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CHAPTER
H
ousehold ammonia, chlorine bleach, and oxygen
bleach can be found in almost everyone's cleaning
closet. But ammonia, chlorine, and oxygen have many
other uses. They are some of the most important reactants
and products of the chemical industry.
When ammonia gas is synthesized, nitrogen gas and
hydrogen gas react by volume in the following way.
1 L Nig)
+ 3 L Hig)
-+
2 L NH 3 (g)
According to the reaction above, 1 L of nitrogen and 3 L of
hydrogen combine to form 2 L of ammonia.
According to Avogadro's principle, the volumes of gas
combine in simple whole-number ratios because 1 L of
nitrogen contains as many particles as 1 L of hydrogen. But
how many particles is that?
The volumes of nitrogen, hydrogen, and ammonia gas
are easily measured, but the particles in a gas are too small
and too numerous to be counted easily. How can you determine the number of particles in a gas without counting?
Concept Check
Review the following concepts
before studying this chapter.
Chapter 4: the types of compounds
Chapter 5: naming compounds; writing chemical formulas
Chapter 6: writing and balancing
chemical reactions
Chapter 10: the factor label method;
volume, pressure, and temperature
relationships of gases
4
Reading Chemistry
Go through the chapter, and write
down the sections heads. As you
read, jot down several key points or
vocabulary under each section head.
Also, for each section, write down
any questions you form about the
topics.
Chemistry Around You
Is it possible to count the atoms,
formula units, and molecules of
substances? When a liter of
methanol, for example, is formed
from hydrogen gas and carbon
dioxide gas, how many molecules
of methanol are formed? How
many molecules of hydrogen gas
and carbon dioxide gas reacted?
To find out more about molar mass,
visit the Glencoe Science Web site.
science.glencoe.com
403
Counting Particles
ofMatter
L
Objectives
Compare and contrast the mole as a
number and the mole
as a mass.
Relate counting particles to weighing samples of substances.
Solve stoichiometric
problems using molar
mass.
Key Terms
stoichiometry
mole
Avogadro constant
molar mass
molecular mass
formula mass
Chapter 12
ouis Staffilino of Dillonvale,
Ohio, saved pennies for 65
years. When he deposited
them in a bank in 1994, he had
40 large drums of pennies. His
wealth represented an enormous counting job for the bank
teller.
Just as a bank teller counts
coins and bills, a chemist counts
atoms, molecules, and formula
units of substances. Unlike a
bank teller, a chemist cannot
count individual items because the particles of matter are too small and just
too numerous. How do you determine the number of particles in a sample of
matter without counting?
Stoichiometry
In Chapter 11, you learned that volumes of gases always combine in
definite ratios. This observation, called the law of combining volumes, is
based on measurements of the gas volumes. When Avogadro suggested
that gases combine in fixed ratios because equal volumes of gases at the
same temperature and pressure contain equal numbers of particles, he
may have been thinking of particles rearranging themselves. Individual
gas particles are so small that their rearranging cannot be observed, but
the volumes of gases can be measured directly. Avogadro's principle is one
of the earliest attempts to relate the number of particles in a sample of a
substance to a direct measurement made on the sample.
Today, by using the methods of stoichiometry> we can measure the
amounts of substances involved in chemical reactions and relate them to
one another. For example, a sample's mass or volume can be converted to
a count of the number of its particles, such as atoms, ions, or molecules.
Pennies are not small, but counting a drum full of pennies, like counting the number of particles in a gas, is a formidable task. Can we measure
the pennies by grouping them in some conveniently large quantity and
then counting the number of groups to find the total number of pennies
in the drum? Can methods of stoichiometry help? Figure 12.1 presents
two ways to count large numbers of pennies. To count the pennies directly, you would have to handle everyone-all 24 216 pennies. But you
Chemical Quantities
could also just make three measurements: the weight of the drum of pennies) the weight of the empty drum, and the weight of a group of 1000
pennies. Now, if you want to count atoms, what size group is most suitable? You will need a larger group than you used for the pennies-much
larger than a thousand or even a million.
Atoms are so tiny that an ordinary-sized sample of a substance contains so
many of these submicroscopic particles that counting them by grouping
them in thousands would be unmanageable. Even
grouping them by millions would not help. The group
or unit of measure used to count numbers of atoms,
molecules) or formula units of substances is the mole
(abbreviated mol). The number of things in one mole
is 6.02 X 1023 , This big number has a short name: the
Avogadro constant, as illustrated in Figure 12.2.
The most precise value
of the Avogadro constant is
6.0221367 x 1023 •
For most purposes,
rounding to
6.02 x 1023 is sufficient.
Figure 12.1
Two Ways to
Count Pennies
You could count
the pennies
directly. ~
•
You could calculate the number of pennies in a
drum by using the mass of all the pennies,
70125 g, and the mass of 1000 pennies. The
mass of 1000 pennies is 2890.7 g. Use the ratio
1000 pennies to find the total number of pennies.
2890.7 g
70 125 Jt X
1000 pennies
2890.7 g'
=
.
24259 pennies
The drum contains 24259 pennies, worth
$242.59.
Figure 12.2
One Mole Is a BIG Number.
If 6.02 X 1023 sheets of paper were placed in a stack, they
would reach from Earth to the sun-more than a million times.
The thickness of a sheet of paper is small, but an atom is much
smaller. One mole of magnesium atoms is hardly a handful.
12.1
Counting Particles of Matter
405
All kinds of submicroscopic particles can be conveniently counted
using the Avogadro constant. There are 6.02 X 10 23 carbon atoms in a
mole of carbon, and there are 6.02 X 10 23 carbon dioxide molecules in a
mole of carbon dioxide. There are 6.02 X 1023 sodium ions in a mole of
sodium ions. For that matter) there are 6.02 X 1023 eggs in a mole of eggs,
but eggs are so large compared with the particles that interact in chemistry that you would have no need to estimate how many there are in a
typical sample. Can you see a relationship here? For counting many small
things, use a large unit of measure; for counting fewer things or larger
things, use a small unit of measure, as shown in Figure 12.3.
Molar Mass
stoichiometry
stoichen (G K)
element or part
metreon (GK)
measure
Stoichiometry
allows you to
determine the
number of atoms
in a substance by
relating other measurable quantities
of that substance.
,
How many moles of methanol are in 500 g of methanol? Methanol is
formed from CO 2 gas and hydrogen gas according to the balanced chemical equation below.
CO 2 (g)
~
CH 30H(g)
+ H 20(g)
Suppose you wanted to produce 500 g of methanol. How many grams of
CO 2 gas and H 2 gas would you need? How many grams of water would be
produced as a by-product? Those are questions about the masses of reactants and products. But the balanced chemical equation shows that three
molecules of hydrogen gas react with one molecule of carbon dioxide gas.
The equation relates molecules, not masses, of reactants and products.
Like Avogadro, you need to relate the macroscopic measurements-the
masses of carbon dioxide and hydrogen-to the number of molecules of
methanol. To find the mass of carbon dioxide and the mass of hydrogen
needed to produce 500 g of methanol, you first need to know how many
molecules of methanol are in 500 g of methanol.
•
figure 12.3
Grouping Items for Counting
The size of the group depends
upon how many items we
would buy or use at one
time. Would you rather
count the sheets of
paper or the reams?
