000 g x
= 0.01529 mol Zn
1.000 g x
65.39 g
(
)
1 mol
= 0.003897 mol S8
(256.6
g)
Next, we calculate the amount of product formed if we use all the zinc and have an excess
of sulfur.
8 mol ZnS
0.01529 mol Zn x
= 0.01529 mol ZnS
8 mol Zn
(
)
Next, we calculate the amount of product formed if we use all the sulfur and have an excess
of zinc.
8 mol ZnS
0.003897 mol S8 x
= 0.03118 mol ZnS
1 mol S8
(
)
Now we look at the amount of product formed in each reaction. The calculation that gives the least
amount of product is the line that indicates the limiting reactant. In this problem, zinc is the limit ing reactant. If we use all of the zinc, we will have extra sulfur. To do any more calculations, we
must use the number of moles of zinc to do all subsequent calculations. The calculation indicates
that we can only produce 0.01529 mol ZnS. The molar mass of ZnS is 97.46 g/mol.
0.01529 mol ZnS x
g
= 1.490 g ZnS are produced.
( 97.46
1 mol )
7. If the theoretical yield is 10.5 g and your actual yield is 9.80 g, what is your percent yield?
The calculation for percent yield is the actual yield divided by the theoretical yield.
This answer is multiplied by 100 to give a percent.
9.80 g
x 100 = 93.3 % yield
10.5 g
24
Suggestions for Instructors
1. Many students have a difficulty grasping the concept of limiting reactants. I use recipes as
an example in the classroom. Isn’t that what reactions really are? Whenever we cook, some
ingredient often limits the amount of food we make. I usually run out of meat. An experiment I have used is referred to as “the s’mores” lab. If you have a box of Teddy Grahams, a
bag of mini marshmallows, and a bag of chocolate chips, how many “mini-s’mores” can you
make? The recipe is one mini-marshmallow, four chocolate chips, and two Teddy Grahams.
The students (even in college) like this lab.
2. One point I stress to the students is that the stoichiometric calculations are actually theore t i c a l
yields. The calculations are what you should produce. In reality, many reactions have side
reactions that do not give desired products, so yields are reduced.
3. Molarity, like density, molar mass, etc., can be used as a conversion factor. Molarity is probably one of the most important units of concentrations that can be used in stoichiometry,
titrations, and acid-base calculations.
4. I will often demonstrate dilutions by using a known concentration and volume of copper (II)
sulfate and dilute it in a volumetric flask. The students see the solution get lighter—a good
visual connection.
5. Do a titration of acetic acid and sodium hydroxide to a phenolphthalein endpoint. Color
changes are always a good attention-getter. You can also use a pH meter with the indicator
to show that the neutralization reaction does not always give a pH neutral solution.
25
PROGRAM 5:
THERMOCHEMISTRY
Key Terms and Formulas
Thermochemistry measures the heat changes associated with chemical reactions.
A system is all the substances taking part in a reaction, plus the reaction vessel.
Enthalpy, or ΔH, is the amount of heat absorbed or released in a system at constant pressure
during a reaction.
In an exothermic reaction, heat flows out of the system.
In an endothermic reaction, heat flows into the system.
The standard enthalpy of form a t i o n, or ΔH°f , is the enthalpy change during a reaction that
f o rms a compound from elements in their standard states at 25 degrees Celsius and 1 atmosphere
pressure.
According to Hess’s Law, the total enthalpy change for a reaction is the sum of the enthalpy
changes of the little reactions that make up the total reaction.
Calorimetry measures the amount of heat absorbed or created in reactions.
Heat capacity is the amount of heat re q u i red to raise the temperature of an object or substance
by one Kelvin.
Molar heat capacity is the amount of heat required to raise the temperature of one mole of
a substance by one Kelvin.
Specific heat is the amount of heat required to raise the temperature of one gram of a substance
by one Kelvin. You can look up the specific heat of substances in your chemistry text.
26
Quiz
1. Define an endothermic reaction.
2. Define an exothermic reaction.
3. What is the ΔH° for the combustion of propane, C3H8(g), that forms gaseous carbon
dioxide and gaseous water? Use the heat-of-formation table in your textbook.
4. Use the following reactions:
2 Cu2S(s) + 3 O2(g) <=> 2 Cu2O(s) + 2 SO2(g)
ΔH° = -771.8 kJ
Cu2O(s) + C(s) <=> 2 Cu(s) + CO(g)
ΔH° = +58.1 kJ
to find the ΔH° of:
2 Cu2S(s) + 3 O2(g) + 2 C(s) <=> 4 Cu(s) + 2 SO2(g) + 2 CO(g)
5. Calculate the amount of energy required to raise the temperature of 240.0 grams of water
from 298.2 K to 373.2 K. The specific heat of water is 4.18 J/mol K.
27
Solutions to Quiz
1. Define an endothermic reaction.
An endothermic reaction is a reaction that absorbs heat.
2. Define an exothermic reaction.
An exothermic reaction is a reaction that produces heat.
3. What is the ΔH° for the combustion of propane, C3H8(g), that forms gaseous carbon dioxide
and gaseous water? Use the heat-of-formation table in your textbook.
The first step is to write the balanced equation.
C3H8(g) + 5 O2(g) <=> 3 CO2(g) + 4 H2O(g)
Use a textbook to find the heat of formation of each of the species in the reaction.
