Chemistry 11 Solutions
Chapter 5
The Mole: A Chemist’s Counter
Section 5.1 The Mole and the Avogadro Constant
Solutions for Practice Problems
Student Edition page 228
1. Practice Problem (page 228)
An average refrigerator has a volume of 0.6 m3. If a grain of salt has a volume
of 9.39 × 10−11 m3, how any refrigerators would 1 mol of salt grains fill?
What Is Required?
You need to determine the number of refrigerators that would be filled by 1
mol of grains of sand.
What Is Given?
You know the volume of one refrigerator: 0.6 m3
You know the volume of one grain of sand: 9.39 × 10–11 m3
Plan Your Strategy
The Avogadro constant, NA, is the number of particles in 1 mol: 6.02 × 1023
Multiply the volume of one grain of sand by the Avogadro constant to
determine the total volume of 6.02 × 1023 grains of sand. Then divide the total
volume of the grains of sand by the volume of one refrigerator.
Act on Your Strategy
total volume of 1 mol of grains of
sand = 6.02 × 1023 grains × 9.39 × 10−11 m3 / grain
= 5.5986 × 1013 m3
number of refrigerators =
5.5986 × 1013 m3
0.6 m3 /refrigerator
= 9.331 × 1013 refrigerators
= 9 × 1013 refrigerators
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 1 Chemistry 11 Solutions
Therefore, 6.02 × 1023 grains of sand would fill 9 × 1013 refrigerators.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
2. Practice Problem (page 228)
If you drove for 6.02 × 1023 days at a speed of 110 km/h, how far would you
travel?
What Is Required?
You need to calculate the total distance travelled.
What Is Given?
You know the speed you are travelling at: 110 km/h
You know the total time elapsed during the trip: 6.02 × 1023 days
Plan Your Strategy
Use the conversion factor of 24 h/day to determine the total time for the trip in
hours.
Calculate the distance travelled using the following relationship: distance =
speed × time.
Act on Your Strategy
time elapsed (in hours) = 6.02 ×1023 days × 24 h/ day
= 1.4448 × 1025 h
distance travelled = speed × time
= 110 km/ h × 1.4448 × 1025 h
= 1.58928 × 1027 km
= 1.6 × 1027 km
The distance travelled in 6.02 × 1023 days is 1.6 × 1027 km.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 2 Chemistry 11 Solutions
3. Practice Problem (page 228)
How long would it take to count 6.02 × 1023 raisins, if you counted at a rate of
one raisin per second?
What Is Required?
You need to calculate the time needed to count 6.02 × 1023 of raisins.
What Is Given?
You know the total number of raisins to be counted: 6.02 × 1023
You know the rate at which the raisins are counted: 1 raisin per second
Plan Your Strategy
Divide the total number of raisins by the rate of counting. To express the
answer in days or years, multiply the time in seconds by the appropriate
1 year
1h
1 day
conversion factors:
,
, and
.
365 days
3600 s 24 h
Act on Your Strategy
time (in seconds) to count raisins =
number of raisins
rate of counting
6.02 ×1023 raisins
=
1 raisin/s
= 6.02 × 1023 s
To convert the time from seconds to days:
1 day
1h
×
time (in days) = 6.02 × 1023 s ×
24 h
3600 s
= 6.9676 × 1018 days
To convert the time from days to years:
time (in years) = 6.9676 × 1018 days ×
1 year
365 days
= 1.90892 × 1016 years
= 1.9 × 1016 years
It would take 6.97 × 1018 days, or 1.9 × 1016 years, to count 6.02 × 1023 raisins.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 3 Chemistry 11 Solutions
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
4. Practice Problem (page 228)
A ream of paper (500 sheets) is 4.8 cm high. If you stacked 6.02 × 1023 sheets
of paper on top of each other, how high would the stack be, in kilometres?
What Is Required?
You need to calculate the height (in kilometres) of 6.02 × 1023 sheets of paper.
What Is Given?
You know the total number of sheets of paper: 6.02 × 1023 sheets
You know the height of 500 sheets of paper: 4.8 cm
Plan Your Strategy
Divide the total number of sheets by 500.
Multiply this amount by the height (in centimetres) of 500 sheets.
Convert this height to kilometres: 1 cm = 1 × 10–5 km
Act on Your Strategy
total height (in centimetres) = 6.02 × 1023 sheets ×
4.8 cm
500 sheets
= 5.779 × 1021 cm
total height (in kilometres) = 5.779 × 1021 cm × 1 × 10 –5 km/ cm
= 5.779 × 1016 km
= 5.8 × 1016 km
The height of 6.02 × 1023 sheets of paper is 5.8 × 1016 km.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 4 Chemistry 11 Solutions
5. Practice Problem (page 228)
The total volume of the Rogers Centre is 1.6 × 10 m3. If the volume of 100
peas is about 55 cm3, how many Rogers Centres would 6.02 × 1023 peas fill?
What Is Required?
You need to determine the number of Rogers Centres that would be filled by
6.02 × 1023 peas.
What Is Given?
You know the volume of one Rogers Centre: 1.6 × 106 m3
You know the volume of 100 peas: 55 cm3
You know the total number of peas: 6.02 × 1023
Plan Your Strategy
Divide the total number of peas by 100. Multiply this answer by 55 to calculate
the total volume of the peas in cm3.
Convert the total volume of peas to m3: 1 m3 = 1 × 106 cm3
Divide the total volume of peas in m3 by the volume of one Rogers Centre.
Act on Your Strategy
total volume (in cm3) of 6.02 × 1023 peas = 6.02 × 10 23 peas ×
55 cm 3
100 peas
= 3.311 × 1023 cm 3
total volume (in km3) of 6.02 × 1023 peas = 3.311 × 1023 cm3 ×
1 m3
1 × 106 cm3
= 3.311 × 1017 m3
number of Rogers Centres =
3.311 × 1017 m3
1.6 × 106 m3
= 2.069 × 1011
= 2.1 × 1011
Therefore, 6.02 × 1023 peas would fill 2.1 × 1011 Rogers Centres.
Check Your Solution
The units cancel properly. Comparing the powers of ten in the calculation
indicates that the answer is reasonable. The number of significant digits in the
answer agrees with the given data having the least number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 5 Chemistry 11 Solutions
6. Practice Problem (page 228)
Canada’s coastline is 243 042 km long. If you laid 6.02 × 1023 metre sticks end
to end along the coast of Canada, how many rows of sticks would there be?
What Is Required?
You need to determine the number of rows of metre sticks that would line the
coastline.
What Is Given?
You know the length of the coastline (which is one row of sticks): 243 042 km
You know the total number of metre sticks: 6.02 × 1023
Plan Your Strategy
Convert the length of the coastline to metres: 1 km = 1 × 103 m
Calculate the number of rows of metre sticks by dividing the total number of
metre sticks by the length of the coastline.
Act on Your Strategy
length of coastline (in metres) = 243 042 km × 1 × 103 m/ km
= 2.43042 × 108 m
6.02 × 1023 metre sticks
number of rows of metre sticks =
2.43042 × 108 metre sticks /row
= 2.4769 × 1015 rows
= 2.48 × 1015 rows
There would be 2.48 × 1015 rows of metre sticks.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
7. Practice Problem (page 228)
Earth’s oceans contain about 1.31 × 109 km3 of water. One tablespoon is equal
to 15 cm3. If you could remove 6.02 × 1023 tablespoons of water from Earth’s
oceans, would you completely drain them?
What Is Required?
You need to determine if the removal of 6.02 × 1023 tablespoons of water
would drain the oceans of water.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 6 Chemistry 11 Solutions
What Is Given?
You know the total volume of water in the ocean: 1.31 × 109 m3
You know the volume of one tablespoon: 15 cm3
Plan Your Strategy
Convert the volume of water in the oceans from km3 to cm3: 1 km3 = 1 × 1015
cm3
Divide the total volume of water in the oceans by the volume of one
tablespoon.
Compare the number of tablespoons of water in the ocean with 6.02 × 1023
tablespoons.
Act on Your Strategy
volume (in cm3) of water in the ocean = 1.31 × 109 km3 ×
1 × 1015 cm3
1 km3
= 1.31 ×1024 cm3
total number of tablespoons of water in the oceans =
1.31 × 1024 cm3
15 cm3 /tablespoon
= 8.73 × 1022 tablespoons
Since Earth’s oceans contain only 8.73 × 1022 tablespoons of water, removing
any more than that would completely drain the oceans, which would be the
case if removing 6.02 × 1023 tablespoons.
Alternative Solution
Plan Your Strategy
The 6.02 × 1023 tablespoons equates to 1 mol of tablespoons. Calculate the
volume (in cm3) of 1 mol of teaspoons.
Convert the volume of 1 mol of tablespoons to km3: 1 cm3 = 1 × 10–15 km3
Compare the volume of 1 mol of tablespoons to the total volume of the oceans.
Act on Your Strategy
Total volume, V, of 1 mol of tablespoons:
V = 6.02 × 1023 tablespoons × 15 cm 3 / tablespoons
= 9.03 × 1024 cm 3
= 9.03 × 1024 cm3 × 1 × 10 –15 km3 / cm3
= 9.03 × 109 km3
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 7 Chemistry 11 Solutions
This is greater than the total volume of water in the oceans. The oceans would
be drained by the removal of 1 mol of tablespoons.
Check Your Solution
The units cancel properly. A check of the powers of ten indicates that the
answer is reasonable.
8. Practice Problem (page 228)
Suppose that you were given 6.02 × 1023 pennies when you were born and you
lived for 100 years. How much money, in dollars, would you need to spend
each second if you wanted to spend all this money in your lifetime?
What Is Required?
You need to determine how much money (in dollars) to spend per second.
What Is Given?
You know the total amount of money to be spent: 6.02 × 1023 pennies
You know that the money must be spent in 100 years.
Plan Your Strategy
1 year 1 day
1h
,
, and
.
365 days 24 h
3600 s
Divide the number of pennies by the time in seconds to determine the rate of
spending.
Divide the rate of spending in pennies by 100 to determine the rate of spending
in dollars.
Convert the time of 100 years to seconds using
Act on Your Strategy
time (in seconds) of 100 years = 100 years ×
365 days
1 year
×
24 h
3600 s
×
1 day
h
= 3.1536 × 109 s
6.02 × 1023 pennies
rate of spending (in pennies) =
3.1536 × 109 s
= 1.9089 × 1014 pennies/s
= 1.91 × 1014 pennies/s
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 8 Chemistry 11 Solutions
rate of spending (in dollars) = 1.91 × 1014 pennies /s ×
1 dollar
100 pennies
= 1.91 × 1012 dollars/s
The 6.02 × 1023 pennies must be spent at a rate of 1.91 × 1012 dollars per
second to be used up in 100 years.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
9. Practice Problem (page 228)
How would the mass of pennies you were given in question 8 compare with
the mass of Earth? The mass of 10 pennies is about 24 g. The mass of Earth is
5.98 × 1024 kg.
What Is Required?
You need to compare the mass of Earth with the mass of 6.02 × 1023 of
pennies.
What Is Given?
You know the total number of pennies: 6.02 × 1023
You know the mass of 10 pennies: 24 g
You know the mass of Earth: 5.98 × 1024 kg
Plan Your Strategy
Calculate the mass of 6.02×1023 pennies (in grams) by dividing the number of
pennies by 10. Then multiply this number by 24.
Convert the mass of pennies to kilograms: 1 g = 1 × 10–3 kg
Compare the mass of Earth with the mass of 6.02 × 1023 pennies, expressed in
kilograms.
Act on Your Strategy
Mass, m, (in grams) of 6.02 × 1023 pennies:
24 g
m = 6.02 × 1023 pennies ×
10 pennies
= 1.4448 × 1024 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 9 Chemistry 11 Solutions
Mass, m, (in kilograms) of 6.02 × 1023 pennies:
m = 1.4448 × 1024 g × 1 × 10 –3 kg/ g
= 1.4448 × 1021 kg
Ratio of mass of Earth to mass of pennies:
5.98 × 1024 kg
mass of Earth
=
mass of 6.02 × 1023 pennies 1.4448 × 1021 kg
4.139 × 103
=
1
= 4.139 × 103
= 4.1 × 103
The mass of Earth is about 4.1 × 102 times the mass of 6.02 × 1023 pennies.
Check Your Solution
The units cancel properly. The answer seems reasonable.
10. Practice Problem (page 228)
If a row of approximately 5.0 × 107 atoms measured 1.0 cm, how long would a
row of 6.02 × 1023 atoms be?
What Is Required?
You need to determine the length of a row containing 6.02 × 1023 atoms.
What Is Given?
You know the total number of atoms: 6.02 × 1023
You know the number of atoms in a length of 1 cm: 5.0 × 107
Plan Your Strategy
Divide the total number of atoms by the number of atoms in a 1 cm length.
Act on Your Strategy
6.02 × 10 23 atoms
length of 6.02 × 10 atoms =
5.0 × 107 atoms /cm
23
= 1.2 × 1016 cm
The length of a row of 6.02 × 1023 of atoms would be 1.2 × 1016 cm.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 10 Chemistry 11 Solutions
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
Section 5.1 The Mole and the Avogadro Constant
Solutions for Practice Problems
Student Text page 230
11. Practice Problem (page 230)
A Canadian penny contains 0.106 mol of copper. How many atoms of copper
are in a Canadian penny?
What Is Required?
You need to calculate the number of atoms, N, of copper in a Canadian penny.
What Is Given?
You know that the penny contains 0.106 mol of copper, Cu(s), atoms.
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of atoms of Cu(s) using the relationship N = n × N A .
Act on Your Strategy
Number of atoms, N, of Cu(s):
N = n × NA
= 0.106 mol × 6.02 × 1023 atoms/ mol
= 6.3812 × 1022 atoms
= 6.38 × 1022 atoms
There are 6.38 × 1022 atoms of Cu(s) in the Canadian penny.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 11 Chemistry 11 Solutions
12. Practice Problem (page 230)
The head of a small pin contains about 8 × 10–3 mol of iron. How many iron
atoms are in the head of the pin?
What Is Required?
You need to determine the number of atoms, N, of iron in the head of a pin.
What Is Given?
You know that the head of the pin contains 8 × 10–3 mol of iron, Fe(s), atoms.
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of atoms of Fe(s) using the relationship N = n × N A .
Act on Your Strategy
Number of atoms, N, of Fe(s):
N = n × NA
= 8 × 10−3 mol × 6.02 × 1023 atoms/ mol
= 4.816 × 1021 atoms
= 5 × 1021 atoms
There are 5 × 1021 atoms of Fe(s) in the head of a pin.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
13. Practice Problem (page 230)
How many molecules of oxygen gas are in a room that contains 8.5 × 103 mol
of oxygen gas?
What Is Required?
You need to determine the number of molecules, N, of oxygen gas, O2(g), in a
room.
What Is Given?
You know that there is 8.5 × 103 mol of oxygen molecules in the room.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 12 Chemistry 11 Solutions
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules of O2(g) using the relationship N = n × N A .
Act on Your Strategy
Number of molecules, N, of O2(g):
N = n × NA
= 8.5 × 103 mol × 6.02 × 1023 molecules/ mol
= 5.117 × 1027 molecules
= 5.1 × 1027 molecules
There are 5.1 × 1027 molecules of O2(g) in the room.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
14. Practice Problem (page 230)
If a marble countertop contains 849 mol of calcium carbonate, CaCO3(s), how
many formula units of calcium carbonate are in the countertop?
What Is Required?
You need to determine the number of formula units, N, of calcium carbonate in
a countertop.
What Is Given?
You know that the countertop contains 849 mol of calcium carbonate.
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of CaCO3(s) using the
relationship N = n × N A .
Act on Your Strategy
Number of formula units, N, of CaCO3(s):
N = n × NA
= 849 mol × 6.02 × 1023 formula units/ mol
= 5.11 × 1026 formula units
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 13 Chemistry 11 Solutions
There are 5.11 × 1026 formula units of CaCO3(s) in the countertop.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
15. Practice Problem (page 230)
A recipe calls for half a teaspoon of salt, which contains 5.23 × 10–2 mol of
sodium chloride. How many formula units of sodium chloride are needed?
What Is Required?
You need to determine the number of formula units, N, of sodium chloride,
NaCl(s), in half a teaspoon of salt.
What Is Given?
You know that half a teaspoon contains 5.23 × 10–2 mol of sodium chloride.
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of NaCl(s) using the
relationship N = n × N A .