406
+ 3H 2 (g)
Chapter 12 Chemical Quantities
Remember the drum of pennies?
By knowing the mass of one group of
1000 pennies and knowing the mass
of all the pennies, you can find the
number of groups of 1000 pennies.
Then, finding the total number of
pennies is an easy matter. You have a
suitable unit of measure, the mole,
and you know the mass of methanol
you want to produce. But, you still
need to know the mass of 1 mol of
methanol molecules.
Six iron atoms
Six carbon atoms
Molar Mass of an Element
Figure 12.4
You know from Chapter 2 that average atomic masses of the elements
are given on the periodic table. For example, the average mass of one iron
atom is 55.8 u, where u means "atomic mass units." The atomic mass unit
is defined so that the atomic mass of an atom of the most common carbon isotope is exactly 12 u, and the mass of 1 mol of the most common
isotope of carbon atoms is exactly 12 g. The mass of 1 mol of a pure substance is called its molar mass. For example, the molar mass of iron is
55.847 g, and the molar mass of platinum is 195.08 g. Relative masses of
elements are demonstrated in Figure 12.4. The molar mass is the mass in
grams of the average atomic mass.
If an element exists as a molecule,
remember that the particles in 1 mol of
that element are themselves composed
of atoms. For example, the element
oxygen exists as molecules composed
of two oxygen atoms, so a mole of
oxygen molecules contains 2 mol of
oxygen atoms. Therefore, the molar
mass of oxygen molecules is twice the
molar mass of oxygen atoms:
2 X 16.00 g = 32.00 g. Figure 12.5
shows the molar masses of some elements.
Relative Masses of Iron and
Carbon
An average iron atom is 4.65
times as heavy as an average
carbon atom. 5ix iron atoms
are 4.65 times as heavy as six
carbon atoms. One mole of
iron atoms is 4.65 times as
heavy as 1 mol of carbon
atoms.
Figure 12.5
Molar Masses of Some Elements
Each sample contains 6.02 x 1023 atoms. Note
that the masses of the samples are all different.
Moles relate counts of atoms, molecules, or ions
to mass because a molar mass always
contains an Avogadro constant
of particles.
12.1
Counting Particles of Matter
407
Determining Number Without Counting
Chemists and chemical engineers usually need to control the number
of atoms, molecules, and ions in their reactions carefully. These particles are too small to be seen and too numerous to count, but their
numbers can be determined by measuring their masses. Simulate this
process by determining the approximate number of small, identical
items in a large bag by weighing instead of counting.
Procedure
~;
1. Count out ten items and weigh
them. Record the mass.
2. Weigh and record the mass of
an empty, self-sealing plastic
storage bag.
3. Completely fill the bag with
items and seal it.
4. Weigh and record its mass.
5. Develop and carry out an experi-
ment asking yourself the following question, "How can I determine the number of buttons in
the bag without opening it?"
Analysis
1. How many items are in the
bag?
2. Explain how you determined
that number.
Number of Atoms in a Sample of an Element
SAMPLE PROBLEM
The mass of an iron bar is 16.8 g. How many Fe atoms are in the
sample?
Problem-Solving
HI NT
• Use the periodic table to find the molar
mass of iron. The average mass of an
Remember that the units of molar mass
iron atom is 55.8 u. Then the mass of
ore grams per mole, which con be used
1 mol of iron atoms is 55.8 g.
as a conversion factor.
Set Up
• To convert the mass of the iron bar to
the number of moles of iron, use the mass of 1 mol of iron atoms as a
conversion factor.
••
••
16.8 g Fe 1 mol Fe
•
55.8 g Fe
•
•
•
•
Now, use the number of atoms in a mole to find the number of iron
•
•
atoms in the bar.
•
•
23
•
16.8 g Fe 11 mol Fe 16.02 X 10 Fe atoms
•
1 mol Fe
55.8 g Fe
•
•
•
Solve : • Simplify the expression above.
23
•
16.8.g-Fe'11~ 16.02 X 10 Fe atoms _
•
55.8.g-Fe'
l~
•••
23
•
16.8 X 6.02 X 10 Fe atoms
23
•
•
= 1.81 X 10 Fe atoms
55.8
•
•
•
Check : • Notice how the units of measure cancel to leave only Fe atoms.
Analyze
t
·
·
4D8
Chapter 12 Chemical Quantities
.-.
Molar Mass of aCompound
As you learned in Chapter 4, covalent compounds are composed of
molecules, and ionic compounds are composed of formula units. The
molecular mass of a covalent compound is the mass in atomic mass units
of one molecule. Its molar mass is the mass in grams of 1 mol of its molecules. The formula mass of an ionic compound is the mass in atomic
mass unlts of one formula unit. Its molar mass is the mass in grams of
1 mol of its formula units. How to calculate the molar mass for ethanol, a
covalent compound, and for calcium chloride, an ionic compound, is
shown below.
Ethanol, C 2H 60, a covalent compound
2 C atoms
6 H atoms
10 atom
molecular mass of C2H 60
2 X 12.0 u =
6 X 1.00 u =
1 X 16.0 u =
24.0 u
6.00 u
+ 16.0 u
46.0 u
46.0 g
46.0 glmol
mass of 1 mol C 2H 6 0 molecules
molar mass of ethanol
Calcium chloride, CaC1 2 , an ionic compound
40.1 u
1 Ca atom
1 X 40.1 u =
2 Cl atoms
2 X 35.5 u = +71.0 u
111.1 u
formula mass of CaCl 2
=
mass of 1 mol CaC1 2 formula units
molar mass of calcium chloride
IlLIg
111.1 glmol
By making calculations with molar masses, you can find the number of
molecules of methanol in your 500 g; the number of molecules of carbon
dioxide and hydrogen that reacted; and the masses of carbon dioxide,
hydrogen, methanol, and water in grams. Figure 12.6 shows the molar
masses of some compounds.
Figure 12.6
Molar Masses of Some
Compounds
Each sample contains
6.02 X 1023 molecules of a
covalent compound or
6.02 x 1023 formula units of
an ionic compound. Each
compound has its own molar
mass.
12.1
-
..-
Counting Particles of Matter
409
SAMPLE PROBLEM
Number of Formula Units in a Sample of a Compound
•
•
•
•
'
The mass of a quantity of iron(III) oxide is 16.8 g. How many
formula units are in the sample?
•
••
Analyze : • Use the periodic table to calculate the mass of one formula unit of Fe 20 3.
2 X 55.8 u =
3 X 16.0 u =
2 Fe atoms
:
3 0 atoms
formula mass of Fe 2 0 3
111.6 u
+ 48.0 u
159.6 u
Therefore, the molar mass of Fe203 (rounded off) is 160 g.
•
•
•
•
•
••
••
Solve
16.8 g FeZ03 1 mol FeZ03
160 g Fe203
•
Set Up
•
:
•
••
•
•
•
•
•
•
•
·
Now, multiply the number of moles of iron oxide by the number in a
mole.
23
16.8 g FeZ03 1 mol Fe203 6.02 X 10 Fe203 formula units
160 g Fe Z03
mol FeZ03
.
