An Introduction to Chemistry

12.2 Drawing Lewis Structures 459

Step 3 We started with 14 total valence electrons for CH3Br, and we have used eight
of them for the four bonds in the skeleton we drew in Step 2.
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Step 4 After Step 3, we have the following skeleton for CH3Br and six valence
electrons still to distribute.
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Hydrogen atoms never have lone pairs, and the carbon atom has an octet of
electrons around it from its four bonds. In contrast, the bromine atom needs
six more electrons to obtain an octet, so we put the remaining six electrons
around the bromine atom as three lone pairs.

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Step 5 The structure drawn in Step 4 for CH3Br has an octet of electrons around the
carbon and bromine atoms. Hydrogen has its one bond. We have also used all
of the valence electrons. Thus we move on to Step 6.

Step 6 All of the atoms in the structure drawn for CH3Br have their most common
bonding pattern, so we have a reasonable Lewis structure.
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Step 7 Because the atoms in our structure for CH3Br have their most common
bonding pattern, we skip this step.

Shortcut The shortcut to drawing Lewis structures described in Section 3.3 can
often be used for uncharged molecules such as CH3Br. Carbon atoms usually have four
bonds and no lone pairs, hydrogen atoms always have one bond, and bromine atoms
most commonly have one bond and three lone pairs. The only way to give these atoms
their most common bonding patterns is with the following Lewis structure, which is
the same Lewis structure we arrived at with the stepwise procedure.

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460 Chapter 12 Molecular Structure

Example 12.2 - Drawing Lewis Structures

Objective 7 Formaldehyde, CH2O, has many uses, including the preservation of biological
specimens. Draw a reasonable Lewis structure for formaldehyde.

Solution
Step 1 Carbon is in group 4A, so its atoms have four valence electrons. Hydrogen

has one valence electron. Oxygen is in group 6A, so its atoms have six valence
electrons.

CH O
CH2O Number of valence e- = 1(4) + 2(1) + 1(6) = 12

Step 2 There are two possible skeletons for this structure. We will work with skeleton
1 first.
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Skeleton 1 Skeleton 2

Step 3 (Skeleton 1) Number of e- remaining = 12 - 3(2) = 6

Step 4 (Skeleton 1) The most common bonding pattern for oxygen atoms is two
bonds and two lone pairs. It is possible for carbon atoms to have one lone pair,
although that is rare. Thus we might use our six remaining electrons as two
lone pairs on the oxygen atom and one lone pair on the carbon atom.

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Step 5 (Skeleton 1) The structure above leaves the carbon atom without its octet.
Because we are short two electrons for octets, we convert one lone pair to
another bond. Converting the lone pair on the carbon to a multiple bond
would still leave the carbon with only six electrons around it and would put
ten electrons around the oxygen atom. Carbon and oxygen atoms always have
eight electrons around them in a reasonable Lewis structure. Thus we instead
try converting one of the lone pairs on the oxygen atom to another C-O
bond.

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Structure 1
Step 6 In structure 1, the carbon atom and the oxygen atom have rare bonding

patterns. This suggests that there might be a better way to arrange the atoms.
We proceed to Step 7.
Step 7 To attempt to give each atom its most common bonding pattern, we return to
Step 2 and try another skeleton.
Step 2 The only alternative to skeleton 1 for CH2O is

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Skeleton 2

Step 3 (Skeleton 2)

12.2 Drawing Lewis Structures 461

Step 4 (Skeleton 2) Oxygen atoms commonly have one bond and three lone pairs,
so we might use our remaining electrons as three lone pairs on the oxygen
atom.
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Step 5 (Skeleton 2) In Step 4, we used all of the remaining electrons but left the

carbon atom with only six electrons around it. Because we are short two
electrons needed to obtain octets of electrons around each atom, we convert
one lone pair into another bond.

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Structure 2

Step 6 (Skeleton 2) All of the atoms have their most common bonding pattern, so we
have a reasonable Lewis structure.

Step 7 (Skeleton 2) Because each atom in Structure 2 has its most common bonding
pattern, we skip this step.

Because Structure 2 is the only arrangement that gives each atom its most common
bonding pattern, we could have skipped the stepwise procedure and used the shortcut
for this molecule.

Example 12.3 - Drawing Lewis Structures Objective 7

The cyanide polyatomic ion, CN- is similar in structure to carbon monoxide, CO.
Although they work by different mechanisms, they are both poisons that can disrupt
the use of oxygen, O2, in organisms. Draw a reasonable Lewis structure for the cyanide
ion.

Solution
The shortcut does not work for polyatomic ions, so we use the stepwise procedure for
CN-.
Step 1 Carbon is in group 4A, so its atoms have four valence electrons. Nitrogen is

in group 5A, so its atoms have five valence electrons. Remember to add one
electron for the -1 charge.

C N [-]
CN- #valence e- = 1(4) + 1(5) + 1 = 10
Step 2 There is only one way to arrange two atoms.
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Step 3 Number e- remaining = 10 - 1(2) = 8
Step 4 Both the carbon atom and the nitrogen atom need six more electrons. We do
not have enough electrons to provide octets by the formation of lone pairs.
You will see in the next step that we will make up for this lack of electrons
with multiple bonds. For now, we might put two lone pairs on each atom.
$/

462 Chapter 12 Molecular Structure

Step 5 Because we are short four electrons (or two pairs) in order to obtain octets,
we convert two lone pairs into bonds. If we convert one lone pair from each
atom into another bond, we get octets of electrons around each atom.

$ / UP $ /


Step 6 The nitrogen atom now has its most common bonding pattern, three bonds
and one lone pair. The carbon atom has a rare bonding pattern, so we
proceed to Step 7.

Step 7 There is no other way to arrange the structure and still have an octet of
electrons around each atom, so the Lewis structure for CN- is

$ /


Remember to put the Lewis structures for polyatomic ions in brackets and show the
charge on the outside upper right.

Objective 7 Example 12.4 - Drawing Lewis Structures

Draw a reasonable Lewis structure for CF3CHCl2, the molecular formula for
HCFC-123, which is one of the hydrochlorofluorocarbons used as a replacement for
more damaging chlorofluorocarbons. (See the Special Topic 7.2: Other Ozone Depleting
Chemicals.)

Solution

Step 1 Carbon is in group 4A, so its atoms have four valence electrons. Chlorine

and fluorine are in group 7A, so their atoms have seven valence electrons.

Hydrogen has one valence electron.

CF H Cl

CF3CHCl2 Number valence e- = 2(4) + 3(7) + 1(1) + 2(7) = 44

Step 2 The best way to start our skeleton is to remember that carbon atoms often
bond to other carbon atoms. Thus we link the two carbon atoms together in
the center of the skeleton. We expect hydrogen, fluorine, and chlorine atoms to
form one bond, so we attach all of them to the carbon atoms. There are many
ways that they could be arranged, but the way the formula has been written
(especially the separation of the two carbons) gives us clues: CF3CHCl2, tells
us that the three fluorine atoms are on one carbon atom, and the hydrogen
and chlorine atoms are on the second carbon.

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CF3CHCl2
Step 3 Number of e- remaining = 44 - 7(2) = 30

12.2 Drawing Lewis Structures 463

Step 4 We expect halogen atoms to have three lone pairs, so we can use the 30
remaining electrons to give us three lone pairs for each fluorine and chlorine
atom.

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Structure 1

We could also represent CF3CHCl2, with the following Lewis structures:

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Structure 2 Structure 3

Structures 2 and 3 might look like they represent different molecules, but they
actually represent the same molecule as Structure 1. To confirm that this is true,
picture yourself sitting on the hydrogen atom in either the space filling or the
ball-and-stick model shown in Figure 12.3 for CF3CHCl2. Atoms connected
to each other by single bonds, such as the two carbon atoms in this molecule,
are constantly rotating with respect to each other. Thus your hydrogen atom is
sometimes turned toward the top of the molecule (somewhat like in Structure
1), sometimes toward the bottom of the structure (somewhat like Structure
3), and sometimes in one of the many possible positions in between. (Section
12.4 describes how you can predict molecular shapes. If you do not see at this
point why Structures 1, 2, and 3 all represent the same molecule, you might
return to this example after reading that section.)
Step 5 We have used all of the valence electrons, and we have obtained octets of
electrons around each atom other than hydrogen. Thus we move to Step 6.
Step 6 All of the atoms in our structure have their most common bonding pattern, so
we have a reasonable Lewis structure.
Step 7 Because each atom in our structure has its most common bonding pattern, we
skip this step.

Figure 12.3
Models of CF3CHCl2

464 Chapter 12 Molecular Structure

More Than One Possible Structure

Objective 7 It is possible to generate more than one reasonable Lewis structure for some formulas.
Objective 7
When this happens, remember that the more common the bonding pattern is (as

summarized in Table 12.1), the more stable the structure. Even after you apply this

criterion, you may still be left with two or more Lewis structures that are equally

reasonable. For example, the following Lewis structures for C2H6O both have the
most common bonding pattern for all of their atoms.

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In fact, both these structures describe actual compounds. The first structure is dimethyl
ether, and the second is ethanol. Substances that have the same molecular formula but
different structural formulas are called isomers. We can write the formulas for these
two isomers so as to distinguish between them: CH3OCH3 represents dimethyl ether,
and CH3CH2OH represents ethanol.

Example 12.5 - Drawing Lewis Structures

Acetaldehyde can be converted into the sedative chloral hydrate (the “Mickey Finn”
or knockout drops often mentioned in detective stories). In the first step of the
reaction that forms chloral hydrate, acetaldehyde, CH3CHO, changes to its isomer,
CH2CHOH. Draw a reasonable Lewis structures for each of these isomers.

Solution
Step 1 Both molecules have the molecular formula C2H4O.

C HO
Number valence e- = 2(4) + 4(1) + 1(6) = 18
Step 2 Because we expect carbon atoms to bond to other carbon atoms, we can link
the two carbon atoms together to start each skeleton, which is then completed
to try to match each formula given.

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CH3CHO CH2CHOH
Skeleton 1 Skeleton 2

Step 3 Number of e- remaining = 18 - 6(2) = 6

Step 4 For Skeleton 1, we can add three lone pairs to the oxygen atom to give it its
octet. For Skeleton 2, we can add two lone pairs to the oxygen atom to get its
most common bonding pattern of two bonds and two lone pairs. We can place

12.3 Resonance 465

the remaining two electrons on one of the carbon atoms as a lone pair.

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Step 5 In each case, we have one carbon atom with only six electrons around it, so we
convert one lone pair into another bond.

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Structure 1

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Structure 2

Step 6 All of the atoms in both structures have their most common bonding pattern,
so we have two reasonable Lewis structures representing isomers. Structure 1
is acetaldehyde (or ethanal) and structure 2 is ethenol.

