Calculations and the Chemical Equation
Chapter Outline
CHEMISTRY CONNECTION: The Chemistry of Automobile Air Bags
5.1 The Mole Concept and Atoms
The Mole and Avogadro's Number
Calculating Atoms, Moles, and Mass
5.2 Compounds
The Chemical Formula
5.3 The Mole Concept Applied to Compounds
5.4 The Chemical Equation and The Information It Conveys
A Recipe for Chemical Change
Features of a Chemical Equation
The Experimental Basis of a Chemical Equation
5.5 Balancing Chemical Equations
A CLINICAL PERSPECTIVE: Carbon Monoxide Poisoning: A Case of Combining Ratios
5.6 Calculations Using the Chemical Equation
General Principles
Use of Conversion Factors
Theoretical and Percent Yield
A MEDICAL PERSPECTIVE: Pharmaceutical Chemistry: The Practical Significance of Percent
Yield
Summary
Key Terms
Questions and Problems
Critical Thinking Problems
Instructional Objectives
Conceptual Objectives
• Know the major function served by the chemical equation, the basis for chemical calculations.
• Know the relationship between the mole and Avogadro's number, and the usefulness of these
quantities.
Performance Objectives
• Perform calculations using Avogadro's number and the mole.
• Write chemical formulas for common inorganic substances.
• Calculate the formula weight and molar mass of a compound.
• Balance chemical equations given the identity of products and reactants.
• Calculate the number of moles of product resulting from a given number of moles of reactants or
the number of moles of reactant needed to produce a certain number of moles of product.
• Calculate theoretical and percent yield.
Health Applications
• Recognize the importance of combining ratios in the formation of carbon monoxide, a toxic
substance, during combustion.
• Remember that the fundamentals of chemical change have been very practical applications in the
pharmaceutical industry.
In-Chapter Examples
Example 5.1:
Example 5.2:
Example 5.3:
Example 5.4:
Example 5.5:
Example 5.6:
Example 5.7:
Example 5.8:
Example 5.9:
Example 5.10:
Example 5.11:
Example 5.12:
Example 5.13:
Example 5.14:
Example 5.15:
Example 5.16:
Example 5.17:
Example 5.18:
Example 5.19:
Example 5.20:
Relating Avogadro's number to molar mass: calculation of the mass of Avogadro's
number of sodium atoms
Converting moles to atoms.
Converting atoms to moles.
Converting moles of a substance to mass in grams.
Converting kilograms to moles.
Converting grams to number of atoms.
Calculating formula weight and molar mass (of water).
Calculating formula weight and molar mass (of sodium sulfate).
Calculating formula weight and molar mass (of calcium phosphate)
Balancing equations: the reaction of hydrogen and oxygen.
Balancing equations: the combustion of propane.
Balancing equations: the combustion of butane.
Balancing equations: the reaction of aqueous ammonium sulfate and aqueous lead
nitrate.
Converting between moles and grams.
Calculating reacting quantities: calculation of the mass of oxygen reacting with
one mole of propane.
Calculating grams of product from moles of reactant: calculation of the mass of
carbon dioxide produced from the combustion of one mole of propane.
Relating masses of reactants and products: calculation of the mass of propane
needed to produce a given amount of water.
Calculating a quantity of reactant: the reaction of hydrochloric acid with calcium
hydroxide.
Calculating reactant quantities.
Calculation of percent yield.
Chapter Overview
The Mole Concept and Atoms
Atoms are exceedingly small, yet their masses have been experimentally determined for each of the
elements. The periodic table provides atomic masses in atomic mass units (amu). A more practical unit
for defining a "collection" of atoms is the mole, Avogadro's number of particles.
Calculations based on the chemical equation relate the number of atoms, moles and their
corresponding mass. Conversion factors are used to relate the information provided in the problem to the
information requested by the problem. It is often useful to map a pattern for the required conversion
before beginning the problem.
Compounds
Compounds are pure substances that are composed of two or more elements that are chemically
combined. They are represented by their chemical formula, a combination of symbols of the various
elements which make up the compounds. The chemical formula is based upon the formula unit. This is
the smallest collection of atoms that provides the identity of the atoms present in the compound and the
relative numbers of each type of atom.
The Mole Concept Applied to Compounds
Just as a mole of atoms is based on the atomic mass or atomic weight, a mole of a compound is based
upon the formula mass or formula weight. To calculate the formula weight, the formula unit must be
known.
The Chemical Equation and the Information it Conveys
In a chemical equation, the identity of reactants and products must be specified. Reactants are written
to the left of the reaction arrow () and products to the right. The physical states of reactants and
products are shown in parentheses. The symbol  over the reaction arrow means that heat energy is
necessary for the reaction to occur. The equation must be balanced to reflect the law of conservation of
mass, which states that matter can neither be gained nor lost in the process of a chemical reaction.