23
16.8 ~ 1 ~ 6.02 X 10 FeZ03 formula units
160~
l~
23
16.8 X 6.02 X 10 FeZ03 formula units
=
160
22
6.32 X 10 FeZ03 formula units
•
•
Check •• • The units of measure show that the calculation is set up correctly. The
•
•
•
•
•
•
•
•
•
•
•
•
•
••
•
•
•
•
••
•
••
•
•
••
·
••
•
•
•
multiplication also checks.
1. Without calculating, decide whether 50.0 g of sulfur or 50.0 g of
tin represents the greater number of atoms. Verify your answer by
calculating.
2. Determine the number of atoms in each sample below.
a)
b)
c)
d)
98.3 g mercury, Hg
45.6 g gold, Au
10.7 g lithium, Li
144'.6 g tungsten, W
3. Determine the number of moles in each sample below.
a)
b)
c)
d)
6.84
16.0
68.0
17.5
g sucrose, C 12 H 22 0 11
g sulfur dioxide, S02
g ammonia, NH 3
g copper(II) oxide, CuO
In the previous problems, you used the molar mass to convert a mass
measurement to a number of moles. Now, you will learn to convert a
number of moles to a mass measurement.
41Z
Chapter 12 Chemical Quantities
Mass of a Number of Moles of a Compound
SAMPLE PROBLEM
.
•
•
What mass of water must be weighed to obtain 7.50 mol of H 20?
•
•
Analyze : • The molar mass of water is obtained from its molecular mass.
•
•
2 H atoms
2 X 1.00 u = 2.00 u
•
•
1 X 16.0 u = 16.0 u
1 0 atom
•
•
•
molecular
mass
of H 20
18.0 u
•
•
The molar mass of water is 18.0 g/mo1.
••
Set Up •• • Use the molar mass to convert the number of moles to a mass mea•
surement.
•
•
•
7.50 mol H20 118.0 g H 20
•
•
1 mol H 20
•
•
:
Solve : • 7.5~118.0gH20
1~ = 7.50 X 18.0gH20 = 135gH20
•
•
Check : • The final units are grams of H 20, as expected. The multiplication also
·
:
•
••
•
•
•
•
•
•
•
•••
checks.
4. Determine the mass of the following molar quantities.
a) 3.52 mol 5i
b) 1.25 mol aspirin, C9 H s0 4
c) 0.550 mol F2
d) 2.35 mol barium iodide, Ba1 2
The concept of molar mass makes it easy to determine the number of
particles in a sample of a substance by simply measuring the mass of the
sample. The concept is also useful in relating masses of reactants and
products in chemical reactions.
rpR.lemental
Problems
For more practice
with solving problems,
see Supplemental Practice
Problems, Appendix B.
Understanding Concepts
1. How is counting a truckload of pennies like
counting the atoms in 10.0 g of aluminum?
How is it different?
2. The average atomic mass of nitrogen is 14 times
greater than the average atomic mass of hydrogen. Would you expect the molar mass of nitrogen gas, N 2, to be 14 times greater than the
molar mass of hydrogen gas, H/ Why?
3. Why is the following statement meaningless?
An industrial process requires 2.15 mol of a
sugar-salt mixture.
Thinking Critically
4. Analyzing Determine the mass in grams of one
average atomic mass unit.
Applying Chemistry
5. Balances If the mole is truly central to quantitative work in chemistry, why do balances in
chemistry labs measure mass and not moles?
12.1
Counting Particles of Matter
413
Using Moles
H~·'llI;.'r"
T
he rose's distinctive scent is
mostly the odor of one
compound, geraniol. The
balanced chemical equation for
the formation of geraniol shows
the ratios of carbon, hydrogen,
and oxygen that interact.
Although the particles are too
small to be seen or weighed, a
mole of carbon, hydrogen, or
oxygen is large enough. Can you use the chemical equation to predict the
masses of carbon, hydrogen, and oxygen that react and the mass of geraniol
that is formed?
Objectives
Predict quantities of
reactants and prod·
I ~·;~::. ucts in chemical reac,
,.... tions.
I·,,..••~,- Determine mole
ratios from formulas
for compounds.
Identify formulas of
compounds by using
mass ratios.
Using Molar Masses in Stoichiometric Problems
Key Terms
molar volume
ideal gas law
empirical formula
In Chapter 11, you saw how balanced chemical equations indicate the
volume of gas required for a reaction or the volume of gas produced. Similarly, you can use balanced chemical equations and numbers of moles of
each substance to predict the masses of reactants or products.
Predicting Mass of a Reactant
SAMPLE PROBLEM
I
As you review this problem, note that nitrogen and hydrogen
are related in terms of moles, not mass. In chemical reactions,
when substances react, their particles react. Ammonia gas is
synthesized from nitrogen gas and hydrogen gas according to
the balanced chemical equation below.
N 2 (g)
+
3 H 2 (g)
-+
2 NH 3 (g)
How many grams of hydrogen gas are required for 3.75 g of
nitrogen gas to react completely?
• The amount of hydrogen needed depends upon the number of
Analyze
nitrogen molecules present in 3.75 g and the mole ratio of
hydrogen gas to nitrogen gas in the balanced chemical equation.
•
•
Find
the number of moles of N 2 molecules by using the molar
Set Up
:
mass of nitrogen.
•
••
3.75 g N211 mol N 2
•
28.0 g N2
·
·
Chapter 1Z Chemical Quantities
To find the mass of hydrogen needed, first find the number of moles
of H 2 molecules needed to react with all the moles of N 2 molecules.
The balanced chemical equation shows that 3 mol of H 2 molecules
react with 1 mol of N 2 molecules. Multiply the number of moles of N 2
molecules, as shown in Set Up, by this ratio.
••
••
•
•
•
•
•
•
•
•
•
•
The units in the expression above simplify to moles of H 2 molecules.
To find the mass of hydrogen, multiply the number of moles of hydrogen molecules by the mass of 1 mol of H 2 molecules, which is 2.00 g.
•
••
·••• •
Solve •
•
•
••
••
•
3.75.g-N211~13~12.00gH2
28.0.g.Ni l~ l~
=
3.75 X 1 X 3 X 2.00 g H 2
28.0
= 0.804 g H 2
•
Check •• • Check the multiplication to verify that the result is correct.
Predicting Mass of a Product
•
:
:
I
What mass of ammonia is formed when 3.75 g of nitrogen gas react
with hydrogen gas according to the balanced chemical equation below?
•
:
N 2 (g) + 3 H 2 (g) ~ 2 NH 3(g)
•
Analyze : • The amount of ammonia formed
'"
Problem-Solving
:
depends upon the number of nitrogen
HI NT
•
:
molecules present and the mole ratio of
:
nitrogen and ammonia in the balanced
The balanced chemical equation is the
:
chemical equation.
source of conversion factors relating
•
Set Up • • As in Sample Problem 4, the number of moles of one substance to moles of
another substance.
moles of nitrogen molecules is given by
the expression below.
3.75 g N 2 1 mol N 2
28.0 g N2
••
••
To find the mass of ammonia produced, first find the number of
moles of ammonia molecules that form from 3.75 g of nitrogen. Use
the mole ratio of ammonia molecules to nitrogen molecules to find
the number of moles of ammonia formed.
•
••
•
·•••
•
•
•
•
•
•
•
•
Use the molar mass of ammonia, 17.0 g, to find the mass of ammonia
formed.
17.0 g NH 3
I mol NH 3
12.2 Using Moles
415
,
-•
•
Solve •
•
..