Act on Your Strategy
Number of formula units, N, of NaCl(s):
N = n × NA
= 5.23 × 10–2 mol × 6.02 × 1023 formula units/ mol
= 3.1485 × 1022 formula units
= 3.15 × 1022 formula units
A total of 3.15 × 1022 formula units of NaCl(s) are needed.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 14 Chemistry 11 Solutions
16. Practice Problem (page 230)
A window-cleaning solution contains 3.86 mol of acetic acid, CH3COOH(ℓ).
How many molecules of acetic acid are in the solution?
What Is Required?
You need to determine the number of molecules, N, of acetic acid in a
window- cleaning solution.
What Is Given?
You know that there are 3.86 mol of acetic acid in the solution.
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules of CH3COOH(ℓ) using the
relationship N = n × N A .
Act on Your Strategy
Number of molecules, N, of CH3COOH(ℓ):
N = n × NA
= 3.86 mol × 6.02 × 1023 molecules/ mol
= 2.32 × 1024 molecules
There are 2.32 × 1024 molecules of CH3COOH(ℓ) in the window-cleaning
solution.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
17. Practice Problem (page 230)
A fuel tank used in a barbecue contains 2.0 × 102 mol of propane, C3H8(g).
What is the total number of atoms in the tank?
What Is Required?
You need to determine the total number of atoms in a fuel tank containing
propane.
What Is Given?
You know that there are 2.0 × 102 mol of propane molecules in the fuel tank.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 15 Chemistry 11 Solutions
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules, N, of C3H8(g) using the
relationship N = n × N A .
The chemical formula for propane shows that one molecule of C3H8(g)
contains 3 carbon atoms and 8 hydrogen atoms for a total of 11 atoms.
Multiply the number of molecules of C3H8(g) by the number of atoms per
molecule.
Act on Your Strategy
Number of molecules, N, of C3H8(g):
N = n × NA
= 2.0 × 102 mol × 6.02 × 1023 molecules/ mol
= 1.204 × 1026 molecules
total number of atoms = 1.204 × 1026 molecules × 11 atoms/ molecule
= 1.3 × 1027 atoms
There are 1.3 × 1025 atoms in the fuel tank.
Check Your Solution
The units cancel properly. The answer seems reasonable. The number of atoms
is about 10 times the number of molecules. The number of significant digits in
the answer agrees with the given data having the least number of significant
digits.
18. Practice Problem (page 230)
Freon™, CCl2F2 (g), is a refrigerant that is no longer used in car air
conditioners because it damages the ozone layer. A sample contains 4.82 mol
of Freon™.
a. How many molecules of Freon™ are in the sample?
b. How many atoms, in total, are in the sample?
What Is Required?
a. You need to determine the total number of molecules, N, of Freon™ in a
sample.
b. You need to determine the total number of atoms in the sample of Freon™.
What Is Given?
You know that there are 4.82 mol of Freon™ molecules in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 16 Chemistry 11 Solutions
Plan Your Strategy
a. number of molecules
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules of CCl2F2(g) using the
relationship N = n × N A .
b. number of atoms
The chemical formula, CCl2F2(g), shows that one molecule of Freon™ contains
1 carbon atom, 2 chlorine atoms, and 2 fluorine atoms for a total of 5 atoms.
Multiply the number of molecules of CCl2F2(g) by the number of atoms per
molecule.
Act on Your Strategy
a. number of molecules
Number of molecules, N, of CCl2F2(g):
N = n × NA
= 4.82 mol × 6.02 × 1023 molecules/ mol
= 2.90 × 1024 molecules
There are 2.90 × 1024 molecules of Freon™ in the sample.
b. number of atoms
number of atoms = 2.9016 × 1024 molecules ×
5 atoms
1 molecule
= 1.45 × 1025 atoms
There are 1.45 × 1025 atoms in the Freon™ sample.
Check Your Solution
The units cancel properly. The answers seem reasonable. The total number of
atoms is about 5 times the number of molecules. The number of significant
digits in the answers agrees with the given data having the least number of
significant digits.
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19. Practice Problem (page 230)
Glauber’s salt is a common name for sodium sulfate decahydrate,
Na2SO4•10H2O(s). It is used in the manufacture of detergents. Suppose that a
sample of 36.2 mol of sodium sulfate decahydrate is required.
a. What number of sodium atoms would be in the sample?
b. What number of water molecules would be in the sample?
What Is Required?
a. You need to determine the total number of sodium, Na(s), atoms in a sample
of sodium sulfate decahydrate.
b. You need to determine the total number of molecules of water in a sample
of sodium sulfate decahydrate.
What Is Given?
You know that there are 36.2 mol of sodium sulfate decahydrate in the sample.
Plan Your Strategy
a. number of sodium atoms
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of Na2SO4•10H2O using the relationship
N = n × N A.
Determine the number of atoms of Na(s) per formula unit of Na2SO4•10H2O.
Calculate the number of Na(s) atoms by multiplying the total number of
formula units by the number of Na(s) atoms per formula unit.
b. number of water molecules
Calculate the number of water molecules by multiplying the total number of
formula units by the number of water molecules per formula unit.
Act on Your Strategy
a. number of sodium atoms
Number of formula units of Na2SO4•10H2O(s):
N = n × NA
= 36.2 mol × 6.02 × 1023 formula units/ mol
= 2.179 × 1025 formula units
From the chemical formula, there are 2 sodium atoms in one formula unit of
Na2SO4•10H2O.
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number of Na(s) atoms sample = 2.179 × 1025 formula units
⎛ Na atoms ⎞
× 2⎜
⎟
⎝ formula unit ⎠
= 4.36 × 1025 atoms
There are 4.36 × 1025 sodium atoms in the sample of sodium sulfate
decahydrate.
b. number of water molecules
From the chemical formula, there are 10 water molecules in one formula unit
of Na2SO4•10H2O.
number of water molecules in sample = 2.179 × 1025 formula units
⎛ H O molecules ⎞
× 10 ⎜ 2
⎟
⎝ formula unit ⎠
= 2.18 × 1026 H 2 O molecules
There are 2.18×1026 water molecules in the sample of sodium sulfate
decahydrate.
Check Your Solution
The units cancel properly. The answers seem reasonable. The number of water
molecules is about 5 times the number of sodium atoms. The number of
significant digits in the answers agrees with the given data having the least
number of significant digits.
20. Practice Problem (page 230)
A sample of sucrose, C12H22O11 (s), contains 0.16 mol of oxygen atoms.
a. What amount in moles of sucrose is in the sample?
b. How many atoms of carbon are in the sample?
What Is Required?
a. You need to determine the total amount in moles (N) of molecules of
sucrose in a sample.
b. You need to determine the total number of atoms of carbon in the sucrose
sample.
What Is Given?
You know there are 0.16 moles of oxygen atoms in the sample of sucrose.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 19 Chemistry 11 Solutions
Plan Your Strategy
a. moles of sucrose
Determine the amount in moles of atoms of oxygen per mole of C12H22O11(s).
Calculate the number of moles of sucrose using the
1 mol C12 H 22 O11
.
ratio
11 mol O per mol C12 H 22 O11
b. atoms of carbon
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules, N, of C12H22O11 using the relationship
N = n × N A.
Determine the number of carbon atoms per molecule of C12H22O11(s).
Calculate the number of carbon atoms in the sample of C12H22O11(s).
Act on Your Strategy
a. moles of sucrose
From the chemical formula for sucrose, there are 11 moles of oxygen atoms
per mole of C12H22O11(s).
Amount in moles, n, of C12H22O11(s):
1 mol C12 H 22 O11
n
=
0.16 mol oxygen atoms 11 mol oxygen atoms
n = 0.16 mol oxygen atoms ×
1 mol C12 H 22 O11
11 mol oxygen atoms
= 0.014545 mol C12 H 22 O11
= 0.015 mol C12 H 22 O11
There is 0.015 mol of C12H22O11(s) in the sample.
b. atoms of carbon
Number of molecules, N, of C12H22O11:
N = n × NA
= 0.014545 mol × 6.02 × 10 23 molecules/ mol
= 8.756 × 10 21 molecules C12 H 22 O11 (s)
From the chemical formula for sucrose, there are 12 carbon atoms per
molecule of C12H22O11(s).
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 20 Chemistry 11 Solutions
Number of carbon atoms, N, in the sample:
N
12 C atoms
=
21
8.756 × 10 molecules 1 molecule
N = 8.756 × 10 21 molecules ×
12 C atoms
1 molecule
= 1.051 × 1023 C atoms
= 1.05 × 10 23 C atoms
There are 1.04 × 1023 atoms of carbon in the sample.
Check Your Solution
The units cancel properly. The answers seem reasonable and the number of
significant digits in the answers agrees with the given data having the least
number of significant digits. Changing the number of atoms of carbon to
moles gives approximately the same number as the original number of moles
oxygen. This is reasonable since the number of C atoms and the number of O
atoms in C12H22O11(s) are approximately the same.
Section 5.1 The Mole and the Avogadro Constant
Solutions for Practice Problems
Student Edition page 231
21. Practice Problem (page 231)
A gold coin contains 9.51 × 1022 atoms of gold. What amount in moles of gold
is in the coin?
What Is Required?
You need to determine the amount in moles of gold, Au(s), is in a gold coin.
What Is Given?
You know the number of atoms, N, of gold in a gold coin: 9.51 × 1022
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of gold using the relationship n =
978‐0‐07‐105107‐1 N
.
NA
Chapter 5 The Mole: A Chemist’s Counter • MHR | 21 Chemistry 11 Solutions
Act on Your Strategy
Amount in moles, n, of Au(s):
N
nAu =
NA
9.51 × 1022 atoms
6.02 × 1023 atoms /mol
= 0.15797 mol
=
= 0.158 mol
There is 0.158 mol of gold in the coin.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
22. Practice Problem (page 231)
A patient in a dentist’s office is given 1.67 × 1023 molecules of dinitrogen
monoxide (laughing gas), N2O(g), during a procedure. What amount in moles
of dinitrogen monoxide is the patient given?
What Is Required?
You need to determine the amount in moles of dinitrogen monoxide a patient
receives at the dentist’s office.
What Is Given?
You know the number of molecules, N, of dinitrogen monoxide: 1.67 × 1023
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of N2O(g) using the relationship n =
N
.
NA
Act on Your Strategy
Amount in moles, n, of N2O(g):
N
nN2O =
NA
1.67 × 1023 molecules
6.02 × 10 23 molecules /mol
= 0.277 mol
=
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 22 Chemistry 11 Solutions
There is 0.277 mol of dinitrogen monoxide given to the patient at the dentist’s
office.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
23. Practice Problem (page 231)
A sheet of drywall contains 1.2 × 1026 formula units of gypsum (calcium
sulfate dihydrate), CaSO4•2H2O(s). What amount in moles of gypsum is in the
sheet of drywall?
What Is Required?
You need to determine the amount in moles of calcium sulfate dihydrate in a
sheet of drywall.
What Is Given?
You know the number of formula units, N, of calcium sulfate dihydrate:
1.2 × 1026
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of CaSO4•2H2O(s) using the relationship
N
n=
.
NA
Act on Your Strategy
Amount in moles, n, of CaSO4•2H2O(s):
N
nCaSO4 •2H2O =
NA
1.2 × 1026 forumula units
=
6.02 × 1023 formula units /mol
= 199.34 mol
= 2.0 × 102 mol
The sheet of drywall contains 2.0 × 102 mol of calcium sulfate dihydrate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 23 Chemistry 11 Solutions
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
24. Practice Problem (page 231)
Limewater, a weak solution of calcium hydroxide, Ca(OH)2(s), is used to
detect the presence of carbon dioxide gas. Suppose that you are given a
solution that contains 8.7 × 1019 formula units of calcium hydroxide. What
amount in moles of calcium hydroxide is in the solution?
What Is Required?
You need to determine the amount in moles of calcium hydroxide in a solution
of limewater.
What Is Given?
You know the number of formula units, N, of calcium hydroxide: 8.7 × 1019
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of Ca(OH)2(s) using the relationship n =
N
.
NA
Act on Your Strategy
Amount in moles, n, of Ca(OH)2(s):
N
nCa ( OH ) =
2
NA
=
8.7 × 1019 formula units
6.02 × 1023 formula units /mol
= 1.445 × 10 –4 mol
= 1.4 × 10 –4 mol
The limewater solution contains 1.4 × 10–4 mol of calcium hydroxide.
Check Your Solution
The units cancel properly. The answer seems reasonable and the number of
significant digits in the answer agrees with the given data having the least
number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 24 Chemistry 11 Solutions
25. Practice Problem (page 231)
If there are a total of 7.3 × 1029 atoms in a sample of glucose, C6H12O6(s), what
amount in moles of glucose is in the sample?
What Is Required?
You need to determine the amount in moles of glucose in a given sample.
What Is Given?
You know the total number of atoms in the sample of glucose: 7.3 × 1029
Plan Your Strategy
Determine the total number of atoms in one molecule of C6H12O6(s).
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of C6H12O6(s) using the relationship n =
N
.
NA
Act on Your Strategy
From the chemical formula for glucose, there are 6 carbon atoms, 12 hydrogen
atoms, and 6 oxygen atoms for a total of 24 atoms in one molecule of
C6H12O6(s).
Total number of molecules, NT, of C6H12O6(s):
NT
1 molecule
=
29
7.3 × 10 atoms
24 atoms
1 molecule
N T = 7.3 × 10 29 atoms ×
24 atoms
= 3.0416 × 10 28 molecules
Amount in moles, n, of C6H12O6(s):
N
nC6H12O6 = T
NA
=
3.0416 × 1028 molecules
6.02 × 1023 molecules /mol
= 5.05249 × 104 mol
= 5.1 × 104 mol
There is 5.1 × 104 mol of glucose in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 25 Chemistry 11 Solutions
Check Your Solution
The units cancel properly. Estimating using the powers of ten in the
calculation, the answer seems reasonable:
1
1
7 × 10 29 ×
×
= 4.8 × 10 4
24
6 × 1023
The number of significant digits in the answer agrees with the given data
having the least number of significant digits.
26. Practice Problem (page 231)
A sample of aluminum oxide, Al2O3(s), contains 8.29 × 1025 total atoms.
Calculate the amount in moles of aluminum oxide in the sample.
Hint: This is a two-step problem. Calculate the number of formula units first.
What Is Required?
You must determine the amount in moles of aluminum oxide in a sample.
What Is Given?
You know the total number of atoms, N, in the sample of aluminum oxide:
8.29 × 1025
Plan Your Strategy
Determine the number of atoms in one formula unit of Al2O3(s).
Calculate the number of formula units of Al2O3(s).
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of Al2O3(s) using the relationship n =
N
.
NA
Act on Your Strategy
From the chemical formula for aluminum oxide, there are 2 atoms of
aluminum and 3 atoms of oxygen for a total of 5 atoms in one formula unit of
Al2O3(s).
Number of formula units, N, of Al2O3(s):
N
1 formula unit
=
25
8.29 × 10 total atoms 5 total atoms
N = 8.29 × 1025 total atoms ×
1 formula unit
5 total atoms
= 1.658 × 1025 formula units
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 26 Chemistry 11 Solutions
Amount in moles, n, of Al2O3(s):
N
nAl2O3 =
NA
1.658 × 1025 formula units
6.02 × 1023 formula units /mol
= 27.541 mol
=
= 27.5 mol
There is 27.5 mol of aluminum oxide in the sample.
Check Your Solution
The units cancel properly. Estimating using the powers of ten in the
1
1
calculation, the answer seems reasonable: 8 × 10 25 × ×
= 27 mol
5 6 × 10 23
The number of significant digits in the answer agrees with the given data
having the least number of significant digits.
27. Practice Problem (page 231)
Trinitrotoluene, or TNT for short, has the chemical formula C7H5N3O6(s). If a
stick of dynamite is pure TNT and it contains 2.5 × 1025 atoms in total, what
amount in moles of TNT does it contain?
What Is Required?
You must determine the amount in moles of trinitrotoluene that the stick of
dynamite contains.
What Is Given?
You know the total number of atoms in the sample of trinitrotoluene: 2.5×1025
Plan Your Strategy
Determine the total number of atoms in one molecule of C7H5N3O6(s).
Calculate the number of molecules, N, of C7H5N3O6(s).
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of C7H5N3O6(s) using the relationship n =
978‐0‐07‐105107‐1 N
.
NA
Chapter 5 The Mole: A Chemist’s Counter • MHR | 27 Chemistry 11 Solutions
Act on Your Strategy
From the chemical formula for trinitrotoluene, there are 7 carbon atoms, 5
hydrogen atoms, 3 nitrogen atoms, and 6 oxygen atoms for a total of 21 atoms
in one molecule of C7H5N3O6(s).