Balancing Chemical Equations
The chemical equation enables the determination of the quantity of reactants needed to produce a
certain quantity of products, and expresses these quantities in terms of moles. The number of moles of
each product and reactant is indicated by placing a whole-number coefficient before the formula of each
substance in the chemical equation. Although many equations are balanced by trial and error, a method
for correctly balancing a chemical equation is presented.
Calculations Involving Chemical Equations
A properly balanced equation for a chemical reaction is the basis for the calculation of quantities of
products or reactants participating in the reaction. The mole is the basis for calculations, but since masses
are generally measured in grams (or kg), the interconversion between grams and moles are often
necessary to perform chemical arithmetic. The law of conservation of mass must be obeyed.
Hints for Faster Coverage
The chapter emphasizes the basics of chemical calculations. Depending on the purpose of the course
and the backgrounds of the students, the instructor may choose to de-emphasize or eliminate the
calculations of moles of atoms and ions. If the students have sufficient preparation, Sections 1-4 may be
omitted and Sections 5-6 may be covered in detail.
Suggested Problem Sets
The Mole Concept and Atoms - 19, 21, 23
Chemical Compounds - 25, 27, 29
The Mole Concept Applied to Compounds - 31, 33, 35, 37, 39
The Chemical Equation and the Information it Conveys - 41, 43
Balancing Chemical Equations - 45, 47, 49
Calculations Using the Chemical Equation - 51, 53, 55, 57, 59, 61, 63, 65, 67, 69
In-Chapter Perspectives
A CLINICAL PERSPECTIVE: Carbon Monoxide Poisoning: A Case of Combining Ratios. This wellknown phenomenon serves as a practical example of the importance of combining ratios in determining
the identity of the product. Additionally, it points out the potentia danger inherent in any combustion
reaction.
A MEDICAL PERSPECTIVE: Pharmaceutical Chemistry: The Practical Significance of Percent Yield.
Students often have difficulty grasping the meaning of incomplete reactions. This discussion shows the
practical economic, as well as scientific ramifications of reactions that have low yields.
Additional Perspectives
The cost to operate an energy delivery system is directly related to its efficiency. Systems that are
combustion-based, automobiles and furnaces, rely on a suitable fuel-to-oxygen (air) ratio for optimum
performance, and that is a direct application of the principles outlined in this chapter.
Critical Thinking Problems
1. and 2.
Quantities expressed in different units must be converted to a common set of units prior to
comparison. The smaller number is not necessarily the correct answer. The importance of
units is reinforced.
The student uses a variety of units of quantity, both English and metric, to obtain a molar unit,
the number of molecules. [Note that in 4.b., SiO2 does not exist as discrete molecules, but
rather as an extended 3-dimensional structure; this explains its appearance of sand in its
natural setting.]
3. and 4.
Chapter 5
Calculations and the Chemical Equation
Solutions to the Even-Numbered Problems
In-Chapter Questions and Problems
5.2
200.59 amu Hg
1.661 x 10 -24 g Hg
6.022 x 10 23 atoms Hg
x
x
1 atom Hg
1 amu Hg
1 mol Hg
5.4
a.
mol Na = 9.03 x 1023 atoms Na x
= 200.59
6.022 x 10 23 atoms Na
= 1.50 mol Na
1 mol Na
g Hg
mol Hg
5.6
O atoms = 40.0 g O x
5.8
a.
1 mol O
6.022 x 10 23 atoms O
x
= 1.51 x 1024 O atoms
16.00 g O
1 mol O
2 atoms of carbon x 12.01 amu/atom =
2 atoms of fluorine x 19.00 amu/atom =
4 atoms of chlorine x 35.45 amu/atom =
203.82 amu
24.02 amu
38.00 amu
141.80 amu
The average mass of a single unit of C2F2Cl4 is 203.82 amu/formula unit. Therefore, the
mass of 1 mole of formula units is 203.82 grams or 203.82 g/mol.
b.
3 atoms of carbon x 12.01 amu/atom =
8 atoms of hydrogen x 1.01 amu/atom =
1 atoms of oxygen x 16.00 amu/atoms =
60.11 amu
36.03 amu
8.08 amu
16.00 amu
The average mass of a single unit of C3H7OH is 60.11 amu/formula unit. Therefore, the
mass of 1 mole of formula units is 60.11 grams, or 60.11 g/mol.
c.
1 atom of carbon x 12.01 amu/atom =
3 atoms of hydrogen x 1.01 amu/atom =
1 atom of bromine x 79.91 amu/atom =
94.95 amu
12.01 amu
3.03 amu
79.91 amu
The average mass of a single unit of CH3Br is 94.95 amu/formula unit. Therefore, the mass
of 1 mole of formula units is 94.95 grams or 94.95 g/mol.
5.10
5.12
a.