· • 3.75..g.Nzll~12~117.0 g NH
3
28.0~
•
•
•
1~
=
1~
3.75 X 1 X 2 X 17.0gNH3
•
28.0
= 4.55 g NH3
•
•
Check : • Convince yourself that the factors are set up correctly and that the
•
multiplication is correct.
•
•
Supplemental
problems
••
••
•
•
5. The combustion of propane, C 3H s ) a fuel used in backyard grills
and camp stoves, produces carbon dioxide and water vapor.
••
C3H s(g)
+
+ 4H 2 0(g)
50 2 (g) -+ 3COig)
W'hat mass of carbon dioxide forms when 95.6 g of propane burns?
For more practice with solving
problems, see Supplemental
Practice Problems,
Appendix B.
6. Solid xenon hexafluoride is prepared by allowing xenon gas and
fluorine gas to react.
Xe(g)
••
•
•
•
+ 3F2 (g)
-+
XeF 6 (s)
How many grams of fluorine are required to produce 10.0 g of XeF 6?
7. Using the reaction in Practice Problem 6, how many grams of
xenon are required to produce 10.0 g of XeF 6 ?
Using Molar Volumes in Stoichiometric Problems
r
Figure 12.7
Molar Volumes of Gases
One mole of any gas at STP
occupies 22.4 L. How large is
that? It is the volume of a
cube that is 28.2 cm on each
edge. Each such volume contains 6.02 X 1023 atoms of a
gaseous element or 6.02 X
10 23 molecules of a molecular
element or compound, but
the mass of 1 mol is different
for each element. One mole
of helium floats because its
mass is less than the mass of
22.4 L of air.
416
Chapter 12
In Chapter 11, you used the law of combining volumes. Avogadro
inferred from that law that equal volumes of gases contain equal numbers
of particles. In terms of moles, Avogadro's principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles of gases.
The molar volume of a gas is the volume that a mole of a gas occupies
at a pressure of one atmosphere (equal to 101 kPa) and a temperature of
O.Oo°e. Under these conditions of STP, the volume of 1 mol of any gas is
22.4 L, as shown in Figure 12.7. Like the molar mass, the molar volume is
used in stoichiometric calculations.
He
22.4 L
Chemical Quantities
6.02 x 10
23
Particles
One
E
u
N
00
22.4
Liters
N
28.2 cm
mole
Air Bags
In 1990, on a Virginia hilltop, two cars collided
in a head-on crash. Both drivers walked away with
only minor injuries. A chemical reaction, along
with seat belts, saved their lives. This was the first
recorded head-on collision between two cars with
air bags.
How air bags function Although air bags seem to
inflate instantaneously, the process occurs in steps.
1. When a car collides with a rigid barrier at a
speed of 12 mph or greater, two or three sensors
on the front of the car send an electric current
to fire the control unit 0.01 s after impact.
Reliability During a ten-year period, one car in a
million may have an air-bag defect. Why are they so
successful? There are no moving parts to wear out,
all exposed components are tightly sealed, and the
gold-plated electrical connectors corrode slowly.
Gas-volume relationships The driver's air bag
requires 0.0650 m 3 of nitrogen to inflate-no
more, no less. The passenger's air bag needs
0.1340 m 3 • The pellet must have the exact amount
of sodium azide needed to produce the correct
amount of nitrogen. As with all expanding gases,
pressure and temperature affect the amount of
sodium azide needed. Because the nitrogen gas is
formed in an explosion, it has to be cooled before
it goes into the air bag.
2. After 0.05 s, a chemical reaction in the stored
air bag creates a gaseous product that inflates it
and pushes open the cover on the steering
wheel or on the passenger-side dash.
3. The driver or passenger strikes the inflated bag.
4. The bag deflates in 0.045 s as the gas escapes
through holes at the base of the bag.
The chemical reactions Sodium azide (NaN 3) is
the chemical that produces nitrogen gas to inflate
the air bag. Sodium azide pellets, an igniter, inflator, and a tightly folded nylon air bag are stored
under a breakaway cover in the steering wheel or
dash. The igniter provides a current to decompose
the sodium azide into nitrogen gas and sodium.
2NaN 3 (s)
electricity
---+
3N 2(g) + 2Na(s)
The sodium immediately reacts with iron(llI)
oxide in the pellet to form sodium oxide and iron.
6Na(s)
+ Fe 20 3 (s)
-+ 3Na20(S)
+ 2Fe(s)
The sodium oxide reacts with carbon dioxide and
water vapor in the air to form sodium hydrogen
carbonate.
Na 20(s)
+ 2C0 2(g) + H 20(g)
-+ 2NaHC0 3 (s)
1. Applying If 130 g of sodium azide are needed
for the driver's air bag, how much is needed for
the passenger's bag? Explain.
2. Inferring What effect does the heat from the
sodium azide reaction have on the pressure and
volume of the nitrogen gas formed?
To find out more about
·~~lIJmtflJ"'~~~air bags, visit the
,
Glencoe Science Web site.
science.glencoe.com
12.2
Using Moles
417
Using Molar Volume
SAMPLE PROBLEM
c
In the space shuttle, exhaled carbon dioxide gas is removed from the
air by passing it through canisters of lithium hydroxide. The following
reaction takes place.
CO 2(g)
+ 2LiOH(s)
~
Li 2C0 3(S)
+ H 20(g)
How many grams of lithium hydroxide are required to remove 500.0 L
•
of carbon dioxide gas at 101 kPa pressure and 25.0°C?
Analyze : • In this problem, you use the molar volume to find the number of
:
moles~
•
Set Up : • The volume of gas at 25°C must be converted to a volume at STP.
•
:•
V = 500.0 L CO 2 (273K)
298K = 458 L CO 2
•
••
Now, find the number of moles of CO 2 gas as below.
CD-ROM
•
•
Use the Interac••
458 L CO 2 1 mol CO 2
tive Virtual Exploration
••
22.4 L CO2
Predicting Mass ofProduct
•
•
The chemical equation shows that the ratio of moles of LiOH to
•
found on Disc 2 of
•
•
CO
the CD-ROM.
2 is 2 to 1. Therefore, the number of moles of lithium hydroxide is
•
given by the expression below.
••
•
458 L CO 2 1 mol CO 2 2 mol LiOH
••
•
22.4 L CO 2 1 mol CO 2
•
•
•
To convert the number of moles of LiOH to mass, use its molar mass,
•
•
23.9 g/mo!.
•
•
•
458 L CO 2 1 mol CO 2 2 mol LiOR 23.9 g LiOH
•
•
22.4 L CO 2 1 mol CO 2 1 mol LiOH
•
•
Solve : • 458J...GE1211~ 12~123.9gLiOH _
•
22.4.J-GB2 1 jP.Ol-€e)i 1 ~ •
•
•
458 x 2 X 23.9 g LiOH
.
•
•
=
977
g
LtOH
22.4
•
•
•
Check : • A mass of 1000 g of lithium hydroxide is about 40 mol of the com:•
pound. According to the chemical equation, about half as much, or
20 mol of CO 2, will be removed from the air. At STP, 20 mol of any gas
:
:
occupy about 450 L, which will expand to about 500 L at 25°C.
·
·
·
8. What mass of sulfur must burn to produce 3.42 L of S02 at 273°C
and 101 kPa? The reaction is S(s) + 02(g) ~ S02(g).