Number of molecules, N, of C7H5N3O6(s):
N
1 molecule
=
25
2.5 × 10 total atoms 21 total atoms
N = 2.5 × 1025 total atoms ×
1 molecule
21 total atoms
= 1.1905 × 1024 molecules
Amount in moles, n, of C7H5N3O6(s):
N
nC7 H5 N3O6 =
NA
1.1905 × 1024 molecules
6.02 × 1023 molecules /mol
= 1.977 mol
=
= 2.0 mol
There is 2.0 mol of trinitrotoluene in the stick of dynamite.
Check Your Solution
The units cancel properly. Estimating using the powers of ten in the
calculation, the answer seems reasonable:
1
1
2.5 × 1025 ×
×
= 2 mol
21 6 × 1023
The number of significant digits in the answer agrees with the given data
having the least number of significant digits.
28. Practice Problem (page 231)
A sample of rubbing alcohol solution contains ethanol, C2H5OH(ℓ). If the
sample contains 1.25 × 1023 atoms of hydrogen in the ethanol, what amount in
moles of ethanol is in the sample?
What Is Required?
You must determine the amount in moles of ethanol in a sample of rubbing
alcohol.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 28 Chemistry 11 Solutions
What Is Given?
You know the total number of hydrogen atoms in the sample (N) of ethanol:
1.25×1023
Plan Your Strategy
Determine the number of hydrogen atoms in one molecule of C2H5OH(ℓ).
Calculate the number of molecules of C2H5OH(ℓ).
Use the Avogadro constant: N A = 6.02 × 10 23
N
.
Calculate the amount in moles of C2H5OH(ℓ) using the relationship n =
NA
Act on Your Strategy
From the chemical formula for ethanol, there are 6 atoms of hydrogen in one
molecule of C2H5OH(ℓ).
Number of molecules, N, of C2H5OH(ℓ):
1 molecule C 2 H 5OH
N
=
23
1.25 × 10 H atoms
6 H atoms
N = 1.25 × 10 23 H atoms ×
1 molecule C 2 H 5OH
6 H atoms
= 2.0833 × 10 22 molecules C 2 H 5OH
Amount in moles, n, of C2H5OH(ℓ):
N
nC2 H5OH =
NA
2.0833 × 1022 molecules
6.02 × 1023 molecules /mol
= 0.0346 mol
=
There is 0.0346 mol of ethanol in the sample.
Check Your Solution
The units cancel properly. Estimating using the powers of ten in the
calculation, the answer seems reasonable:
1
1
1.3 × 10 23 ×
×
= 0.36 mol
6 6 × 1023
The number of significant digits in the answer agrees with the given data
having the least number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 29 Chemistry 11 Solutions
29. Practice Problem (page 231)
A cleaning solution contains 7.9 × 1026 molecules of ammonia, NH3(aq). What
amount in moles of ammonia is in the solution?
What Is Required?
You need to determine the amount in moles of ammonia in a cleaning solution.
What Is Given?
You know the total number of molecules, N, of ammonia: 7.9 × 1026
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of ammonia using the relationship n =
N
.
NA
Act on Your Strategy
Amount in moles, n, of NH3(aq):
N
nNH3 =
NA
=
7.9 × 1026 molecules
6.02 × 1023 molecules /mol
= 1.312 × 103 mol
= 1.3 × 103 mol
There is 1.3 × 103 mol of ammonia in the cleaning solution.
Check Your Solution
The units cancel properly. Comparing the powers of ten in the solution, the
answer seems reasonable. The number of significant digits in the answer
agrees with the given data having the least number of significant digits.
30. Practice Problem (page 231)
A muffin recipe calls for cream of tartar, or potassium hydrogen tartrate,
KHC4H4O6(s). The amount of cream of tartar that is required contains
2.56 × 1023 atoms of carbon. What amount in moles of potassium hydrogen
tartrate is required?
What Is Required?
You need to determine the amount in moles of potassium hydrogen tartrate
required in a recipe.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 30 Chemistry 11 Solutions
What Is Given?
You know the number of atoms, N, of carbon in the sample: 2.56 × 1023
Plan Your Strategy
Determine the number of carbon atoms in one molecule of KHC4H4O6(s).
Calculate the number of molecules of KHC4H4O6(s).
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of KHC4H4O6(s) using the relationship n =
N
.
NA
Act on Your Strategy
From the chemical formula of potassium hydrogen tartrate, there are 4 carbon
atoms in one molecule of KHC4H4O6(s).
Number of molecules, N, of KHC4H4O6(s):
1 molecule KHC4 H 4O
N
=
23
2.56 × 10 C atoms
4 C atoms
N = 2.56 × 1023 C atoms ×
1 molecule KHC4 H 4 O
4 C atoms
= 6.40 × 1022 molecules KHC4 H 4O
Amount in moles, n, of KHC4H4O6(s):
N
nKHC4 H4O6 =
NA
6.40 × 10 22 molecules
6.02 × 1023 molecules /mol
= 0.106 mol
=
There is 0.106 mol of potassium hydrogen tartrate in the sample.
Check Your Solution
1
the number
4
of moles of C, which is consistent. The number of significant digits in the
answer agrees with the given data having the least number of significant digits.
The units cancel properly. The 0.106 mol of KHC4H4O6 is about
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 31 Chemistry 11 Solutions
Section 5.1 The Mole and the Avogadro
Solutions for Selected Review Questions
Student Edition page 232
4. Review Question (page 232)
If the volume of one mole of water at room temperature is 18.02 mL, what is
the volume of one molecule of water, in litres?
What Is Required?
You need to determine the volume (in litres) of one molecule of water.
What Is Given?
You know the volume of 1 mol of water molecules: 18.02 mL
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Divide the volume of 1 mol of molecules by the number of molecules in 1 mol
to determine the volume of one molecule.
Convert the volume from millilitres to litres: 1 mL = 1 × 10–3 L
Act on Your Strategy
volume, V, (in mL) of 1 molecule of H2O(ℓ):
18.02 mL
V=
6.02 × 1023 molecules
= 2.993 × 10−23 mL/molecule
volume, V, (in L) of 1 molecule of H2O(ℓ):
V = 2.993 ×10 –23 mL /molecule × 1 × 10 –3 L/ mL
= 2.993 ×10 –26 L/molecule
The volume of one molecule of water is 2.993×10–26 L.
Check Your Solution
The units cancel properly. Estimating using rounded numbers gives an answer
that is close to the calculated answer:
20
× 10–3 = 3 × 10−26 L
23
6 × 10
The answer is reasonable and has the correct number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 32 Chemistry 11 Solutions
6. Review Question (page 232)
Calculate the number of formula units of sodium chloride in 0.0578 mol.
What is Required?
You need to determine the number of formula units of sodium chloride,
NaCl(s).
What is Given?
You know the amount in moles of sodium chloride: 0578 mol
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of NaCl(s) using the relationship
N = n × NA .
Act on Your Strategy
Number of formula units, N, of NaCl(s):
N = n × NA
= 0.0578 mol ×
6.02 × 1023 formula units
1 mol
= 3.4796 × 10 22 formula units
= 3.48 × 10 22 formula units
There are 3.48 × 1022 formula units of sodium chloride in the sample.
Check Your Solution
The units cancel properly. The answer has the correct number of significant
digits. Converting the answer from formula units to moles gives the original
amount in moles.
7. Review Question (page 232)
Calculate the number of particles in each sample. Indicate the correct type of
particle (atom, molecule, ion, or formula unit).
a. 0.156 mol Au(s)
b. 7.8 mol MgCl2(s)
c. 15.2 mol H2O2(ℓ)
What Is Required?
You need to determine the number of particles and indicate the correct type of
particle in each sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 33 Chemistry 11 Solutions
What Is Given?
You know the chemical formula for each sample.
You know the amount of each substance:
a. 0.156 mol of Au(s)
b. 7.8 mol of MgCl2(s)
c. 15.2 mol of H2O2(ℓ)
Plan Your Strategy
Identify the type of particle that makes up each sample.
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of particle units of each sample using the relationship
N = n × NA .
Act on Your Strategy
a. Au(s) is gold, and is made up of atoms
Number of atoms, N, of Au(s):
N = n × NA
= 0.156 mol ×
6.02 × 1023 atoms
1 mol
= 9.39 × 1022 atoms
There are 9.39× 1022 atoms of Au(s) in the sample.
b. MgCl2(s) is the ionic compound magnesium chloride, and the solid form is
made up of formula units
Number of formula units, N, of MgCl2(s):
N = n × NA
= 7.8 mol ×
6.02 × 1023 formula units
1 mol
= 4.7 × 1024 formula units
There are 4.7 × 1024 formula units of MgCl2(s) in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 34 Chemistry 11 Solutions
c. H2O2(ℓ) is hydrogen peroxide, and is made up of molecules.
Number of molecules, N, of H2O2(ℓ):
N = n × NA
= 15.2 mol ×
6.02 × 1023 molecules
1 mol
= 9.15 × 1024 molecules
There are 9.15 × 1024 molecules of H2O2(ℓ) in the sample.
Check Your Solution
In each case, the units cancel properly. The answers seem reasonable and have
the correct number of significant digits.
8. Review Question (page 232)
Calculate the amount in moles of molecules for each substance listed.
a. a sample of ammonia, NH3(g), containing 8.1 × 1020 atoms of hydrogen
b. a sample of diphosphorus pentoxide, P2O5(s), containing a total of 4.91 ×
1022 atoms of phosphorus and oxygen
What Is Required?
You need to determine the amount in moles of molecules of each sample.
What Is Given?
a. You know the sample of ammonia contains 8.1 × 1020 atoms of hydrogen.
b. You know the sample of diphosphorus pentoxide contains a total of 4.91 ×
1022 atoms of phosphorus and oxygen.
Plan Your Strategy
a. ammonia
From the chemical formula, NH3, determine the number of hydrogen atoms per
molecule of ammonia.
Calculate the number of molecules of the NH3(g).
Use the Avogadro constant: N A = 6.02 × 10 23
N
.
Calculate the amount in moles of NH3(g) using the relationship n =
NA
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 35 Chemistry 11 Solutions
b. diphosphorus pentoxide
Calculate the number of molecules of P2O5(s) using the information that there
is a total of 7 atoms per molecule.
Use the Avogadro constant: N A = 6.02 × 10 23
N
Calculate the amount in moles of the P2O5(s) using the relationship n =
.
NA
Act on Your Strategy
a. ammonia
From the chemical formula, NH3, there are 3 atoms of hydrogen per molecule
of ammonia.
Number of molecules, N, of NH3(g):
1 molecule
N
=
20
8.1 × 10 H atoms 3 H atoms
N = 8.1 × 10 20 H atoms ×
1 molecule
3 H atoms
= 2.7 × 1020 molecules
Amount in moles, n, of the NH3(g):
N
nNH3 =
NA
=
2.7 × 10 20 molecules
6.02 × 1023 molecules /mol
= 4.5 × 10 –4 mol
There is 4.5 × 10–4 mol of NH3(g) in the sample.
b. diphosphorus pentoxide
From the chemical formula, P2O5, there are 2 phosphorus atoms and 5 oxygen
atoms for a total of 7 atoms per molecule.
Number of molecules, N, of the P2O5(s):
N
1 molecule
=
22
4.91 × 10 atoms
7 atoms
N = 4.91 × 1022 atoms ×
1 molecule
7 atoms
= 7.014 × 1022 molecules
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 36 Chemistry 11 Solutions
Amount in moles, n, of P2O5(s):
N
nP2O5 =
NA
=
7.014 × 1022 molecules
6.02 × 1023 molecules /mol
= 1.165 × 10 –2 mol
= 1.17 × 10 –2 mol
There is 1.17 × 10–2 mol of P2O5(s) in the sample.
Check Your Solution
In each case, the units cancel properly. Using rounded numbers to estimate the
answer gives a result that is close to the calculated answers.
1
1
a. 8 × 1020 × ×
= 4.4 × 10–4
23
3
6 × 10
20
1
b. 5 × 1022 ×
×
= 1.2 × 10–2
23
23
6 × 10
6 × 10
The answers are reasonable.
9. Review Question (page 232)
A sample of ethanol, C2H5OH(ℓ), contains 2.49 × 1023 molecules of ethanol. A
sample of carbon contains 1.65 mol of carbon atoms. Which sample, the
compound or the element, contains the greater amount in moles of carbon?
What Is Required?
You need to compare the amount in moles of carbon in samples of the
compound ethanol and element carbon.
What Is Given?
You know the amount in moles of carbon, C, atoms in the carbon sample:
1.65 mol
You know the number of molecules of C2H5OH(ℓ) in the ethanol sample:
2.49 × 1023
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of C2H5OH(ℓ) in the sample of ethanol using
N
the relationship n =
.
NA
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 37 Chemistry 11 Solutions
From the chemical formula, C2H5OH, determine the amount in moles of
carbon atoms per mole of C2H5OH(ℓ) molecules.
Calculate the amount in moles of C atoms in the C2H5OH(ℓ).
Compare this amount in moles of C atoms to the given amount in the sample
of the element.
Act on Your Strategy
Amount in moles, n, of C2H5OH(ℓ):
N
nC2 H5OH =
NA
2.49 × 1023 molecules
6.02 × 1023 molecules /mol
= 0.4136 mol
=
From the chemical formula, C2H5OH, there are 2 mol of carbon atoms per
mole of ethanol.
Amount in moles, n, of C atoms in C2H5OH(ℓ):
nC
2 mol C atoms
=
0.4136 mol C 2 H 5OH 1 mol C 2 H 5OH
nC = 0.4136 mol C 2 H 5OH ×
2 mol C atoms
1 mol C2 H 5OH
= 0.8272 mol C atoms
= 0.827 mol C atoms
The sample of ethanol contains 0.827 mol of carbon atoms.
The sample of the element carbon containing 1.65 mol of carbon atoms has the
greater amount of carbon atoms.
Check Your Solution
The units cancel properly and the answers have the correct number of
significant digits. The answer is reasonable.
10. Review Question (page 232)
If there are 4.28 × 1021 atoms of hydrogen and oxygen in a sample of water,
what amount in moles of water is in the sample?
What is Required?
You need to determine the amount in moles of water in a sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 38 Chemistry 11 Solutions
What Is Given?
You know the total number of atoms in the sample of water: 4.28 × 1021
Plan Your Strategy
Use the total number of atoms per molecule of H2O(ℓ) to calculate the number
of molecules.
Use the Avogadro constant: N A = 6.02 × 10 23
N
Calculate the amount in moles of water using the relationship n =
.
NA
Act on Your Strategy
From the chemical formula, H2O, there are 2 hydrogen atoms and 1 oxygen
atom for a total of 3 atoms per molecule.
Number of molecules, N, of H2O(ℓ)
N
1 molecule
=
21
4.28 × 10 atoms
3 atoms
N = 4.28 × 1021 atoms ×
1 molecule
3 atoms
= 1.4266 × 1021 molecules
Amount in moles, n, of H2O(ℓ):
N
nH2O =
NA
=
1.4266 × 1021 molecules
6.02 × 1023 molecules /mol
= 2.369 × 10 –3 mol
= 2.37 × 10 –3 mol
There is 2.37 × 10–3 mol of water molecules.
Check Your Solution
The units cancel properly. Using rounded numbers to estimate the answer:
1
1
= 2 × 10–3 mol
4 × 1021 × ×
3
6 × 1023
The estimated answer is close to the calculated answer. The answer is
reasonable and has the correct number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 39 Chemistry 11 Solutions
11. Review Question (page 232)
Aluminum oxide, Al2O3(s), forms a thin coating on an aluminum surface, such
as the body of an airplane, when aluminum is exposed to oxygen in air. This
coating helps to reduce further corrosion of the aluminum. Suppose that a
sample contains 2.6 mol of aluminum oxide.
a. How many formula units are in the sample?
b. How many atoms, in total, are in the sample?
c. How many aluminum atoms are in the sample?
What Is Required?
a. You need to determine the number of formula units of Al2O3(s).
b. You need to determine the total number of atoms in the sample of Al2O3(s).
c. You need to determine the number of aluminum atoms in the sample.
What Is Given?
You know the amount in moles of the Al2O3(s): 2.6 mol
Plan Your Strategy
a. number of formula units
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of the Al2O3(s) using the
relationship N = n × N A .
b. total number of atoms
From the chemical formula, Al2O3, determine the number of atoms in one
formula unit.