6S2Cl2(s) + 16NH3(g)  N4S4(s) + 12NH4Cl(s) + S8(s)
b.
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
180 g C 6H12O6
106 g C 6H12O6
a. 1.00 x 10 mol C6H12O6 x
x
1 mol C 6H12O6
1 g C 6H12O 6
–5
= 1.80 x 103 g C6H12O6
b.
35.0 g MgCl2 x
1 mol MgCl 2
= 0.368 mol MgCl2
95.21 g MgCl 2
2 mol CO 2
44.0 g CO 2
= 88.0 g CO2
x
1 mol C 2H5OH
1 mol CO 2
5.14
1 mol C2H5OH x
5.16
Barium carbonate is BaCO3
5.18

a.
BaCO 3(s)  BaO(s) + CO 2(g)
b.
50.0 g BaCO3 x
a.
Step 1.
1 mol BaCO 3
1 mol CO 2
44.0 g CO 2
= 11.2 g CO2
x
x
197 g BaCO 3
1 mol BaCO 3
1 mol CO 2
Write down information about the reaction:
CH4(g) + 3 Cl2(g)  3 HCl(g) + CHCl3(g)
(excess) 105 g
Step 2.
Convert the mass of Cl2 to moles of Cl2:
105 g Cl2 x
Step 3.
1 mol Cl 2
= 1.48 mol Cl2
70.90 g Cl2
The reaction states that 3 moles of Cl2 will react to form one mole of CHCl3, so
the mole ratio is 3:1. Use this conversion factor to calculate the mass of product:
1.48 mol Cl2 x
b.
3 mol CHCl 3
119.37 g CHCl 3
= 58.9 CHCl3
x
1 mol Cl 2
1 mol CHCl 3
% yield =
actual yield
x 100 %
theoretical yield
% yield =
10.0 g
x 100 % = 17.0% yield
58.9 g
End-of-Chapter Questions and Problems
5.20
1.66 x 10 -24 g N
14.01 amu N
6.022 x 10 23 atoms N
x
x
1 amu N
1 atom N
1 mol N
5.22
a.
0.10 lb Ca x
b.
4.00 g Fe x
= 14.0
gN
mol N
454 g Ca
1 mol Ca
x
= 1.1 mol Ca
1 lb Ca
40.08 g Ca
1 mol Fe
= 7.16 x 10–2 mol Fe
55.85 g Fe
10 3g N 2
1 mol N 2
c. 2.00 kg N2 x
= 71.4 mol N2
x
1 kg N 2
28.02 g N 2
12.01 g C
= 1.80 x 102 g C
1 mol C
5.24
15.0 mol C x
5.26
The term formula weight may be used in a general way to describe the mass of ions, ion-pairs, or
molecules. The term molecular weight is reserved specifically for molecules.
5.28
a.
32.06 amu S
= 256.5 amu S
1 atom S
8 atoms S x
The average mass of a single unit of S8 is 256.5 amu/formula unit. Therefore the mass of a
mole of S8 units is 256.5 g/mol.
b.
2 atoms N x
14.01 amu N
=
1 atom N
28.02 amu
8 atoms H x
1.008 amu H
=
1 atom H
8.064 amu
32.06 amu S
=
1 atom S
1 atom S x
4 atoms O x
16.00 amu O
=
1 atom O
32.06 amu
64.00 amu
132.14 amu
The average mass of a single unit of (NH4)2SO4 is 132.14 amu/formula unit. Therefore the
mass of a mole of (NH4)2SO4 units is 132.14 g/mol.
c.
1 atom C x
12.01 amu C
=
1 atom C
2 atoms O x
16.00 amu O
=
1 atom O
12.01 amu
32.01 amu
44.01 amu
The average mass of a single unit of CO2 is 44.01 amu/formula unit. Therefore the mass of a
mole of CO2 units is 44.01 g/mol.
5.30
3 atoms O x
16.00 amu O
= 48.00 amu/molecule O3
1 atom O
The average mass of a single unit of O3 is 48.00 amu/formula unit. Therefore the mass of a
mole of O3 units is 48.00 g/mol.
5.32
a.
The formula weight of NH3 is 17.03 g/mol.
15.0 g NH3 x
b.
The formula weight of O2 is 32.00 g/mol.
16.0 g O2 x
5.34
a.
1 mol NH 3
= 0.881 mol NH3
17.03 g NH 3
1 mol O 2
= 0.500 mol O2
32.00 g O 2
The formula weight of NH3 is 17.03 g/mol.
0.400 mol NH3 x
b.
17.03 g NH 3
= 6.81 g NH3
1 mol NH 3
The formula weight of BaCO3 is 197.35 g/mol.
0.800 mol BaCO3 x
5.36
a.
The formula weight of CH4 is 16.04 g/mol.
2.00 mol CH4 x
b.