ΔH°f [C3H8(g)] = -103.85 kJ/mol
ΔH°f [O2(g)] = 0.0 kJ/mol (by definition)
ΔH°f [CO2(g)] = -393.5 kJ/mol
ΔH°f [H2O(g)] = -241.82 kJ/mol
The formula for the calculation:
Δ Η° = ΣnΔΗ°f (products) – ΣmΔ Η°f (reactants)
In the formula, “n” and “m” are the coefficients (in moles) in the balanced equation.
Δ Η° = [(3 mol x -393.5
kJ
kJ
kJ
) + (4 mol x -241.82
)] – [(1mol x -103.85
) + 0]
mol
mol
mol
ΔΗ° = -2043.9 kJ
4. Use the following reactions:
2 Cu2S(s) + 3 O2(g) <=> 2 Cu2O(s) + 2 SO2(g)
ΔH° = -771.8 kJ
Cu2O(s) + C(s) <=> 2 Cu(s) + CO(g)
ΔH° = +58.1 kJ
to find the ΔH° of:
2 Cu2S(s) + 3 O2(g) + 2 C(s) <=> 4 Cu(s) + 2 SO2(g) + 2 CO(g)
28
This is a Hess’s Law problem. You must add the equations to get the answer. You will notice
that the copper (I) oxide will not cancel because there are two in the first equation and one
in the second equation. You must multiply everything in the second equation by two. When
you multiply everything by two, the equation remains balanced.
2 Cu2S(s) + 3 O2(g) <=> 2 Cu2O(s) + 2 SO2(g)
ΔΗ° = -771.8 kJ
2 {Cu2O(s) + C(s) <=> 2 Cu(s) + CO(g)
ΔΗ° = +58.1 kJ }
This gives:
2 Cu2S(s) + 3 O2(g)
2 Cu2O(s) + 2 C(s)
<=> 2 Cu2O(s) + 2 SO2(g)
<=> 4 Cu(s) + 2 CO(g)
2 Cu2S(s) + 3 O2(g) + 2 C(s) <=> 4 Cu(s) + 2 SO2(g) + 2 CO(g)
ΔH° = -771.8 kJ
ΔH° = +116.2 kJ
ΔH° = -655.6 kJ
By adding the two equations, we get the final answer. Remember, whatever factor you multi ply the equation by, you must also do it to the heat of reaction. The answer is -655.6 kJ.
5. Calculate the amount of energy required to raise the temperature of 240.0 grams of water
from 298.2 K to 373.2 K. The specific heat of water is 4.18 J/mol K.
This problem can be treated like a conversion (dimensional analysis) problem or just use the
formula. The formula is:
Q = ΔT (temperature change) x mass (in grams) x specific heat
We need to calculate the ΔT first. The delta (Δ) here means change in. This is the final
temperature minus the initial temperature.
ΔT = 373.2 K – 298.2 K = 75.0 K
Now plug in the numbers.
Q = 75 K x 240.0 g x 4.18 J/mol K = 75200 J
This is the amount of heat required to raise the temperature of a cup of water at room temper ature to its boiling point.
29
Suggestions for Instructors
1. Sometime the specific heat is given in the units of J/mol °C. This may cause students some
distress if the temperature is in Kelvin. This may also occur if the units are in J/mol K and
the temperature is in degrees Celsius. However, the calculation using the formula
Q = ΔT x mass x specific heat
will not cause a problem with temperature. Since we are using a change in temperature, the
amount of change in Celsius is the same amount of change in Kelvin. A one degree Celsius
change equals a one Kelvin change.
2. Students often confuse heat and temperature. Heat is the energy flow from a hot to a cold
body. Temperature is the measure of the average kinetic energy.
3. When using Q = ΔT x mass x specific heat, making sure the students get the sign of ΔT
correct will help them understand the concept of endothermic and exothermic. If the change
in temperature is negative, the heat will be negative. In chemistry sign-conventions, this is
exothermic (heat leaving the system). If the change in temperature is positive, the heat will
be positive. This is endothermic (heat entering the system).
4. In dealing with heat-of-formation problems, students tend to write the balanced chemical
equations but forget to write the state of the substance as well. This can be a problem. For
example, the heats of formation of liquid water and gaseous water differ by 44 kJ. What
number from the heats of formation table do you use if you do not know the state of the
material? It can make a big difference in the calculation.
5. Try some demonstrations of endothermic and exothermic reactions. I like to use barium
hydroxide octahydrate and ammonium thiocyanate for an endothermic reaction. Ice forms
on the outside of the beaker. Be careful—ammonia is formed. For an exothermic reaction,
I like to do the dehydration of sugar with sulfuric acid. It is very hot! As with any demo,
wear proper safety gear and follow proper disposal methods.
30
PROGRAM 6:
ATOMIC STRUCTURE
Key Terms and Formulas
The word orbital refers to the general region in an atom where an electron exists. An electron’s orbital describes its average distance from the nucleus of the atom, and what shape its
possible path forms, or distribution—and how that distribution is oriented in space—in other
words, where it is in relation to the atom’s nucleus and the other orbitals.
Electron orbitals are described using three quantum numbers:
The first, or principal quantum number, is represented by a lowercase n.
It describes how far the electron is from the nucleus of the atom.
The second quantum number is represented by a lowercase l. It describes the shape
formed by the electron distribution, or the shape of the possible path the electron
can take.