9. What volume of hydrogen gas can be produced by reacting 4.20 g of
sodium in excess water at 50.0°C and 106 kPa? The reaction is
2 Na + 2 H 20 ~ 2 NaOH + H 2•
418
Chapter 12
Chemical Quantities
Gases are much less dense than solids, as illustrated by Figure 12.8.
Ideal Gas Law
Exactly how the pressure p, volume V, temperature T, and number of
particles n of gas are related is given by the ideal gas law shown here.
PV= nRT
The value of the constant R can be determined using the definition of
molar volume. At STP, 1 mol of gas occupies 22.4 L. Therefore, when P =
101.3 kPa, V = 22.4 L, n = 1 mol, and T = 273.15 K, the equation for the
ideal gas law can be shown as follows.
Figure 12.8
Explosives
Gases produced in powerful
chemical reactions of explosives expand with great
energy. One mole of a gas
occupies much more space
than 1 mol of any solid.
101.3 kPa X 22.4 L = 1 mol X R X 273.15 K
Now, we can solve for R.
R = 101.3 kPa 1 22 .4 L 11
1 mol -2-7-3.-1-5=K
8.31 kPa·L
mol·K
Now, you can find the volume, pressure, temperature, and number of
moles of a gas.
Using the Ideal Gas Law
SAMPLE PROBLEM
•
••
•
••
How many moles are contained in a 2.44-L sample of gas at 25.0°C
and 202 kPa?
•
Analyze : • Solve the ideal gas law for n, the number of moles.
•
•
PV
•
n=•
RT
•
•
•
202 kPa X 2.44 L
•
Set Up •• • n =
8.31 kPa· L) X 298 K
•
( mol·K
•
·
·••
•
Solve •• • n
•
••
•
·•••
•
••
202 kPa 2.44 L
=
1
8.31 L· kPa
mol·K
202 J>Pa X 2.44X X 1 mol· jt'
8.31X· ~ X 298 jt'
1
298 K
=
202 mol X 2.44
8.31 X 298
= 0.199 mol
Check • • First, find the volume that 2.44 L of a gas would occupy at STP.
•
•
•
•
•
•
•
•
••
•
••
•
•
•••
•
Then, find the number of moles in this volume.
4.47 ~ 11 mol
22.4)':'
= 0.200 mol
0.200 mol is close to the calculated value.
1Z.Z
Using Moles
419
Bagging the Gas
Chemists and chemical engineers often need to determine amounts
of reactants and products that will react efficiently and cost effectively.
Use the molar volume to determine the amount of baking soda
required to react with vinegar to yield just enough carbon dioxide to
fill a one-quart, self-sealing plastic bag.
n
Procedure
~;
Squeeze the air from the bag
and seal the zipper top.
1. Write the balanced equation for
the reaction of baking soda
7. Place the bag in a trash can
(sodium hydrogen carbonate)
or behind an explosion shield.
and vinegar (acetic acid) that
8. Release the twist tie, quickly
produces sodium acetate, water,
mix the reactants, and allow the
and carbon dioxide.
reaction to proceed.
2. Find the volume of a I-quart,
Analysis
self-sealing plastic bag by filling
1. Show and explain the calculait with water and then pouring
tions that you used to deterthe water into a graduated
mine the required mass of sodicylinder or measuring cup.
um hydrogen carbonate.
3. Calculate the mass of sodium
2. What mass of sodium hydrogen
hydrogen carbonate that will
carbonate would be required to
fill the bag with CO 2 gas when
react with excess acetic acid to
the compound reacts with
produce 20 000 L of carbon
excess acetic acid.
dioxide gas at STP for a water4. Put on an apron and goggles.
treatment plant?
5. Weigh the calculated amount of 3. What would have happened in
sodium hydrogen carbonate
step 8 if the amount of acetic
and place it in a bottom corner
acid was insufficient to react
of the bag. Use a plastic-coated
with all of the baking soda?
twist tie to seal off this corner.
4. Suppose the pressure of the gas
6. Pour about 60 mL of 1M acetic
in the bag was measured to be
acid into the other bottom cor101.5 kPa. Is this pressure conner of the bag. Be careful not to
sistent with the ideal gas law?
allow the reactants to mix.
Assume T = 20°C and
P = 101 kPa.
••
Supplemental
prDblems
For more practice with solving
problems, see Supplemental
-Practice Problems,
Appendix B.
420
Chapter 12
·••
·•••
•
•
•
•
••
•
10. How many moles of helium are contained in a 5.00-L canister at
101 kPa and 30.0°C?
11. What is the volume of 0.020 mol Ne at 0.505 kPa and 27.0°C?
12. How much zinc must react in order to form 15.5 L of hydrogen,
H 2 (g), at 32.0°C and 115 kPa?
Chemical Quantities
Zn(s)
+ H 2S0 4
-+
ZnS0 4 (aq)
+ H 2 (g)
Theoretical Yield and Actual Yield
The amount of product of a chemical reaction predicted by stoichiometry is called the theoretical yield. As shown earlier, if 3.75 g of nitrogen
completely react, a theoretical yield of 4.55 g of ammonia would be produced. The actual yield of a chemical reaction is usually less than predicted. The collection techniques and apparatus used, time, and the skills of
the chemist may affect the actual yield.
When actual yield is less than theoretical yield, you express the efficiency of the reaction as percent yield. The percent yield of a reaction is the
ratio of the actual yield to the theoretical yield expressed as a percent.
Suppose the actual yield was 3.86 g of ammonia. Then, the percent yield
could be calculated by using the following equation.
. ld actual yield
percen t Yle - theoretical yield X 100%
3.86~
4.55..g..Ntr; X 100% = 84.8%
This means that 84.8 percent of the possible yield of ammonia was
obtained from the reaction. Calculating percentage yield is similar to calculating a baseball player's batting average, as shown in Figure 12.9.
A manufacturer is interested in producing chemicals as efficiently and
inexpensively as possible. High yields make commercial manufacturing of
substances possible. For example, in human experiments, taxol, a naturally occurring complex compound, is a strong agent against cancer. For ten
years, chemists tried to synthesize this compound in the lab. In 1994, two
independent academic research groups succeeded. However, the process is
so complicated and time-consuming that the percent yield is probably not
even one percent.
CD·ROM
Use the Interactive Virtual Experiment
How much oxygen is available? found on Disc 2 of
the CD-ROM.
Figure 12.9
Batting Average and
Percent Yield
In 1995, Alfredo Spinelli
batted .352, which means:
hits
352
---=-attempts
1000 .
Just as batting averages measure a hitter's efficiency, percent yield measures a reaction's efficiency.
12.2 Using Moles
421
Improving Percent Yield in
Chemical Synthesis
The single most important industrial chemical in
the world is probably sulfuric acid. In the United
States, its production exceeds 40 million tons per
year. The strength of the U.S. economy can be
gauged by the amount ofsulfuric acid produced
annually. Over the years, the manufacturing process
ofsulfuric acid has been improved to provide a higher and more economical yield.
Step 1
Sulfur is burned in air to produce sulfur dioxide, a
stable compound. This reaction takes place quickly
and readily.