197.35 g BaCO 3
= 158 g BaCO3
1 mol BaCO 3
16.04 g CH 4
= 32.1 g CH4
1 mol CH 4
The formula weight of Ca(NO3)2 is 164.10 g/mol.
0.400 mol Ca(NO3)2 x
5.38
a.
The formula weight of NaOH is 40.00 g/mol.
0.100 mol NaOH x
b.
40.00 g NaOH
= 4.00 g NaOH
1 mol NaOH
The formula weight of H2SO4 is 98.08 g/mol.
0.100 mol H2SO4 x
c.
164.10 g Ca(NO 3 )2
= 65.6 g Ca(NO3)2
1 mol Ca(NO 3 )2
98.08 g H 2SO4
= 9.81 g H2SO4
1 mol H 2SO4
The formula weight of C2H5OH is 46.07 g/mol.
0.100 mol C2H5OH x
d.
180.16 g C 6H12O 6
= 4.61 g C2H5OH
1 mol C 6H12O6
The formula weight of Ca3(PO4)2 is 310.18 g/mol.
0.100 mol Ca3(PO4)2 x
5.40
a.
The formula weight of CS2 is 76.13 g/mol.
50.0 g CS2 x
b.
1 mol CS 2
= 0.657 mol CS2
76.13 g CS2
The formula weight of Al2(CO3)3 is 233.99 g/mol.
50.0 g Al2(CO3)3 x
c.
1 mol Al 2 (CO3 )3
= 0.214 mol Al2(CO3)3
233.99 g Al 2 (CO3 )3
The formula weight of Sr(OH) 2 is 121.64 g/mol.
50.0 g Sr(OH)2 x
d.
310.18 g Ca 3(PO 4 ) 2
= 31.0 g Ca3(PO4)2
1 mol Ca 3(PO 4 ) 2
1 mol Sr(OH) 2
= 0.411 mol Sr(OH)2
121.64 g Sr(OH) 2
The formula weight of LiNO3 is 68.95 g/mol.
50.0 g LiNO3 x
1 mol LiNO 3
= 0.725 mol LiNO3
68.95 g LiNO 3
5.42
A chemical equation provides the identity of the products and reactants, the physical state of the
products and reactants, reaction conditions, such as temperature or the presence of a catalyst, and
most importantly, the molar ratio of products and reactants, based on the law of conservation of
mass (the balanced equation).
5.44
The coefficients in a chemical equation show the relative number of moles of reactants and
products involved in the reaction.
5.46
a.
3Fe2O3(s) + CO(g)  2Fe3O42(g)
b.
2C6H6(l) + 1502(g)  12CO2(g) + 6H2O(g)
c.
2I4O9(s) + I2O6(s)  SI2(s) + 12O2(g)
d.
2KClO3 2KCl(s) + 3O2(g)
e.
C6H12O6(s)  2C2H6O(l) + 2CO2(g)
a.
2HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2H2O(l)
b.
2C4H10(g) + 13O2(g)  10H2O(g) + 8CO2(g)
a.
S(s) + O 2(g) 

 SO2(g)
5.48
5.50

b.
5.52
4HF(aq) + SiO2(s)  SiF4(aq) + 2H2O(l)
The formula weight of Al2O3 is 101.96 g/mol.
15.0 g Al x
5.54
1 mol Al
2 mol Al 2O3
101.96 g Al 2O 3
= 28.3 g Al2O3
x
x
26.98 g Al
4 mol Al
1 mol Al 2O3
The formula weight of H2O is 18.02 g/mol and that of H3PO3 is 82.00 g/mol.
3.50 g H2O x
5.56
1 mol H 2O
1 mol H 3PO3
82.00 g H 3PO 3
= 5.31 g H3PO3
x
x
18.02 g H 2O
3 mol H 2O
1 mol H 3PO3
a.
C7H6O3 + C2H4O2  C9H8O4 + H2O is balanced as written.
b.
1.00 x 102 mol salicylic acid x
c.
The formula weight of aspirin is 180.17 g/mol.
Using the answer from (b) above, 1.00 x 10 2 mol aspirin,
1.00 x 102 mol aspirin x
d.
1 mol aspirin
1 mol salicylic acid
= 1.00 x 102 mol aspirin
180.17 g aspirin
= 1.80 x 104 g aspirin
1 mol aspirin
The formula weight of acetic acid is 60.05 g/mol.
1.00 x 102 mol salicylic acid x
1 mol acetic acid
1 mol salicylic acid
x
60.05 g acetic acid
1 mol acetic acid
= 6.01 x 103 g acetic acid
5.58
a.
3 C atoms x 12.01 amu/atom C =
8 H atoms x 1.008 amu/atom H =
3 O atoms x 16.00 amu/atom O =
36.03 amu
8.064 amu
48.00 amu
92.09 amu
The mass of a single unit of C3H8O3 is 92.09 amu/formula unit. Therefore the mass of a
mole of C3H8O3 formula units is 92.09 g/mol.
b.
