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10.2 Mole-Mass and MoleVolume Relationships
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Chapter 10
Chemical Quantities
10.1 The Mole: A Measurement of
Matter
10.2 Mole-Mass and MoleVolume Relationships
10.3 Percent Composition and
Chemical Formulas
1
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10.2 Mole-Mass and MoleVolume Relationships
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CHEMISTRY
& YOU
How can you calculate the pieces of a
substance in a given mass
or volume?
Guess how many pennies are
in the container. In a similar
way, chemists use the
relationships between the
mole and quantities such as
mass, volume, and number of
particles to solve problems in
chemistry.
2
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Mass
Relationship
The Mole-Mass Relationship
How do you convert the mass of a
substance to the number of moles of
the substance?
3
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Mass
Relationship
Use the molar mass of an element or
compound to convert between the mass
of a substance and the moles of the
substance.
4
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Mass
Relationship
Use the molar mass of an element or
compound to convert between the mass
of a substance and the moles of the
substance.
• The conversion factors for these calculations
are based on these relationships.
molar mass
1 mol
5
or
1 mol
molar mass
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Mass
Relationship
In some situations the term molar mass may be
unclear.
• What is the molar mass of nitrogen?
6
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Mass
Relationship
In some situations the term molar mass may be
unclear.
• What is the molar mass of nitrogen?
– If you assume elemental nitrogen (N), then the
answer is 14.0 g/mol.
– If you assume molecular nitrogen (N2), then the
molar mass is 28.0 g/mol (2 × 14.0 g/mol).
7
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Mass
Relationship
In some situations the term molar mass may be
unclear.
• What is the molar mass of nitrogen?
– If you assume elemental nitrogen (N), then the
answer is 14.0 g/mol.
– If you assume molecular nitrogen (N2), then the
molar mass is 28.0 g/mol (2 × 14.0 g/mol).
• You can avoid confusion by using the formula of
the substance you want. ( N or N2 )
8
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
Converting Moles to Mass
Items made out of aluminum, such
as aircraft parts and cookware, are
resistant to corrosion because the
aluminum reacts with oxygen in the
air to form a coating of aluminum
oxide (Aℓ2O3). This tough, resistant
coating prevents any further
corrosion. What is the mass, in
grams, of 12.51 mol of aluminum
oxide?
9
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
1 Analyze List the known and the unknown.
The mass of the compound is calculated from
the known number of moles of the compound.
The desired conversion is moles → mass.
KNOWN
number of moles = 12.51 mol Aℓ2O3
UNKNOWN
mass = ? g Aℓ2O3
10
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10.2 Mole-Mass and MoleVolume Relationships
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Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Aℓ2O3.
2 mol Aℓ × 27.0 g Aℓ
1
1 mol Aℓ
3 mol O
1
11
= 54.0 g Aℓ
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Aℓ2O3.
2 mol Aℓ × 27.0 g Aℓ
1
1 mol Aℓ
3 mol O × 16.0 g O
1
1 mol O
12
=
=
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Aℓ2O3.
2 mol Aℓ × 27.0 g Aℓ
1
1 mol Aℓ
3 mol O × 16.0 g O
1
1 mol O
= 54.0 g Aℓ
= 48.0 g O
1 mol Aℓ2O3 = 54.0 g Aℓ + 48.0 g O
13
= 102.0 g of Aℓ2O3
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Identify the conversion factor needed.
102.0 g Aℓ2O3
1 mol Aℓ2O3
or
1 mol Aℓ2O3
102.0 g Aℓ2O3
14
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Multiply the given number of moles
by the conversion factor.
12.51 mol Aℓ2O3 × 102.0 g Aℓ2O3
1
1 mol Aℓ2O3
= 1276.02 g of Aℓ2O3
= 1276 g of Aℓ2O3
15
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Sample Problem
3 Evaluate Does the result make sense?
• The number of moles of Aℓ2O3 is
approximately 12, and each has a mass
of approximately 100 g.
• The answer should be close to 1200 g.
• The answer has been rounded to the
correct number of significant figures.
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10.2 Mole-Mass and MoleVolume Relationships
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Sample Problem
Converting Mass to Moles
When iron is exposed to air,
it corrodes to form a redbrown rust. Rust is iron(III)
oxide (Fe2O3). How many
moles of iron(III) oxide are
contained in 33.3 g of pure
Fe2O3?
17
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10.2 Mole-Mass and MoleVolume Relationships
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Sample Problem
1 Analyze List the known and the unknown.
The number of moles of the compound is
calculated from the known mass of the
compound. The conversion is mass → moles.
KNOWN
mass = 33.3 g Fe2O3
UNKNOWN
number of moles = ? mol Fe2O3
18
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10.2 Mole-Mass and MoleVolume Relationships
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Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Fe2O3.