Use the total number of atoms in one formula unit to find the number of atoms
in the sample.
c. number of aluminum atoms
Use the number of Al atoms in one formula unit of the Al2O3(s) to find the
number of Al atoms in the sample.
Act on Your Strategy
a. number of formula units
Number of formula units, N, of the Al2O3(s):
N = n × NA
= 2.6 mol × 6.02 × 1023 formula units/ mol
= 1.5652 × 1024 formula units
= 1.6 × 1024 formula units
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 40 Chemistry 11 Solutions
There are 1.6 × 1024 formula units of aluminum oxide in the sample.
b. total number of atoms
From the chemical formula, Al2O3, there are 2 aluminum atoms and 3 oxygen
atoms for a total of 5 atoms in one formula unit of Al2O3(s).
Total number of atoms, N, in the Al2O3(s):
N
5 atoms
=
24
1.5652 × 10 formula units 1 formula unit
N = 1.5652 × 10 24 formula units ×
5 atoms
1 formula unit
= 7.826 × 1024 atoms
= 7.8 × 1024 atoms
There are 7.8 × 1024 atoms in total in the sample of aluminum oxide.
c. number of aluminum atoms
From the chemical formula, Al2O3, there are 2 aluminum atoms in one formula
unit of Al2O3(s).
Number of Al atoms, N, in the Al2O3(s):
2 Al atoms
N
=
24
1.5652 × 10 formula units 1 formula unit
N = 1.5652 × 1024 formula units ×
2 Al atoms
1 formula unit
= 3.1304 × 1024 Al atoms
= 3.1 × 1024 Al atoms
There are 3.1 × 1024 Al atoms in the sample of aluminum oxide.
Check Your Solution
The units cancel properly. The answer is reasonable and has the correct
number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 41 Chemistry 11 Solutions
12. Review Question (page 232)
Zinc chloride hexahydrate, ZnCl2•6 H2O(s), is used in the textile industry,
often to prepare fireproofing agents. In a particular fireproofing solution,
0.46 mol of zinc chloride hexahydrate is used.
a. How many formula units of zinc chloride hexahydrate are in the solution?
b. How many atoms of chlorine are in the solution?
What Is Required?
a. You need to determine the number of formula units of ZnCl2•6H2O(s).
b. You need to determine the number of chlorine, Cl, atoms in the sample of
ZnCl2•6H2O(s).
What Is Given?
You know the amount in moles of the ZnCl2•6H2O(s): 0.46 mol
Plan Your Strategy
a. number of formula units
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of the ZnCl2•6H2O(s) using the
relationship N = n × N A .
b. number of chlorine atoms
From the chemical formula, ZnCl2•6H2O, determine the number of chlorine
atoms per formula unit of zinc chloride hexahydrate.
Calculate the number of chlorine atoms in the sample.
Act on Your Strategy
a. number of formula units
Number of formula units, N, of the ZnCl2•6H2O(s):
N = n × NA
= 0.46 mol × 6.02 × 1023 formula units/ mol
= 2.769 × 1023 formula units
= 2.8 × 1023 formula units
There are 2.8 × 1023 formula units of ZnCl2•6H2O(s) in the sample.
b. number of chlorine atoms
From the chemical formula, ZnCl2•6H2O, there are 2 chlorine atoms in one
formula unit of zinc chloride hexahydrate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 42 Chemistry 11 Solutions
Number of Cl atoms, N, in ZnCl2•6H2O(s):
N
2 Cl atoms
=
23
2.769 × 10 formula units 1 formula unit
N = 2.769 × 1023 formula units ×
2 Cl atoms
1 formula unit
= 5.538 × 1023 Cl atoms
There are 5.5 × 1023 chlorine atoms in the sample.
Check Your Solution
a. Estimating, you have approximately
1
mol (0.46 mol) of ZnCl2•6H2O that
2
1
of the Avogadro constant of formula units, or
2
2.8 × 1023. The units cancel properly. The answer is reasonable and has the
correct number of significant digits.
would be equivalent to about
b. An estimate of the answer is approximately 3 × 2 atoms, which is close to
the answer that was calculated. The units cancel properly. The answer is
reasonable and has the correct number of significant digits.
14. Review Question (page 232)
Arrange the following three samples from largest to smallest in terms of their
numbers of representative particles:
• 3.92 mol of octane, C8H18(ℓ)
• 6.52 × 1023 atoms of copper, Cu(s)
• 1.25 × 1024 formula units of sodium hydrogen carbonate, NaHCO3(s)
What Is Required?
You need to arrange the number of particles of three different samples in order
from the largest number to the smallest.
What Is Given?
You know the amount in moles of octane: 3.92 mol
You know the number of atoms of copper: 6.52 × 1023
You know the number of formula units of sodium hydrogen carbonate:
1.25 × 1024
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 43 Chemistry 11 Solutions
Plan Your Strategy
Compare the number of atoms in each sample.
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules in 3.92 mol of C8H18(ℓ) using the
relationship N = n × N A .
Use the chemical formulas for the compounds octane and sodium hydrogen
carbonate to determine the number of atoms per unit.
List the compounds from the largest number of atoms to the smallest.
Act on Your Strategy
• octane
Number of molecules, N, of the C8H18(ℓ):
N = n × NA
= 3.92 mol × 6.02 × 1023 molecules/ mol
= 2.3598 × 1024 molecules
From the chemical formula, C8H18, there are 8 carbon atoms and 18 hydrogen
atoms for a total of 26 atoms per molecule.
Total number of atoms, N, in the C8H18(ℓ):
N
26 atoms
=
24
2.3598 × 10 molecules 1 molecule
N = 2.3598 × 1024 molecules ×
26 atoms
1 molecule
= 6.14 × 1025 atoms
The sample of octane contains 6.14 × 1025 atoms.
• sodium hydrogen carbonate
From the chemical formula, NaHCO3, there are 1 sodium atom, 1 hydrogen
atom, 1 carbon atom, and 3 oxygen atoms for a total of 6 atoms per molecule.
Total number of atoms, N, in the NaHCO3(s):
6 atoms
N
=
24
1.25 × 10 formula units 1 formula unit
N = 1.25 × 10 24 formula units ×
6 atoms
1 formula unit
= 7.50 × 1024 atoms
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 44 Chemistry 11 Solutions
The sample of sodium hydrogen carbonate contains 7.50 × 1024 atoms.
• copper
Number of atoms in the Cu(s): 6.52 × 1023 atoms (given)
Samples arranged from largest number of representative particles (atoms) to
the smallest:
C8H18(ℓ) (6.14 × 1025 atoms) > NaHCO3(s) (7.50 × 1024 atoms ) > Cu(s)
(6.52 × 1023 atoms)
Check Your Solution
In each case, the units cancel properly. Using rounded numbers to estimate the
calculated amounts:
• octane: 4 × 6 × 1023 × 26 = 6 × 1025 atoms
• sodium hydrogen carbonate 1 × 1024 × 6 = 6 × 1024 atoms
The estimated answers are close to the calculated answers. The answers are
reasonable and have the correct number of significant digits.
15. Calculate the number of atoms in 6.0 mol of fluorine, F2(g), molecules.
What Is Required?
You need to determine the number of atoms of fluorine, F, in a sample of
F2(g).
What Is Given?
You know the amount in moles of the F2(g): 6.0 mol
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules of F2(g) using the relationship N = n × NA.
Use the number of F atoms (2) in one molecule of F2(g) to find the number of
F atoms in the sample.
Act on Your Strategy
Number of molecules, N, of the F2(g):
N = n × NA
= 6.0 mol × 6.02 × 1023 molecules/ mol
= 3.612 × 1024 molecules
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 45 Chemistry 11 Solutions
Number of F atoms, N, in the F2(g):
2 F atoms
N
=
24
3.612 × 10 molecules 1 molecule
N = 3.612 × 1024 molecules ×
2 F atoms
1 molecule
= 7.224 × 1024 atoms
= 7.2 × 1024 atoms
There are 7.2 × 1023 F atoms in the sample.
Check Your Solution
The units cancel properly. Using rounded numbers to estimate the answer:
6 × 6 × 1023 × 2 = 7.2 × 1024 atoms
This estimate agrees with the calculated answer. The answer is reasonable and
has the correct number of significant digits.
Section 5.2 Mass and the Mole
Solutions for Practice Problems
Student Edition page 235
31. Practice Problem (page 235)
State the molar mass of each element.
a. sodium
b. tungsten
c. xenon
d. nickel
What Is Required?
You need to determine the molar mass of
a. sodium.
b. tungsten.
c. zenon.
d. nickel.
What Is Given?
You are the given the names of the elements.
Plan Your Strategy
The molar mass of any element is numerically equal to the atomic mass of the
element found on the periodic table.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 46 Chemistry 11 Solutions
Act on Your Strategy
a. MNa = 22.99 g/mol
b. MW = 183.84 g/mol
c. MXe = 131.29 g/mol
d. MNi = 58.69 g/mol
Check Your Solution
Check to see that the atomic masses have been recorded correctly.
32. Practice Problem (page 235)
Calculate the molar mass of phosphorus, P4(s).
What Is Required?
You need to determine the mass of 1 mol of phosphorus.
What Is Given?
You know the chemical formula for phosphorus: P4
Plan Your Strategy
Use the periodic table to determine the total mass contributed by 4 mol of
phosphorus atoms.
Act on Your Strategy
Molar mass, M, of P4(s):
M P4 = 4M P
= 4 ( 30.97 g/mol )
= 123.88 g/mol
The molar mass of phosphorus is 123.88 g/mol.
Check Your Solution
Check that the atomic mass has been recorded and multiplied correctly.
33. Practice Problem (page 235)
Determine the molar mass of calcium phosphate, Ca3(PO4)2(s).
What Is Required?
You need to determine the mass of 1 mol of calcium phosphate.
What Is Given?
You know the chemical formula for calcium phosphate: Ca3(PO4)2
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 47 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of each element in
Ca3(PO4)2.
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to get the molar mass of
the compound.
Act on Your Strategy
Molar mass, M, of Ca3(PO4)2(s):
M Ca 3 (PO4 )2 = 3M Ca + 2M P + 8M O
= 3 ( 40.08 g/mol ) + 2 ( 30.97 g/mol ) + 8 (16.00 g/mol )
= 120.24 g/mol + 61.94 g/mol + 128 g/mol
= 310.18 g/mol
The molar mass of calcium phosphate is 310.18 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(3 × 40) + (2 × 30) + (8 × 16) = 308 g/mol
34. Practice Problem (page 235)
Calculate the molar mass of lead(II) nitrate, Pb(NO3)2(s).
What Is Required?
You need to determine the mass of 1 mol of lead nitrate.
What Is Given?
You know the chemical formula for lead nitrate: Pb(NO3)2
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of each element in
Pb(NO3)2.
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to get the molar mass of
the compound.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 48 Chemistry 11 Solutions
Act on Your Strategy
Molar mass, M, of Pb(NO3)2:
M Pb(NO3 )2 = 1M Pb + 2M N + 6M O
= 1(207.2 g/mol) + 2(14.01 g/mol) + 6(16.00 g/mol)
= 207.2 g/mol + 28.02 g/mol + 96.00 g/mol
= 331.2 g/mol
The molar mass of lead nitrate is 331.2 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(1 × 210) + (2 × 14) + (6 × 16) = 334 g/mol
35. Practice Problem (page 235)
Determine the molar mass of iron(III) thiocyanate ion, FeSCN2+(aq).
What Is Required?
You need to determine the mass of 1 mol of iron(III) thiocyanate ion.
What Is Given?
You know the chemical formula for iron(III) thiocyanate ion: FeSCN2+
Plan Your Strategy
Use the periodic table to find the atomic molar mass of each element in
FeSCN2+(aq).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to get the molar mass of
the compound.
Note: The fact that this is an ion rather than a neutral species has insignificant
impact on the molar mass.
Act on Your Strategy
Molar mass, M, of FeSCN2+(aq):
M FeSCN2+ = 1M Fe + 1M S + 1M C + 1M N
= 1(55.85 g/mol) + 1(32.07 g/mol) + 1(12.01 g/mol) + 1(14.01 g/mol)
= 55.85 g/mol + 32.07 g/mol + 12.01 g/mol + 14.01 g/mol
=113.94 g/mol
The molar mass of FeSCN2+(aq) is 113.94 g/mol.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 49 Chemistry 11 Solutions
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(1 × 55) + (1 × 32) + (1 × 12) + (1 × 14) = 113 g/mol
36. Practice Problem (page 235)
Calculate the molar mass of sodium stearate, NaC17H35COO(s).
What Is Required?
You need to determine the mass of 1 mol of sodium stearate.
What Is Given?
You know the chemical formula for sodium stearate: NaC17H35COO
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of each element in
NaC17H35COO(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to get the molar mass of
the compound.
Act on Your Strategy
Molar mass, M, of NaC17H35COO(s):
M NaC17 H35COO = 1M Na + 18M C + 35M H + 2M O
= 1( 22.99 g/mol ) + 18 (12.01 g/mol ) + 35 (1.01 g/mol ) + 2 (16.00 g/mol )
= 22.99 g/mol + 216.18 g/mol + 35.35 g/mol + 32.00 g/mol
= 306.52 g/mol
The molar mass of sodium stearate is 306.52 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(1 × 23) + (18 × 12) + (35 × 1) + (2 × 16) = 306 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 50 Chemistry 11 Solutions
37. Practice Problem (page 235)
Calculate the molar mass of barium hydroxide octahydrate, Ba(OH)2•8H2O(s).
Hint: The molar mass of a hydrate must include the water component.
What Is Required?
You need to determine the mass of 1 mol of barium hydroxide octahydrate.
What Is Given?
You know the chemical formula for barium hydroxide octahydrate:
Ba(OH)2•8H2O
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of each element in
Ba(OH)2•8H2O(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to get the molar mass of
the compound.
Act on Your Strategy
Molar mass, M, of Ba(OH)2•8H2O(s):
M Ba ( OH ) •8H2O = 1M Ba ( OH ) + 8M H2O
2
2
= 1[1M Ba + 2M O + 2M H ] + 8[2M H + 1M O ]
= 1[1(137.33 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol)]
+ 8[2(1.01 g/mol) + 1(16.00 g/mol)]
= 137.33 g/mol + 32.00 g/mol + 2.02 g/mol + 8 (18.02 g/mol )
= 137.33 g/mol + 32.00 g/mol + 2.02 g/mol + 144.16 g/mol
= 315.51 g/mol
The molar mass of barium hydroxide octahydrate is 315.51 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(1 × 140) + (2 × 16) + (2 × 1) + (8 × 18) = 318 g
38. Practice Problem (page 235)
Determine the molar mass of tetraphosphorus decoxide, P4O10(s).
What Is Required?
You need to determine the molar mass of 1 mol of tetraphosphorus decoxide.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 51 Chemistry 11 Solutions
What Is Given?
You know the chemical formula for tetraphosphorus decoxide: P4O10
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of each element in
P4O10(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to get the molar mass of
the compound.
Act on Your Strategy
Molar mass, M, of P4O10(s):
M P4O10 = 4 M P + 10M O
= 4(30.97 g/mol) + 10(16.00 g/mol)
= 123.88 g/mol + 160.00 g/mol
= 283.88 g/mol
The molar mass of tetraphosphorus decoxide is 283.88 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(4 × 30) + (10 × 16) = 280 g/mol
39. Practice Problem (page 235)
Calculate the molar mass of iron(II) ammonium sulfate hexahydrate,
(NH4)2Fe(SO4)2•6H2O(s).
What Is Required?
You need to determine the mass of 1 mol of iron(II) ammonium sulfate
octahydrate.
What Is Given?
You know the chemical formula for iron(II) ammonium sulfate hexahydrate:
(NH4)2Fe(SO4)2•6H2O(s)
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 52 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of each element in
(NH4)2Fe(SO4)2•6H2O(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of the atoms of each element to find the molar mass of
the compound.
Act on Your Strategy
Molar mass, M, of (NH4)2Fe(SO4)2•6H2O(s):
M ( NH4 ) Fe(SO4 ) •6H2O = 1M ( NH4 ) Fe(SO4 ) + 6M H2O
2
2
2
2
= 1[2M N + 8M H + 1M Fe + 2M S + 8M O ] + 6[2M H + 1M O ]
= 1[2 (14.01 g/mol ) + 8 (1.01 g/mol ) + 1( 55.85 g/mol )
+ 2 ( 32.07 g/mol ) + 8 (16.00 g/mol ) ]
+ 6 ⎡⎣ 2 (1.01 g/mol ) + 1(16.00 g/mol ) ⎤⎦
= 28.02 g/mol + 8.08 g/mol + 55.85 g/mol + 64.14 g/mol
+ 128.00 g/mol + 6 (18.02 g/mol )
= 28.02 g/mol + 8.08 g/mol + 55.85 g/mol + 64.14 g/mol
+ 128.00 g/mol + 108.12 g/mol
= 392.21 g/mol
The molar mass of ammonium iron(II) sulfate hexahydrate is 392.21 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. Rounded values
for the atomic masses can be used to estimate the answer:
(2 × 14) + (8 × 1) + (1 × 55) + (2× 30) + (8 × 16) + (6 × 18) = 387 g/mol
40. Practice Problem (page 235)
The formula for a compound that contains an unknown element, A, is A2SO4.