1 mol C3H8O3 x
3 mol O atoms
6.02 x 10 23 O atoms
= 1.81 x 1024 O atoms
x
1 mol C 3 H8 O3
1 mol O atoms
c.
1 mol C3H8O3 x
3 mol O atoms
16.00 g O
= 48.00 g O
x
1 mol C 3H8O3
1 mol O atoms
d.
50.0 g C3H8O3 x
1 mol C 3H 8O 3
3 mol O atoms
x
x
92.09 g C 3H 8O 3
1 mol C 3H8O3
16.00 g O
= 26.1 g O
1 mol O atoms
5.60
The formula weight of N2O is 44.02 g/mol. The formula weight of NH4NO3 is 80.04 g/mol.
1.00 x 102 g NH4NO3 x
1 mol NH 4NO 3
1 mol N 2O
44.02 g N 2O
x
x
80.05 g NH 4NO3
1 mol NH 4NO3
1 mol N 2O
= 55.0 g N2O
5.62
The formula weight of CaH2 is 42.10 g/mol.
1.00 x 102 g CaH2 x
5.64
Step 1.
1 mol CaH 2
1 mol H 2
2.02 g H 2
= 4.80 g H2
x
x
42.10 g CaH2
1 mol CaH 2
1 mol H 2
Write down information about the reaction:
16H+ + 2Cr2O72– + 3CH3CH2OH  3CH3COOH + 4Cr3+ + 11H2O
(excess)
0.100 g
Step 2.
Convert the mass of CH3CH2OH to moles of CH3CH2OH:
0.100 g CH3CH2OH x
Step 3.
1 mol CH 3CH2OH
= 2.17 x 10–3 mol CH3CH2OH
46.07 g CH3CH2OH
The reaction states that ethanol and acetic acid react in a 3:3 mole ratio. Use this
conversion factor to calculate the mass of product:
2.17 x 10–3 mol CH3CH2OH x
3 mol CH 3COOH
60.05 g CH 3COOH
x
3 mol CH 3CH 2OH
3 mol CH 3COOH
= 0.130 g CH3COOH
5.66
Step 1.
Write down information about the reaction:
4Fe3O4(s) + O2(g)  6Fe2O3
4.00 g
(excess)
Step 2.
Convert the mass of Fe3O4 to moles of Fe3O4:
4.00 g Fe3O4 x
Step 3.
1 mol Fe 3O4
= 1.73 x 10–2 mol Fe3O4
231.6 g Fe 3O 4
The reaction states that Fe2O3 and Fe3O4 react in a 6:4 mole ratio. Use this conversion
factor to calculate the mass of product:
1.73 x 10–2 mol Fe3O4 x
5.68
6 mol Fe 2O 3
159.7 g Fe 2O3
= 4.14 g Fe2O3
x
4 mol Fe 3O 4
1 mol Fe 2O 3
% yield =
actual yield
x 100%
theoretical yield
% yield =
0.110 g
x 100% = 84.6%
0.130 g
5.70
% yield =
actual yield
x 100%
theoretical yield
Rearranging, and solving for the actual yield:
actual yield = (% yield)(theoretical yield)/(100%)
= (90.0%)(4.14g)/(100%)
= 3.73 g
States of Matter: Gases, Liquids, and Solids
Chapter Outline
CHEMISTRY CONNECTION: The Demise of the Hindenburg
6.1 The Gaseous State
Ideal Gas Concept
Measurement of Gases
Boyle's Law
Charles's Law
Combined Gas Law
A CLINICAL PERSPECTIVE: Autoclaves and the Gas Laws
Avogadro's Law
Molar Volume a Gas
Gas Densities
The Ideal Gas Law
AN ENVIRONMENTAL PERSPECTIVE: The Greenhouse Effect and Global Warming
Dalton's Law of Partial Pressures
Kinetic Molecular Theory of Gases
Ideal Gases Versus Real Gases
6.2 The Liquid State
Compressibility
Viscosity
A CLINICAL PERSPECTIVE: Blood Gases and Respiration
Surface Tension
Vapor Pressure of a Liquid
Van der Waals Forces
Hydrogen Bonding
6.3 The Solid State
Properties of Solids
Types of Crystalline Solids
Summary
Key Terms
Questions and Problems
Critical Thinking Problems
Instructional Objectives
Conceptual Objectives
• Know the three common states of matter and their general properties.
• Understand the concept of the various gas laws: Boyle’s Law, Charles’ Law, Avogadro’s Law,
the ideal gas law, and Dalton’s Law.
• Describe the major points of the kinetic molecular theory of gases.