Step 7 Because each atom in our structures has its most common bonding pattern, Web
we skip this step. Molecules

Exercise 12.1 - Lewis Structures Objective 7

Draw a reasonable Lewis structure for each of the following formulas.

a. CCl4 f. N2H4
b. Cl2O g. H2O2

c. COF2 h. NH2OH
d. C2Cl6 i. NCl3
e. BCl3

12.3 Resonance

As the valence-bond model was being developed, chemists came to recognize that it
did not always describe all molecules and polyatomic ions adequately. For example, if
we follow the procedures in Section 12.2 to draw a Lewis structure for the nitrate ion,
NO3-, we would obtain a structure that chemists have discovered is not an accurate
description of the ion’s bonds:

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0/ 0

466 Chapter 12 Molecular Structure

This Lewis structure shows two different types of bonds, single and double. Because
more energy is required to break a double bond than to break a single bond, we say
that a double bond is stronger than a single bond. Double bonds also have a shorter
bond length (the distance between the nuclei of the two atoms in the bond) than
single bonds do. Thus, if the above Lewis structure for nitrate were correct, the nitrate
polyatomic ion would have one bond that is shorter and stronger than the other two.

This is not the case. Laboratory analyses show all three of the bonds in the nitrate
ion to be the same strength and the same length. Interestingly, analysis of the bonds
suggests they are longer than double bonds and shorter than single bonds. They are also
stronger than single bonds but not as strong as double bonds. In order to explain how
these characteristics are possible for the nitrate ion and for molecules and polyatomic
ions like it, the valence-bond model had to be expanded.

The model now allows us to view certain molecules and polyatomic ions as if they
were able to resonate—to switch back and forth—between two or more different
structures. For example, the nitrate ion can be viewed as if it resonates between the three
different structures below. Each of these structures is called a resonance structure. The
hypothetical switching from one resonance structure to another is called resonance. In
chemical notation, the convention is to separate the resonance structures with double
headed arrows.

0 0 0
0/ 0 0/0 0/0

Objective 8 It is important to stress that the nitrate ion is not really changing from one resonance
structure to another, but chemists find it useful, as an intermediate stage in the process
of developing a better description of the nitrate ion, to think of it as if it were doing so.
In actuality, the ion behaves as if it were a blend of the three resonance structures.

We can draw a Lewis-like structure that provides a better description of the actual
character of the nitrate ion by blending the resonance structures into a single resonance
hybrid:

Step 1 Draw the skeletal structure, using solid lines for the bonds that are
found in all of the resonance structures.

Step 2 Where there is sometimes a bond and sometimes not, draw a dotted
line.

Step 3 Draw only those lone pairs that are found on every one of the
resonance structures. (Leave off the lone pairs that are on one or more
resonance structure but not on all of them.)

The resonance hybrid for the nitrate polyatomic ion is
" CPOE GPVOE JO BU MFBTU POF
CVU OPU BMM UIF SFTPOBODF TUSVDUVSFT

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" CPOE GPVOE JO BMM " MPOF QBJS GPVOE JO
UIF SFTPOBODF TUSVDUVSFT BMM UIF SFTPOBODF TUSVDUVSFT

12.4 Molecular Geometry from Lewis Structures 467

Resonance is possible whenever a Lewis structure has a multiple bond and an Objective 8
adjacent atom with at least one lone pair. The arrows in the following generalized Objective 8
structures show how you can think of the electrons shifting as one resonance structure
changes to another.

9:; 9: ;

For example, the two resonance structures for the formate ion, HCO2- are

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To generate the second resonance structure from the first, we imagine one lone pair
dropping down to form another bond, and pushing an adjacent bond off to form
a lone pair. The arrows show this hypothetical shift of electrons, which leads to the
resonance hybrid below.

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)$ 0

Exercise 12.2 - Resonance

Draw all of the reasonable resonance structures and the resonance hybrid for the
carbonate ion, CO32-. A reasonable Lewis structure for the carbonate ion is

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You can find a more comprehensive description of resonance at the
textbook’s Web site.

12.4 Molecular Geometry from Lewis Structures

The shapes of molecules play a major role in determining their function. For example,
Special Topic 3.1 describes how the shape of ethanol molecules allows them to attach
to specific sites on nerve cell membranes and slow the transfer of information from
one neuron to another. Special Topic 5.2 describes how the shapes of the molecules
in our food determine whether they taste sweet or bitter. You will find out in Special
Topic 17.2 that the fat substitute Olestra is indigestible because it does not fit into
the enzyme that digests natural fat. The purpose of this section is to show you how
to use Lewis structures to predict three-dimensional shapes of simple molecules and
polyatomic ions. Let’s start with a review of some of the information from Section 3.1,
where this topic was first introduced.

468 Chapter 12 Molecular Structure

Objective 9 The carbon atoms in methane, CH4, molecules are surrounded by four electron
groups, each composed of a two electron covalent bond to a hydrogen atom.

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Because each electron group is negatively charged and because negative charges repel
each other, the most stable arrangement for the electron groups is one in which they
are as far apart as possible. In Chapter 3, we saw that the geometric arrangement in
which four electron groups are as far apart as possible is a tetrahedral arrangement,
with bond angles of 109.5°. A tetrahedron is a four sided solid for which each side is an
equilateral triangle, and a bond angle is the angle formed by any three adjacent atoms
in a molecule. The carbon in a methane molecule can be viewed as sitting in the center
of a tetrahedron with a hydrogen atom at each of the four corners.

H

HC H
H

Figure 12.4 shows four ways to describe the methane molecule. The space filling
model in that figure is probably the best representation of the actual structure of the
CH4 molecule, but when scientists are focusing on the bonds in a molecule, they often
prefer to think of the structure as represented by the ball-and-stick model, the Lewis
structure, and the geometric sketch. Notice that the Lewis structure for CH4 does not
show the correct bond angles. In order to display bond angles correctly with a Lewis-
like structure, chemists use geometric sketches, such as the last image in Figure 12.4.
Remember that in a geometric sketch, a solid line represents a bond whose central
axis lies in the plane of the paper, a solid wedge shape represents a bond whose axis
is extending out of the plane of the paper toward you, and a dashed wedge shape
represents a bond whose axis extends out behind the plane of the paper, away from
you.

Figure 12.4
Four Ways to Describe a Methane, CH4, Molecule

Objective 9 ) )

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The tetrahedral arrangement is very common and important in chemistry, so be sure
that you have a clear image in your mind of the relationship between the geometric
sketch in Figure 12.4 and the three-dimensional shape of a molecule such as methane.
You can make a ball-and-stick model of this molecule using a firm piece of fruit such as
an apple or a green tomato for the carbon atom, four toothpicks to represent the axes
through each of the four bonds, and four grapes to represent the hydrogen atoms. Try

12.4 Molecular Geometry from Lewis Structures 469

to arrange the toothpicks and grapes so that each grape is an equal distance from the Figure 12.5
other three. When you think you have successfully constructed a CH4 fruit molecule, A Green Tomato-Grape
turn your model over and over, so that a different grape is on top each time. If your
model is really tetrahedral, it will always look the same, no matter which grape is on Ball-and-Stick Model for
top (Figure 12.5).
the CH4 Molecule
The nitrogen atom in each ammonia molecule, NH3, has four electron groups around
it (three bond groups and one lone pair), and, like with methane molecules, the best
way to get these four groups as far apart as possible is in a tetrahedral arrangement. The
electrons in the lone pair exert a greater repulsive force than the electrons in the bonds,
however, so the bond groups are pushed a little closer together. As a result, the bond
angles for ammonia are 107°, rather than the 109.5° found in the methane molecule.

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There are two ways to describe the geometry of the ammonia molecule. We can Objective 9
describe the arrangement of all the electron groups, including the lone pair, and call
the shape tetrahedral. Or we can describe the arrangement of the atoms only, without
considering the lone pair, and call ammonia’s shape a trigonal pyramid (Figure 12.6).
The three hydrogen atoms represent the corners of the pyramid’s base, and the nitrogen
atom forms the peak. In this book, we will call the geometry that describes all of the
electron groups, including the lone pairs, the electron group geometry. The shape that
describes the arrangement of the atoms only—treating the lone pairs as invisible—will
be called the molecular geometry.

)/ ) ) /) Figure 12.6
) ) Geometry of an
¡ Ammonia Molecule

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The water molecule’s electron group geometry, which is determined by the
arrangement of its four electron groups around the oxygen atom, is tetrahedral (Figure
12.7).

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Geometry of a Water
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Its molecular geometry, which describes the arrangement of the atoms only, is called
bent. The two lone pairs on the oxygen atom exert a stronger repulsion than the two
bond groups and push the bond groups closer together than they would be if all of the
electron groups around the oxygen atom were identical. The bond angle becomes 105°
instead of the 109.5° predicted for four identical electron groups.

470 Chapter 12 Molecular Structure

Objective 9 The boron atom in the center of a BF3 molecule is surrounded by three electron
groups, which are each composed of a two-electron covalent bond to a fluorine atom.

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The geometric arrangement that keeps three electron groups as far apart as possible
is called trigonal planar (often called triangular planar) and leads to angles of 120°
between the groups. Therefore, the B–F bonds in BF3 molecules have a trigonal planar
arrangement with 120° angles between any two fluorine atoms (Figure 12.8).

Figure 12.8
Four Ways to Describe a Boron Trifluoride, BF3, Molecule

Objective 9 ' '
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Molecules quite often have more than one central atom (an atom with two or more
atoms bonded to it). To describe the arrangement of atoms in such molecules, we need
to consider the central atoms one at a time. Let’s look at the ethene, C2H4, molecule:

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There are three electron groups around each of ethene’s carbon atoms (two single bonds
to hydrogen atoms and the double bond to the other carbon). Because these groups
repel each other, the most stable arrangement around each carbon atom—with the
electron groups as far apart as possible—is a triangle with all three groups in the same
plane and with angles of about 120° between the groups. That is, ethene has a trigonal
planar arrangement of atoms around each carbon.

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Figure 12.9 shows four ways to describe a C2H4 molecule. Because the bond groups
around each carbon are not identical (two are single bonds and one is a double bond),
the angles are only approximately 120°.

12.4 Molecular Geometry from Lewis Structures 471

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Figure 12.9
Four Ways to Describe an

Ethene (Ethylene), C2H4,
Molecule

When atoms have only two electron groups around them, the groups are arranged
in a line, with bond angles of 180°. This arrangement is called linear. For example,
the carbon atom in HCN has two electron groups around it (one single bond and one
triple bond), and the best way to keep these groups apart is in a linear shape.

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There are two central atoms in an acetylene, C2H2, molecule.
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Because each of acetylene’s central atoms has two electron groups around it, the
arrangement of atoms around each is linear. Figure 12.10 shows four ways to describe
a C2H2 molecule.

Objective 9 Figure 12.10
Four Ways to Describe an

Ethyne (Acetylene), C2H2,
Molecule

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The following sample study sheet summarizes a systematic procedure for predicting
electron group geometries, drawing geometric sketches, and describing molecular
geometries.