If the molar mass of the compound is 361.89 g/mol, what is the atomic molar
mass of A?
What Is Required?
You need to determine the atomic molar mass of element A in 1 mol of
A2SO4(s).
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 53 Chemistry 11 Solutions
What Is Given?
You know the chemical formula of the compound: A2SO4
You know the molar mass of the compound: 361.89 g/mol
Plan Your Strategy
Let y be the atomic molar mass of element A.
Use the periodic table to determine the atomic molar mass of the two named
elements in A2SO4(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Subtract the molar mass of each of the named elements from the total molar
mass of the compound to determine the molar mass of element A.
Solve for y by dividing the molar mass of element A by 2.
Act on Your Strategy
Molar mass, M, of A2SO4(s):
M A2SO4 = 2M A + 1M S + 4M O
= 2 ( y ) + 1( 32.07 g/mol ) + 4 (16.00 g/mol )
361.89 g/mol = 2 ( y ) + 1( 32.07 g/mol ) + 4 (16.00 g/mol )
2 ( y ) = 361.89 g/mol − [1(32.07 g/mol ) + 4 (16.00 g/mol )]
2 ( y ) = 361.89 g/mol − 96.01 g/mol
2 ( y ) = 265.82 g/mol
2( y)
265.82 g/mol
2
2
y = 132.91 g/mol
=
The atomic molar mass of element A is 132.91 g/mol.
Check Your Solution
Check that the atomic masses have been recorded correctly. The atomic molar
mass of y corresponds to the element cesium in the periodic table.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 54 Chemistry 11 Solutions
Section 5.2 Mass and the Mole
Solutions for Practice Problems
Student Edition page 237
41. Practice Problem (page 237)
Calculate the mass of 3.57 mol of vanadium.
What Is Required?
You need to determine the mass of a sample of vanadium, V(s).
What Is Given?
You know the amount in moles of vanadium: 3.57 mol
Plan Your Strategy
Use the periodic table to determine the atomic molar mass of V(s).
Use the relationship m = n × M .
Multiply the amount in moles of the V(s) by its molar mass.
Act on Your Strategy
Molar mass of V(s): M = 50.94 g/mol (from the periodic table)
Mass, m, of V(s):
mV = n × M
= 3.57 mol × 50.94 g/ mol
= 181.8558 g
= 182 g
The mass of the sample of vanadium is 182 g.
Check Your Solution
The units cancel properly. Using rounded values to estimate the answer:
3.5 × 50 = 175 g
This value is close to the calculated answer.
42. Practice Problem (page 237)
Calculate the mass of 0.24 mol of carbon dioxide.
What Is Required?
You need to determine the mass of carbon dioxide.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 55 Chemistry 11 Solutions
What Is Given?
You know the amount in moles of the carbon dioxide: 0.24 mol
You know the chemical formula for carbon dioxide: CO2
Plan Your Strategy
Use the periodic table to find the atomic molar masses of carbon and oxygen.
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of each element to find the molar mass of CO2(g).
Use the relationship m = n × M .
Multiply the amount in moles of CO2(g) by its molar mass.
Act on Your Strategy
Molar mass, M, of CO2(g):
M CO2 = 1M C + 2M O
= 1(12.01 g/mol) + 2(16.00 g/mol)
= 12.01 g/mol + 32.00 g/mol
= 44.01 g/mol
Mass, m, of the CO2(g):
mCO2 = n × M
= 0.24 mol × 44.01 g/ mol
= 10.56 g
= 11 g
The mass of the carbon dioxide is 11 g.
Check Your Solution
The units cancel properly. Estimating, 0.24 is approximately
approximately
1
mol, and 11 is
4
1
the molar mass. The answer is reasonable.
4
43. Practice Problem (page 237)
Calculate the mass of 1.28 × 10–3 mol of glucose, C6H12O6(s).
What Is Required?
You need to determine the mass of a sample of glucose.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 56 Chemistry 11 Solutions
What Is Given?
You know the amount in moles of glucose: 1.28 × 10–3 mol
You know the chemical formula for glucose: C6H12O6
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
C6H12O6(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of each element to find the molar mass of the
compound.
Use the relationship m = n × M .
Multiply the amount in moles of C6H12O6(s) by its molar mass.
Act on Your Strategy
Molar mass, M, of C6H12O6(s):
M C6 H12O6 = 6 M C + 12M H + 6M O
= 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
= 180.18 g/mol
Mass, m, of the C6H12O6(s):
mC6 H12O6 = n × M
= 1.28 × 10 –3 mol × 180.18 g/ mol
= 0.2306 g
= 0.231 g
The mass of the glucose is 0.231 g.
Check Your Solution
The units cancel properly.
Use rounded values for the atomic molar masses to determine an estimate of
the answer:
0.001 × 200 = 0.200 g
This estimate is close to the answer that was calculated.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 57 Chemistry 11 Solutions
44. Practice Problem (page 237)
Calculate the mass of 0.0029 mol of magnesium bromide, MgBr2(s), in
milligrams.
What Is Required?
You need to determine the mass (in milligrams) of magnesium bromide.
What Is Given?
You know the amount in moles of magnesium bromide: 0.0029 mol
You know the chemical formula for magnesium bromide: MgBr2
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
MgBr2(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of each element to find the molar mass of the
compound.
Use the relationship m = n × M .
Multiply the amount in moles of MgBr2(s) by its molar mass.
Convert the mass from grams to milligrams: 1 g = 1 × 103 mg
Act on Your Strategy
Molar mass, M, of MgBr2(s):
M MgBr2 = 1M Mg + 2M Br
= 1(24.31 g/mol) + 2(79.90 g/mol)
= 24.31 g/mol + 159.80 g/mol
= 184.11 g/mol
Mass, m, of the MgBr2(s):
mMgBr2 = n × M
= 0.0029 mol × 184.11 g/ mol
= 0.533919 g
= 0.533919 g × 1 × 103 mg/ g
= 533.919 mg
= 5.3 × 102 mg
The mass of the magnesium bromide is 5.3 × 102 mg.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 58 Chemistry 11 Solutions
Check Your Solution
The units cancel properly.
Use rounded values for the atomic molar masses to get an estimate of the
answer:
0.003 × 180 × 103 = 540 mg
This estimate is close to the calculated answer.
45. Practice Problem (page 237)
Name each compound, and then calculate its mass. Express this value in
scientific notation.
a. 4.5 × 10–3 mol of Co(NO3)2(s)
b. 29.6 mol of Pb(S2O3)2(s)
What Is Required?
a. You need to name and determine the mass of Co(NO3)2(s).
b. You need to name and determine the mass of Pb(S2O3)2(s).
What Is Given?
a. You know the amount in moles of the Co(NO3)2(s): 4.5 × 10–3 mol
b. You know the amount in moles of the Pb(S2O3)2(s): 29.6 mol
Plan Your Strategy
Refer to Chapter 2, Section 2.2 (beginning at page 64) to name the compounds.
Use the periodic table to find the atomic molar masses of the elements in each
compound.
For each compound, multiply each element’s atomic molar mass by the
number of atoms of the element in the compound.
Add the molar masses of each element to find the molar mass of the
compound.
Use the relationship m = n × M .
Multiply the amount in moles of each compound by the compound’s molar
mass.
Act on Your Strategy
a. Co(NO3)2(s)
The name of the compound is cobalt(II) nitrate.
Molar mass, M, of Co(NO3)2(s):
M Co(NO3 )2 = 1M Co + 2M N + 6M O
= 1( 58.93 g/mol ) + 2 (14.01 g/mol ) + 6 (16.00 g/mol )
= 58.93 g/mol + 28.02 g/mol + 96.00 g/mol
= 182.95 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 59 Chemistry 11 Solutions
Mass, m, of the Co(NO3)2(s):
mCo( NO3 ) = n × M
2
= 4.5 × 10 –3 mol × 182.95 g/ mol
= 0.82328 g
= 8.2 × 10 –1 g
The mass of the cobalt(II) nitrate is 8.2 × 10–1 g.
b. Pb(S2O3)2(s)
The name of the compound is lead(IV) thiosulfate.
Molar mass, M, of Pb(S2O3)2(s):
M Pb(S2O3 )2 = 1M Pb + 4M S + 6M O
= 1( 207.2 g/mol ) + 4 ( 32.07 g/mol ) + 6 (16.00 g/mol )
= 207.2 g/mol + 128.28 g/mol + 96.00 g/mol
= 431.48 g/mol
Mass, m, of the Pb(S2O3)2(s):
mPb(S2O3 ) = n × M
2
= 29.6 mol × 431.48 g/ mol
=1.2772 × 10 4 g
=1.28 × 10 4 g
The mass of the lead(IV) thiosulfate is 1.28 × 104 g.
Check Your Solution
The units cancel properly. Check that the correct atomic molar masses have
been used.
Use rounded values for the atomic molar masses to get an estimate of the
answers:
a. 5 × 10–3 × 200 = 1 g
b. 30 × 400 = 1.2 × 104 g
These estimates are reasonably close to the answers that were calculated.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 60 Chemistry 11 Solutions
46. Practice Problem (page 237)
Determine the chemical formula for each compound, and then calculate its
mass.
a. 4.9 mol of ammonium nitrate
b. 16.2 mol of iron(III) oxide
What Is Required?
a. You need to determine the chemical formula for ammonium nitrate, and
then calculate its mass.
b. You need to determine the chemical formula for iron(III) oxide, and then
calculate its mass.
What Is Given?
a. You know the amount in moles of ammonium nitrate: 4.9 mol
b. You know the amount in moles of iron(III) oxide: 16.2 mol
Plan Your Strategy
Refer to Chapter 2, Section 2.2 (beginning at page 64) to write the chemical
formulas.
Use the periodic table to find the atomic molar masses of the elements in each
compound.
For each compound, multiply each element’s atomic molar mass by the
number of atoms of the element in the compound.
Add the molar masses of each element to find the molar mass of the
compound.
Use the relationship m = n × M .
Multiply the amount in moles of each compound by the compound’s molar
mass.
Act on Your Strategy
a. ammonium nitrate
The chemical formula for ammonium nitrate is NH4NO3.
Molar mass, M, of NH4NO3(s):
M NH4 NO3 = 2M N + 4M H + 3M O
= 2 (14.01 g/mol ) + 4 (1.01 g/mol ) + 3 (16.00 g/mol )
= 28.02 g/mol + 4.04 g/mol + 48.00 g/mol
= 80.06 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 61 Chemistry 11 Solutions
Mass, m, of the NH4NO3(s):
mNH 4 NO3 = n × M
= 4.9 mol × 80.06 g/ mol
= 392.294 g
= 3.9 × 102 g
The mass of the ammonium nitrate is 3.9 × 102 g.
b. iron(III) oxide
The chemical formula for iron(III) oxide is Fe2O3.
Molar mass, M, of Fe2O3(s):
M Fe2O3 = 2M Fe + 3M O
= 2 ( 55.85 g/mol ) + 3 (16.00 g/mol )
= 111.7 g/mol + 48.00 g/mol
= 159.7 g/mol
Mass, m, of the Fe2O3(s):
mFe2O3 = n × M
= 16.2 mol × 159.7 g/ mol
= 2.587 × 103 g
= 2.59 × 103 g
The mass of the iron(III) oxide is 2.59 × 103 g.
Check Your Solution
The units cancel properly. Check that the correct atomic molar masses have
been used.
Use rounded values for the atomic molar masses to get an estimate of the
answers:
a. 80 × 5 = 400 g
b. 16 × 160 = 2.6 × 103 g
These estimates are reasonably close to the answers that were calculated.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 62 Chemistry 11 Solutions
47. Practice Problem (page 237)
What is the mass of 1.6 × 10–3 mol of calcium chloride dihydrate,
CaCl2•2H2O(s), in milligrams?
What Is Required?
You need to determine the mass (in milligrams) of calcium chloride dihydrate.
What Is Given?
You know the chemical formula for calcium chloride dihydrate: CaCl2•2H2O
You know the amount in moles of CaCl2•2H2O(s): 1.6 × 10–3 mol
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
CaCl2•2H2O(s).
Multiply each element’s atomic molar mass by the number of atoms of the
element in the compound.
Add the molar masses of each element to find the molar mass of the
compound.
Use the relationship m = n × M .
Multiply the amount in moles of CaCl2•2H2O(s) by its molar mass.
Convert the mass from grams to milligrams: 1 g = 1 × 103 mg
Act on Your Strategy
Molar mass, M, of CaCl2•2H2O(s):
M CaCl2 •2H 2O = 1M CaCl2 + 2 M H 2O
= 1[1M Ca + 2M Cl ] + 2[2M H + 1M O ]
= 1[1(40.08 g/mol) + 2(35.45 g/mol)] + 2 ⎡⎣ 2 (1.01 g/mol ) + 1(16.00 g/mol ) ⎤⎦
= 40.08 g/mol + 70.90 g/mol + 2 (18.02 g/mol )
= 40.08 g/mol + 70.90 g/mol + 36.04 g/mol
= 147.02 g/mol
Mass, m, of the CaCl2•2H2O(s):
mCaCl2 •2H2O = n × M
= 1.6 × 10 –3 mol × 147.02 g/ mol
= 0.23523 g
= 0.23523 g × 1 × 103 mg/ g
= 235.23 mg
= 2.4 × 102 mg
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 63 Chemistry 11 Solutions
The mass of calcium chloride dihydrate is 2.4 × 102 mg.
Check Your Solution
The units cancel properly.
Use rounded values for the atomic molar masses to get an estimate of the
answer:
0.0016 × 150 × 103 = 240 mg
This estimate is close to the answer that was calculated.
48. Practice Problem (page 237)
A litre of water contains 55.56 mol of water molecules. What is the mass of a
litre of water, in kilograms?
What Is Required?
You need to determine the mass (in kilograms) of 1 L of water.
What Is Given?
You know the amount in moles of 1 L of water: 55.56 mol
You know the chemical formula for water: H2O
Plan Your Strategy
Use the periodic table to find the atomic molar masses of hydrogen and
oxygen.
Multiply each element’s atomic molar mass by the number of atoms of the
element in H2O(ℓ).
Add the molar masses of each element to find the molar mass of H2O(ℓ).
Use the relationship m = n × M .
Multiply the amount in moles of H2O(ℓ) by its molar mass.
Convert the mass from grams to milligrams: 1 g = 1 × 103 mg
Act on Your Strategy
Molar mass, M, of H2O(ℓ):
M H 2 O = 2M H + 1 M O
= 2(1.01 g/mol) + 1(16.00 g/mol)
= 2.02 g/mol + 16.00 g/mol
= 18.02 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 64 Chemistry 11 Solutions
Mass, m, of H2O(ℓ):
mH 2 O = n × M
= 55.56 mol × 18.02 g/ mol
= 1.001 × 103 g
= 1.001 × 103 g × 1 × 10 –3 kg/ g
=1.001 kg
The mass of 1 L of water is 1.001 kg.
Check Your Solution
The units cancel properly. Check that the correct atomic molar masses have
been used. Use rounded values for the atomic molar masses to get an estimate
of the answer:
55 × 20 × 10–3 = 1.1 kg
This is a reasonably good estimate of the calculated answer.
49. Practice Problem (page 237)
For each group of three samples, determine the sample with the largest mass.
a. 2.34 mol of bromine, Br2(ℓ); 9.80 mol of hydrogen sulfide, H2S(g);
0.568 mol of potassium permanganate, KMnO4(s)
b. 13.7 mol of strontium iodate, Sr(IO3)2(s); 15.9 mol of gold(III) chloride,
AuCl3(s); 8.61 mol of bismuth silicate, Bi2(SiO3)3(s)
What Is Required?
a. You need to determine the masses of samples of bromine, hydrogen sulfide,
and potassium permanganate and identify which has the largest mass.
b. You need to determine the masses of samples of strontium iodate, gold(III)
chloride, and bismuth silicate and identify which has the largest mass.