5(s) + Oig) --+ 50 2(g)
Impurities produced
in combustion are
removed from the
Sulfur
sulfur dioxide so they
melter
will not react with
and impede the catalyst in the next step.
Melted
sulfur
Uses of Sulfuric Acid
Sixty percent of all manufactured sulfuric acid
is used to make fertilizers. It is also used in manufacturing detergents, photographic film, synthetic fibers, pigments, paints, drugs, and other
acids. It is the electrolyte in some batteries, acts as
a catalyst and a dehydrating agent, and is a component in refining petroleum and metals.
The Lead·Chamber Process-A Low·Yield
Process
The first industrial method, the lead-chamber
process, is not commonly used now because its
purity is low and its percent yield is only 60 to 80
percent. But it is much cheaper than the later and
more productive contact process. The lead-chamber process is used for manufacturing sulfuric acid
for applications that do not demand high purity.
The Contact Process-The High-Yield
Process
The contact process is the most widely used
commercial method. It is more expensive than
the lead-chamber process, but it is simple and it
produces high-purity sulfuric acid at a high percent yield-about 98 percent. In addition, it creates no by-products that pollute the atmosphere.
The contact process has four steps.
l:=====Ai=r-:,-:,-:,:
[
Water --=::..:....--=----+
~--
~::±======-
~=~=====:::J
Maximizing Percent Yield
Reaction yields in the contact process were
increased in several ways-by selecting an optimum temperature, using an efficient catalyst,
removing a product from a reaction that does not
go to completion, and by controlling the rate of
reaction of S03 with water. Keeping operating
pressures at the correct values also increases yield.
Using Scientific Methods to Attain the
Best Yields
To develop higher yields that are closer to the
theoretical values, chemists adjust temperatures,
pressures, or other conditions in their industrial
processes. They look for better catalysts or new
ways to deal with undesirable side reactions.
Unwanted by-products are serious concerns to
the chemist. Because they may cause environmental damage or be expensive to dispose of, byproducts may raise production costs.
Step 3
Step 4
Because sulfur trioxide reacts violently with water, it is
bubbled through 98 percent concentrated sulfuric
acid and forms pyrosulfuric acid, H2 S20 7 •
When water is added to the pyrosulfuric
acid, high-quality, 98 percent concentrated
sulfuric acid is formed.
50 3(g)
+ H2S04 (1)
---*
H2S20 7(1)
H2 S2 0 7 (1)
+ H20(I)
---*
2 H1SOil)
Step 1
Sulfur burner
) S02
S + O2
t
)
Step 2
Step 2
Because sulfur dioxide reacts slowly with
excess oxygen, a catalyst, either vanadium
pentoxide (V 20 S) or finely divided platinum at a temperature of 400°C, is used.
The reaction produces sulfur trioxide.
2S0 ig)
+ Gig) c~t 250 3(g)
Sulfur trioxide is quickly removed from the
contact chamber because it tends to form
sulfur dioxide and oxygen again. Removing sulfur trioxide promotes the production of more sulfur trioxide.
1. Hypothesizing Roasting iron pyrite, FeS 2> can
replace burning sulfur in Step 1 to produce
sulfur dioxide. Write an equation to show
what you think the reaction with pyrite is.
Green Chemistry
The contact process is economically sound
because the reactants are abundant and inexpensive and because it produces no unwanted byproducts, which may be expensive to store or dispose of and which may pollute the environment.
Ecologically sound chemical manufacturing is
called green chemistry or green technology.
2. Acquiring Information Find out how sulfuric
acid is used to manufacture hydrochloric acid
and write the equation(s) of the reactions.
Determining Mass Percents
As you know, the chemical formula of a compound tells you the elements
that comprise it. For example, the formula for geraniol (the main compound
that gives a rose its scent) is CIOH1SO. The formula shows that geraniol is
comprised of carbon, hydrogen, and oxygen. Because all these elements are
nonmetals, geraniol is probably covalent and comprised of molecules.
In addition, the formula CIQH1SO tells you that each molecule of geraniol contains ten carbon atoms, 18 hydrogen atoms, and one oxygen
atom. In terms of numbers of atoms, hydrogen is the major element in
geraniol. How can you tell whether it is the major element by mass? You
can answer this question by determining the mass percents of each element in geraniol, which are shown in Figure 12.10.
o
Figure 12.10
10.4%
Mass Percents of Elements In
Geraniol
This pie graph shows the composition
of geraniol in terms of mass percents
of the elements.
C
77.9%
Suppose you have a mole of geraniol. Its molar mass is 154 glmol. Of this
mass, how many grams do the carbon atoms contribute? The formula shows
that one molecule of geraniol includes ten atoms of carbon. Therefore, 1 mol
of geraniol contains 10 mol of carbon. Multiply the mass of 1 mol of carbon
by 10 to get the mass of carbon in 1 mol of geraniol.
10~112.0gC
j:OOt = 120 g C
Now, use this mass of carbon to find the mass percent of carbon in geraniol.
%C =
120 g C
154 g CIQH 18 0
X 100% = 77.9%
The mass percents of the other elements are calculated below in a similar
fashion.
Mass of hydrogen in 1 mol geraniol:
18~X 1.~H
Mass percent of H =
=
18.0 gH
mass ofH
f
. 1 X 100%
mass a geramo
18.0 g H
426
Chapter 12
Chemical Quantities
X 100% = 11.7% H
Mass of oxygen in 1 mol geraniol:
1~ 116.0g0
~
Mass percent of 0 =
= 16.0g0
mass of 0
f
. 1 X 100%
mass 0 geramo
=
16.0 g 0
154 g C LO HUIO
X 100% =
10.4% 0
In the example above, you learned how to use the chemical formula
and the molar masses to find the mass percent of a compound. You can
also solve the reverse problem. You can use the mass percent to find the
chemical formula of an unknown compound.
Determining Chemical Formulas
Suppose you analyzed an unknown compound and found that, by
mass, it was 18.8 percent sodium, 29.0 percent chlorine, and 52.2 percent
oxygen. Because this compound contains some metal and nonmetal elements, it may be an ionic compound. To determine its chemical formula,
find the relative numbers of sodium, chlorine, and oxygen atoms in the
formula unit of the compound.
Suppose you have 100.0 g of the unknown compound. Because you
know the sample includes 18.8 g of sodium, 29.0 g of chlorine, and 52.2 g of
oxygen, use the molar mass to find the number of moles of each element.
18.8~ 11 mol Na
.
= 0.817 mol Na
23.0~
29.0~ll mol CI
-
52.2~11 malO
-
35.5~ - 0.817 rna
16.0~ - 3.26 mo
I Cl
10
You know that atoms combine in ratios of small whole numbers to
form compounds. To find the whole-number ratios, divide the mole
numbers by the smallest one.
0.817 mol Na
= 1.00 mol Na
0.817
0.817 mol Cl
= 1.00 mol CI
0.817
3.26 malO = 3.99 malO
0.817
These values are whole numbers or very close to whole numbers. Therefore, the mole ratio for this compound is 1 mol Na :1 mol Cl :4 mol O.
Similarly, the ratio of atoms in this compound is 1 Na :1 Cl :4 O.
12.2
... -
Using Moles
421
The formula of a compound having the smallest whole-number ratio
of atoms in the compound is called the empirical formula. The empirical
formula of this unknown compound is NaCI0 4 •
What is the chemical formula for this compound? You have learned
that the formula for an ionic compound represents the simplest possible
ratio of the ions present and is called a formula unit. Chemical formulas
for most ionic compounds are the same as their empirical formulas.