2 mol Fe ×
1
3 mol O ×
1
19
56 g Fe
=
1 mol Fe
16.0 g O
=
1 mol O
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Fe2O3.
20
2 mol Fe ×
1
56 g Fe =
1 mol Fe
3 mol O ×
1
16.0 g O =
1 mol O
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Fe2O3.
21
2 mol Fe ×
1
56 g Fe = 112 g Fe
1 mol Fe
3 mol O ×
1
16.0 g O = 48.0 g O
1 mol O
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
First determine the mass of 1 mol of Fe2O3.
2 mol Fe ×
1
56 g Fe = 112 g Fe
1 mol Fe
3 mol O ×
1
16.0 g O = 48.0 g O
1 mol O
1 mol Fe2O3 = 112 g Fe + 48.0 g O
22
= 160 g Fe2O3
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10.2 Mole-Mass and MoleVolume Relationships
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Sample Problem
2 Calculate Solve for the unknown.
Identify the conversion factor relating
grams of Fe2O3 to moles of Fe2O3.
1 mol Fe2O3
160 g Fe2O3
Note that the known unit (g) is in the denominator
and the unknown unit (mol) is in the numerator.
23
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Multiply the given mass by the
conversion factor.
33.3 g Fe2O3 ×
1
1 mol Fe2O3
160 g Fe2O3
= 0.208125 mol of Fe2O3
= 0.208 mol of Fe2O3
24
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Sample Problem
3 Evaluate Does the result make sense?
The given mass (about 33 g) is slightly less
than the mass of one-quarter mole of Fe2O3
(about 40 g), so the answer should be
slightly less than one-quarter (0.25) mol.
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You know how many grams of a
substance you have and want to find
out how many moles this is. What other
information do you need to know, and
where can you find it?
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You know how many grams of a
substance you have and want to find
out how many moles this is. What other
information do you need to know, and
where can you find it?
You also need to know the molar mass, which
you can calculate using the chemical formula
based on the atomic mass listed in the periodic
table of elements.
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Notice that the volumes of one mole of
different solid and liquid substances are not
the same.
• The volumes of one
mole of glucose (blood
sugar) and one mole of
parachlorobenzene
(moth crystals) are much
larger than the volume of
one mole of liquid water.
28
moth crystals
glucose
water
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
The Mole-Volume Relationship
How does the relationship between
the volume of a gas at STP and the
number of moles of that gas compare
to solids and liquids?
STP stands for STANDARD TEMPERATURE and PRESSURE
29
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10.2 Mole-Mass and MoleVolume Relationships
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The Mole-Volume
Relationship
The Mole-Volume Relationship
Unlike liquids and solids, the volumes of
moles of gases, are much more
predictable measured under the same
physical conditions.
30
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Avogadro’s Hypothesis
Avogadro’s hypothesis states that
equal volumes of gases at STP contain
equal numbers of particles.
* Whether the particles are big or small, there
are large amounts of empty space
between individual particles of gas.
31
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The Mole-Volume
Relationship
Avogadro’s Hypothesis
Same amount of space and same amount of particles.
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10.2 Mole-Mass and MoleVolume Relationships
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The Mole-Volume
Relationship
Avogadro’s Hypothesis
The volume of a gas varies with a change
in temperature and/or a change in pressure.
33
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10.2 Mole-Mass and MoleVolume Relationships
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The Mole-Volume
Relationship
Avogadro’s Hypothesis
The volume of a gas varies with a change
in temperature and/or a change in pressure.
• Due to variations in temperature and pressure,
the volume of a gas is usually measured at STP.
34
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Avogadro’s Hypothesis
The volume of a gas varies with a change
in temperature and/or a change in pressure.
• Due to variations in temperature and pressure,
the volume of a gas is usually measured at STP.
• Standard temperature and pressure (STP)
means a temperature of 0°C and a pressure of
101.3 kPa, or 1 atmosphere (atm).
35
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10.2 Mole-Mass and MoleVolume Relationships
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The Mole-Volume
Relationship
Avogadro’s Hypothesis
At STP, 1 mol or 6.02 × 1023 representative
particles, of any gas occupies a volume of
22.4 L.
• The quantity, 22.4 L, is called the molar
volume of a gas.
36
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Calculating the Volume and Moles of a
Gas at STP
The molar volume is used to convert between
the number of moles of gas and the volume
of the gas at STP.
• The conversion factors for these calculations are
based on the relationship 22.4 L = 1 mol at STP.
22.4 L
and
1 mol
37
1 mol
22.4 L
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
Calculating Gas Quantities at STP
Carbon dioxide (CO2) is a gas
produced by burning coal. It is
the leading pollutant that adds
to the “Greenhouse Effect”.