472 Chapter 12 Molecular Structure

Sample Study Tip-off In this chapter, you are given a Lewis structure for a molecule or polyatomic
Sheet 12.2 ion (or a formula from which you can draw a Lewis structure), and asked to (1) name
Predicting the electron group geometry around one or more atoms in the structure, (2) draw a
Molecular geometric sketch of the structure, and/or (3) name the molecular geometry around one
Geometry or more of the atoms in the structure. You will find in later chapters that there are other
tip-offs for these tasks.
Objective 9
General Steps Follow these steps.
Web Step 1 To determine the name of the electron group geometry around each atom that
Molecules
is attached to two or more other atoms, count the number of electron groups
around each “central” atom and apply the guidelines found on Table 12.3. An
electron group can be a single bond, a lone pair, or a multiple bond. (Double
and triple bonds count as one group.)
Step 2 Use one or more of the geometric sketches shown on Table 12.3 as models for
the geometric sketch of your molecule. If the groups around the central atom
are identical, the bond angle is exact. If the groups are not identical, the angles
are approximate.
Step 3 To determine the name of the molecular geometry around each atom that
has two or more atoms attached to it, count the number of bond groups and
lone pairs, and then apply the guidelines found on Table 12.3. Single, double,
and triple bonds all count as one bond group. Notice that if all of the electron
groups attached to the atom are bond groups (no lone pairs), the name of the
molecular geometry is the same as the name of the electron group geometry.
Example See Examples 12.6 and 12.7.

Table 12.3 Electron Group and Molecular Geometry

e- e- group General Bond Bond Lone molecular
groups geometry geometric sketch angles groups pairs geometry

2 linear ¡ 180° 2 0 linear

3 trigonal planar ¡ 120° 3 0 trigonal planar

21 bent

4 tetrahedral ¡
109.5° 4 0 tetrahedral

31 trigonal
22 pyramid

bent

12.4 Molecular Geometry from Lewis Structures 473

Example 12.6 - Predicting Molecular Geometry Objective 9

Nitrosyl fluoride, NOF, is used as an oxidizer in rocket fuels.

'/0

Using its Lewis structure, (a) write the name of the electron group geometry around
the nitrogen atom, (b) draw a geometric sketch of the molecule, including bond angles,
and (c) write the name of the molecular geometry around the nitrogen atom.

Solution

1. Table 12.3 tells us that because there are three electron groups around the
nitrogen in nitrosyl fluoride (two bond groups and one lone pair), the electron
group geometry around the nitrogen atom is trigonal planar.

2. The geometric sketch for NOF is

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3. The nitrogen atom in NOF has one single bond, one double bond, and one
lone pair. According to Table 12.3, an atom with two bond groups and one
lone pair has the molecular geometry called bent.

Example 12.7 - Predicting Molecular Geometry Objective 9

Methyl cyanoacrylate is the main ingredient in Super Glue.

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Using the Lewis structure above, (a) write the name of the electron group geometry
around each atom that has two or more atoms attached to it, (b) draw a geometric
sketch of the molecule, including bond angles, and (c) write the name of the
molecular geometry around each atom that has two or more atoms attached to it.
(This example was chosen because it contains several different types of electron
groups and molecular shapes. The structures in the exercises and problems that
accompany this chapter are much simpler.)

474 Chapter 12 Molecular Structure

Solution FMFDUSPO HSPVQT

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b and c. There are many ways to sketch the molecule and still show the correct
geometry. The following is just one example. (Remember that your assigned exercises
will be simpler than this one.)

CPOE HSPVQT
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Exercise 12.3 - Molecular Geometry

Objective 9 For each of the Lewis structures given below, (a) write the name of the electron group
geometry around each atom that has two or more atoms attached to it, (b) draw a
geometric sketch of the molecule, including bond angles, and (c) write the name of the
molecular geometry around each atom that has two or more atoms attached to it.

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Chapter Glossary 475

Valence electrons The highest energy s and p electrons for an atom. Chapter
Isomers Compounds that have the same molecular formula but different molecular Glossary

structures.
Resonance structures Two or more Lewis structures for a single molecule or polyatomic

ion that differ in the positions of lone pairs and multiple bonds but not in the
positions of the atoms in the structure. It is as if the molecule or ion were able to
shift from one of these structures to another by shifting pairs of electrons from one
position to another.
Resonance The hypothetical switching from one resonance structure to another.
Resonance hybrid A structure that represents the average of the resonance structures
for a molecule or polyatomic ion.
Bond angle The angle between any three adjacent atoms in a molecule.
Trigonal pyramid The molecular geometry formed around an atom with three bonds
and one lone pair.
Electron group geometry A description of the arrangement of all the electron groups
around a central atom in a molecule or polyatomic ion, including the lone pairs.
Molecular geometry The description of the arrangement of all the atoms around a
central atom in a molecule or polyatomic ion. This description does not consider
lone pairs.
Bent The molecular geometry formed around an atom with two bond groups and two
lone pairs or two bond groups and one lone pair.
Trigonal planar (often called triangular planar) The geometric arrangement that keeps
three electron groups as far apart as possible. It leads to angles of 120° between the
groups.
Linear geometry The geometric arrangement that keeps two electron groups as far
apart as possible. It leads to angles of 180° between the groups.

You can test yourself on the glossary terms at the textbook’s Web site.

The goal of this chapter is to teach you to do the following. Chapter
Objectives
1. Define all of the terms in the Chapter Glossary.

Section 12.1 A New Look at Molecules and the Formation of Covalent Bonds

2. Describe the advantages and disadvantages of using models to describe physical
reality.

3. Describe the assumptions that lie at the core of the valence-bond model.
4. Write or identify the number of valence electrons in an atom of any

representative element.

476 Chapter 12 Molecular Structure

5. Use the valence-bond model to explain how (a) fluorine atoms form one
covalent bond and have three lone pairs in molecules such as F2, (b) hydrogen
atoms can form one covalent bond and have no lone pairs in molecules such as
HF, (c) carbon atoms can form four covalent bonds and have no lone pairs in
molecules such as CH4, (d) nitrogen, phosphorus, and arsenic atoms can form
three covalent bonds and have one lone pair in molecules such as NH3, PH3, and
AsH3, (e) nitrogen atoms can form four covalent bonds and have no lone pairs
in structures such as NH4+, (f ) oxygen, sulfur, and selenium atoms can form two
covalent bonds and have two lone pairs in molecules such as H2O, H2S, and
H2Se (g) oxygen atoms can form one covalent bond and have three lone pairs in
structures such as OH-, (h) carbon and oxygen atoms can form three covalent
bonds and have one lone pair in molecules such as CO, (i) boron atoms can form
three covalent bonds and have no lone pairs in molecules such as BF3, (j) halogen
atoms can form one covalent bond and have three lone pairs in molecules such as
HF, HCl, HBr, and HI.

6. Write electron-dot symbols for the representative elements.

Section 12.2 Drawing Lewis Structures
7. Given a formula for a molecule or polyatomic ion, draw a reasonable Lewis

structure for it.

Section 12.3 Resonance
8. Given a Lewis structure or enough information to draw one, predict whether

it is best described in terms of resonance, and if it is, draw all of the reasonable
resonance structures and the resonance hybrid for the structure.

Section 12.4 Molecular Geometry from Lewis Structures
9. Given a Lewis structure or enough information to write one, do the following.

a. Write the name of the electron group geometry around each atom that has
two or more atoms attached to it.

b. Draw a geometric sketch of the molecule, including bond angles (or
approximate bond angles).

c. Write the name of the molecular geometry around each atom that has two
or more atoms attached to it.

Review 1. Using the A-group convention, what is the group number of the column on the
Questions periodic table that includes the element chlorine, Cl?

2. Draw Lewis structures for CH4, NH3, and H2O.
3. Write the definition of orbital.
4. Write a complete electron configuration and an orbital diagram for each of the

following.
a. oxygen, O
b. phosphorus, P

Key Ideas 477

Complete the following statements by writing one of these words or phrases in each Key Ideas

blank.

109.5° most
A-group most common
anions never
as far apart oxygen
as if pair unpaired
cations polyatomic ions
change predict
eight rarely
electron group s and p
explain structural
fewest triple
four true
function two
highest-energy uncharged
lone pairs useful
molecular visualize
more

5. When developing a model of physical reality, scientists take what they think is
_____________ and simplify it enough to make it _____________.

6. Models have advantages and disadvantages. They help us to _____________,
_____________, and _____________ chemical changes, but we need to remind
ourselves now and then that they are only models and that as models, they have
their limitations.

7. One characteristic of models is that they _____________ with time.
8. The valence-bond model, which is commonly used to describe the formation

of covalent bonds, is based on the assumptions that only the _____________
electrons participate in bonding, and covalent bonds usually form to
_____________ electrons.
9. Valence electrons are the highest-energy _____________ electrons in an atom.
10. When the columns in the periodic table are numbered by the _____________
convention, the number of valence electrons in each atom of a representative
element is equal to the element’s group number in the periodic table.
11. Paired valence electrons are called _____________.
12. When forming four bonds to hydrogen atoms in a methane molecule, each
carbon atom behaves _____________ one electron is promoted from the 2s
orbital to the 2p orbital.
13. Carbon atoms frequently form double bonds, in which they share
_____________ electrons with another atom, often another carbon atom.
14. A(n) _____________ bond can be viewed as the sharing of six electrons between
two atoms.

478 Chapter 12 Molecular Structure

15. The shortcut for drawing Lewis structures for which we try to give each atom
its _____________ bonding pattern works well for many simple uncharged
molecules, but it does not work reliably for molecules that are more complex or
for _____________.

16. For _____________ molecules, the total number of valence electrons is the sum
of the valence electrons of each atom.

17. For polyatomic _____________, the total number of valence electrons is the
sum of the valence electrons for each atom minus the charge.

18. For polyatomic _____________, the total number of valence electrons is the
sum of the valence electrons for each atom plus the charge.

19. Hydrogen and fluorine atoms are _____________ in the center of a Lewis
structure.

20. Oxygen atoms are _____________ in the center of a Lewis structure.
21. The element with the _____________ atoms in the formula is often in the

center of a Lewis structure.
22. The atom that is capable of making the _____________ bonds is often in the

center of a Lewis structure.
23. Oxygen atoms rarely bond to other _____________ atoms.
24. In a reasonable Lewis structure, carbon, nitrogen, oxygen, and fluorine always

have _____________ electrons around them.
25. In a reasonable Lewis structure, hydrogen will always have a total of

_____________ electrons from its one bond.
26. Boron can have fewer than eight electrons around it in a reasonable Lewis

structure but never _____________ than eight.
27. Substances that have the same molecular formula but different _____________

formulas are called isomers.
28. The shapes of molecules play a major role in determining their _____________.
29. The most stable arrangement for electron groups is one in which they are

_____________ as possible.
30. The geometric arrangement in which four electron groups are as far apart as

possible is a tetrahedral arrangement, with bond angles of _____________.
31. In this book, we call the geometry that describes all of the electron groups,

including the lone pairs, the _____________ geometry. The shape that describes
the arrangement of the atoms only—treating the lone pairs as invisible—is called
the _____________ geometry.