Because the unknown compound is ionic, the chemical formula for a formula unit of the compound' is the same as its empirical formula, NaCI0 4 .
The compound is called sodium perchlorate.
As another example, suppose the mass percents of a compound are
40.0 percent carbon, 6.70 percent hydrogen, and 53.3 percent oxygen.
Because all its elements are nonmetals, the compound is covalent. Imagine you have 100 g of the compound. Then you have 40.0 g of carbon,
6.70 g of hydrogen, and 53.3 g of oxygen. Use the molar masses of these
elements to find the number of moles of each element.
empirical
empeirikos (GK)
working from
experience
From experimental
work, the empirical
formula of an
unknown compound can be
derived.
4o.0~11 mol C _
12.0,..g.e- - 3.33 mol C
I'
6.70,.g.-H' 11 mol H
1.00~ = 6.70 mol H
J
°
53.3,g.e--\1 mol
16.0,g.e-- = 3.33 malO
~~~~,~
Now, divide all the mole numbers by the smallest one.
3.33 mol C
3.33
- - - - = 1.00 mol C
6.70 mol H
= 2.01 mol H
3.33
3.33 mol 0
3.33
=
1.00 mol 0
Because the ratio of moles is the same as the ratio of atoms, CH 20 is the
empirical formula for this compound. But, the empirical formula is not
always the chemical formula. Many different covalent compounds have
the same empirical formula, as demonstrated in Figure 12.11, because
atoms can share electrons in different ways.
Figure 12.11
Compounds Having the
Empirical Formula CH20
For each of these compounds, the ratio of atoms
is 1C:2H:l0. Because one
molecule of each compound has a different
number of each atom, it is
a different compound with
its own molecular formula.
428
Chapter 12
Chemical Quantities
To decide which multiple of the empirical formula is the correct molecular formula, you need the molar mass of the compound. Suppose a separate
analysis shows that the molar mass of the compound is 90.0 glmol. Therefore, the molecular mass of the compound is 90.0 u. The molecular mass of
a CH 20 molecule is 30.0 u. By dividing the molecular mass of the compound by 30.0 u, you find the multiple.
90.0 u
--=3
30.0 u
The molecular formula of the compound contains three empirical formula units. The molecular formula is C3 H 6 0 3 • The compound is lactic
acid, the sour-tasting substance in spoiled milk. Lactic acid is also produced by active muscles, causing them to feel sore after strenuous activity.
Connecting Ideas
In this chapter, you learned to solve several kinds of stoichiometric
problems. For each kind, you used the mole concept because when substances react, their particles interact. The number of particles at this submicroscopic level controls what happens macroscopically. In the next
chapter, you will use the mole concept and the particle nature of matter to
study the mixtures of substances called solutions.
Understanding Concepts
1. Octane, C sH 1s ' is one of the many components
of gasoline. Write the balanced chemical equation for the combustion of octane to form carbon dioxide gas and water vapor. Identify as
many mole ratios among the substances in the
combustion as you can.
If 20.0 g of
each reactant
are used,
which is the
limiting
reactant?
2. If 25.0 g of octane burn as in Problem 1, how
many grams of water will be produced? How
many grams of carbon dioxide?
3. A manufacturer advertises a new synthesis reaction for methane with a percent yield of 110 percent. Comment on this claim.
Thinking Critically
4. Predicting The reaction of iron(III) oxide and
aluminum is called the thermite reaction
because of its intense heat. The iron produced
is molten and was formerly used to weld railroad tracks in remote areas. Its balanced chemical equation is shown below.
Applying Chemistry
5. Blue Jeans Indigo, the dye used to color blue
jeans, is prepared using sodium amide. Sodium
amide contains the following mass percents of
elements: hydrogen, 5.17 percent; nitrogen, 35.9
percent; and sodium, 58.9 percent. Find an
empirical formula for sodium amide.
12.2
Using Moles
429
12.1 Counting Particles of Matter
The number of particles (atoms, molecules,
or ions) in macroscopic matter controls the
consumption and formation of substances in
chemical reactions.
• One mole equals 6.02 X 10
23
•
• Use the molar mass to convert mass to moles
or moles to mass.
12.2 Using Moles
• A balanced chemical equation provides mole
ratios of the substances in the reaction.
• The mole is a central concept in making
chemical calculations.
• The ideal gas law is expressed in the following
equation.
PV= nRT
• Percent yield measures the efficiency of a
chemical reaction.
actual yield
X 100%
theoretical yield
• Percent composition can be determined from
the chemical formula of a compound.
Percent yield =
• The empirical formula of a compound can be
determined from its percent composition.
• The chemical formula of a compound can be
determined if the molar mass and the empirical formula are known.
Key Terms
For each of the following terms, write a sentence that shows
your understanding of its meaning.
Avogadro constant
empirical formula
formula mass
ideal gas law
molar mass
molar volume
mole
molecular mass
stoichiometry
UNDERSTANDING CONCEPTS
3. Explain how you would use weighing to count
40 000 washers.
1. Your uncle leaves you a barrel of nickels. You
determine that the nickels weigh 1345 pounds
and that 50 nickels weigh 232 g. How many
dollars is the barrel of nickels worth?
4. What is the molar mass of the following sub-
2. A manufacturer must supply 20 000 connector
units, each consisting of a bolt, two washers, and
three nuts. How many of each part are needed?
stances?
a) bromine
b) argon
c) lead(II) nitrate
d) dinitrogen tetroxide
5. A reaction requires 0.498 mol of CU2S04' How
many grams must you weigh to obtain this
number of formula units?
6. Which has the largest mass?
a) three atoms of magnesium
b) one molecule of sucrose, C 12 H n O Il
c) ten atoms of helium
7. Which has the largest mass?
a) 3 mol of magnesium
b) 1 mol of sucrose, C 12 H n O\l
c) 10 mol of helium
8. What is the molecular mass of UF 6? What is
the molar mass of UF 6 ?
430
Chapter 12
Chemical Quantities
I
-
, • ,~CHAPTER 12 ASSESSMENT
~.""'.I.:'
__ J~~'_-----=-=---.
9. What mass of copper contains the same number of atoms as 68.7 g of iron?
10. Explain why the mass of copper is not equal to
the mass of iron in question 9.
11. Determine the molar mass of each of the following compounds.
a) C 6H sBr
b) K 2Cr 20 7
c) (NH4)3P04
d) Fe(N0 3)3
12. Which molar mass in question 11 contains the
largest number of atoms? Largest number of
ions? Largest number of formula units?
13. Calculate the mass of 0.345 mol of sodium
nitrite, NaN0 2 •
14. Calculate each of the following:
a) the mass in grams of 0.254 mol of calcium
sulfate
b) the number of atoms in 2.0 g of helium
c) the number of moles in 198 g of glucose,
C6H 12 0 6
d) the total number of ions in 10.8 g of magnesium bromide
15. What are the molar masses of the following
elements?
a) nitrogen
b) iodine
c) oxygen
d) nickel
16. Sodium nitrate decomposes upon heating to
form sodium nitrite and oxygen gas. This reaction is sometimes used to produce small quantities of oxygen in the lab. How many grams of
sodium nitrate must be heated to produce
128 g of oxygen?