•
•
•
•
•
•
•
Explain the relationship between the kinetic molecular theory and the physical properties of
macroscopic quantities of gases.
Recognize the differences between real gases and ideal gases.
Understand the properties of the liquid state: compressibility, viscosity, surface tension, and
vapor pressure.
Describe the processes of melting, boiling, evaporation, and condensation.
Describe the dipolar attractions known collectively as van der Waals forces.
Describe hydrogen bonding and its relationship to boiling and melting temperatures.
Know the properties of the various classes of solids: ionic, covalent, molecular, and metallic.
Performance Objectives
• Perform ideal gas calculations, using each of the gas laws.
Health Applications
• Understand the chemical processes involved in the use of autoclaves.
• Recognize the essential role of that blood gases play in the process of respiration.
In-Chapter Examples
Example 6.1:
Example 6.2:
Example 6.3:
Example 6.4:
Example 6.5:
Example 6.6:
Example 6.7:
Example 6.8:
Calculating a final pressure (Boyle's Law).
Calculating a final volume (Charles' Law).
Using the combined gas law.
Using the combined gas law.
Using Avogadro's law.
Calculating a molar volume.
Calculating the number of moles of a gas.
Converting mass to volume.
Chapter Overview
The Gaseous State
The gaseous state is characterized by particles widely separated and weakly attracted to each other.
The kinetic molecular theory describes an ideal gas in which gas particles exhibit no interactive or
repulsive forces and the volumes of the individual gas particles are assumed to be negligible.
The behavior of an ideal gas is described by Boyle's law (relationship of volume and pressure),
Charles' law (relating volume to temperature), Avogadro's law (relating volume and number of moles),
and the combined gas law which relates the four variables noted above. Mixtures of gases are described
by Dalton's law of partial pressures.
The Liquid State
The liquid state is more ordered than the gaseous state, with strong attractive forces between
particles. Liquids are practically incompressible and may be described in terms of their viscosity and
surface tension. Surfactants decrease surface tension.
Condensation is the conversion of a gas to its liquid state; evaporation is the conversion of liquid to
vapor at a temperature below its boiling point. Both phenomena, as well as the boiling points, are
explained in terms of a liquid's vapor pressure.
Hydrogen bonding in liquids is responsible for lower than expected vapor pressures and higher than
expected boiling points. The presence of van der Waals forces and hydrogen bonds significantly affects
the boiling points of liquids as well as the melting points of solids.
The Solid State
The solid state is the most highly ordered state of matter; the strong attractive forces between
particles impart a definite shape to the solid. Solids are incompressible. Solids may be crystalline
(regular, repeating structure) or amorphous (without organized structure). The crystalline solids may be
classified into four major groups: ionic solids, covalent solids, molecular solids, or metallic solids.
Electrons in metallic solids are extremely mobile, resulting in the high conductivity exhibited by many
metallic solids.
Hints for Faster Coverage
If the course is designed to be more qualitative, a discussion of the gas phase without excessive gas
law calculations is possible. The instructor may choose to omit discussion of the various types of solids.
Suggested Problem Sets
Boyle's Law: 15, 17, 19
Charles's Law: 21, 23, 25
Combined Gas Law: 27, 29, 31
Avogadro's Law: 33, 35
Molar Volume and the Ideal Gas Law: 37, 39, 41
Kinetic Molecular Theory, Ideal and Real Gases: 45, 47
Dalton's Law: 49, 51
The Liquid State: 53, 55, 57, 59
The Solid State: 61, 63, 65
In-Chapter Perspectives
A CLINICAL PERSPECTIVE: Autoclaves and the Gas Laws. This discussion shows the interrelationship
between temperature and pressure and does so in the context of a medically related application. The
instructor may wish to expand on the topic of sterilization and the current controversy involving the reuse
of surgical implements on patients in surgery.
AN ENVIRONMENTAL PERSPECTIVE: The Greenhouse Effect and Global Warming. This topic is
one of international concern. It has broad implications that involve politics, economics, and international
law. Students, and all concerned citizens should be aware of the basic causes of this problem; in the
future they will certainly have some say in its resolution.
A CLINICAL PERSPECTIVE: Blood Gases and Respiration. This perspective affords the instructor
another opportunity to apply the basic concepts, developed early in the book, to the more career-oriented
biochemical material.
Additional Perspectives
Mention of liquid crystals and their applications meshes well with a discussion of liquids and solids.
Liquid crystals are synthesized to emphasize order in one or two dimensions. They, and cathode ray
devices, represent the principal approaches to display technology ... refer to your students' calculators as
common examples.
Critical Thinking Problems
1.
This problem accomplishes several objectives:
- it provides practice in the use of conversion factors and the ideal gas law.
- it deals with a practical ecological problem: plants generating oxygen for animals.
- it forces the students to make some assumptions to solve the problem (the assumptions to solve the
problem—the pressure and temperature).