What Is Given?
a. You know the chemical formulas for the samples and the amount in moles
of each:
bromine, Br2; n = 2.34 mol
hydrogen sulfide, H2S; n = 9.80 mol
potassium permanganate, KMnO4; n = 0.568 mol
b. You know the chemical formulas for the samples and the amount in moles
of each:
strontium iodate, Sr(IO3)2; n = 13.7 mol
gold(III) chloride, AuCl3; n = 15.9
bismuth silicate, Bi2(SiO3)3; n = 8.61 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 65 Chemistry 11 Solutions
Plan Your Strategy
For each of a. and b.:
Use the periodic table to find the atomic molar masses of the elements in each
compound.
Multiply each element’s atomic molar mass by the number of atoms of the
element in each compound.
For each compound, add the molar masses of each element to find the molar
mass.
Use the relationship m = n × M .
Multiply the amount in moles of each compound by the compound’s molar
mass.
Identify the sample in each group that is the largest in mass.
Act on Your Strategy
a. samples of bromine, hydrogen sulfide, and potassium permanganate
Molar mass, M, of Br2(ℓ):
M Br2 = 2 M Br
= 2(79.90 g/mol)
=159.8 g/mol
Mass, m, of the Br2(ℓ):
mBr2 = n × M
= 2.34 mol × 159.8 g/ mol
= 3.74 × 102 g
The mass of the sample of bromine is 3.74 × 102 g.
Molar mass, M, of H2S(g):
M H2S = 2 M H + 1M S
= 2 (1.01 g/mol ) + 1( 32.07 g/mol )
= 2.02 g/mol + 32.07 g/mol
= 34.09 g/mol
Mass, m, of the H2S(g):
mH 2S = n × M
= 9.8 mol × 34.09 g/mol
= 3.3 × 102 g
The mass of the sample of hydrogen sulfide is 3.74 × 102 g.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 66 Chemistry 11 Solutions
Molar mass, M, of KMnO4(s):
M KMnO4 = 1M K + 1M Mn + 4M O
= 1( 39.1 g/mol ) + 1( 54.94 g/mol ) + 4 (16.00 g/mol )
= 39.10 g/mol + 54.94 g/mol + 64.00 g/mol
= 158.04 g/mol
Mass, m, of the KMnO4(s):
mKMnO4 = n × M
= 0.568 mol × 158.04 g/mol
= 89.8 g
The mass of the sample of potassium permanganate is 89.8 g.
The sample of bromine has the largest mass.
b. samples of strontium iodate, gold(III) chloride, and bismuth silicate:
Molar mass, M, of Sr(IO3)2(s):
M Sr(IO3 )2 = 1M Sr + 2 M I + 6M O
= 1(87.62 g/mol) + 2(126.9 g/mol) + 6(16.00 g/mol)
= 87.62 g/mol + 253.8 g/mol + 96.00 g/mol
= 437.42 g/mol
Mass, m, of the Sr(IO3)2(s):
mSr ( IO3 ) = n × M
2
= 13.7 mol × 437.42 g/ mol
= 5.99 × 103 g
The mass of the sample of strontium iodate is 5.99 × 103 g.
Molar mass, M, of AuCl3(s):
M AuCl3 = 1M Au + 3M Cl
= 1(196.97 g/mol ) + 3 ( 35.45 g/mol )
= 196.97 g/mol + 106.35 g/mol
= 303.32 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 67 Chemistry 11 Solutions
Mass, m, of the AuCl3(s):
mAuCl3 = n × M
= 15.9 mol × 303.32 g/ mol
= 4.82 × 103 g
The mass of the sample of gold(III) chloride is 4.82 × 103 g.
Molar mass, M, of Bi2(SiO3)3(s):
M Bi2 (SiO3 )3 = 2 M Bi + 3M Si + 9M O
= 2 ( 208.98 g/mol ) + 3 ( 28.09 g/mol ) + 9 (16.00 g/mol )
= 417.96 g/mol + 84.27 g/mol + 144.00 g/mol
= 646.23 g/mol
Mass, m, of the Bi2(SiO3)3(s):
mBi2 (SiO3 ) = n × M
3
= 8.61 mol × 646.23 g/ mol
= 5.56 × 103 g
The mass of the sample of bismuth silicate is 5.56 × 103 g.
The sample of strontium iodate has the largest mass.
Check Your Solution
The units cancel properly. Check that the correct atomic molar masses have
been used.
The answers seem reasonable. However, the comparisons are too close to
estimate using rounded numbers. The calculations should be rechecked.
50. Practice Problem (page 237)
Which has the smallest mass: 0.215 mol of potassium hydrogen sulfite,
KHSO3(s); 1.62 mol of sodium hydrogen sulfite, NaHSO3(s); or 0.0182 mol of
aluminum iodate, Al(IO3)3(s)?
What Is Required?
You need to determine the masses of samples of potassium hydrogen sulfite,
sodium hydrogen sulfite, and aluminum iodate and identify which has the
smallest mass.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 68 Chemistry 11 Solutions
What Is Given?
You know the chemical formulas for the compounds and their amounts in
moles:
potassium hydrogen sulfite, KHSO3; n = 0.215 mol
sodium hydrogen sulfite, NaHSO3; n = 1.62 mol
aluminum iodate, Al(IO3)3; n = 0.0182 mol
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in each
compound.
Multiply each element’s atomic molar mass by the number of atoms of the
element in each compound.
For each compound, add the molar masses of each element to find the molar
mass.
Use the relationship m = n × M .
Multiply the amount in moles of each compound by the compound’s molar
mass.
Identify the sample that has the smallest mass.
Act on Your Strategy
Molar mass, M, of KHSO3(s):
M KHSO3 = 1M K + 1M H + 1M S + 3M O
= 1(39.10 g/mol) + 1(1.01 g/mol) + 1(32.07 g/mol) + 3(16.00 g/mol)
= 39.10 g/mol + 1.01 g/mol + 32.07 g/mol + 48.00 g/mol
= 120.18 g/mol
Mass, m, of the KHSO3(s):
mKHSO3 = n × M
= 0.215 mol × 120.18 g/ mol
= 25.8 g
The mass of the sample of potassium hydrogen sulfite is 25.8 g.
Molar mass, M, of NaHSO3(s):
M NaHSO3 = 1M Na + 1M H + 1M S + 3M O
= 1( 22.99 g/mol ) + 1(1.01 g/mol ) + 1( 32.07 g/mol ) + 3 (16.00 g/mol )
= 22.99 g/mol + 1.01 g/mol + 32.07 g/mol + 48.00 g/mol
= 104.07 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 69 Chemistry 11 Solutions
Mass, m, of the NaHSO3(s):
mNaHSO3 = n × M
= 1.62 mol × 104.07 g/ mol
= 168 g
The mass of the sample of sodium hydrogen sulfite is 168 g.
Molar mass, M, of Al(IO3)3(s):
M Al(IO3 )3 = 1M Al + 3M I + 9M O
= 1( 26.98 g/mol ) + 3 (126.90 g/mol ) + 9 (16.00 g/mol )
= 26.98 g/mol + 380.7 g/mol + 144.00 g/mol
= 551.68 g/mol
Mass, m, of the Al(IO3)3(s):
mAl( IO3 ) = n × M
3
= 0.0182 mol × 551.68 g/ mol
= 10.0 g
The mass of the sample of aluminum iodate is 10.0 g.
The sample of aluminum iodate has the smallest mass.
Check Your Solution
The units cancel properly. Check that the correct atomic molar masses have
been used.
The answers seem reasonable. However, the comparisons are too close to
estimate using rounded numbers. The calculations should be rechecked.
Section 5.2 Mass and the Mole
Solutions for Practice Problems
Student Edition page 239
51. Practice Problem (page 239)
Convert 29.5 g of ammonia to the amount in moles.
What Is Required?
You need to convert 29.5 g of ammonia to the amount in moles.
What Is Given?
You know the mass of the ammonia: 29.5 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 70 Chemistry 11 Solutions
Plan Your Strategy
Determine the chemical formula for ammonia.
Use the periodic table to find the atomic molar masses of the elements in
ammonia.
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of ammonia.
Divide the mass by the molar mass to determine the amount in moles of
m
ammonia using the relationship n = .
M
Act on Your Strategy
The chemical formula for ammonia is NH3.
Molar mass, M, of NH3(g):
M NH3 = 1M N + 3M H
= 1(14.01 g/mol ) + 3 (1.01 g/mol )
= 17.04 g/mol
Amount in moles, n, of the NH3(g):
m
nNH3 =
M
29.5 g
=
17.04 g /mol
= 1.73 mol
There is 1.73 mol of ammonia in the sample.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
52. Practice Problem (page 239)
Determine the amount in moles of potassium thiocyanate, KSCN(s), in
13.5 kg.
What Is Required?
You need to determine the amount in moles of 13.5 kg of potassium
thiocyanate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 71 Chemistry 11 Solutions
What Is Given?
You know the chemical formula for potassium thiocyanate: KSCN
You know the mass of the KSCN(s): 13.5 kg
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
KSCN(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of KSCN(s).
Convert the mass of the KSCN(s) from kilograms to grams: 1 kg = 1 × 103 g
Divide the mass by the molar mass to determine the amount in moles, n, of
m
KSCN(s) using the relationship n = .
M
Act on Your Strategy
Molar mass, M, of KSCN(s):
M KSCN = 1M K + 1M S + 1M C + 1M N
= 1( 39.10 g/mol ) + 1( 32.07 g/mol ) + 1( 2.01 g/mol ) + 1(14.01 g/mol )
= 97.19 g/mol
Mass (in grams), m, of the KSCN(s):
mKSCN = 13.5 kg × 1 × 103 g/ kg
= 13 500 g
Amount in moles, n, of KSCN(s):
m
nKSCN =
M
13 500 g
=
97.19 g /mol
= 138.9 mol
= 139 mol
There is 139 mol of potassium thiocyanate in the sample.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 72 Chemistry 11 Solutions
53. Practice Problem (page 239)
Determine the amount in moles of sodium dihydrogen phosphate, NaH2PO4(s),
in 105 mg.
What Is Required?
You need to determine the amount in moles of 105 mg of sodium dihydrogen
phosphate.
What Is Given?
You know the chemical formula for sodium dihydrogen phosphate: NaH2PO4
You know the mass of the NaH2PO4(s): 105 mg
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
NaH2PO4(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of NaH2PO4(s).
Convert the mass of the NaH2PO4(s) from kilograms to grams: 1 kg = 1 × 103 g
Divide the mass by the molar mass to determine the amount in moles, n, of
m
NaH2PO4(s) using the relationship n =
.
M
Act on Your Strategy
Molar mass, M, of NaH2PO4(s):
M NaH2 PO4 = 1M Na + 2M H + 1M P + 4M O
= 1( 22.99 g/mol ) + 2 (1.01 g/mol ) + 1( 30.97 g/mol ) + 4 (16.00 g/mol )
= 119.98 g/mol
Mass (in grams), m, of the NaH2PO4(s):
mNaH2 PO4 = 105 mg × 1 × 10 –3 g/ mg
= 0.105 g
Amount in moles, n, of NaH2PO4(s):
m
nNaH2 PO4 =
M
0.105 g
=
119.98 g /mol
= 8.7514 × 10 –4 mol
= 8.75 × 10 –4 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 73 Chemistry 11 Solutions
There is 8.75 × 10–4 mol of sodium dihydrogen phosphate in the sample.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
54. Practice Problem (page 239)
Determine the amount in moles of xenon tetrafluoride, XeF4(s), in 22 mg.
What Is Required?
You need to determine the amount in moles of 22 mg of xenon tetrafluoride.
What Is Given?
You know the chemical formula for xenon tetrafluoride: XeF4
You know the mass of the XeF4(s): 22 mg
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
XeF4(s). Multiply the atomic molar masses by the number of atoms of each
element in the compound.
Add these values to calculate the molar mass of XeF4(s).
Convert the mass of the XeF4(s) from kilograms to grams: 1 kg = 1 × 103 g
Divide the mass by the molar mass to determine the amount in moles, n, of
m
.
XeF4(s) using the relationship n =
M
Act on Your Strategy
Molar mass, M, of XeF4(s):
M XeF4 = 1M Xe + 4M F
= 1(131.29 g/mol ) + 4 (19.00 g/mol )
= 207.29 g/mol
Mass (in grams), m, of the XeF4(s):
mXeF4 = 22 mg × 1 × 10 –3 g/ mg
= 0.022 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 74 Chemistry 11 Solutions
Amount in moles, n, of XeF4(s):
m
nXeF4 =
M
0.022 g
=
207.29 g /mol
= 1.0613 × 10 –4 mol
= 1.1 × 10 –4 mol
There is 1.1 × 10–4 mol of xenon tetrafluoride in the sample.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
55. Practice Problem (page 239)
Write the chemical formula for each compound, and then calculate the amount
in moles in each sample.
a. 3.7 × 10–3 g of silicon dioxide
b. 25.38 g of titanium(IV) nitrate
c. 19.2 mg of indium carbonate
d. 78.1 kg of copper(II) sulfate pentahydrate
What Is Required?
You need to write the chemical formula for, and determine the amount in
moles of, samples of
a. silicon dioxide.
b. titanium(IV) nitrate.
c. indium carbonate.
d. copper(II) sulfate pentahydrate.
What Is Given?
a. You know the mass of the silicon dioxide: 3.7 × 10–3 g
b. You know the mass of the titanium(IV) nitrate: 25.38 g
c. You know the mass of the indium carbonate: 19.2 mg
d. You know the mass of the copper(II) sulfate pentahydrate: 78.1 kg
Plan Your Strategy
For each compound:
Write the chemical formula.
Use the periodic table to find the atomic molar masses of the elements in the
compound.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 75 Chemistry 11 Solutions
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Convert the mass in milligrams to grams: 1 mg = 1 × 10–3 g
Convert the mass in kilograms to grams: 1 kg = 1 × 103 g
Divide the mass by the molar mass to determine the amount in moles of each
m
compound using the relationship n = .
M
Act on Your Strategy
a. silicon dioxide
The chemical formula for silicon dioxide is SiO2.
Molar mass, M, of SiO2(s):
M SiO2 = 1M S + 2M O
= 1( 28.09 g/mol ) + 2 (16.00 g/mol )
= 60.09 g/mol
Amount in moles, n, of SiO2(s):
m
nSiO2 =
M
3.7 ×10 –3 g
=
60.09 g /mol
= 6.157 × 10 –5 mol
= 6.2 × 10 –5 mol
There is 6.2 × 10–5 mol of silicone dioxide in the sample.
b. titanium(IV) nitrate
The chemical formula for titanium(IV) nitrate is Ti(NO3)4.
Molar mass, M, of Ti(NO3)4(s):
M Ti(NO3 )4 = 1M Ti + 4M N + 12M O
= 1( 47.87 g/mol ) + 4 (14.01 g/mol ) + 12 (16.00 g/mol )
= 295.91 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 76 Chemistry 11 Solutions
Amount in moles, n, of the Ti(NO3)4(s):
m
nTi( NO3 ) =
4
M
25.38 g
=
295.91 g /mol
= 8.5769 × 10 –2 mol
= 8.577 × 10 –2 mol
There is 8.577 × 10–2 mol of titanium(IV) nitrate in the sample.
c. indium carbonate
The chemical formula for indium carbonate is In2(CO3)3.
Molar mass, M, of In2(CO3)3(s):
M In 2 (CO3 )3 = 2 M In + 3M C + 9M O
= 2 (114.82 g/mol ) + 3 (12.01 g/mol ) + 9 (16.00 g/mol )
= 409.67 g/mol
Mass (in grams), m, of In2(CO3)3(s):
mIn 2 ( CO3 ) = 19.2 mg × 1 × 10 –3 g/ mg
3
= 0.0192 g
Amount in moles, n, of In2(CO3)3(s):
m
nIn 2 ( CO3 ) =
3
M
0.0192 g
=
409.67 g /mol
= 4.68669 × 10 –5 mol
= 4.69 × 10 –5 mol
There is 4.69 × 10–5 mol of indium carbonate in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 77 Chemistry 11 Solutions
d. copper(II) sulfate pentahydrate
The chemical formula for copper(II) sulfate pentahydrate is CuSO4•5H2O.