Finally, ml, the magnetic quantum number, describes where the orbital is in relation
to the nucleus and the other orbitals.
An electron shell is a group of orbitals with the same value for n.
A subshell is a group of orbitals that have the same values for n and l.
Electron diagrams are used to represent electron configurations, or the way electrons are
distributed among an atom’s orbitals.
An ion is an atom with a charge. When atoms gain electrons they acquire a negative charge
and become negative ions. When atoms loses electrons they acquire a positive charge and
become positive ions.
31
Quiz
1. What is an orbital?
2. What are the general shapes of the “s”, “p”, and “d” orbitals?
3. Which of the following sets of quantum numbers is possible? If not possible, explain why.
n
l
ml
a.
0
3
-2
b.
2
0
1
c.
1
2
0
d.
3
1
-3
e.
3
2
2
4. What is the difference between a “3p” orbital and a “2p” orbital?
5. Give the electron configurations for the following:
a. P
b. O
c. Na
d. Zn
e. K+
f. F–
6. What is the set of quantum numbers (n, l, ml) for the outermost electron in the potassium
atom?
32
Solutions to Quiz
1. What is an orbital?
An orbital is a representation of where we will probably find an electron 90% of the time.
2. What are the general shapes of the “s”, “p”, and “d” orbitals?
d
p
d
s
3. Which of the following sets of quantum numbers is possible? If not possible, explain why.
a. Not possible—the first energy level is 1.
b. Possible.
c. Not possible—the range for “l” is 0 to (n –1). Since n=1, l must equal 0.
The numbers indicate that this orbital is a d orbital in the first energy level.
d. Not possible—n and l are fine, but the ml can be an integer ranging from -l to +l. The
possible values for l=1 are -1, 0, 1.
e. Possible.
4. What is the difference between a “3p” orbital and a “2p” orbital?
The “3p” orbital is bigger than the “2p” orbital. The “3p” electrons are generally farther
away from the nucleus than the “2p” electrons.
5. Give the electron configurations for the following:
a. P
1s2 2s2 2p6 3s2 3p3
b. O
1s2 2s2 2p4
c. Na
1s2 2s2 2p6 3s1
d. Zn
1s2 2s2 2p6 3s2 3p6 4s2 3d10
+
e. K
1s2 2s2 2p6 3s2 3p6
f. F–
1s2 2s2 2p6
6. What is the set of quantum numbers (n, l, ml) for the outermost electron in the potassium
atom?
The outermost electron of the potassium atom occupies the 4s orbital. The principal quan tum number (n) is 4. The “s” orbital has an l value of 0. The “ml” value for an l = 0 is 0.
The quantum numbers are (4, 0, 0).
33
Suggestions for Instructors
1. Students tend to think that electrons move along the surface of an orbital. In truth, the electron moves all over the interior of the orbital 90% of the time. The concept of nodal regions
inside the orbital is also confusing. If an electron can’t be in a node, how can it get from
one region to another? Remember that electrons can act as a wave as well as a particle.
Waves can pass through a nodal region, a particle cannot. “Beam me up!”
2. Students will need to memorize the electron configurations for chromium and copper.
These elements are exceptions to “aufbau”.
Cr: 1s2 2s2 2p6 3s2 3p6 4s1 3d5
Cu: 1s2 2s2 2p6 3s2 3p6 4s1 3d10
3. Try some flame tests on alkali salts. This is a good introduction to quantum mechanics.
The light given off is due to the electrons returning to the ground state after being excited
in the flame. You can use this to explain fireworks as well.
4. The students need to realize that we have never seen an orbital. Orbitals arise from calculations on the wave equations. The energy results match the experimental data from emission
spectra. Orbitals are theoretical entities that are supported by experimental evidence.
34
PROGRAM 7:
CHEMICAL BONDING
Key Terms and Formulas
Lewis structures, also called electron dot structures, are diagrams that use dots to show the
number of electrons in an atom’s outer shell, indicating the bonding pattern between the atoms
that form a compound.
The valence shell is the outermost shell of an atom.
According to the octet rule, when a compound forms, an atom gains or loses electrons, or
shares pairs of electrons, until it has eight electrons in its valence shell. An atom with eight
electrons in its valence shell is a happy atom. A couple exceptions to the octet rule are
hydrogen and boron.
Ionic bonds are formed by the electrostatic attraction between positive and negative ions.
Covalent bonds are formed between two atoms that share a pair of electrons.
Bond energy, represented by the letters BE, refers to the energy needed to break or form
chemical bonds. You can use bond energy to estimate how much heat will be gained or lost
during a reaction. BE values can be found in your textbook.
35
Quiz
1. Draw the Lewis structures for the following molecules or ions:
a. NO3–
b. NO+
c. F2
d. N2
e. O2
f. NCl3
g. CH4
h. O3
2. Based on periodic trends, which element is more electronegative?
a. O or N
b. F or K
c. Cl or Al
d. C or O
3. When a compound forms, the component atoms will gain, lose, or share electrons to meet
what rule?
4. Using your textbook to determine bond energies, find the reaction enthalpies for the
following reactions:
a. 2 H2(g) + O2(g) <=> 2 H2O(g)
b. N2(g) + 2 H2(g) <=> N2H4(g)
c. CH4(g) + 2 O2(g) <=> CO2(g) + 2 H2O(g)
d. C2H6(g) + Cl2(g) <=> C2H5Cl(g) + HCl(g)
5. Define the following terms:
a. covalent bond
b. ionic bond
c. polar covalent bond
36
Solutions to Quiz
1. Draw the Lewis structures for the following molecules or ions:
a. NO3– (resonance structures)
..