Chapter Section 12.1 A New Look at Molecules and the Formation of Covalent Bonds
Problems
32. Describe the advantages and disadvantages of using models to describe physical
Objective 2
reality.
Objective 3
Objective 4 33. Describe the assumptions that lie at the core of the valence-bond model.

34. How many valence electrons do the atoms of each of the following elements

have? Write the electron configuration for these electrons. (For example, fluorine
has seven valence electrons, which can be described as 2s 2 2p 5.)

a. nitrogen, N b. sulfur, S c. iodine, I d. argon, Ar

Chapter Problems 479

35. How many valence electrons do the atoms of each of the following elements Objective 4
Objective 6
have? Write the electron configuration for these electrons. (For example, fluorine Objective 6
has seven valence electrons, which can be described as 2s 2 2p 5.)
Objective 5
a. oxygen, O b. boron, B c. neon, Ne d. phosphorus, P e. carbon, C Objective 5

36. Draw electron-dot symbols for each of the following elements.

a. nitrogen, N b. sulfur, S c. iodine, I d. argon, Ar

37. Draw electron-dot symbols for each of the following elements.

a. oxygen, O b. boron, B c. neon, Ne d. phosphorus, P e. carbon, C

38. To which group on the periodic table would atoms with the following electron-

dot symbols belong? List the group numbers using the 1-18 convention and

using the A-group convention.

a. 9 b. 9 c. 9

39. To which group on the periodic table would atoms with the following electron-
dot symbols belong? List the group numbers using the 1-18 convention and
using the A-group convention.

a. 9 b. 9 c. 9

40. For each of the following elements, sketch all of the ways mentioned in Section
12.1 that their atoms could look in a Lewis structure. For example, fluorine has

only one bonding pattern, and it looks like ' .

a. nitrogen, N b. boron, B c. carbon, C

41. For each of the following elements, sketch all of the ways mentioned in Section

12.1 that their atoms could look in a Lewis structure. For example, fluorine has

only one bonding pattern, and it looks like ' .

a. hydrogen, H b. oxygen, O c. chlorine, Cl

42. Use the valence-bond model to explain the following.

a. Fluorine atoms have one bond and three lone pairs in F2.
b. Carbon atoms have four bonds and no lone pairs in CH4.
c. Nitrogen atoms have three bonds and one lone pair in NH3.
d. Sulfur atoms have two bonds and two lone pairs in H2S.
e. Oxygen atoms have one bond and three lone pairs in OH-.

43. Use the valence-bond model to explain the following.

a. Phosphorus atoms have three bonds and one lone pair in PH3.
b. Nitrogen atoms have four bonds and no lone pairs in NH4+.
c. Oxygen atoms have two bonds and two lone pairs in H2O.
d. Boron atoms have three bonds and no lone pair in BF3.
e. Chlorine atoms have one bond and three lone pairs in HCl.

44. Based on your knowledge of the most common bonding patterns for the

nonmetallic elements, predict the formulas with the lowest subscripts for the

compounds that would form from the following pairs of elements. (For example,

hydrogen and oxygen can combine to form H2O and H2O2, but H2O has lower
subscripts.)

a. C and H b. S and H c. B and F

480 Chapter 12 Molecular Structure

45. Based on your knowledge of the most common bonding patterns for the
nonmetallic elements, predict the formulas with the lowest subscripts for the
compounds that would form from the following pairs of elements. (For example,
hydrogen and oxygen can combine to form H2O and H2O2, but H2O has lower
subscripts.)
a. P and I b. O and Br c. N and Cl

Section 12.2 Drawing Lewis Structures

46. Copy the following Lewis structure and identify the single bonds, the double
bond, and the lone pairs.
))

)$ $'

47. Copy the following Lewis structure and identify the single bonds, the triple
bond, and the lone pairs.

)

) $ $ $ #S

)

48. For each of the following molecular compounds, identify the atom that is most
likely to be found in the center of its Lewis structure. Explain why.
a. CBr4 b. SO2 c. H2S d. NOF
49. For each of the following molecular compounds, identify the atom that is most
likely to be found in the center of its Lewis structure. Explain why.
a. BI3 b. SO3 c. AsH3 d. HCN
50. Calculate the total number of valence electrons for each of the following
formulas.
a. HNO3 b. CH2CHF
51. Calculate the total number of valence electrons for each of the following
formulas.
a. H3PO4 b. HC2H3O2
52. Draw a reasonable Lewis structure for each of the following formulas.
Objective 7 a. CI4 b. O2F2 c. HC2F d. NH2Cl
Objective 7 e. PH3 f. S2F2 g. HNO2 h. N2F4
i. CH2CHCH3
Objective 8 53. Draw a reasonable Lewis structure for each of the following formulas.
a. H2S b. CHBr3 c. NF3 d. Br2O
e. H2CO3 f. H2S2 g. HOCl h. BBr3
i. CH3CH2CHCH2

Section 12.3 Resonance
54. Draw a reasonable Lewis structure for the ozone molecule, O3, using the skeleton

below. The structure is best described in terms of resonance, so draw all of its
reasonable resonance structures and the resonance hybrid that summarizes these
structures.

000

Chapter Problems 481

55. Draw a reasonable Lewis structure for a nitric acid, HNO3, using the skeleton Objective 8
below. The structure is best described in terms of resonance, so draw all of its
reasonable resonance structures and the resonance hybrid that summarizes these Objective 9
structures. Objective 9

0

)0 /0

Section 12.4 Molecular Geometry from Lewis Structures

56. Although both CO2 molecules and H2O molecules have three atoms, CO2
molecules are linear, and H2O molecules are bent. Why?

57. Although both BF3 molecules and NH3 molecules have four atoms, the BF3
molecules are planar, and NH3 molecules are pyramidal. Why?

58. Using the symbol X for the central atom and Y for the outer atoms, draw the
general geometric sketch for a three-atom molecule with linear geometry.

59. Using the symbol X for the central atom and Y for the outer atoms, draw the
general geometric sketch for a molecule with trigonal planar geometry.

60. Using the symbol X for the central atom and Y for the outer atoms, draw the
general geometric sketch for a molecule with tetrahedral geometry.

61. For each of the Lewis structures given below, (1) write the name of the electron
group geometry around each atom that has two or more atoms attached to it, (2)
draw a geometric sketch of the molecule, including bond angles (or approximate
bond angles), and (3) write the name of the molecular geometry around each
atom that has two or more atoms attached to it.
'*

a. ' $ ' b. 0 / $M c. #S # #S

'

#S 0

d. #S "T #S e. #S $ #S f. 0 $ 4

62. For each of the Lewis structures given below, (1) write the name of the electron
group geometry around each atom that has two or more atoms attached to it, (2)
draw a geometric sketch of the molecule, including bond angles (or approximate
bond angles), and (3) write the name of the molecular geometry around each
atom that has two or more atoms attached to it.

#S $M '

a. * # * b. $M "T $M c. $M $ $M

$M

d. 4F $ 4 e. ) 0 0 )

Chapter 13

Gases

t’s Monday morning, and Lilia is walking out of the chemistry building, thinking 13.1 Gases and Their
about the introductory lecture on gases that her instructor just presented. Dr. Properties
Scanlon challenged the class to try to visualize gases in terms of the model she
described, so Lilia looks at her hand and tries to picture the particles in the air 13.2 Ideal Gas
bombarding her skin at a rate of 1023 collisions per second. Lilia has high hopes that a Calculations
week of studying gases will provide her with answers to the questions her older brothers
and sisters posed to her the night before at a family dinner. 13.3 Equation
When Ted, who is a mechanic for a Formula One racing team, learned that Lilia Stoichiometry and
was going to be studying gases in her chemistry class, he asked her to find out how to Ideal Gases
calculate gas density. He knows that when the density of the air changes, he needs to
adjust the car’s brakes and other components to improve its safety and performance. 13.4 Dalton’s Law of
John, who is an environmental scientist, wanted to be reminded why balloons that Partial Pressures
carry his scientific instruments into the upper atmosphere expand as they rise. Amelia
is an artist who recently began to add neon lights to her work. After bending the tubes The gas particles in the air
into the desired shape, she fills them with gas from a high pressure cylinder. She wanted around us are constantly
to know how to determine the number of tubes she can fill with one cylinder. colliding with our skin.
Lilia’s sister Rebecca, the oldest, is a chemical engineer who could answer Ted’s,
John’s, and Amelia’s questions, but to give Lilia an opportunity to use her new
knowledge, she keeps quiet except to describe a gas-related issue of her own. Rebecca
is helping to design an apparatus in which two gases will react at high temperature,
and her responsibility is to equip the reaction vessel with a valve that will keep the
pressure from rising to dangerous levels. She started to explain to Lilia why increased
temperature leads to increased pressure, but when Lilia asked what gas pressure was
and what caused it, Rebecca realized that she had better save her explanation for the
next family dinner. Lilia (like yourself ) will learn about gas pressures and many other
gas related topics by reading Chapter 13 of her textbook carefully and listening closely
in lecture.

Review Skills

The presentation of information in this chapter assumes that you can already perform
the tasks listed below. You can test your readiness to proceed by answering the Review
Questions at the end of the chapter. This might also be a good time to read the Chapter
Objectives, which precede the Review Questions.

Describe the particle nature of gases. (Section substance or volume of a solution
2.1) containing a given molarity of one of the
Convert between temperatures in the Celsius substances. (Sections 10.1 and 10.3)
and Kelvin scales. (Section 8.6) Given an actual yield and a theoretical
Convert the amount of one substance in yield for a chemical reaction (or enough
a given reaction to the amount of another information to calculate a theoretical
substance in the same reaction, whether the yield), calculate the percent yield for the
amounts are described by mass of pure reaction. (Section 10.2)

483

484 Chapter 13 Gases

13.1 Gases and Their Properties

Objective 2 If you want to understand how gases behave—such as why fresh air rushes into your
Objective 3 lungs when certain chest muscles contract or how gases in a car’s engine move the
pistons and power the car—you need a clear mental image of the model chemists
use to explain the properties of gases and the relationships between them. The model
was introduced in Section 2.1, but we’ll be adding some new components to it in the
review presented here.

Gases consist of tiny particles widely spaced (Figure 13.1). Under typical conditions,
the average distance between gas particles is about ten times their diameter. Because
of these large distances, the volume occupied by the particles themselves is very small
compared to the volume of the empty space around them. For a gas at room temperature
and pressure, the gas particles themselves occupy about 0.1% of the total volume. The
other 99.9% of the total volume is empty space (whereas in liquids and solids, about
70% of the volume is occupied by particles). Because of the large distances between gas
particles, the attractions or repulsions among them are weak.