Molar mass, M, of CuSO4•5H2O(s):
M CuSO4 •5H2O = 1M CuSO4 + 5M H2O
= 1[1M Cu + 1M S + 4M O ] + 5[2M H + 1M O ]
= 1[1( 63.55 g/mol ) + 1( 32.7 g/mol ) + 4 (16.00 g/mol )] + 5 (18.02 g/mol )
= 249.62 g/mol
Mass (in grams), m, of CuSO4•5H2O(s):
mCuSO4 •5H2O = 78.1 kg × 1 × 103 g/ kg
= 7.81 × 104 g
Amount in moles, n, of CuSO4•5H2O(s):
m
nCuSO4 •5H2O =
M
7.81 × 104 g
=
249.62 g /mol
= 3.12875 × 102 mol
= 3.13 × 102 mol
There is 3.13 × 102 mol of copper(II) sulfate pentahydrate in the sample.
Check Your Solution
Check that the correct atomic molar masses have been used and substitutions
have been made correctly. The units are correct. The answers seem reasonable
and show the correct number of significant digits.
56. Practice Problem (page 239)
The characteristic odour of garlic comes from allyl sulfide, (C3H5)2S(ℓ).
Determine the amount in moles of allyl sulfide in 168 g.
What Is Required?
You need to determine the amount in moles of 168 g of allyl sulfide.
What Is Given?
You know the chemical formula for allyl sulfide: (C3H5)2S
You know the mass of the (C3H5)2S(ℓ): 168 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 78 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
(C3H5)2S(ℓ).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of (C3H5)2S(ℓ).
Divide the mass by the molar mass to determine the amount in moles of
m
.
(C3H5)2S(ℓ) using the relationship n =
M
Act on Your Strategy
Molar mass, M, of (C3H5)2S(ℓ):
M (C3H5 )2 S = 6 M C + 10M H + 1M S
= 6 (12.01 g/mol ) + 10 (1.01 g/mol ) + 1( 32.07 g/mol )
= 114.23 g/mol
Amount in moles, n, of (C3H5)2S(ℓ):
m
n( C3H5 ) S =
2
M
168 g
=
114.23 g /mol
= 1.47 mol
There is 1.47 mol of allyl sulfide in the sample.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
57. Practice Problem (page 239)
Road salt, CaCl2(s), is often used on roads in the winter to prevent the build-up
of ice. What amount in moles of calcium chloride is in a 20.0 kg bag of road
salt?
What Is Required?
You need to determine the amount in moles of 20.0 kg of calcium chloride.
What Is Given?
You know the chemical formula for calcium chloride: CaCl2
You know the mass of the CaCl2(s): 20.0 kg
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 79 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
CaCl2(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of CaCl2(s).
Convert the mass from kilograms to grams: 1 kg = 1 × 103 g
Divide the mass by the molar mass to determine the amount in moles of
m
CaCl2(s) using the relationship n = .
M
Act on Your Strategy
Molar mass, M, of CaCl2(s):
M CaCl2 = 1M Ca + 2M Cl
= 1( 40.08 g/mol ) + 2 ( 35.45 g/mol )
= 110.98 g/mol
Mass (in grams), m, of CaCl2(s):
mCaCl2 = 20.0 kg × 1 × 103 g/ kg
= 2.00 × 104 kg
Amount in moles, n, of CaCl2(s):
m
nCaCl2 =
M
2.00 × 104 g
=
110.98 g /mol
= 1.80 × 102 mol
There is 1.80 × 102 mol of calcium chloride in the bag.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 80 Chemistry 11 Solutions
58. Practice Problem (page 239)
Calculate the amount in moles of trinitrotoluene, C7H5(NO2)3(s), an explosive,
in 3.45 × 10–3 g.
What Is Required?
You need to determine the amount in moles of 3.45 × 10–3 g of trinitrotoluene.
What Is Given?
You know the chemical formula for trinitrotoluene: C7H5(NO2)3
You know the mass of the C7H5(NO2)3(s): 3.45 × 10–3 g
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
C7H5(NO2)3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of C7H5(NO2)3(s).
Divide the mass by the molar mass to determine the amount in moles of
m
C7H5(NO2)3(s) using the relationship n = .
M
Act on Your Strategy
Molar mass, M, of C7H5(NO2)3(s):
M C7 H5 ( NO2 )3 = 7 M C + 5M H + 3M N + 6M O
= 7 (12.01 g/mol ) + 5 (1.01 g/mol ) + 3 (14.01 g/mol ) + 6 (16.00 g/mol )
= 227.15 g/mol
Amount in moles, n, of the C7H5(NO2)3(s):
m
nC7 H5 ( NO2 ) =
3
M
3.45 × 10 –3 g
=
227.15 g /mol
= 1.51880 × 10 –5 mol
= 1.52 × 10 –5 mol
There is 1.52 × 10–5 mol of trinitrotoluene in the explosive.
Check Your Solution
The units are correct and substitutions have been made correctly. The answer
seems reasonable and shows the correct number of significant digits.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 81 Chemistry 11 Solutions
59. Practice Problem (page 239)
Arrange the following substances in order from largest to smallest amount in
moles:
• 865 mg of Ni(NO3)2(s)
• 9.82 g of Al(OH)3(s)
• 10.4 g of AgCl(s)
What Is Required?
You need to determine the amount in moles of samples of Ni(NO3)2(s),
Al(OH)3(s), and AgCl(s) and arrange these amounts from largest to smallest.
What Is Given?
You know the mass of each substance:
Ni(NO3)2(s) = 865 mg
Al(OH)3(s) = 9.82 g
AgCl(s) =10.4 g
Plan Your Strategy
For each compound:
Use the periodic table to find the atomic molar masses of the elements in the
compound.
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Convert the mass from milligrams to grams: 1 mg = 1 × 10–3 g
Divide the mass by the molar mass to determine the amount in moles of each
m
compound using the relationship n = .
M
Arrange the amounts in moles from largest to smallest.
Act on Your Strategy
Molar mass, M, of Ni(NO3)2(s):
M Ni(NO3 )2 = 1M Ni + 2M N + 6M O
= 1( 58.69 g/mol ) + 2 (14.01 g/mol ) + 6 (16.00 g/mol )
= 182.71 g/mol
Mass (in grams), m, of the Ni(NO3)2(s):
mNi( NO3 ) = 865 mg × 1 × 10 –3 g/ mg
2
= 0.865 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 82 Chemistry 11 Solutions
Amount in moles, n, of the Ni(NO3)2:
m
nNi( NO3 ) =
2
M
0.865 g
=
182.71 g /mol
= 4.734 × 10 –3 mol
= 4.73 × 10 –3 mol
The amount in moles of the Ni(NO3)2 is 4.73 × 10−3 mol.
Molar mass, M, of Al(OH)3(s):
M Al( OH ) = 1M Al + 3M O + 3M H
3
= 1( 26.98 g/mol ) + 3 (16.00 g/mol ) + 3 (1.01 g/mol )
= 78.10 g/mol
Amount in moles, n, of the Al(OH)3(s):
m
nAl( OH ) =
3
M
9.82 g
=
78.10 g /mol
= 0.12574 mol
= 0.126 mol
The amount in moles of the Al(OH)3(s) is 0.126 mol.
Molar mass, M, of AgCl(s):
M AgCl = 1M Ag + 1M Cl
= 1(107.87 g/mol ) + 1( 35.45 g/mol )
= 143.32 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 83 Chemistry 11 Solutions
Amount in moles, n, of the AgCl(s):
m
nAgCl =
M
10.4 g
=
143.32 g /mol
= 7.2565 × 10 –2 mol
= 7.26 × 10 –2 mol
The amount in moles of the AgCl(s) is 7.26 × 10−2 mol.
Amounts in moles from largest to smallest:
Al(OH)3(s) (0.126 mol) > AgCl(s) (0.0726 mol) > Ni(NO3)2(s) (0.00473 mol)
Check Your Solution
Check that the correct atomic molar masses have been used and substitutions
have been made correctly. The units are correct. The answers seem reasonable
and show the correct number of significant digits. The amounts in moles are
arranged from largest to smallest.
60. Practice Problem (page 239)
Place the following substances in order from smallest to largest amount in
moles, given 20.0 g of each:
• glucose, C6H12O6(s)
• barium perchlorate, Ba(ClO4)2(s)
• tin(IV) oxide, SnO2(s)
What Is Required?
You need to determine the amount in moles of samples of glucose, barium
perchlorate, and tin(IV) oxide and arrange these amounts from smallest to
largest.
What Is Given?
You know the chemical formula for each compound: glucose, C6H12O6; barium
perchlorate, Ba(ClO4)2; tin(IV) oxide, SnO2
You know the mass of each compound: 20.0 g
Plan Your Strategy
For each compound:
Use the periodic table to find the atomic molar masses of the elements in the
compound.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 84 Chemistry 11 Solutions
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Divide the mass by the molar mass to determine the amount in moles of each
m
.
compound using the relationship n =
M
Arrange the amounts in moles from smallest to largest.
Act on Your Strategy
Molar mass, M, of C6H12O6(s):
M C6 H12O6 = 6 M C + 12M H + 6M O
= 6 (12.01 g/mol ) + 12 (1.01 g/mol ) + 6 (16.00 g/mol )
= 180.18 g/mol
Amount in moles, n, of the C6H12O6(s):
m
nC6 H12O6 =
M
20.0 g
=
180.18 g / mol
= 0.111 mol
The amount in moles of glucose is 0.111 mol.
Molar mass, M, of Ba(ClO4)2(s):
M Ba(ClO4 )2 = 1M Ba + 2M Cl +8M O
= 1(137.33 g/mol ) + 2 ( 35.45 g/mol ) + 8 (16.00 g/mol )
= 336.23 g/mol
Amount in moles, n, of the Ba(ClO4)2(s):
m
nBa ( ClO4 ) =
2
M
20.0 g
=
336.23 g /mol
= 0.0595 mol
The amount in moles of barium perchlorate is 0.0595 mol.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 85 Chemistry 11 Solutions
Molar mass, M, of SnO2(s):
M SnO2 = 1M Sn + 2M O
= 1(118.71 g/mol ) + 2 (16.00 g/mol )
= 150.71 g/mol
Amount in moles, n, of the Sn2O4(s):
m
nSn 2O4 =
M
20.0 g
=
150.71 g /mol
= 0.133 mol
The amount in moles of tin(IV) oxide is 0.133 mol.
Amounts in moles from smallest to largest:
Ba(ClO4)2(s) (0.0595 mol) < C6H12O6(s) (0.111 mol) < SnO2(s) (0.133 mol)
Check Your Solution
Check that the correct atomic molar masses have been used and substitutions
have been made correctly. The units are correct. The answers seem reasonable
and show the correct number of significant digits. The amounts in moles are
arranged from smallest to largest.
Section 5.2 Mass and the Mole
Solutions for Practice Problems
Student Edition page 242
61. Practice Problem (page 242)
Calculate the mass of each sample.
a. 1.05 × 1026 atoms of neon, Ne(g)
b. 2.7 × 1024 molecules of phosphorus trichloride, PCl3(ℓ)
c. 8.72 × 1021 molecules of karakin, C15H21N3O15(s)
d. 6.7 × 1027 formula units of sodium thiosulfate, Na2S2O3(s)
What Is Required?
You need to determine the mass of a sample of
a. neon.
b. phosphorus trichloride.
c. karakin.
d. sodium thiosulfate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 86 Chemistry 11 Solutions
What Is Given?
a. You know the number of atoms of neon, Ne(g): 1.05 × 1026
b. You know the number of molecules of phosphorus trichloride, PCl3(ℓ): 2.7
× 1024
c. You know the number of molecules of karakin, C15H21N3O15(s): 8.72 × 1021
d. You know the number of formula units of sodium thiosulfate, Na2S2O3(s):
6.7 × 1027
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of each compound using the relationship
N
.
n=
NA
Use the periodic table to find the atomic molar masses of the elements in the
compound.
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Calculate the mass of each compound using the relationship m = n × M .
Alternative Solution
The steps can be completed in one line, using the units to calculate the answer.
Calculate the mass of each compound by multiplying the given amount of the
1 mol
and then by the molar mass, M.
compound by
6.02 × 10 23 units
Act on Your Strategy
a. neon
M Ne = 20.18 g/mol (from the periodic table)
Amount in moles, n, of Ne(g):
N
nNe =
NA
=
1.05 × 1026 atoms
6.02 × 1023 atoms /mol
= 1.7442 × 102 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 87 Chemistry 11 Solutions
Mass, m, of the Ne(g):
mNe = n × M
= 1.7442 × 102 mol × 20.18 g/ mol
= 3.52 × 103 g
The mass of the neon is 3.52 × 103 g.
Alternative solution
M Ne = 20.18 g/mol (from the periodic table)
Mass, m, of the Ne(g):
mNe = 1.05 × 10 26 atoms ×
1 mol
20.18 g
×
23
1 mol
6.02 × 10 atoms
= 3.52 × 103 g
The mass of the neon is 3.52 × 103 g.
b. phosphorus trichloride
Molar mass, M, of PCl3(ℓ):
M PCl 3 = 1M P + 3M Cl
= 1( 30.97 g/mol ) + 3 ( 35.45 g/mol )
= 137.32 g/mol
Amount in moles, n, of PCl3(ℓ):
N
nPCl3 =
NA
2.7 × 1024 molecules
6.02 × 1023 moledules /mol
= 4.485 mol
=
Mass, m, of the PCl3(ℓ):
mPCl3 = n × M
= 4.485 mol × 137.32 g/ mol
= 6.15887 × 102 g
= 6.2 × 102 g
The mass of the phosphorus trichloride is 6.2 × 102 g.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 88 Chemistry 11 Solutions
Alternative Solution
Molar mass, M, of PCl3(ℓ):
M PCl 3 = 1M P + 3M Cl
= 1( 30.97 g/mol ) + 3 ( 35.45 g/mol )
= 137.32 g/mol
Mass, m, of the PCl3(ℓ):
mPCl3 = 2.7 × 1024 molecules ×
137.32 g
1 mol
×
23
1 mol
6.02 × 10 molecules
= 6.15887 × 102 g
= 6.2 × 102 g
The mass of the phosphorus trichloride is 6.2 × 102 g.
c. karakin
Molar mass, M, of C15H21N3O15(s):
M C15 H21N3O15 = 15M C + 21M H + 3M N + 15M O
= 15 (12.01 g/mol ) + 21(1.01 g/mol ) + 3 (14.01 g/mol ) + 15 (16.00 g/mol )
= 483.39 g/mol
Amount in moles, n, of C15H21N3O15(s):
N
nC15 H21N3O15 =
NA
=
8.72 × 1021 molecules
6.02 × 1023 moledules /mol
= 1.4485 × 10 –2 mol
Mass, m, of the C15H21N3O15(s):
mC15 H21N3O15 = n × M
= 1.4485 × 10 –2 mol × 483.39 g/ mol
= 7.0019 g
= 7.00 g
The mass of the karakin is 7.00 g.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 89 Chemistry 11 Solutions
Alternative Solution
Molar mass, M, of C15H21N3O15(s):
M C15 H21N3O15 = 15M C + 21M H + 3M N + 15M O
= 15 (12.01 g/mol ) + 21(1.01 g/mol ) + 3 (14.01 g/mol ) + 15 (16.00 g/mol )
= 483.39 g/mol
Mass, m, of the C15H21N3O15(s):
mC15 H21N3O15 = 8.72 × 1021 molecules ×
483.39 g
1 mol
×
23
6.02 × 10 molecules
1 mol
= 7.0019 g
= 7.00 g
The mass of the karakin is 7.00 g.
d. sodium thiosulfate
Molar mass, M, of Na2S2O3(s):
M Na 2S2O3 = 2 M Na + 2M S + 3M O
= 2 ( 22.99 g/mol ) + 2 ( 32.07 g/mol ) + 3 (16.00 g/mol )
= 158.12 g/mol
Amount in moles, n, of the Na2S2O3(s):
N
nNa 2S2O3 =
NA
=
6.7 × 1027 formula units
6.02 × 1023 formula units / mol
= 1.11295 × 104 mol
Mass m, of the Na2S2O3(s):
mNa 2S2O3 = n × M
= 1.1129 × 104 mol × 158.12 g/ mol
= 1.7598 × 106 g
= 1.8 × 106 g
The mass of the sodium thiosulfate is 1.8 × 106 g.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 90 Chemistry 11 Solutions
Alternative Solution
Molar mass, M, of Na2S2O3(s):
M Na 2S2O3 = 2 M Na + 2M S + 3M O
= 2 ( 22.99 g/mol ) + 2 ( 32.07 g/mol ) + 3 (16.00 g/mol )
= 158.12 g/mol
Mass, m, of the Na2S2O3(s):
mNa 2S2O3 = 6.7 × 1027 formula units ×
158.12 g
1 mol
×
23
1 mol
6.02 × 10 formula units
= 1.7598 × 106 g
= 1.8 × 106 g
The mass of the sodium thiosulfate is 1.8 × 106 g.