:o:
:o:
:o:
..
:o:
..
:N
O: ]+
N:
..
..
O—O
..
..
f. NCl3
..
:Cl:
..
:N — Cl:
..
:Cl:
..
H
H—C—H
H
h. O3
37
..
:o:
..
N o:
d. N2
g. CH4
–
..
N—o:
..
..
..
: F — F:
..
..
e. O2
:o:
..
N—o:
..
b. NO+
[ :N
c. F2
–
.. .. ..
O
.. O—O:
..
–
2. Based on periodic trends, which element is more electronegative?
a. O or N
b. F or K
c. Cl or Al
d. C or O
3. When a compound forms, the component atoms will gain, lose, or share electrons to meet
what rule?
The octet rule.
4. Using your textbook to determine bond energies, find the reaction enthalpies for the
following reactions:
a. 2 H2(g) + O2(g) <=> 2 H2O(g)
b. N2(g) + 2 H2(g) <=> N2H4(g)
c. CH4(g) + 2 O2(g) <=> CO2(g) + 2 H2O(g)
d. C2H6(g) + Cl2(g) <=> C2H5Cl(g) + HCl(g)
The calculations for these problems come from the following equation:
ΔHrxn = Σ (bond enthalpies of bonds broken) – Σ (bond enthalpies of bonds form e d )
You should do the Lewis structures for the compounds if you do not know the structures.
a. ΔHrxn = [ {2 x BE (H-H)} + {BE (O=O)} ] – [ {4 x BE (H-O)} ]
ΔHrxn = [ {2 x 436 kJ/mol} + { 495 kJ/mol} ] – [ 4 x 463 kJ/mol ] = -485 kJ
b. ΔHrxn = [ {BE (N N)} + {2 x BE (H-H)} ] – [ {BE (N N)} + {4 x BE (N-H)} ]
ΔHrxn = [ {941 kJ/mol} + {2 x 436 kJ/mol} ] – [ {163 kJ/mol} + {4 x 391 kJ/mol} ]
= +86 kJ
c. ΔHrxn = [ {4 x BE (C-H)} + {2 x BE (O=O)} ] – [ {2 x BE (C=O)} + {4 x BE (O-H)} ]
ΔHrxn = [ {4 x 413 kJ/mol} + {2 x 495 kJ/mol} ] – [ {2 x 799 kJ/mol} + {4 x 463 kJ/mol}
]
= -808 kJ
d. In this problem, the bonds broken are one C-H bond and one Cl-Cl bond. The bonds
formed are one C-Cl bond and one H-Cl bond.
ΔHrxn = [ {BE (C-H)} + {BE (Cl-Cl}) ] – [ {BE (C-Cl)} + {BE (H-Cl)} ]
ΔHrxn = [ {413 kJ/mol} + {242 kJ/mol} ] – [ {328 kJ/mol)} + {431 kJ/mol} ]
= -104 kJ
38
5. Define the following terms:
a. covalent bond: a bond in which electrons are shared by two nuclei.
b. ionic bond: a bond formed due to the attraction of oppositely charged particle (ions).
c. polar covalent bond: a covalent bond in which the electrons are not shared equally
between two nuclei.
39
Suggestions for Instructors
1. Students often ask, “which element is the central atom?” If you have a binary molecule,
the single atom is usually the central atom.
2. When adding the total number of electrons for Lewis structures, remember to add electrons
to anions and subtract electrons for cations.
3. Show students the common exceptions to the octet rule. Hydrogen follows the “duet” rule.
Boron and beryllium run short and usually will not complete their octets. If your class is an
advanced class, be sure to show the ones that have the expanded octet.
4. Remember that a common pattern of trends on the periodic table is up and to the right.
Ionization energies increase as you go up and to the right. Atomic sizes decrease as you go
up and to the right. Metallic character decreases as you go up and to the right.
5. The number of valence electrons can be determined from the group number on the periodic
table.
40
PROGRAM 8:
MOLECULAR GEOMETRY AND BONDING THEORIES
Key Terms and Formulas
The Valence Shell Electron Pair Repulsion model, or VSEPR model, is a common model
used to predict molecular geometry by describing how electron pairs bond and theorizing how
electron pairs and bonds will affect the shape of the molecule. According to the VSEPR model,
the electron pairs in a molecule’s valence shell are arranged so they are as far away from each
other as possible. In other words, the electrons act like they repel each other, affecting the
shape of a molecule.
Electronegativity is the ability of an atom to pull the electrons being shared in a covalent
bond closer to itself. Strong electronegative atoms pull the shared electrons in a covalent bond
toward themselves, causing the molecule to become polar.
Polar molecules have a slight positive charge on the less electronegative atom, and a slight
negative charge on the more electronegative atom.
A dipole is a molecular arrangement in which a strongly electronegative atom shares a covalent
bond with an atom that has a weaker electronegativity.
In a dipole molecule, the stronger electronegative atom will gain a slight negative charge, and
the weaker atom gains a slightly positive charge, so they tend to line up so that the negative
end of one molecule lines up with the positive end of another molecule. These attractions are
called dipole-dipole forces.