The particles in a gas are in rapid and continuous motion. For example, the average
velocity of nitrogen molecules, N2, at 20 °C is about 500 m/s. As the temperature
of a gas increases, the particles’ velocity increases. The average velocity of nitrogen
molecules at 100 °C is about 575 m/s.

The particles in a gas are constantly colliding with the walls of the container and
with each other. Because of these collisions, the gas particles are constantly changing
their direction of motion and their velocity. In a typical situation, a gas particle moves
a very short distance between collisions. For example, oxygen, O2, molecules at normal
temperatures and pressures move an average of 10-7 m between collisions.

Figure 13.1
Particles of a Gas

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PS SFQVMTJPO CFUXFFO
QBSUJDMFT
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JO EJSFDUJPO BOE WFMPDJUZ

13.1 Gases and Their Properties 485

Ideal Gases Objective 4

The model described above applies to real gases, but chemists often simplify the model Objective 5
further by imagining the behavior of an ideal gas. An ideal gas differs from a real gas Objective 6
in that Objective 7
Objective 8
The particles are assumed to be point masses, that is, particles that have a Objective 9
mass but occupy no volume.
There are no attractive or repulsive forces at all between the particles.
When we add these assumptions to our model for gases, we call it the ideal gas model.
As the name implies, the ideal gas model describes an “ideal” of gas behavior that
is only approximated by reality. Nevertheless, the model succeeds in explaining and
predicting the behavior of typical gases under typical conditions. In fact, some actual
gases do behave very much in accordance with the model, and scientists may call
them ideal gases. The ideal gas assumptions make it easier for chemists to describe the
relationships between the properties of gases and allow us to calculate values for these
properties.

Properties of Gases

The ideal gas model is used to predict changes in four related gas properties: volume,
number of particles, temperature, and pressure. Volumes of gases are usually described
in liters, L, or cubic meters, m3, and numbers of particles are usually described in
moles, mol. Although gas temperatures are often measured with thermometers that
report temperatures in degrees Celsius, °C, scientists generally use Kelvin temperatures
for calculations. Remember that you can convert between degrees Celsius, °C, and
kelvins, K, using the following equations.

? K = °C + 273.15

? °C = K - 273.15

To understand gas pressure, picture a typical gas in a closed container. Each time a
gas particle collides with and ricochets off one of the walls of its container, it exerts a
force against the wall. The sum of the forces of these ongoing collisions of gas particles
against all the container’s interior walls creates a continuous pressure upon those walls.
Pressure is force divided by area.

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"SFB

(BT QSFTTVSF 'PSDF EVF UP QBSUJDMF DPMMJTJPOT XJUI UIF XBMMT
"SFB PG UIF XBMMT

The accepted SI unit for gas pressure is the pascal, Pa. A pascal is a very small amount of
pressure, so the kilopascal, kPa, is more commonly used. Other units used to describe
gas pressure are the atmosphere (atm), torr, millimeter of mercury (mmHg), and bar.
The relationships between these pressure units are

1 atm = 101,325 Pa = 101.325 kPa = 760 mmHg = 760 torr

1 bar = 100 kPa = 0.9869 atm = 750.1 mmHg

The numbers in these relationships come from definitions, so they are all exact. At sea
level on a typical day, the atmospheric pressure is about 101 kPa, or about 1 atm.

In calculations, the variables P, T, V, and n are commonly used to represent pressure,
temperature, volume, and moles of gas.

486 Chapter 13 Gases

Discovering the Relationships Between Properties

If we want to explain why a weather balloon carrying instruments into the upper
atmosphere expands as it rises, we need to consider changes in the properties of the
gases (pressure, volume, temperature, or number of gas particles) inside and outside
the balloon. For example, as the balloon rises, the pressure outside of it, called the
atmospheric pressure, decreases. But, there are also variations in temperature, and the
balloon might have small leaks that change the number of gas particles it contains.

In a real situation, pressure, temperature, and number of gas particles may all be
changing, and predicting the effect of such a blend of changing properties on gas
volume is tricky. Therefore, before we tackle predictions for real world situations, such
as the weather balloon, we will consider simpler systems in which two of the four gas
properties are held constant, a third property is varied, and the effect of this variation on
the fourth property is observed. For example, it is easier to understand the relationship
between volume and pressure if the number of gas particles and temperature are held
constant. The volume can be varied, and the effect this has on the pressure can be
measured. An understanding of the relationships between gas properties in controlled
situations will help us to explain and predict the effects of changing gas properties in
more complicated, real situations.

Figure 13.2 shows a laboratory apparatus that can be used to demonstrate all the
relationships we are going to be discussing. It consists of a cylinder with a movable
piston, a thermometer, a pressure gauge, and a valve through which gas may be added
to the cylinder’s chamber or removed from it.

Figure 13.2
Apparatus Used to Demonstrate Relationships
Between the Properties of Gases

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BOE SFNPWF HBT
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Objective 10a The Relationship Between Volume and Pressure
Objective 10a
Figure 13.3 shows how our demonstration apparatus would be used to determine
the relationship between gas volume and pressure. While holding the number of gas
particles constant (by closing the valve) and holding the temperature constant (by
allowing heat to transfer in or out so that the apparatus remains the same temperature
as the surrounding environment), we move the piston to change the volume, and then
we observe the change in pressure. When we decrease the gas volume, the pressure
gauge on our system shows us that the gas pressure increases. When we increase the gas
volume, the gauge shows that the pressure goes down.

Decreased volume → Increased pressure
Increased volume → Decreased pressure

13.1 Gases and Their Properties 487

For an ideal gas (in which the particles occupy no volume and experience no attractions
or repulsions), gas pressure and volume are inversely proportional. This means that
if the temperature and the number of gas particles are constant and if the volume
is decreased to one-half its original value, the pressure of the gas will double. If the
volume is doubled, the pressure decreases to one-half its original value. The following
expression summarizes this inverse relationship:

1 t

488 Chapter 13 Gases

Objective 10b The Relationship Between Pressure and Temperature
Objective 10b
In order to examine the relationship between pressure and temperature, we must adjust
Objective 10b our demonstration apparatus so that the other two properties (number of gas particles
and volume) are held constant. This can be done by locking the piston so it cannot
move and closing the valve tightly so that no gas leaks in or out (Figure 13.4). When
the temperature of a gas trapped inside the chamber is increased, the measured pressure
increases. When the temperature is decreased, the pressure decreases.

Increased temperature → Increased pressure
Decreased temperature → Decreased pressure
We can explain the relationship between temperature and pressure using our model
for gas. Increased temperature means increased motion of the particles. If the particles
are moving faster in the container, they will collide with the walls more often and with
greater force per collision. This leads to a greater overall force pushing on the walls and
to a greater force per unit area or pressure (Figure 13.4).
If the gas is behaving like an ideal gas, a doubling of the Kelvin temperature doubles
the pressure. If the temperature decreases to 50% of the original Kelvin temperature,
the pressure decreases to 50% of the original pressure. This relationship can be expressed
by saying that the pressure of an ideal gas is directly proportional to the Kelvin temperature
of the gas if the volume and the number of gas particles are constant. This relationship is
sometimes called Gay-Lussac’s Law.
P ∝ T if n and V are constant

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Relationship Between JO QPTJUJPO PG HBT QBSUJDMFT
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Increased temperature leads )FBU BEEFE $POTUBOU WPMVNF
to increased pressure if the
number of gas particles and *ODSFBTFE UFNQFSBUVSF
volume are constant. *ODSFBTFE QSFTTVSF

Objective 10b *ODSFBTFE UFNQFSBUVSF

You can see an *ODSFBTFE BWFSBHF WFMPDJUZ PG UIF HBT QBSUJDMFT
animation that
demonstrates *ODSFBTFE OVNCFS PG *ODSFBTFE GPSDF QFS DPMMJTJPO
this relationship DPMMJTJPOT XJUI UIF XBMMT
at the textbook’s
Web site.

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*ODSFBTFE GPSDF EVF UP DPMMJTJPOT
BSFB PG XBMM

*ODSFBTFE HBT QSFTTVSF

13.1 Gases and Their Properties 489

The Relationship Between Volume and Temperature Objective 10c

Consider the system shown in Figure 13.5. To demonstrate the relationship between Placing a balloon in
temperature and volume of gas, we keep the number of gas particles and gas pressure liquid nitrogen lowers
constant. If our valve is closed and if our system has no leaks, the number of particles the temperature of
is constant. We keep the gas pressure constant by allowing the piston to move freely the gas and causes
throughout our experiment, because then it will adjust to keep the pressure pushing an initial decrease in
on it from the inside equal to the constant external pressure pushing on it due to the pressure. With its internal
weight of the piston and the atmospheric pressure. The atmospheric pressure is the pressure now lower
pressure in the air outside the container, which acts on the top of the piston due to the than the pressure of the
force of collisions between particles in the air and the top of the piston. air outside, the balloon
shrinks to a much smaller
If we increase the temperature, the piston in our apparatus moves up, increasing the volume.
volume occupied by the gas. A decrease in temperature leads to a decrease in volume.
The University of Michigan
Increased temperature → Increased volume Department of Physics Lecture
Decreased temperature → Decreased volume Demonstration Lab
The increase in temperature of the gas leads to an increase in the average velocity of
the gas particles, which leads in turn to more collisions with the walls of the container
and a greater force per collision. This greater force acting on the walls of the container
leads to an initial increase in the gas pressure. Thus the increased temperature of our
gas creates an internal pressure, acting on the bottom of the piston, that is greater
than the external pressure. The greater internal pressure causes the piston to move up,
increasing the volume of the chamber. The increased volume leads to a decrease in gas
pressure in the container, until the internal pressure is once again equal to the constant
external pressure (Figure 13.5). Similar reasoning can be used to explain why decreased
temperature leads to decreased volume when the number of gas particles and pressure
are held constant.
For an ideal gas, volume and temperature described in kelvins are directly proportional
if the number of gas particles and pressure are constant. This is called Charles’ Law.
V ∝ T if n and P are constant

1JTUPO GSFF UP NPWF $POTUBOU OVNCFS Figure 13.5
)FBU BEEFE PG HBT QBSUJDMFT Relationship Between Temperature and
Volume Increased temperature leads to
*ODSFBTFE increased volume if the number of gas
WPMVNF particles and pressure are constant.
*ODSFBTFE
UFNQFSBUVSF Objective 10c

$POTUBOU You can see an animation that
QSFTTVSF demonstrates this relationship at
the textbook’s Web site.