17. Calculate the mass of silver chloride, Agel, and
dihydrogen monosulfide, H 2S, formed when
85.6 g of silver sulfide, Ag 2S, reacts with excess
hydrochloric acid, HCl.
18. Aluminum nitrite and ammonium chloride
react to form aluminum chloride, nitrogen,
and water. What mass of aluminum chloride is
present after 43.0 g of aluminum nitrite and
43.0 g of ammonium chloride have reacted
completely?
-
I'i'
-----~
19. A 1O.O-g sample of magnesium reacted with
excess hydrochloric acid to form magnesium
chloride by the reaction below. When the reaction was complete, 30.8 g of magnesium chloride were recovered. What was the percent
yield?
Mg(s)
+ 2HCI(aq)
-+ MgCI 2(aq)
+ H 2(g)
20. Calcium carbide, CaC2> reacts with water to
form calcium oxide, CaO, and acetylene, C 2H 2.
Acetylene is a fuel that combines with oxygen
in an exothermic reaction to produce water
and carbon dioxide. This combination of reactions was used to produce light in lanterns
known as "carbide lamps." A certain lamp produces 10.0 L of CO 2 (g) at STP. How many
grams of calcium carbide must have reacted in
the lamp?
21. What is the molecular formula for each of the
following compounds?
a) empirical formula: CH 2i
molar mass: 42 glmol
b) empirical formula: CHi
molar mass: 78 g/mol
c) empirical formula: N0 2i
molar mass: 92 glmol
22. An oxide of nitrogen is 26 percent nitrogen by
mass. The molar mass of the oxide is approximately 105 g/mol. What is the formula of the
compound?
23. How many moles are in each of the following
samples?
a) 43.6 g NH 3
b) 5.0 g of aspirin, C 9H 80 4
c) 15.0 g CuO
APPLYING CONCEPTS
24. Hydrogen fuels are rated with respect to their
hydrogen content. Determine the percent
hydrogen for the followir...g fuels.
a) ethane, C 2H 6
b) methane, CH4
c) whale oil, C32H6402
Chapter 12
Assessment
431
25. Oxidizers are generally compounds that contain relatively large mass percentages of oxygen. Hydrogen peroxide and sodium nitrate
are good oxidizing agents. Compare the mass
percent of oxygen in each of these compounds.
THINKING CRITICALLY
Applying Concepts
33. What is the mass in grams of a single molecule
of water?
26. Oleic acid is a component in olive oil. It is 76.5
percent C, 12.1 percent H, and 11.3 percent O.
The molar mass of the compound is approximately 282 g/mol. What is the molecular formula of oleic acid?
34. Explain how a mole is used in chemistry as
both a number and a mass.
27. Serotonin is a compound that conducts nerve
impulses in the brain. Serotonin is 68.2 percent
C, 6.86 percent H, 15.9 percent N, and 9.08
percent 0. The molar mass is 176 g/mol. What
is the molecular formula of serotonin?
36. All of the group 2 metals form metal oxides
with the general formula MO, where M is a
group 2 metal. How do the mass percentages
of oxygen in these compounds compare?
28. Gas masks used by firefighters often contain
potassium superoxide, K0 2• K0 2 reacts with
carbon dioxide and water in exhaled breath to
produce oxygen by the following reaction:
4K0 2(s) + 2H 20(g) + 4C0 2 (g)
4KHC0 3 (s) + 30 2 (g)
~
Assume that a person exhales 14.0 g of CO 2 (g)
over 20 minutes. How many grams of potassium superoxide must the mask contain in order
to react with the carbon dioxide?
29. Suppose that the percentage yield of the reaction in the last problem is 80.0 percent. How
many grams of K0 2 would need to be in the
mask to insure enough oxygen?
Art Connection
30. How does the United States standardize measurements of mass and weight?
Everyday Chemistry
31. How many grams of sodium azide, NaN 31 are
needed to fill a 0.0650-m 3 driver's air bag with
nitrogen if the pressure required is 2.00 atmospheres and the gas temperature is 40.0°C?
(Hint: 1 m 3 = 1000 L)
Chemistry and Technology
32. What are some uses of sulfuric acid?
432
Chapter 12
Comparing and Contrasting
35. Distinguish between molar mass, formula
mass, and molecular mass.
37. ChemLab Suggest a procedure to determine
the mass percent of silver nitrate, AgN0 3, in a
mixture of silver nitrate and sodium nitrate,
NaN0 3 ·
Comparing and Contrasting
38. Mini Lab 1 Explain how your procedure in this
activity is like using the molar mass to determine the number of particles in a sample of a
known substance.
Relating Cause and Effect
39. MiniLab 2 If you used calcium carbonate
instead of baking soda, how would the amount
of reactants needed change?
CUMULATIVE REVIEW
40. Distinguish between atomic number and mass
number. How do these two numbers compare
for isotopes of an element? (Chapter 2)
41. Answer the following questions about sugar
and table salt. (Chapter 4)
a) What type of compound is each substance?
b) Which compound is an electrolyte?
c) Which compound is made up of molecules?
42. Why does the second period of the periodic
table contain eight elements? (Chapter 7)
43. Explain why the metals zinc, aluminum, and
magnesium are resistant to corrosion, while
iron is not. (Chapter 8)
Chemical Quantities
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.
I~
• ~CHAPTER 12 ASSESSMENT '.1
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.
44. The volume of a gas at STP is 8.50 L. What is
its volume if the pressure is 1250 mm Hg and
the temperature is 75.0°C? (Chapter 11)
SKILL REVIEW
45. Using a Table Complete the table below. In
the last column, rank the number of particles
from smallest to largest. Did you use the number of moles or the number of individual particles for your ranking? Explain.
WRITING IN CHEMISTRY
46. A mole is such a large number that it is hard to
envision just how big 6.02 X 1023 is. Write
three of your own examples of the number of
things in a mole. One example must use time,
one must use distance, and the third is up to
you.
PROBLEM SOLVING
47. A student carries out the following sequence of
chemical reactions on a 0.635-g sample of pure
copper.
Cu(s) + 2HN0 3 (aq) ---7 CU(N0 3 )2(aq) + H 2(g)
Cu(N0 3 h(aq) + 2NaOH(aq) ---7
Cu(OHMs) + 2NaN0 3 (aq)
CU(OH)2(S) ---7 CuO(s) + H 20(l)
CuO(s) + H 2S0 4(aq) ---7 CuS04(aq) + H 20(l)
CuS04(aq) + Mg(s) ---7 Cu(s) + MgS0 4(aq)
The student isolates and weighs the product at
the end of each step before proceeding to the
next. What is the theoretical yield of the copper product in each step?
The formula units in
10.0 9 of calcium fluoride
The sodium ions in
10.0 9 of sodium chloride
The water molecules in
10.0 9 of water
The hydrogen atoms
in 10.0 9 of water
The carbon dioxide
molecules in 10.0 9 of
carbon dioxide
The carbon monoxide
molecules in 10.0 9 of
carbon monoxide
The aspirin molecules,
C9 Hs04,o in 10.0 g of aspirin
The carbon atoms contained
within 10.0 g of aspirin
The valence electrons
in 10.0 9 of aspirin
Chapter 12
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Assessment
433