According to the hybrid orbital theory, when molecules form, the orbitals of the component
atoms overlap with each other to form hybrid orbitals, or mixtures of the overlapping atomic
orbitals. These overlapping, or hybrid, orbitals let two electrons with opposite spins come
together to form a covalent bond that helps hold the molecule together.
According to the molecular orbital theory, the orbitals from the atoms in the molecule overlap to form covalent bonds that glue the atoms together.
41
Quiz
1. Give the molecular geometry for the following molecules:
a. BF3
b. CH4
c. H2S
d. CO2
e. PH3
f. O3
g. CCl4
2. For the above question, give the hybridization of the central atom in each molecule.
3. Draw the molecular orbital diagram for the H2 molecule and calculate the bond order.
42
Solutions to Quiz
1. Give the molecular geometry for the following molecules:
In using VSEPR, you must do the Lewis structure first. The next step is to figure how to spread
the number of electron pairs out in three-dimensional space.
a. BF3. The structure fits the AX3 notation. The molecule is trigonal planar. It lies in one
plane.
..
:F:
B
..
:F
..
..
F:
..
b. CH4. The structure fits the AX4 notation. The molecule is tetrahedral.
H
H—C—H
H
c. H2S. This molecule is slightly different. It has nonbonding or lone pairs of electrons.
Having four electron pairs around the central atom would give an electron pair
geometry of tetrahedral (geometry due to all electron pairs). Since many diagrams
do not show the nonbonding pairs, the original tetrahedron now looks like a bent stick.
We call this structure bent.
..
H—S—H
..
d. CO2. This structure is also different. It has four shared pairs of electrons. Instead of each
pair pointing in separate directions, two bonds point in one direction and the other two
bonds point in the opposite direction. This actually fits the AX2 notation and is linear.
..
..
O
C
O
..
..
e. PH3. This molecule has four electron pairs pointing in different directions. Having four
electron pairs around the central atom would give an electron pair a geometry of tetrahe dral. Since many diagrams do not show the nonbonding pairs, the original tetrahedron
now looks like a pyramid with a triangular base. We call this structure trigonal pyramid.
..
H—P—H
H
43
f. O3. This molecule is similar to BF3. Even though it has four electron pairs, two of them
make a double bond. They point in one direction and are treated like one bond for the
purposes of VSEPR. This is the same for CO2. The geometry for all the electron pairs is trigo nal planar. Since many diagrams do not show the nonbonding pairs, the atoms would look
like a bent stick. This is b e n t.
..
..
..
O
O
O:
..
..
g. CCl4. This molecule is like CH4. It has a notation of AX4 and is tetrahedral. It does not
have lone pairs.
..
:Cl:
..
..
:Cl—C—Cl:
..
..
:Cl:
..
2. For the above question, give the hybridization of the central atom in each molecule.
Hybrid orbitals hold the nonbonding electrons and one pair of electrons in a multiple bond
(multiple bonds are like a single bond). Counting the bonds (remember that multiple bonds
count as one bond) and number of lone pairs on the central atom will give the hybrid
orbital number (HON). If the HON is 2, the hybridization is sp. If the HON is 3, the
hybridization is sp2. If the HON is 4, the hybridization is sp4.
a. sp2 HON = 3
c. sp3 HON = 4
e. sp3 HON = 4
g. sp3 HON = 4
b. sp3
d. sp
f. sp2
HON = 4
HON = 2
HON = 3
3. Draw the molecular orbital diagram for the H2 molecule and calculate the bond order.
H–H
antibonding
1s
1s
bonding
To calculate the bond order (B.O.), you take the number of bonding electrons minus the
antibonding electrons and divide the quantity by two.
(bonding electrons – antibonding electrons) (2 – 0)
=
2
2
This indicates a single bond.
B.O. = 1
B.O. =
44
Suggestions for Instructors
1. A common problem that students have is to try to guess a structure without doing a Lewis
structure. For example, a student might think that PH3 has a notation of AX3. It would be
trigonal planar. But since PH3 has lone pairs its geometry is actually tetrahedral.
2. Use molecular models to demonstrate the molecular geometries. They allow students to get
a hands-on and a 3-D visual of the shapes.
3. You can use the hybridization of orbitals to explain molecular geometry.
4. In many general chemistry courses, molecular orbital (M.O.) theory is not covered. If you
choose to cover the topic, make sure you do homonuclear diatomic molecules. The nice
thing about M.O. theory is that it can explain properties not explained by VSEPR, for example, the fact that O2 is attracted to a magnet.
45
PROGRAM 9:
GASES AND STATES OF MATTER
Key Terms and Formulas
We describe gases based on these four properties: mass, volume, pre s s u re, and t e m p e r a t u re.
When using the gas equations, mass is measured in grams and kilograms, volume is measured
in liters and milliliters, pressure is measured in atmospheres, and temperature is measured in
Kelvins.
Kinetic Molecular Theory explains why gases act they way they do, and is based on five
assumptions:
1. A gas is composed of molecules that are far apart from each other in comparison with
their own dimensions.
2. Gas molecules are in constant random motion.
3. Gas molecules exert no force on each other or on the container, until they collide with
each other or with the walls of the container.
4. The average kinetic energy of the molecules of a gas is proportional to the temperature
of the gas.