Determine the volume, in liters,
of 0.71 mol of CO2 gas at STP.
38
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
1 Analyze List the knowns and the unknown.
Since CO2 is a gas, the volume at STP can be
calculated from the known number of moles.
KNOWNS
number of moles = 0.71 mol CO2
1 mol CO2 = 22.4 L CO2 at STP
UNKNOWN
volume = ? L CO2
39
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Sample Problem
2 Calculate Solve for the unknown.
First identify the conversion factor
relating moles of CO2 to volume of CO2
at STP.
22.4 L CO2
1 mol CO2
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Multiply the given number of moles by
the conversion factor.
0.71 mol CO2 × 22.4 L CO2
1 mol CO2
1
= 15.904 L CO2
= 16 L CO2
41
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Sample Problem
3 Evaluate Does the result make sense?
• One mole of any gas at STP has a
volume of 22.4 L, so 0.71 mol should
have a volume slightly less than threequarters of a mole of the gas or 16.8 L.
• The answer should have two significant
figures.
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Calculating Molar Mass and Density
A gas-filled balloon will either sink or
float in the air depending on whether
the density of the gas inside the balloon
is greater or less than the density of the
surrounding air.
• Different gases have different densities.
• Usually the density of a gas is measured in
grams per liter (g/L) and at a specific
temperature.
43
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Calculating Molar Mass and Density
• At STP, the density of a gas and the
molar volume (22.4 L/mol) can be
used to calculate the molar mass of
the gas.
44
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
Calculating Molar Mass and Density
• At STP, the density of a gas and the
molar volume (22.4 L/mol) can be
used to calculate the molar mass of
the gas.
• Similarly, the molar mass of a gas
and the molar volume can be used to
calculate the density of a gas at STP.
45
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
Calculating the Molar Mass of a
Gas at STP
The density of an unknown
Noble gas is found to be
5.862 g/L at STP. What is the
molar mass of this gas?
What is the identity of this
Noble gas?
46
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
1 Analyze List the knowns and the unknown.
The molar mass of a gas is calculated from the known
density of the gas and the molar volume at STP.
KNOWNS
density = 5.862 g/L
1 mol of gas at STP = 22.4 L
UNKNOWN
molar mass = ? g/mol
Noble gas = ?
47
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Sample Problem
2 Calculate Solve for the unknown.
First identify the conversion factor needed
to convert density to molar mass.
22.4 L
1 mol
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Multiply the given density by the
conversion factor.
5.862 g
1L
×
22.4 L
1 mol
= 131.3088 g/mol
= 131.3 g/mol
Noble gas = Xenon
49
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Sample Problem
3 Evaluate Does the result make sense?
• The density of the gas is relatively high so
one would expect the molar mass to be
relatively high.
• The answer should have four significant
figures.
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
Calculating the Molar Mass of a
Gas at STP
What is the density of
fluorine gas at STP?
51
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
1 Analyze List the knowns and the unknown.
The density of a gas is calculated from the known
molar mass of the gas and the molar volume at STP.
KNOWNS
molar mass of F = 19 g/mol
1 mol of gas at STP = 22.4 L
UNKNOWN
density of F2 = ? g/L
52
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
First calculate the molar mass of F2.
2 mol F
1
x
19 g
1 mol F
= 38 g
1 mol F2 = 38 g F
53
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Now identify the conversion factor needed
to convert density to molar mass.
1 mol
22.4 L
54
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10.2 Mole-Mass and MoleVolume Relationships
>
Sample Problem
2 Calculate Solve for the unknown.
Multiply the given density by the
conversion factor.
38 g
1 mol
×
1 mol
22.4 L
= 1.6964285 g/L
= 1.7 g/L
55
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10.2 Mole-Mass and MoleVolume Relationships
>
The Mole-Volume
Relationship
The mole is at the center
of chemical calculations.
• To convert from one unit to
another, you must use the
mole as an intermediate
step.
56
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Why does one mole of any gas always
occupy the same volume (22.4 L) at
standard temperature and pressure?
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Why does one mole of any gas always
occupy the same volume (22.4 L) at
standard temperature and pressure?
Gas molecules vary in size, just like molecules of
solids and liquids, but the particles in a gas are so
far apart that the change in volume (size) of the
molecules doesn’t change the volume the gas
occupies at STP.
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10.2 Mole-Mass and MoleVolume Relationships
> Key Concepts
Use the molar mass of an element or
compound to convert between the
mass of a substance and the moles
of the substance.
The molar volume is used to convert
between the number of moles of gas
and the volume of the gas at STP.
59
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10.2 Mole-Mass and MoleVolume Relationships
>
END OF 10.2
60
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