3.3 Moles,
3.4 Molar Mass,
and
3.5 Percent
Composition
Collection Terms
A collection term states
a specific number of items.
• 1 dozen donuts
= 12 donuts
• 1 ream of paper
= 500 sheets
• 1 case
= 24 cans
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3–2
A Mole of Atoms
A mole is a collection that contains
• the same number of particles as there are pure 12carbon atoms in 12.0 g of carbon.
• 6.022 x 1023 atoms of an element (Avogadro’s number).
1 mole element
Number of Atoms
1 mole C
= 6.022 x 1023 C atoms
1 mole Na
= 6.022 x 1023 Na atoms
1 mole Au
= 6.022 x 1023 Au atoms
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A Mole of a Compound
A mole
• of a covalent compound has Avogadro’s number of
molecules.
1 mole CO2 = 6.022 x 1023 CO2 molecules
1 mole H2O = 6.022 x 1023 H2O molecules
• of an ionic compound contains Avogadro’s number of
formula units.
1 mole NaCl
= 6.022 x 1023 NaCl formula units
1 mole K2SO4 = 6.022 x 1023 K2SO4 formula units
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3–4
Samples of One Mole Quantities
1 mole C
=
6.02 x 1023 C atoms
1 mole Al
=
6.02 x 1023 Al atoms
1 mole S
=
6.02 x 1023 S atoms
1 mole H2O =
6.02 x 1023 H2O molecules
1 mole CCl4 =
6.02 x 1023 CCl4 molecules
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3–5
Avogadro’s Number
Avogadro’s number 6.022 x 1023 can be written as an
equality and two conversion factors.
Equality:
1 mole
= 6.022 x 1023 particles
Conversion Factors:
6.022 x 1023 particles
1 mole
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and
1 mole
6.022 x 1023 particles
3–6
Using Avogadro’s Number
Avogadro’s number is used to convert
moles of a substance to particles.
How many Cu atoms are in 0.50 mole Cu?
0.50 mole Cu x 6.022 x 1023 Cu atoms
1 mole Cu
= 3.0 x 1023 Cu atoms
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Publishing as Benjamin Cummings
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3–7
Using Avogadro’s Number
Avogadro’s number is used to convert
particles of a substance to moles.
How many moles of CO2 are in
2.50 x 1024 molecules CO2?
2.50 x 1024 molecules CO2 x
1 mole CO2
6.02 x 1023 molecules CO2
= 4.15 mole CO2
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3–8
Learning Check
1. The number of atoms in 2.0 mole Al is
A. 2.0 Al atoms.
B. 3.0 x 1023 Al atoms.
C. 1.2 x 1024 Al atoms.
2. The number of moles of S in 1.8 x 1024 atoms S is
A. 1.0 mole S atoms.
B. 3.0 mole S atoms.
C. 1.1 x 1048 mole S atoms.
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3–9
Solution
A. The number of atoms in 2.0 mole Al is
2.0 mole Al x 6.02x1023 Al atoms = 1.2 x 1024Al atoms
1 mole Al
Avogadro’s
Number
B. The number of moles of S in 1.8 x 1024 atoms S is
1.8 x 1024 atoms S x 1 mole S
6.02 x 1023 S atoms
Avogadro’s
Number
= 3.0 moles of S atoms
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3–10
Subscripts and Moles
The subscripts in a formula show
• the relationship of atoms in the formula.
• the moles of each element in 1 mole of compound.
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O
In 1 mole:
6 mole C 12 mole H 6 mole O
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3–11
Subscripts State Atoms and Moles
9 mole C 8 mole H 4 mole O
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Factors from Subscripts
The subscripts are used to write conversion factors for
moles of each element in 1 mole compound. For aspirin
C9H8O4, the following factors can be written:
9 mole C
1 mole C9H8O4
and
1 mole C9H8O4
9 mole C
8 mole H
4 mole O
1 mole C9H8O4 1 mole C9H8O4
1 mole C9H8O4
8 mole H
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1 mole C9H8O4
4 mole O
3–13
Learning Check
A. How many mole O are in 0.150 mole
aspirin C9H8O4?
B. How many O atoms are in 0.150 mole
aspirin C9H8O4?
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3–14
Solution
A. How many mole O are in 0.150 mole aspirin C9H8O4?
0.150 mole C9H8O4 x 4 mole O
= 0.600 mole O
1 mole C9H8O4
subscript factor
B. How many O atoms are in 0.150 mole aspirin C9H8O4?
0.150 mole C9H8O4 x 4 mole O x 6.02 x 1023 O atoms
1 mole C9H8O4
1 mole O
subscript
Avogadro’s
factor
Number
= 3.61 x 1023 O atoms
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3–15
Molar Mass
The molar mass
• is the mass of one mole
of an element or
compound.
• is the atomic mass
expressed in grams.
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3–16
Molar Mass from Periodic Table
Molar mass is
the atomic
mass
expressed in
grams.
1 mole Ag
= 107.9 g
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1 mole C
= 12.01 g
1 mole S
= 32.07 g
3–17
Molar Mass of a Compound
The molar mass of a compound is the sum of the
molar masses of the elements in the formula.
Example: Calculate the molar mass of CaCl2 to the
tenths decimal place.
Element
Ca
Number
of Moles
1
40.1 g/mole
40.1 g
Cl
2
35.5 g/mole
71.0 g
CaCl2
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Atomic Mass Total Mass
111.1 g
3–18
Some One-mole Quantities
One-Mole Quantities
32.1 g
55.9 g
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58.5 g
294.2 g
342.2 g
3–20
Learning Check
What is the molar mass of each of the following?