5. Every time a molecule collides with the wall, it exerts a force on it.
According to Boyle’s Law, the volume of a sample of gas at a given temperature varies
inversely with applied pressure.
Diffusion is the process by which one gas mixes with another.
Effusion is the process by which a gas escapes from a container through a very small hole.
According to Graham’s Law, the rate of diffusion or effusion of a gas is inversely proportional to
the square root of its molar mass—that is, the heavier a gas is, the slower it diffuses or eff u s e s .
The ideal gas equation, PV = nRT, describes the pressure (P), volume (V), and temperature
(T) of any hypothetical gas. The “n” refers to the number of moles of the gas, and the “R”
refers to the gas constant.
R = 0.0821 atm-L
mol-k
The partial pressure of a gas is the pressure exerted by one of the gases in a mixture.
According to Dalton’s Law of partial pressures, the total pressure exerted by a gas mixture
equals the sum of the pressures that each gas would exert if it were alone under the same
conditions.
Intermolecular forces are the forces of attraction between molecules. Ion-dipole forces,
dipole-dipole forces, London dispersion forces, and hydrogen bonds all make molecules stick
together.
In a closed system, vapor pressure is the constant force that a vapor exerts above a liquid
once equilibrium has been established.
46
Quiz
1. What are the five main points of the Kinetic Molecular Theory?
2. Use the Kinetic Molecular Theory to explain Boyle’s Law.
3. Describe Graham’s Law.
4. What volume does 1.60 g of CH4 gas occupy at a pre s s u re of 0.986 atmospheres and 31.0°C?
5. If a 2.00 L container has 1.00 g of H2(g) and 1.00 g O2(g), what is the partial pressure of
each gas at 25.0° C?
6. If 1.65 g of solid potassium chlorate is heated to form solid potassium chloride and gaseous
oxygen, what volume of oxygen is formed at 298 K and a pressure of 1.12 atm?
7. What are the four types of intermolecular forces? Define each force.
8. Which has the higher vapor pre s s u re, hexane (has only London forces) or water (has hydrogen
bonding)?
9. Define the points “a” through “h” on the phase diagram of carbon dioxide.
P
a
b
c
solid
liquid
d
e
f
h
g
gas
T
47
Solutions to Quiz
1. What are the five main points of the Kinetic Molecular Theory?
1. A gas is composed of molecules that are far apart from each other in comparison with
their own dimensions.
2. Gas molecules are in constant random motion.
3. Gas molecules exert no force on each other or on the container, until they collide with
each other or with the walls of the container.
4. The average kinetic energy of the molecules of a gas is proportional to the temperature of
the gas.
5. Every time a molecule collides with the wall, it exerts a force on it.
2. Use the Kinetic Molecular Theory to explain Boyle’s Law.
Every collision causes pressure. If you have more collisions, you get a higher pressure. If the
volume of the container gets smaller, the gas particles hit the walls more often and we get an
increase in pressure. If we have a larger volume, we get fewer collisions and a lower pres sure. This is Boyle’s Law.
3. Describe Graham’s Law.
The rate of diffusion of a gas is inversely proportional to the square root of the molar mass of
the gas. In other words, the bigger the mass, the slower it moves.
4. What volume does 1.60 g of CH4 gas occupy at a pre s s u re of 0.986 atmospheres and 31.0°C?
This problem utilizes PV=nRT. For all gas calculations, we must convert the temperature to
Kelvin. R has a value of 0.0821 L • atm/mol • K.
T = 31.0°C + 273.2 = 304.2 K
We must also convert grams to moles. CH4 has a molar mass of 16.05 g/mol.
1 mol
1.60 g x
= 0.10 mol
16.05 g
(
)
Now we can plug into the equation.
(0.986 atm) x V = (0.10 mol) x (0.0821
Now, solve for V.
V = 2.5 L
48
L • atm
) x (304.2 K)
mol • K
5. If a 2.00 L container has 1.00 g of H2(g) and 1.00 g O2(g), what is the partial pressure of
each gas at 25.0° C?
Ideally, pressure is due to the number of moles of gas. We need to find the number of moles
of each gas.
1.00 g x
1 mol
2.016 g
(
)
= 0.496 mol H2
1.00 g x
1 mol
32.00 g
(
)
= 0.0313 mol O2
As long as the gases do not react with one another, the total pressure is the sum of the individ ual pressures. One thing to keep in mind: two or more gases can occupy the same volume.
The volume for each gas is 2.00 L. After we convert the temperature to Kelvin, we can plug
our numbers into the Ideal Gas Equation and solve for P.
T = 25.0° C + 273.2 = 298.2 K
P x (2.00 L) = (0.496 mol) x (0.0821
L • atm
) x (298.2)
mol • K
P x (2.00 L) = (0.313 mol) x (0.0821
L • atm
) x (298.2)
mol • K
P (H2) = 6.07 atm
P (O2) = 0.383 atm
6. If 1.65 g of solid potassium chlorate is heated to form solid potassium chloride and gaseous
oxygen, what volume of oxygen is formed at 298 K and a pressure of 1.12 atm?
The first thing you need to do is to write a balanced equation.
2 KClO3(s) <=> 2 KCl(s) + 3 O2(g)
We assume that all the potassium chlorate is converted to oxygen and potassium chloride.
We are looking for the number of moles of oxygen (the only gas present). The molar mass of
potassium chlorate is 122.6 g/mol.