*ODSFBTFE UFNQFSBUVSF *ODSFBTFE BWFSBHF WFMPDJUZ PG UIF HBT QBSUJDMFT
*ODSFBTFE OVNCFS PG DPMMJTJPOT XJUI UIF XBMMT *ODSFBTFE GPSDF QFS DPMMJTJPO

*OJUJBM JODSFBTF JO GPSDF QFS BSFBĹUIBU JT
JO QSFTTVSF

*ODSFBTFE WPMVNF *OTJEF QSFTTVSF JT HSFBUFS UIBO FYUFSOBM QSFTTVSF
$POUBJOFS FYQBOET %FDSFBTFE QSFTTVSF VOUJM JOTJEF QSFTTVSF FRVBMT FYUFSOBM QSFTTVSF

490 Chapter 13 Gases

Objective 10d The Relationship Between Number of Gas Particles and Pressure
Objective 10d
Objective 10d To explore the relationship between gas pressure and the number of gas particles, we
could set up our experimental system as shown in Figure 13.6. The volume is held
constant by locking the piston so it cannot move. The temperature is kept constant by
allowing heat to flow in or out of the cylinder in order to keep the temperature of the
gas in the cylinder equal to the external temperature. When the number of gas particles
is increased by adding gas through the valve on the left of the cylinder, the pressure
gauge shows an increase in pressure. When gas is allowed to escape from the valve, the
decrease in the number of gas particles causes a decrease in the pressure of the gas.

Increased number of gas particles → Increased pressure

Decreased number of gas particles → Decreased pressure

The increase in the number of gas particles in the container leads to an increase in the
number of collisions with the walls per unit time. This leads to an increase in the force
per unit area–that is, to an increase in gas pressure.

If the temperature and the volume of an ideal gas are held constant, the number of
gas particles in a container and the gas pressure are directly proportional.

P ∝ n if T and V are constant

Figure 13.6
Relationship Between Number of Gas Particles and Pressure

Increased number of gas particles leads to increased pressure if the temperature and volume are constant.

Objective 10d 1JTUPO MPDLFE (BT BEEFE $POTUBOU
JO QPTJUJPO WPMVNF
You can see an $POTUBOU
animation that UFNQFSBUVSF
demonstrates
this relationship *ODSFBTFE
at the textbook’s QSFTTVSF
Web site.
*ODSFBTFE OVNCFS PG HBT QBSUJDMFT
*ODSFBTFE OVNCFS PG DPMMJTJPOT XJUI UIF XBMMT

*ODSFBTFE UPUBM GPSDF PG DPMMJTJPOT
*ODSFBTFE HBT QSFTTVSF

13.1 Gases and Their Properties 491

The Relationship Between Number of Gas Particles and Volume

Figure 13.7 shows how the relationship between the number of gas particles and volume Objective 10e
can be demonstrated using our apparatus. Temperature is held constant by allowing
heat to move into or out of the system, thus keeping the internal temperature equal
to the constant external temperature. Pressure is held constant by allowing the piston
to move freely to keep the internal pressure equal to the external pressure. When we
increase the number of gas particles in the cylinder by adding gas through the valve on
the left of the apparatus, the piston rises, increasing the volume available to the gas. If
the gas is allowed to escape from the valve, the volume decreases again.

Increased number of gas particles → Increased volume
Decreased number of gas particles → Decreased volume

The explanation for why an increase in the number of gas particles increases When this person blows more air
volume starts with the recognition that the increase in the number of gas into the balloon, the increased
particles results in more collisions per second against the walls of the container. number of gas particles initially
The greater force due to these collisions creates an initial increase in the force leads to an increased pressure.
per unit area—or gas pressure—acting on the walls. This will cause the piston Because the internal pressure is
to rise, increasing the gas volume and decreasing the pressure until the internal now greater than the pressure
and external pressure are once again equal (Figure 13.7). Take a minute or two of the air outside the balloon,
to work out a similar series of steps to explain why decreased number of gas the balloon expands to a larger
particles leads to decreased volume. volume.

The relationship between moles of an ideal gas and volume is summarized
by Avogadro’s Law, which states that the volume and the number of gas particles
are directly proportional if the temperature and pressure are constant.

V ∝ n if T and P are constant

1JTUPO GSFF UP NPWF "U DPOTUBOU QSFTTVSF Figure 13.7
(BT BEEFE Relationship Between
*ODSFBTFE WPMVNF Number of Gas Particles
$POTUBOU and Volume
UFNQFSBUVSF Increased number of gas
particles leads to increased
volume if the temperature
and pressure are constant.

Objective 10e

*ODSFBTFE OVNCFS PG HBT QBSUJDMFT You can see an
animation that
*ODSFBTFE OVNCFS PG DPMMJTJPOT XJUI UIF XBMMT demonstrates
*ODSFBTFE UPUBM GPSDF PG DPMMJTJPOT this relationship
at the textbook’s
*OJUJBM JODSFBTFE JO GPSDF QFS BSFB UIBU JT
JO QSFTTVSF Web site.

*OTJEF QSFTTVSF JT HSFBUFS UIBO FYUFSOBM QSFTTVSF

$POUBJOFS FYQBOET *ODSFBTFE WPMVNF

%FDSFBTFE QSFTTVSF VOUJM UIF JOTJEF QSFTTVSF FRVBMT UIF FYUFSOBM QSFTTVSF

492 Chapter 13 Gases

Objective 11 Gases and the Internal Combustion Engine

Now, let’s apply our model for gases to a real situation. Because the changes that take
place in a typical car engine illustrate many of the characteristics of gases, let’s take a
look at how the internal combustion engine works (Figure 13.8).

Liquid gasoline is a mixture of hydrocarbons, with from five to twelve carbon atoms
in each molecule. It evaporates to form a gas, which is mixed with air and injected into
the engine’s cylinders. (See cylinder 1 in Figure 13.8.) The movement of the engine’s
pistons turns a crankshaft that causes the piston in one of the cylinders containing the
gasoline-air mixture to move up, compressing the gases and increasing the pressure of
the gas mixture in the cylinder. (See cylinder 2 in Figure 13.8.)

A spark ignites the mixture of compressed gases in a cylinder, and the hydrocarbon
compounds in the gasoline react with the oxygen in the air to form carbon dioxide gas
and water vapor. (See cylinder 3 in Figure 13.8.) A typical reaction is

$ ) H
0 H
m $0 H
) 0 H
&OFSHZ

2 mol + 25 mol = 27 mol 16 mol + 18 mol = 34 mol

Objective 11 In this representative reaction, a total of 27 moles of gas are converted into 34 moles of
Objective 11 gas. The increase in moles of gas leads to an increase in number of collisions per second
with the walls of the cylinder, which creates greater force acting against the walls and a
greater gas pressure in the cylinder. The pressure is increased even more by the increase
in the temperature of the gas due to the energy released in the reaction. The increased
temperature increases the average velocity of the gas particles, which leads to more
frequent collisions with the walls and a greater average force per collision.

The increased pressure of the gas pushes the piston down. (See cylinder 4 in Figure
13.8.) This movement of the pistons turns the crankshaft, which, through a series of
mechanical connections, turns the wheels of the car.

ɨF DPNCVTUJPO PG UIF HBTPMJOF MFBET UP BO JODSFBTF JO
NPMFT PG HBT
XIJDI BMTP DBVTFT UIF HBT QSFTTVSF UP JODSFBTF

%FDSFBTFE WPMVNF MFBET UP ɨF SFBDUJPO JT FYPUIFSNJD
TP UIF UFNQFSBUVSF
JODSFBTFE HBT QSFTTVSF JOTJEF PG UIF QSPEVDU HBTFT JODSFBTFT
DPOUSJCVUJOH UP
UIF DZMJOEFS UIF JODSFBTFE HBT QSFTTVSF

(BTFPVT HBTPMJOF ɨF JODSFBTFE QSFTTVSF QVTIFT UIF QJTUPO
NJYFE XJUI BJS EPXO
UVSOJOH UIF DSBOLTIBGU BOE
NPWFT JOUP UIF VMUJNBUFMZ UIF DBShT XIFFMT
DZMJOEFS

$ZMJOEFS

Objective 11 1JTUPO

Figure 13.8 $SBOLTIBGU
Gases and the Internal

Combustion Engine

13.1 Gases and Their Properties 493

Explanations for Other Real World Situations Objective 12

The relationships between gas properties can be used to explain how we breathe in
and out. When the muscles of your diaphragm contract, your chest expands, and the
volume of your lungs increases. This change leads to a decrease in the number of
particles per unit volume inside the lungs, leaving fewer particles near any give area
of the inner surface of the lungs. Fewer particles means fewer collisions per second
per unit area of lungs and a decrease in force per unit area, or gas pressure. During
quiet, normal breathing, this increase in volume decreases the pressure in the lungs to
about 0.4 kilopascals lower than the atmospheric pressure. As a result, air moves into
the lungs faster than it moves out, bringing in fresh oxygen. When the muscles relax,
the lungs return to their original, smaller volume, causing the pressure in the lungs to
increase to about 0.4 kilopascals above atmospheric pressure. Air now goes out of the
lungs faster than it comes in (Figure 13.9).

Objective 12 Figure 13.9
The Mechanics of Breathing

&YUFSOBM
1SFTTVSF L1B
L1B L1B L1B L1B

%JBQISBHN ɨF EJBQISBHN DPOUSBDUT
BOE UIF DIFTU ɨF EJBQISBHN SFMBYFT BOE UIF DIFTU

FYQBOET
DBVTJOH UIF MVOHT UP FYQBOE SFUVSOT UP JUT PSJHJOBM WPMVNF
DBVTJOH
ɨF JODSFBTFE WPMVNF EFDSFBTFT UIF UIF MVOH WPMVNF UP EFDSFBTF
QSFTTVSF JO UIF MVOHT UP CFMPX UIF ɨJT JODSFBTFT UIF QSFTTVSF JO UIF MVOHT

FYUFSOBM QSFTTVSF
DBVTJOH BJS UP NPWF DBVTJOH BJS UP NPWF PVU PG UIF MVOHT
JOUP UIF MVOHT GBTUFS UIBO JU NPWFT PVU GBTUFS UIBO JU NPWFT JO

Let’s return to the gas-related issues discussed at Lilia’s family dinner. When Lilia’s Objective 13
brother John adds helium gas to one of his instrument-carrying balloons, its
volume increases. Increased number of gas particles leads to increased gas pressure,
making the internal pressure greater than the external pressure acting on the outside
surface of the balloon. The increase of internal pressure leads to an expansion of the
balloon. When the balloon is released and rises into the air, the concentration of
gases outside the balloon decreases with increased distance from the earth, leading
to a decrease in the atmospheric pressure acting on the outside of the balloon. The
greater pressure inside the balloon causes the balloon to expand as it rises. This
increase in volume is diminished somewhat by the slight loss of gas from tiny holes
in the balloon and by the general decrease in temperature with increased distance
from the earth.
When the temperature is increased for the gases in the reaction vessel that Lilia’s The volume of this balloon
will increase as it moves up
sister Rebecca is helping design, the gas pressure rises. This is due to an increase in into the atmosphere.

494 Chapter 13 Gases

the average velocity of the particles and therefore an increase in the rate of collisions
of gas particles with the constant area of the walls and an increase in the average force
per collision. Rebecca’s pressure release valve allows gas to escape if the pressure gets to
a certain level. The decrease in the number of gas particles when gas escapes from the
valve keeps the pressure below dangerous levels.