A. K2O
B. Al(OH)3
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3–21
Solution
A. K2O 94.2 g/mole
2 mole K (39.1 g/mole) + 1 mole O (16.0 g/mole)
78.2 g + 16.0 g = 94.2 g
B. Al(OH)3 78.0 g/mole
1 mole Al (27.0 g/mole) + 3 mol O (16.0 g/mole)
+ 3 mol H (1.01 g/mole)
27.0 g + 48.0 g + 3.03 g = 78.0 g
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3–22
Learning Check
Prozac, C17H18F3NO, is an antidepressant that inhibits
the uptake of serotonin by the brain. What is the molar
mass of Prozac?
1) 40.1 g/mole
2) 262 g/mole
3) 309 g/mole
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3–23
Solution
Prozac, C17H18F3NO, is an antidepressant that inhibits
the uptake of serotonin by the brain. What is the molar
mass of Prozac?
3) 309 g/mole
17 C (12.0) + 18 H (1.01) + 3 F (19.0)
+ 1 N (14.0) + 1 O (16.0) =
204 + 18.2 + 57.0 + 14.0 + 16.0 = 309 g/mole
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3–24
Molar Mass Factors
Molar mass conversion factors
• are written from molar mass.
• relate grams and moles of an element or compound.
Example: Write molar mass factors for methane, CH4, used in
gas stoves and gas heaters.
Molar mass:
1 mol CH4 = 16.0 g
Conversion factors:
and
16.0 g CH4
1 mole CH4
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1 mole CH4
16.0 g CH4
3–25
Learning Check
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass conversion factors for acetic
acid.
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3–26
Solution
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass factors for acetic acid.
Calculate molar mass:
24.0 + 4.04 + 32.0 = 60.0 g/mole
1 mole of acetic acid
Molar mass factors
1 mole acetic acid
60.0 g acetic acid
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=
and
60.0 g acetic acid
60.0 g acetic acid
1 mole acetic acid
3–27
Calculations Using Molar Mass
Molar mass factors are used to convert between the
grams of a substance and the number of moles.
Grams
Molar mass factor
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Moles
3–28
Moles to Grams
Aluminum is often used to build lightweight bicycle
frames. How many grams of Al are in 3.00 mole Al?
Molar mass equality: 1 mole Al = 27.0 g Al
Setup with molar mass as a factor:
3.00 mole Al
x
27.0 g Al
1 mole Al
= 81.0 g Al
molar mass factor for Al
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3–29
Learning Check
Allyl sulfide C6H10S is a
compound that has the odor
of garlic. How many moles
of C6H10S are in 225 g?
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3–30
Solution
Calculate the molar mass of C6H10S.
(6 x 12.0) + (10 x 1.01) + (1 x 32.1) = 114.2 g/mole
Set up the calculation using a mole factor.
225 g C6H10S x 1 mole C6H10S
114.2 g C6H10S
molar mass factor (inverted)
= 1.97 mole C6H10S
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3–31
Grams, Moles, and Particles
A molar mass factor and Avogadro’s number convert
• grams to particles.
molar mass
g
mole
Avogadro’s
number
particles
• particles to grams.
Avogadro’s
number
particles
mole
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molar mass
g
3–32
Learning Check
How many H2O molecules are in 24.0 g H2O?
1) 4.52 x 1023
2) 1.44 x 1025
3) 8.03 x 1023
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3–33
Solution
How many H2O molecules are in 24.0 g H2O?
3) 8.02 x 1023
24.0 g H2O x 1 mole H2O x 6.02 x 1023 H2O molecules
18.0 g H2O
1 mole H2O
= 8.03 x 1023 H2O molecules
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3–34
Learning Check
If the odor of C6H10S can be detected from 2 x 10-13
g in one liter of air, how many molecules of C6H10S
are present?
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3–35
Solution
If the odor of C6H10S can be detected from 2 x 10-13 g
in one liter of air, how many molecules of C6H10S are
present?
2 x 10-13 g x
1 mole
114.2 g
x 6.02 x 1023 molecules
1 mole
= 1 x 109 molecules C6H10S
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• Percent Composition of Compounds: What is the mass
percentage of C in CO2?
• The mass percentage is calculated using the
following equation:
• If a sample consisting of 1 mole of CO2 is used, the molebased relationships given earlier show that
1 mole CO2=44.01 g CO2 (12.01 g C + 32.00 g O). Thus,
the mass of C in a specific mass of CO2 is known, and the
problem is solved as follows:
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3–37
• What is the mass percentage of oxygen in CO2?
• The mass percentage is calculated using the following
equation:
• Once again, a sample consisting of 1 mole of CO2 is used
to take advantage of the mole-based relationships given
earlier where:
1 mole CO2 = 44.01g CO2 = 12.01 g C + 32.00g O
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• Thus, the mass of O in a specific mass of CO2 is
known, and the problem is solved as follows:
• We see that the % C + % O = 100% , which
should be the case because C and O are the only
elements present in CO2.
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3–39
QUESTION
Morphine, derived from opium plants, has the potential for use and
abuse. It’s formula is C17H19NO3. What percent, by mass, is the
carbon in this compound?
1.
2.
3.
4.
42.5%
27.9%
71.6%
This cannot be solved until the mass of the sample is given.
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3–40
ANSWER
Choice 3 is correct. First determine the molar mass of the
compound, then divide that into the total mass of carbon present,
and finally, multiply that by 100.
(17 × 12.01) / ((17 × 12.01) + (19 × 1.008) + (1 × 14.01) + 3 ×
16.00)) = 0.716
0.716 × 100 = 71.6 %
Section 3.5: Percent Composition of Compounds
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