2.
The experiment illustrates a method for determining ideality of a gas; it helps to emphasize that real
gases do not exhibit perfect ideal gas behavior.
3.
The students must use their understanding of molecular structure and polarity and relate this to the
behavior of these molecules in the gas phase.
4.
The students recognize that the coefficients in the balanced equation really do predict relative
volumes, hence moles of product. Avogadro's relationship is incorporated in this question.
5.
The relationship between pressure and boiling temperature is illustrated in a practical, medically
related device.
Chapter 6
States of Matter:
Gases, Liquids, and Solids
Solutions to the Even-Numbered Problems
In-Chapter Questions and Problems
6.2
a. 10.0 torr x
1 atm
760 torr
b. 61.0 cm Hg x
c. 2.75 mm Hg x
6.4
a.
b.
= 1.32 x 10 -2 atm
10 mm Hg
1 atm
x
1 cm Hg
760 mm Hg
1 atm
760 mm Hg
= 0.362 atm
PiVi = PfVf
Vi =
Pf Vf
Pi
Vi =
(0.50 atm)(0.30 L)
1.0 atm
= 0.15 L
PiVi = PfVf
Vf =
Pi Vi
Pf
Vf =
(1.0 atm)(0.75 L)
2.0 atm
= 0.38 L
= 0.803 atm
6.6
Initial temperature: 25˚C + 273 = 298 K
a.
Final temperature = 546 K
Vi Vf

Ti
Tf
b.
Vf =
Vi Tf
Ti
Vf =
(3.00 L)(546 K)
= 5.50 L
298 K
Final Temperature = 0.00˚C + 273 = 273 K
Vi Vf

Ti
Tf
c.
Vf =
Vi Tf
Ti
Vf =
(3.00 L)(273 K)
= 2.75 L
298 K
Final Temperature = 373 K
Vi Vf

Ti
Tf
6.8
Vf =
Vi Tf
Ti
Vf =
(3.00 L)(373 K)
= 3.76 L
298 K
PiVi = PfVf
Solving for the final pressure, Pf ,
Pf =
Pi Vi
Vf
Pi = 760 torr x
Vi = 2.00 L
1 atm
760 torr
= 1.00 atm
Vf = 10.0 L
Substituting,
Pf =
6.10
(1.00 atm)(2.00 L)
(10.0 L)
= 0.200 atm
Begin by assuming that Vi is 1.0 L. Consequently, Vf must be 3.0 L.
Vi Vf

ni
nf
nf =
Vf ni
Vi
nf =
(3.0 L)(0.25 mol H 2 )
= 0.75 mol H 2
1.0 L
Note: Any initial volume can be assumed; the same result would be obtained.
6.12
PV = nRT
Solving for pressure, P,
P =
nRT
V
nRT
V
n = 4.80 g H 2 x
R =
P =
(1 mol H 2 )
= 2.38 mol H 2
2.02 g H 2
0.0821 L  atm
K  mol
T = 25˚C + 273 = 298 K
V = 20.0 L
Substituting,
P =
6.14
(2.38 mol H 2 )(0.0821 L  atm/K  mol)(298 K)
 2.91 atm
20.0 L
PV = nRT
Solving for temperature, T,
T
PV
nR
P = 1.00 atm (standard pressure)
V = 2.00 L
n = 2.00 mol He
R=
0.0821 L  atm
K  mol
Substituting,
T =
(1.00 atm)(2.00 L)
 12.2 K
(2.00 mol He) (0.0821 L  atm/K  mol)
End-of-Chapter Questions and Problems
6.16
A pressure of 5 atm (abscissa) corresponds to a volume of 1 L (ordinate).
6.18
A pressure of 2 atm (abscissa) corresponds to a volume of 2.5 L (ordinate).
PV = k
(2 atm)(2.5 L) = 5 L·atm = k
6.20
Pi = 1.00 atm
Vi = 15.6 L
Pf = 0.150 atm
Vf = ? L
PiVi = PfVf
6.22
Vf =
Pi Vi
Pf
Vf =
(1.00 atm)(15.6 L)
0.150 atm
Vf = 104 L
Mathematically, Charles's law states that the ratio of volume (V) and Temperature (T) is a
constant:
V
T
6.24
= 104 L
Vi = 2.00 L
Ti = 250 K
= k
Vf = ? L
Tf = 500 K
Vi Vf

Ti
Tf
Vf =
Vi Tf
Ti
Vf =
(2.00 L)(500 K)
= 4.00 L
250 K
The change in volume,
V = Vf - Vi
V = 4.00 L - 2.00 L
V = 2.00 L
6.26
Vi = 1.25 L
Ti = 20˚C
Ff = ? L
Tf = 39˚F
Vi Vf

Ti
Tf
6.28
6.30
Vf =
Vi Tf
Ti
Vf =
(1.25 L){[(39 F - 32)/1.8] + 273}
(20C + 273)
Vf =
(1.25 L)(277 K)
= 1.18 L
(293 K)
Examine each effect separately:
Volume and temperature are directly proportional; decreasing T increases V.