To answer Ted’s question about how to calculate gas density and to see how Amelia
can estimate the number of tubes she can fill for her neon light sculptures, we need to
continue on to the next section.

13.2 Ideal Gas Calculations

This section shows how to do calculations such as those necessary to answer Ted’s and
Amelia’s questions about gas density and volume, and in addition, it considers some of
the gas-related issues that Lilia’s sister Rebecca and her co-workers need to resolve. For
one thing, the design team needs to know the amount of gas that they can safely add to
their reaction vessel, and then Rebecca needs to determine the maximum temperature
at which the reaction can be run without causing the pressure of that amount of gas to
reach dangerous levels.

All these calculations, and others like them, can be done with the aid of two useful
equations that we will now derive from the relationships described in Section 13.1.

Calculations Using the Ideal Gas Equation

We discovered in Section 13.1 that pressure of an ideal gas is directly proportional to
the number of gas particles (expressed in moles), directly proportional to temperature,
and inversely proportional to the volume of the container.

1 t O JG 5 BOE 7 BSF DPOTUBOU
1 t 5 JG O BOE 7 BSF DPOTUBOU
1 t 7 JG O BOE 5 BSF DPOTUBOU
These three relationships can be summarized in a single equation:

1 t O5
7

Another way to express the same relationship is

1 B DPOTUBOU
O75

The constant in this equation is the same for all ideal gases. It is called the universal
gas constant and is expressed with the symbol R. The value of R depends on the units
of measure one wishes to use in a given calculation. Two choices are given below,
showing R for different pressure units (atmospheres, atm, and kilopascals, kPa).

Objective 14 3 - BUN PS - L1B
, NPM , NPM

13.2 Ideal Gas Calculations 495

Substituting R for “a constant” and rearranging the equation yields the ideal gas Objective 15
equation (often called the ideal gas law) in the form that is most commonly memorized Objective 16
and written:

PV = nRT
Because we will often be interested in the masses of gas samples, it is useful to remember
an expanded form of the ideal gas equation that uses mass in grams (g) divided by
molar mass (M ) instead of moles (n).

O NPMFT HH SS BB NN TT N N B TP T M BJ O S N H S B B TN T T . H
NPMF

17 . H 35

The following sample study sheet describes how calculations can be done using the
two forms of the ideal gas equation.

Tip-off The usual tip-off that you can use the ideal gas equation to answer a question Sample Study
is that you are given three properties of a sample of gas and asked to calculate the Sheet 13.1
fourth. A more general tip-off is that only one gas is mentioned and there are no Using the Ideal
changing properties. Gas Equation

General Steps Follow these steps. Objective 15
Objective 16
Step 1 Assign variables to the values given and the value that is unknown. Use P for
pressure, V for volume, n for moles, T for temperature, g for mass, and M for
molar mass.

Step 2 Write the appropriate form of the ideal gas equation.
If the number of particles in moles is given or desired, use the most
common form of the ideal gas equation.

17 O35 3 - BUN PS - L1B
, NPM , NPM

If mass or molar mass is given or desired, use the expanded form of the ideal
gas equation.

17 . H 35 H NBTT . NPMBS NBTT

Step 3 Rearrange the equation to isolate the unknown.

Step 4 Plug in the known values, including units. Be sure to use Kelvin
temperatures.

Step 5 Make any necessary unit conversions and cancel your units.

Step 6 Calculate your answer and report it to the correct significant figures and with
the correct unit.

Example See Examples 13.1 and 13.2.

496 Chapter 13 Gases

Objective 15 Example 13.1 - Using the Ideal Gas Equation

Incandescent light bulbs “burn out” because their tungsten filament evaporates,
weakening the thin wire until it breaks. Argon gas is added inside the bulbs to reduce
the rate of evaporation. (Argon is chosen because, as a noble gas, it will not react with
the components of the bulb, and because it is easy to obtain in significant quantities.
It is the third most abundant element in air.) What is the pressure in atmospheres of
3.4 × 10-3 moles of argon gas in a 75-mL incandescent light bulb at 20 °C?

Solution
We are given three of the four properties of ideal gases (moles, volume, and temperature),
and we are asked to calculate the fourth (pressure). Therefore, we use the ideal gas
equation to solve this problem.

Step 1 We assign variables to the values that we are given and to the unknown value.
Remember to use Kelvin temperatures in gas calculations.

P = ? n = 3.4 × 10-3 mol V = 75 mL T = 20 °C + 273.15 = 293 K

Step 2 We pick the appropriate form of the ideal gas equation. Because moles of gas
are mentioned in the problem, we use

PV = nRT

Step 3 We rearrange our equation to isolate the unknown property.

1 O35
7

Step 4 We plug in the values that are given, including their units. We use the value
for R that contains the unit we want to calculate, atm.

s
NPM -

13.2 Ideal Gas Calculations 497

When the team of chemists and chemical engineers meets to design the new process,
they begin with some preliminary calculations. First, they want to determine the
amount of gas that they can safely add to their reaction vessel. They know that the
optimum conditions for this reaction are a temperature of about 825 °C and a pressure
of about 700 kPa, and they have been told by the plant architect that the maximum
volume of the reaction vessel will be 2500 m3. Notice that the value for R that contains
the unit kPa is used.

PV = nRT

O 1 7 L1B N
- s NPM
35 N
-

498 Chapter 13 Gases

Objective 17 Example 13.3 - Using the Ideal Gas Equation

In the incandescent light bulb described in Example 13.1, what is the density of argon
gas at 80.8 °C and 131 kPa?

Solution

Because only one gas is mentioned and because there are no changing properties, we
recognize this as an ideal gas equation problem.

Density is mass divided by volume. Using variables from the expanded form of the
ideal gas equation, it can be expressed as g/V.

H =? P = 131 kPa T = 80.8 °C + 273.15 = 354.0 K
7

17 . H 35

To save a little time, we can use the value for R that contains the pressure unit given in
the problem, kPa.

H 1. L1B H H -
7 35 NPM

-

13.2 Ideal Gas Calculations 499

If such a change is anticipated, car designers and mechanics, such as Lilia’s brother Ted,
can make adjustments. For example, because of its altitude, Denver has about 15%
lower air density than does a city at sea level. Therefore, Indy cars that race in Denver
are fitted with larger-than-normal brakes to compensate for the lower air resistance
and greater difficulty stopping. Ted can do calculations similar to Example 13.3 to
determine the density of air on different days and at different racetracks. An example
is the following set of equations calculating the density of air at 762 mmHg and
32 °C and at 758 mmHg and 4 °C. (The average molar mass for the gases in air is 29
g/mol.)

17 . H 35

NN)H H Objective 17
NPM
H 1. BUN H -
7 35 -

500 Chapter 13 Gases

When Properties Change

Another useful equation, derived from the ideal gas equation, can be used to calculate
changes in the properties of a gas. In the first step of the derivation, the ideal gas
equation is rearranged to isolate R.

17 3
O5

In this form, the equation shows that no matter how the pressure, volume, moles of
gas, or temperature of an ideal gas may change, the other properties will adjust so that
the ratio of PV /nT always remains the same. In the equations below, P1, V1, n1, and
T1 are the initial pressure, volume, moles of gas, and temperature, and P2, V2, n2, and
T2 are the final pressure, volume, moles of gas, and temperature.

1 7 O 35 TP 1 7 3
O 5

1 7 O 35 TP 1 7 3
O 5

Therefore,

Objective 18 1 7 1 7
O 5 O 5

The equation above is often called the combined gas law equation. It can be used to do
calculations such as Example 13.4 and Exercise 13.2, which follow the sample study
sheet below.

Sample Study Tip-off – The problem requires calculating a value for a gas property that has
Sheet 13.2
Using the changed. In other words, you are asked to calculate a new pressure, temperature,
Combined Gas moles (or mass), or volume of gas given sufficient information about the initial and
Law Equation other final properties.

Objective 18 General Steps

Step 1 Assign the variables P, T, n, and V to the values you are given and to
the unknown value. Use the subscripts 1 and 2 to show initial or final
conditions.

Step 2 Write out the combined gas law equation, but eliminate the variables for any
constant properties. (You can assume that the properties not mentioned in the
problem remain constant.)

Step 3 Rearrange the equation to isolate the unknown property.
Step 4 Plug in the values for the given properties.
Step 5 Make any necessary unit conversions and cancel your units.
Step 6 Calculate your answer and report it with the correct units and with the

correct significant figures.

Example See Example 13.4.

13.2 Ideal Gas Calculations 501

Example 13.4 - Using the Combined Gas Law Equation

Neon gas in luminous tubes radiates red light—the original “neon light.” The standard Objective 18
gas containers used to fill the tubes have a volume of 1.0 L and store neon gas at a Neon tubes
pressure of 101 kPa at 22 °C. A typical luminous neon tube contains enough neon gas
to exert a pressure of 1.3 kPa at 19 °C. If all the gas from a standard container is allowed
to expand until it exerts a pressure of 1.3 kPa at 19 °C, what will its final volume be?
If Lilia’s sister Amelia is adding this gas to luminous tubes that have an average volume
of 500 mL, what is the approximate number of tubes she can fill?

Solution

We recognize this as a combined gas law problem because it requires calculating a value
for a gas property that has changed. In this case, that property is volume.

Step 1 We assign variables to the given values and to the unknown.

V1 = 1.0 L P1 = 101 kPa T1 = 22 °C + 273.15 = 295 K

V2 = ? P2 = 1.3 kPa T2 = 19 °C + 273.15 = 292 K

Step 2 We write the combined gas law equation, eliminating variables for properties
that are constant. Because moles of gas are not mentioned, we assume that
they are constant (n1 = n2).

1 7 1 7 becomes 1 7 1 7
O 5 O 5 5 5

Steps 3 and 4 We rearrange the equation to solve for our unknown and plug in the
given values.

7 7 5 1 - , L1B
5 1 , L1B

Step 5 Our units cancel to yield liters, which is a reasonable volume unit, so we do
not need to make any unit conversions.

Step 6 We finish the problem by calculating our answer and reporting it with the
correct significant figures.

7 7 5 1 - , L1B -
5 1 , L1B

We can now answer Amelia’s question about how many neon tubes she can expect to
fill from each neon cylinder.

UVCFT - N- UVCF s UVCFT
PS BCPVU UVCFT
- N-

Let’s look at another gas-related issue that Lilia’s sister Rebecca needs to consider in
designing the pressure valve for the reaction vessel at her chemical plant. She knows
that, for safety reasons, the overall pressure must be kept below 1000 kPa, and she
knows that the most likely cause of increased pressure is increased temperature. To
get an idea of how high the temperature can go safely, she could use the combined gas
law equation to calculate the temperature at which the pressure of the gas will reach
1000 kPa if the initial temperature was 825 °C (1098 K) and the initial pressure was

502 Chapter 13 Gases

700.0 kPa. The reaction vessel has a constant volume, and we can assume that it has no
leaks. Therefore, the volume and moles of gas remain constant.