1.65 g x
(
1 mol
122.6 g
)
= 0.0135 mol KClO3
3 mol O2
0.0135 mol KClO3 x
2 mol KClO3
(
)
= 0.0203 mol O2
Now plug the numbers into the gas equation and solve for V.
L • atm
(1.12 atm) x V = (0.0203 mol) x (0.0821
) x (298 K)
mol • K
V = 0.443 L
49
7. What are the four types of intermolecular forces? Define each force.
a. Dipole-dipole forces: forces that result from slight charges that occur at the ends
of polar molecules and cause the molecules to stick together.
b. Ion-Dipole forces: forces that result from the slight charges that occur at the ends
of polar molecules. These slight charges are attracted to ions in a substance. This is the
reason salt dissolves in water.
b. Hydrogen bonding: force that results from a hydrogen atom that is attached to a nitro gen, oxygen, or fluorine atom that is attracted to another nitrogen, oxygen, or fluorine.
d. London dispersion forces: forces that result from fluctuating, instantaneous dipoles.
These dipoles are not permanent.
8. Which has the higher vapor pressure, hexane (has only London forces) or water (has
hydrogen bonding)?
Vapor pressure is inversely proportional to the strength of the intermolecular forces.
Hydrogen bonding is stronger than London dispersion forces. The weaker the forces,
the higher the vapor pressure. Hexane will have the higher vapor pressure.
9. Define the points “a” through “h” on the phase diagram of carbon dioxide.
P
a
b
c
solid
liquid
d
e
f
h
g
gas
T
50
a. melting
b. critical point
c. freezing
d. boiling
e. condensation
f. sublimation
g. triple point
h. deposition
Suggestions for Instructors
1. Because models need to be useful, the Kinetic Molecular Theory would have been tossed
out if it had not explained some gas behavior.
2. The hardest part of the gas equations for some students is solving the equation for the
unknown after plugging in the numbers. Practice (plenty of homework) helps. Checking to
make sure the units work out is also extremely useful.
3. The “collapsing can” demo is a good one for showing the “strength” of the air.
4. Many students think that hydrogen bonding is a true bond. It is an intermolecular force.
True bonds are stronger and are called intramolecular forces.
5. All molecules have London dispersion forces.
6. Hydrogen bonding is what gives water its high specific heat capacity and allows its solid
form to float on its liquid form.
51
PROGRAM 10:
PROPERTIES OF SOLUTIONS
Key Terms and Formulas
Liquids that can dissolve in each other are called miscible, while liquids that are insoluble in
each other are called immiscible.
The general rule for solutions is ‘like dissolves like.’ Ionic and polar solutes are soluble in
polar solvents, and non-polar solutes are soluble only in non-polar solvents.
The solubility of a substance is the maximum amount of that substance that can dissolve in a
particular solvent at a particular temperature.
A saturated solution is a solution at equilibrium with a solvent that has accepted the maximum amount of a solute that it can.
The composition of a solution consists of the components that make up the solution as well
as the concentrations and ratios of the components in the solution. Composition can be
expressed four ways: mass percent, mole fraction, molarity, and molality.
The mass percentage composition of a solution measures the number of grams of solute in
each 100 grams of solution.
mass percent of solute =
mass of solute
x 100
mass of solution
The mole fraction of a compound, also known as X, is equal to the number of moles of the
compound you’re measuring, divided by the total number of moles of all components of the
system.
The molarity of a solution is the number of moles of a solute in one liter of solution.
The molality of a solution is the number of moles of solute per kilogram of solvent.
Colligative properties of solutions are those that are determined by the concentration of solute
in a solvent, like freezing-point depression, boiling-point elevation, and osmosis.
Freezing-point depression calculates how much a solute interferes with a solvent’s ability to
freeze—in other words, how much colder the solvent has to get before it will freeze.
ΔTf = Kf m
Boiling-point elevation calculates how much a solute can interfere with a solvent’s ability to
boil—in other words, how much hotter the solvent has to be before it will boil.
ΔTb = Kbm
Osmosis is the movement of solvent through a semipermeable membrane from a solution of
lower concentration, to a solution of higher concentration.
52
Quiz
1. When ethanol is added to water, they dissolve and mix together. The liquids are
__________________.
2. Oil and water do not dissolve and mix. The liquids are ______________________.
3. Define each of the following
a. solubility
b. saturated
c. electrolyte
d. nonelectrolyte
4. A chemist has mixed a solution of sodium chloride by dissolving 4.5 g of NaCl and 495.5 g
of water. What is the mass percent and mole fraction of each component? What is the molarity of the solution? The density of solution is 1.00 g/ml.
5. What is the molarity of a 355 ml solution that contains 1.56 g of NaOH?
6. What are the boiling point and melting point of an aqueous ethanol solution that contains
25 g of C2H5OH with 0.750 kg of water? (For water, Kf = 1.86 and Kb = 0.52)
7. Define “osmosis.”
53
Solutions to Quiz
1. When ethanol is added to water and they dissolve and mix together. The liquids are miscible.
2. Oil and water do not dissolve and mix. The liquids are immiscible.
3. Define each of the following:
a. solubility: the maximum amount of a solute that can be dissolved in a solvent.
b. saturated: a solution that has accepted as much solute as it can at a particular temperature.
c. electrolyte: a solute that produces ions in solution and conducts electricity.
d. nonelectrolyte: a solute that does not produces ions in solution.