Volume and pressure are inversely proportional; increasing P increases V.
Therefore, both variables work together to decrease the volume.
Pi Vi Pf Vf

Ti
Tf
Pi Vi Tf = Pf Vf Ti
Tf =
6.32
Pf Vf Ti
Pi Vi
Pi = 1.00 atm
Pf = 0.800 atm
Vi = 2.30 L
Ti = 25˚C
Vf = ? L
Tf = 20˚C
Using the equation derived in question 6.29,
Vf =
Pi Vi Tf
Pf Ti
and substituting:
6.34
Vf =
(1.00 atm)(2.50 L)(20 C + 273)
(0.800 atm)(25C + 273)
Vf =
(1.00 atm)(2.50 L)(293 K)
 3.07 L
(0.800 atm)(289 K)
Doubling the number of grams of helium is identical to doubling the number of moles of helium:
8.00 g He x
1 mol He
4.00 g He
2 x 8.00 g Hex
= 2.00 mol He
1 mol He
4.00 g He
= 4.00 mol He
If the volume is directly proportional to the number of moles of the gas, doubling the number of
moles of helium will double the volume.
6.36
6.38
Mathematically, the ratio of volume (V) and number of moles (n) is a constant:
V
n
= k
Pi = 1.20 atm
Vi = 257 mL
Ti = 20˚C
Pf = 1.00 atm
Vf = ? L
Tf = 273 K
Pi Vi Pf Vf

Ti
Tf
Vf =
Vf
Pi Vi Tf
Pf Ti

1 L 
(1.20 atm) 257 mL x
(273 K)
3


10
mL
=
(1.00 atm)(20 C + 273)
Vf =
(1.20 atm)(0.257 L)(273 K)
 0.287 L
(1.00 atm)(293 K)
6.40
6.42
PV = nRT
V =
nRT
P
V =
(1.00 mol)(0.0821 L  atm/K  mol)(273 K)
(1.00 atm)
PV = nRT
Convert F to K :
T =
Convert cm Hg to atm :
6.44
= 22.4 L
72F - 32
 273 = 295 K
1.8
P = 30 cm Hg x
1 atm
76 cm Hg
V =
nRT
P
V =
(6.00 mol)(0.0821 L  atm/K  mol)(295 K)
(0.39 atm)
= 0.39 atm
V = 368 L  3.7 x 102 L
PV = nRT
n =
PV
RT
n =
(1.00 atm)(10.0 L)
(0.0821 L  atm/K  mol)(273 K)
0.446 mol O
2
x
32.0 g O 2
1 mol O 2
= 0.446 mol O
2
= 14.3 g O 2
6.46
Gases exhibit more ideal behavior at high temperatures. At high temperatures, the gas particles
have greater energy to overcome attractive forces between particles. The ideal gas model
assumes negligible attractive forces between gas particles.
6.48
The kinetic molecular theory states that the average kinetic energy of the gas particles increases
as the temperature increases. Kinetic energy is proportional to (velocity)2. therefore, as the
temperature increases the gas particle velocity increases and the force per unit area (pressure)
increases as well. This pressure could cause the container to rupture explosively.
6.50
Pt = P1 + P2 + P3 + ….
6.52
Pt = PHe + PNe
PNe = Pt – PHe
PNe = 0.56 atm – 0.27 atm
PNe = 0.29 atm
6.54
Intermolecular forces in solids are greater than those in liquids. This is evident because solids
have a network of forces that allow them to retain their own shape while liquids will assume the
shape of their containers.
6.56
Liquids with strong intermolecular attractive forces have low vapor pressures; vapor pressure
increases as the strength of intermolecular attractive forces decreases.
6.58
Evaporation is the conversion of a liquid to a gas below the boiling temperature. Boiling occurs
at a temperature (the boiling point) at which the vapor pressure of the liquid becomes equal to
the atmospheric pressure.
6.60
Surface tension of a liquid is a measure of the attractive forces exerted among molecules on the
surface of a liquid. The net force on each surface molecule is greater than that exerted on
molecules in the bulk solution. This is a result of the fact that each surface molecule interacts
with fewer molecules; hence, each interaction is stronger.
6.62
Crystalline solids have a regular, repeating structure. Amorphous solids have no organized
structure.
6.64
a.
b.
6.66
Diamond is a covalent solid. As such, it has a very high melting point (it is unaffected by the
frictional heat generated in the cutting process) and it is extremely hard (it must be harder than
the material it cuts).
molecular solid - soft, low melting point
metallic solid - good conductor of heat and electricity