1 7 1 7 PS 1 1 CFDBVTF 7 7 BOE O O
O 5 O 5 5 5

5 5 1 , L1B , PS o$
1 L1B

Although Rebecca designs her pressure release valve to allow gas to escape when the
pressure reaches 1000 kPa, she hopes it will never need to be used. To avoid the release
of gases, she informs the rest of her team that they should be careful to design the
process to keep the temperature well below 1295 °C.

Objective 18 Exercise 13.2 - Using the Combined Gas Law Equation

A helium weather balloon is filled in Monterey, California, on a day when the
atmospheric pressure is 102 kPa and the temperature is 18 °C. Its volume under
these conditions is 1.6 × 104 L. Upon being released, it rises to an altitude where the
temperature is -8.6 °C, and its volume increases to 4.7 × 104 L. Assuming that the
internal pressure of the balloon equals the atmospheric pressure, what is the pressure
at this altitude?

13.3 Equation Stoichiometry and Ideal Gases

Chemists commonly want to convert from the amount of one substance in a chemical
reaction to an amount of another substance in that reaction. For example, they might
want to calculate the maximum amount of a product that can be formed from a certain
amount of a reactant, or they might want to calculate the minimum amount of one
reactant that would be necessary to use up a known amount of another reactant.
Chapter 10 showed how to do conversions like these. In Example 10.1, we calculated
the minimum mass of water in kilograms necessary to react with 2.50 × 104 kg of
tetraphosphorus decoxide, P4O10, in the reaction

P4O10(s) + 6H2O(l ) → 4H3PO4(aq)
This kind of problem, an equation stoichiometry problem, is generally solved in the
following steps:

Measurable property of 1 → moles 1 → moles 2 → measurable property of 2
For substances that are pure solids and liquids, such as P4O10 and H2O, mass is an
easily measured property, so the specific steps used in solving Example 10.1 were

Mass P4O10 → moles P4O10 → moles H2O → mass H2O
Amounts of gas, however, are more commonly described in terms of volume at a
particular temperature and pressure. (It is far easier to measure a gas’s volume than
to measure its mass.) Therefore, to solve equation stoichiometry problems involving
gases, we must be able to convert between the volume of a gaseous substance and
the corresponding number of moles. For example, if both substances involved in the
equation stoichiometry problem are gases, the general steps would be

Volume of gas 1 → moles 1 → moles 2 → volume of gas 2

13.2 Equation Stoichiometry and Ideal Gases 503

There are several ways to convert between moles of a gaseous substance and its
volume. One approach is to use as a conversion factor the molar volume at STP. STP
stands for “standard temperature and pressure.” Standard temperature is 0 °C, or 273.15
K, and standard pressure is 1 atm, or 101.325 kPa, or 760 mmHg. The molar volume, or
liters per mole of an ideal gas, at STP can be calculated from the ideal gas equation.

PV = nRT

7 35 ,

504 Chapter 13 Gases

the volume of a mole of gas will also change, so 22.414 L/mole cannot be used as a
conversion factor for conditions other than STP. One way to convert between volume
of gas and moles of gas at temperatures and pressures other than 0 °C and 1 atm is to
use the ideal gas equation. We convert from volume to moles by solving the ideal gas
equation for n and plugging in the given values for V, T, and P :

17 O35 MFBET UP O 17
35
We convert from moles to volume by solving the ideal gas equation for V and plugging
in the given values for n, T, and P :

17 O35 MFBET UP 7 O35
1
This technique is demonstrated in Example 13.6.

Example 13.6 - Gas Stoichiometry for Conditions Other Than STP

Objective 19 Ammonia is produced when nitrogen and hydrogen gases react at high pressures and
temperatures:

N2( g ) + 3H2( g ) 2NH3( g )
At intervals, the system is cooled to between -10 °C and -20 °C, causing some of
the ammonia to liquefy so that it can be separated from the remaining nitrogen and
hydrogen gases. The gases are then recycled to make more ammonia. An average
ammonia plant might make 1000 metric tons of ammonia per day. When
4.0 × 107 L of hydrogen gas at 503 °C and 155 atm reacts with an excess of nitrogen,
what is the maximum volume of gaseous ammonia that can be formed at 20.6 °C
and 1.007 atm?

Solution

This equation stoichiometry problem can be solved in three steps:

Volume of H2(g) → moles H2 → moles NH3 → volume of NH3(g)

The first step can be accomplished using the ideal gas equation to convert from volume
of H2 to moles of H2 under the initial conditions.

PV = nRT BUN s -

-

13.3 Equation Stoichiometry and Ideal Gases 505

An alternative technique allows us to work Example 13.6 and problems like it
using a single unit analysis setup. This technique, illustrated in Example 13.7, uses the
universal gas constant, R, as a conversion factor.

Example 13.7 - Gas Stoichiometry for Conditions Other Than Objective 19
STP (alternative technique)

The question is the same as in Example 13.6: When 4.0 × 107 L of hydrogen gas at
503 °C and 155 atm reacts with an excess of nitrogen, what is the maximum volume
of gaseous ammonia that can be formed at 20.6 °C and 1.007 atm?

Solution

We can use R as a conversion factor to convert from volume in liters of H2 to moles of
H2. We use it in the inverted form, with liters (L) on the bottom, so that liters will be
canceled out.

- /) s - ) ,

506 Chapter 13 Gases

The textbook’s Web site has a section that describes a shortcut for working
equation stoichiometry problems in which volume of one gas is converted
into volume of another gas at the same temperature and pressure.

Between this chapter and Chapter 10, we have now seen three different ways to
convert between a measurable property and moles in equation stoichiometry problems.
The different paths are summarized in Figure 13.10 in the sample study sheet on the
next two pages. For pure liquids and solids, we can convert between mass and moles,
using the molar mass as a conversion factor. For gases, we can convert between volume
of gas and moles using the methods described above. For solutions, molarity provides
a conversion factor that enables us to convert between moles of solute and volume of
solution. Equation stoichiometry problems can contain any combination of two of
these conversions, such as we see in Example 13.8.

Objective 19 Example 13.8 - Equation Stoichiometry

The gastric glands of the stomach secrete enough hydrochloric acid to give the
stomach juices a hydrochloric acid concentration of over 0.01 M. Some of this
stomach acid can be neutralized by an antacid that contains calcium carbonate. The
CaCO3 reacts with HCl in the stomach to form aqueous calcium chloride, carbon
dioxide gas, and water. A regular antacid tablet will react with about 80 mL of 0.050
M HCl. How many milliliters of CO2 gas at 37 °C and 1.02 atm will form from the
complete reaction of 80 mL of 0.050 M HCl?

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2( g ) + H2O(l )

Solution
The general steps for this conversion are

Volume of HCl solution → moles HCl → moles CO2 → volume of CO2 gas

N- $0 N- )$M TPMO N - N )P$M )M T$PMM O NPM $0 -

13.3 Equation Stoichiometry and Ideal Gases 507

Tip-off – You are asked to convert an amount of one substance to an amount of Sample Study
Sheet 13.3
another substance, both involved in a chemical reaction. Equation
Stoichiometry
General Steps Problems

Step 1 Convert the amount of substance 1 to moles 1.

For a pure solid or liquid, this will be a molar mass conversion from mass to
moles.

H NPM
NPMBS NBTT
H

For solutions, molarity can be used to convert from volume of solution to
moles.

- TPMVUJPO OVNCFS G S-P NTP MNVUPJPMBOS J UZ
NPM

or N- TPMVUJPO OVNC F S G SNPN- TNPMPVMUBJSP JOUZ
NPM

For a gas at STP, the molar volume at STP can be used as a conversion factor to Objective 19
convert from volume of gas at STP to moles.

- NPM
-

For a gas under conditions other than STP, the ideal gas equation can be used Objective 19
to convert volume to moles, or the universal gas constant, R, can be used as a
conversion factor.

O 1 7
35

or - ,

508 Chapter 13 Gases

For solutions, molarity can be used to convert from moles to volume of
solution.

(coefficient 2) mol 2 1 L solution 2

(coefficient 1) mol 1 (number from molarity) mol 2

or (coefficient 2) mol 2 103 mL solution 2

(coefficient 1) mol 1 (number from molarity) mol 2

Objective 19 For a gas at STP, the molar volume at STP can be used as a conversion factor to
Objective 19 convert from moles to volume of gas at STP.

(coefficient 2) mol 2 22.414 L 2
(coefficient 1) mol 1 1 mol 2

For a gas under conditions other than STP, the ideal gas equation or the universal
gas constant, R, can be used to convert from moles to volume of gas.

7 O 35
1

or (coefficient 2) mol 2 0.082058 L

13.4 Dalton’s Law of Partial Pressures 509

Exercise 13.3 - Equation Stoichiometry Objective 19

Iron is combined with carbon in a series of reactions to form pig iron, which is about
4.3% carbon.

2C + O2 → 2CO
Fe2O3 + 3CO → 2Fe + 3CO2
2CO → C(in iron) + CO2
Pig iron is easier to shape than pure iron, and the presence of carbon lowers its
melting point from the 1539 °C required to melt pure iron to 1130 °C.

a. In the first reaction, what minimum volume of oxygen at STP is necessary
to convert 125 Mg of carbon to carbon monoxide?

b. In the first reaction, what is the maximum volume of carbon monoxide at
1.05 atm and 35 °C that could form from the conversion of 8.74 × 105 L
of oxygen at 0.994 atm and 27 °C?

Exercise 13.4 - Equation Stoichiometry Objective 19

Sodium hypochlorite, NaOCl, found in household bleaches, can be made from a
reaction using chlorine gas and aqueous sodium hydroxide:

Cl2( g ) + 2NaOH(aq) → NaOCl(aq) + NaCl(aq) + H2O(l )
What minimum volume of chlorine gas at 101.4 kPa and 18.0 °C must be used to
react with all the sodium hydroxide in 3525 L of 12.5 M NaOH?

13.4 Dalton’s Law of Partial Pressures

Most gaseous systems contain a mixture of gases. Air is a mixture of nitrogen gas,
oxygen gas, xenon gas, carbon dioxide gas, and many others. A typical “neon” light on
a Las Vegas marquee contains argon gas as well as neon. The industrial reaction that
forms nitric acid requires the mixing of ammonia gas and oxygen gas.

When working with a mixture of gases, we are sometimes interested in the total
pressure exerted by all of the gases together, and sometimes we are interested in the
portion of the total pressure that is exerted by only one of the gases in the mixture.
The portion of the total pressure that one gas in a mixture of gases contributes is called
the partial pressure of the gas. Distance runners find it harder to run at high altitudes
because the partial pressure of oxygen in the air they breathe at sea level is about 21
kPa, but at 6000 ft above sea level, it drops to about 17 kPa. Table 13.1 shows the
partial pressures of various gases in dry air. (The percentage of water vapor in air varies
from place to place and day to day.)