4. A chemist has mixed a solution of sodium chloride by dissolving 4.5 g of NaCl and 495.5 g
of water. What is the mass percent and mole fraction of each component? What is the molarity
of the solution? The density of solution is 1.00 g/ml.
The mass percent is calculated by the following formula:
mass of the part
x 100
mass of the solution
(
(
)
4.5 g
4.5 g + 495.5 g
)
(
x 100 = 0.90% NaCl
495.5 g
4.5 g + 495.5 g
)
x 100 = 99.10% H2O
To find the mole fraction, calculate the number of moles of each component.
4.5 g NaCl x
(
)
1 mol
58.44 g
= 0.077 mol NaCl; 495.5g H2O x
(
)
1 mol
= 27.50 mol H2O
18.02 g
The total number of moles of solution is (0.077 mol + 27.50 mol) or 27.6 moles of solution.
The mole fraction formula is similar to the mass percent formula.
(
(
number moles of the part
total number of moles of the solution
0.077 mol
27.6 mol
)
x 100 = 0.28% NaCl
(
)
x 100
27.5 mol
27.6 mol
)
x 100 = 99.7% H2O
To calculate molarity, you need the volume of the solution. Because you know the total mass
of the solution and its density, you can calculate the volume. Remember that the molarity
refers to the solute.
1 ml
500.0 g solution x
= 500.0 ml = .500 L
1.00 g
(
M=
54
(
)
number of moles of solute
volume of solution in liters
) (
=
0.077 mol
0.500 L
)
= 0.15 M NaCl
5. What is the molarity of a 355 ml solution that contains 1.56 g of NaOH?
Find the number of moles of NaOH. Remember to convert the volume to liters.
1 mol
1.56 g NaOH x 40.00 g = 0.0390 mol NaOH
(
M=
(
)
number of moles of solute
volume of solution in liters
) (
=
)
0.0390 mol
0.355 L
= 0.110 M NaOH
6. What are the boiling point and melting point of an aqueous ethanol solution that contains
25 g of C2H5OH with 0.750 kg of water? (For water, Kf = 1.86 and Kb = 0.52)
The formulas for the temperature changes are as follows:
ΔT (°C) = Kf • m, where m is molality
ΔT (°C) = Kb • m, where m is molality
Molality is calculated by dividing the number of moles of particles of solute by the mass of the
solvent (in kilograms). We need to calculate the number of moles of ethanol. Its molar mass is
46.08 g/mol.
25 g x
(
)
1 mol
46.08 g = 0.54 mol C2H5OH
(
0.54 mol
0.750 kg
)
= .72 m C2H5OH
Now we can plug the numbers into the temperature change formulas. Remember that the
boiling point of water is 100.00°C and its melting point is 0.00°C.
ΔT (°C) = Kf • m
ΔT (°C) = 1.86
( )
°C
m
x 0.72 m = 1.3°C
This is how much the melting point will drop if 25 g of ethanol is added to the water.
The new melting point is (0.00° C – 1.3°C), or –1.3°C.
ΔT (°C) = Kb • m
ΔT (°C) = 0.52
( )
°C
m
x 0.72 m = 0.37°C
This is how much the boiling point will rise if 25 g of ethanol is added to the water.
The new boiling point is (100.00°C + 0.37°C), or 100.37°C.
7. Define “osmosis.”
Osmosis is the flow of solvent through a semipermeable membrane. The flow of solvent is
from a dilute solution to a more concentrated solution.
55
Suggestions for Instructors
1. When dealing with colligative properties, students tend to think that the molality is of the
solution. It is actually the molality of the particles. If the solute is table salt (NaCl), it will
break into sodium ions and chloride ions in water. For every one unit of NaCl, we get two
kinds of particles. The molality will depend on how many ions the ionic compound breaks
into. In the quiz, ethanol was used. It does not break into ions, so we use the number of
moles of ethanol in the equation to calculate molality. Sugar behaves the same way.
2. Remember “like dissolves like.” Polar compounds dissolve polar compounds. Nonpolar compounds dissolve nonpolar compounds. Table salt is an ionic compound. Ionic compounds fit
into the polar category. They are an extreme of polarity. The reason some salts do not dissolve
is that the ions are attracted to each other so strongly that water cannot pull the ions apart.
3. I like to use a demo of water with food coloring and add vegetable oil to see them not mix.
4. It is easy to demonstrate saturated and unsaturated solutions. Take about 250 ml of water
and add some salt. The salt will dissolve. Add some more salt. It will dissolve. Keep doing
this until some solid does not dissolve. It is now saturated.
5. Students think that the volume of solution is the same as the volume of water added.
The volume of the solution is the combined volume of solute and solvent.
56
Contributors
The Core Curriculum Series: Chemistry was designed by the Standard Deviants Academic Team,
including:
David Rowley, Ph.D., George Washington University
David Ramaker, Ph.D., George Washington University
Instructor’s Guide prepared by: William J. Peacy, M.S., Triton College
Headed by:
David Sturdevant, Director of Writing
MFA, University of Louisville
With assistance from Margaret Packard, Films for the Humanities & Sciences.
57
Films for the Humanities & Sciences
A Wealth of Information. A World of Ideas.
PO Box 2053, Princeton, NJ 08543-2053
CALL 800-257-5126 • FAX 609-671-0266
10548
for use with
1 1744, 10122