Check Your Solutions
In each case, check that the correct atomic molar masses and formulas have
been used. Using rounded numbers, estimate the answers:
1
a. 1 × 1026 ×
× 20 = 3 × 103 g
6.02 × 1023
1
b. 3 × 1024 ×
× 140 = 7 × 102 g
23
6.02 × 10
1
c. 9 × 1021 ×
× 480 = 7.2 g
6.02 × 1023
1
d. 7 × 1027 ×
× 160 = 1.9 × 106 g
6.02 × 1023
All of the estimated answers are close to the calculated answers. The answers
are reasonable.
62. Practice Problem (page 242)
Determine the number of molecules or formula units in each sample.
a. 32.4 g of lead(II) phosphate, Pb3(PO4)2(s)
b. 8.62 × 10–3 g of dinitrogen pentoxide, N2O5(s)
c. 48 kg of molybdenum(VI) oxide, MoO3(s)
d. 567 g of tin(IV) fluoride, SnF4(s)
What Is Required?
You need to determine the number of
a. formula units of lead(II) phosphate.
b. molecules of dinitrogen pentoxide.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 91 Chemistry 11 Solutions
c. formula units of molybdenum(VI) oxide .
d. formula units of tin(IV) fluoride.
What Is Given?
You know the mass of each of the samples:
a. Pb3(PO4)2(s) = 32.4 g
b. N2O5(s) = 8.62 × 10–3 g
c. MoO3(s) = 48 kg
d. SnF4(s) = 567 g
Plan Your Strategy
For each compound:
Use the periodic table to find the atomic molar masses of the elements in the
compound.
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Solution 1:
Convert the mass in kilograms to grams: 1 kg = 1 × 103 g
Calculate the amount in moles of each compound using the relationship
m
.
n=
M
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules or formula units of each compound using
N
the relationship n =
.
NA
Solution 2:
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the number of formula units or molecules of the compounds by
1
multiplying the mass by
and then by the Avogadro constant, NA.
molar mass
Act on Your Strategy
a. lead(II) phosphate
Molar mass, M, of Pb3(PO4)2(s):
M Pb3 (PO4 )2 = 3M Pb + 2M P + 8M O
= 3 ( 207.2 g/mol ) + 2 ( 30.97 g/mol ) + 8 (16.00 g/mol )
= 811.54 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 92 Chemistry 11 Solutions
Solution 1
Amount in moles, n, of the Pb3(PO4)2(s):
m
nPb3 ( PO4 ) =
2
M
32.4 g
=
811.54 g /mol
= 3.99 × 10 –2 mol
Number of formula units, N, of Pb3(PO4)2(s):
N = n × NA
= 3.99 × 10 –2 mol ×
6.02 × 1023 formula units
1 mol
= 2.40 × 1022 formula units
There are 2.41 × 1022 formula units of lead(II) phosphate in the sample.
Solution 2
Number of formula units, N, of Pb3(PO4)2(s):
1 mol
6.02 × 1023 formula units
N = 32.4 g ×
×
1 mol
811.54 g
= 2.40 × 10 22 formula units
There are 2.41 × 1022 formula units of lead(II) phosphate in the sample.
b. dinitrogen pentoxide
Molar mass, M, of N2O5(s):
M N2O5 = 2 M N + 5M O
= 2 (14.01 g/mol ) + 5 (16.00 g/mol )
= 108.02 g/mol
Solution 1
Amount in moles, n, of the N2O5(s):
m
nN2O5 =
M
8.64 × 10 –3 g
=
108.02 g /mol
= 7.995 × 10 –5 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 93 Chemistry 11 Solutions
Number of molecules, N, of N2O5(s):
N = n × NA
= 7.995 × 10 –5 mol ×
6.02 ×1023 molecules
1 mol
= 4.82 × 1019 molecules
There are 4.82 × 1019 molecules of dinitrogen pentoxide.
Solution 2
Number of molecules, N, of N2O5(s):
1 mol
6.02 ×1023 molecules
N = 8.64 × 10 –3 g ×
×
1 mol
108.02 g
= 4.82 × 1019 molecules
There are 4.82 × 1019 molecules of dinitrogen pentoxide in the sample.
c. molybdenum(VI) oxide
Molar mass, M, of MoO3(s):
M MoO 3 = 1M Mo + 3M O
= 1(95.96 g/mol) + 1(16.00 g/mol)
= 143.96 g/mol
Solution 1
Mass (in grams), m, of the MoO3(s):
mMoO3 = 48 kg × 1 × 103 g/ kg
= 48 000 g
Amount in moles, n, of the MoO3(s):
m
nMoO3 =
M
48 000 g
=
143.96 g /mol
= 3.3343 × 102 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 94 Chemistry 11 Solutions
Number of formula units, N, of the MoO3(s):
N = n × NA
= 3.3343 × 102 mol ×
6.02 ×1023 formula units
1 mol
= 2.0 × 1026 formula units
There are 2.0 × 1026 formula units of molybdenum(VI) oxide in the sample.
Solution 2
Number of formula units, N, of the MoO3(s):
1 mol
6.02 ×1023 formula units
N = 48 000 g ×
×
1 mol
143.96 g
= 2.0 × 10 26 formula units
There are 2.0 × 1026 formula units of molybdenum(VI) oxide in the sample.
d. tin(IV) fluoride
Molar mass, M, of the SnF4(s)
M SnF4 = 1M Sn + 4M F
= 1(118.71 g/mol ) + 4 (19.00 g/mol )
= 194.71 g/mol
Solution 1
Amount in moles, n, of the SnF4(s):
m
nSnF4 =
M
567 g
=
194.71 g /mol
= 2.912 mol
Number of formula units, N, SnF4(s):
N = n × NA
= 2.912 mol ×
6.02 ×1023 formula units
1 mol
= 1.75 × 1024 formula units
There are 1.75 × 1024 formula units of tin(IV) fluoride in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 95 Chemistry 11 Solutions
Solution 2
Number of formula units, N, of the SnF4(s):
1 mol
6.02 × 10 23 formula units
N = 567 g ×
×
1 mol
194.71 g
= 1.75 × 10 24 formula units
There are 1.75 × 1024 formula units of tin(IV) fluoride in the sample.
Check Your Solutions
In all cases, check that the correct atomic molar masses have been used.
Using rounded numbers, estimate the answers:
1
a. 30 ×
× 6 × 1023 = 2 × 1022 formula units of Pb3(PO4)2(s)
800
1
b. 9 × 10−3 ×
× 6 × 1023 = 5 × 1019 molecules of N2O5(s)
100
1
c. 5 × 104 ×
× 6 × 1023 = 2 × 1026 formula units of MoO3(s)
150
1
d. 560 ×
× 6 × 1023 = 1.7 × 1024 formula units of SnF4(s)
200
All of the estimated answers are close to the calculated answers. The answers
are reasonable.
63. Practice Problem (page 242)
Sodium hydrogen carbonate, NaHCO3(s), is the principal ingredient in many
stomach-relief medicines.
a. A teaspoon of a particular brand of stomach-relief medicine contains
6.82 × 1022 formula units of sodium hydrogen carbonate. What mass of sodium
hydrogen carbonate is in the teaspoon?
b. The bottle of this stomach-relief medicine contains 350 g of sodium
hydrogen carbonate.
How many formula units of sodium hydrogen carbonate are in the bottle?
a. mass of sodium hydrogen carbonate
What Is Required?
You have to determine the mass of sodium hydrogen carbonate in a teaspoon
of stomach-relief medicine.
What Is Given?
You know there are 6.82 × 1022 formula units of sodium hydrogen carbonate.
You know the chemical formula for sodium hydrogen carbonate: NaHCO3
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 96 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
NaHCO3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Solution 1
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of moles of NaHCO3(s) using the relationship n =
N
.
NA
Calculate the mass of NaHCO3(s) using the relationship m = n × M .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the mass by multiplying the given amount of formula units of
1 mol
NaHCO3(s) by the ratio
and then by the molar
6.02 × 10 23 formula units
mass, M.
Act on Your Strategy
Molar mass, M, of NaHCO3(s):
M NaHCO3 = 1M Na + 1M H + 1M C + 3M O
= 1( 22.99 g/mol ) + 1(1.01 g/mol ) + 1(12.01 g/mol ) + 3 (16.00 g/mol )
= 84.01 g/mol
Solution 1
Amount in moles, n, of NaHCO3(s):
N
nNaHCO3 =
NA
6.82 × 1022 formula units
=
6.02 × 10 23 formula units /mol
= 0.113289 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 97 Chemistry 11 Solutions
Mass, m, of the NaHCO3(s):
mNaHCO3 = n × M
= 0.113289 mol × 84.01 g/ mol
= 9.5174 g
= 9.52 g
The mass of the sodium carbonate is 9.52 g.
Solution 2
Mass, m, of the NaHCO3(s):
mNaHCO3 = 6.82 × 1022 formula units ×
1 mol
× 84.01 g/ mol
6.02 × 1023 formula units
= 9.52 g
The mass of the sodium carbonate is 9.52 g.
Check Your Solution
Using rounded numbers to estimate the answer:
1
7 × 1022 ×
× 80 = 9 g
6.02 × 1023
The answer is reasonable.
b. formula units of sodium hydrogen carbonate
What Is Required?
You have to determine the number of formula units of sodium hydrogen
carbonate in the bottle of stomach-relief medicine.
What Is Given?
You know the mass of the sodium hydrogen carbonate: 350 g
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
NaHCO3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Solution 1
Calculate the amount in moles of NaHCO3(s) using the relationship n =
978‐0‐07‐105107‐1 m
.
N
Chapter 5 The Mole: A Chemist’s Counter • MHR | 98 Chemistry 11 Solutions
Calculate the mass of the NaHCO3(s) using the relationship m = n × M .
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units of NaHCO3(s) using the relationship
N = n × NA .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the number of formula units by multiplying the mass by
1
and then by the Avogadro constant, NA.
molar mass
Act on Your Strategy
Molar mass, M, of NaHCO3(s):
M NaHCO3 = 1M Na + 1M H + 1M C + 3M O
= 1( 22.99 g/mol ) + 1(1.01 g/mol ) + 1(12.01 g/mol ) + 3 (16.00 g/mol )
= 84.01 g/mol
Solution 1
Amount in moles, n, of the (s):
m
nNaHCO3 =
N
350 g
=
84.01 g /mol
= 4.166 mol
Number of formula units, N, of the NaHCO3(s):
N = n × NA
6.02 × 10 23 formula units
= 4.166 mol ×
1 mol
= 2.51 × 1024 formula units
There are 2.51 × 1024 formula units of sodium hydrogen carbonate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 99 Chemistry 11 Solutions
Solution 2
Number of formula units, N, of NaHCO3(s):
1 mol
6.02 × 1023 formula units
×
N = 350 g ×
1 mol
84.01 g
= 2.51 × 1024 formula units
There are 2.51 × 1024 formula units of sodium hydrogen carbonate.
Check Your Solution
Check to see that the correct atomic molar masses have been used.
Use rounded numbers to estimate the answer:
1
× 6 × 1023 = 2.6 × 1024 formula units of NaHCO3(s)
350 ×
80
The answer is reasonable.
64. Practice Problem (page 242)
Riboflavin, C17H20N4O6(s), is an important vitamin in the metabolism of fats,
carbohydrates, and proteins in your body.
a. The current recommended dietary allowance (RDA) of riboflavin for adult
men is 1.3 mg/day. How many riboflavin molecules are in this RDA?
b. The RDA of riboflavin for adult women contains 1.8 × 1018 molecules of
riboflavin. What is the RDA for adult women, in milligrams?
a. RDA for men
What Is Required?
You need to determine the number of molecules of riboflavin in a
recommended dietary allowance.
What Is Given?
You know the RDA: 1.3 mg/day
You know the chemical formula for riboflavin: C17H20N4O6
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
C17H20N4O6(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of C17H20N4O6(s).
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 100 Chemistry 11 Solutions
Solution 1
Convert the mass in a daily dosage of C17H20N4O6(s) from milligrams to
grams: 1 mg = 1× 10–3 g
Calculate the number of moles of C17H20N4O6(s) using the relationship n =
m
.
N
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of molecules of riboflavin using the relationship
N = n × NA .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the number of molecules of C17H20N4O6(s) by multiplying the mass
1
and then by the Avogadro constant, NA.
by
molar mass
Act on Your Strategy
Molar mass, M, of C17H20N4O6(s):
M C17 H2 0 N4O6 = 17M C + 20M H + 4M N + 6M O
= 1(12.01 g/mol ) + 20 (1.01 g/mol ) + 4 (14.01 g/mol ) + 6 (16.00 g/mol )
= 376.41 g/mol
Solution 1
Mass (in grams), m, of the C17H20N4O6(s):
mC17 H20 N4O6 = 1.3 mg × 1 × 10 –3 g/ mg
= 1.3 × 10 –3 g
Amount in moles, n, of the C17H20N4O6(s):
m
nC17 H 20 N 4O6 =
N
1.3 × 10 –3 g
=
376.41 g /mol
= 3.4536 × 10 –6 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 101 Chemistry 11 Solutions
Number of molecules, N, of the C17H20N4O6(s):
N = n × NA
= 3.4536 × 10 –6 mol ×
6.02 × 1023 molecules
1 mol
= 2.079 × 1018 molecules
= 2.1 × 1018 molecules
There are 2.1 × 1018 molecules of riboflavin in the RDA for men.
Solution 2
Number of molecules, N, of C17H20N4O6(s):
1 mol
6.02 × 1023 molecules
×
N = 1.3 × 10 –3 g ×
1 mol
376.41 g
= 2.1 × 1018 molecules
There are 2.1 × 1018 molecules of riboflavin in the RDA for men.
Check Your Solution
Check to see that the correct atomic molar masses have been used.
Use rounded numbers to estimate the answer:
1
× 6 × 1023 = 1.6 × 1018
1 × 10–3 ×
380
The estimate is close to the calculated answer. The answer is reasonable.
b. RDA for women
What Is Required?
You need to determine the recommended dietary allowance of riboflavin (in
milligrams) for adult women.
What Is Given?
You know there are 1.8 × 1018 molecules of riboflavin in the RDA.
You know the chemical formula for riboflavin: C17H20N4O6
Plan Your Strategy
Solution 1
Use the periodic table to find the atomic molar masses of the elements
inC17H20N4O6(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of C17H20N4O6(s).
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 102 Chemistry 11 Solutions
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of moles of C17H20N4O6(s) using the relationship
N
.
n=
NA
Calculate the mass of the C17H20N4O6(s) using the relationship m = n × M .
Convert the mass from grams to milligrams: 1 g = 1 × 103 mg
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the mass of C17H20N4O6(s) by multiplying the number of molecules
1 mol
and then by the molar mass, M.
by
6.02 × 1023 molecules
Convert the mass from grams to milligrams: 1 g = 1 × 103 mg
Act on Your Strategy
Molar mass, M, of riboflavin, C17H20N4O6(s):
M C17 H2 0 N4O6 = 17M C + 20M H + 4M N + 6M O
= 1(12.01 g/mol ) + 20 (1.01 g/mol ) + 4 (14.01 g/mol ) + 6 (16.00 g/mol )
= 376.41 g / mol
Amount in moles, n, of C17H20N4O6(s):
1.8 × 1018 molecules
nC17 H 2 0 N 4O6 =
6.02 × 10 23 molecules /mol
= 2.990 × 10 –6 mol
Mass, m, of C17H20N4O6(s):
mC17 H 20 N4O6 = n × M
= 2.990 × 10 –6 mol × 376.41 g/ mol
= 1.12547 × 10 –3 g
= 1.1 × 10 –3 g
= 1.1 × 10 –3 g × 1 × 103 mg/ g
= 1.1 mg
The daily RDA of riboflavin for adult women is 1.1 mg.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 103 Chemistry 11 Solutions
Solution 2
Mass, m, of the C17H20N4O6(s):
mC17 H20 N4O6 = 1.8 × 1018 molecules ×
1 mol
× 376.41 g/ mol
6.02 × 1023 molecules
= 1.1 × 10 –3 g
= 1.1 × 10 –3 g × 1 × 103 mg/ g
= 1.1 mg
The daily RDA of riboflavin for adult women is 1.1 mg.
Check Your Solution
Using rounded numbers to estimate the answer:
1
× 380 × 103 = 1.3 mg
2 × 1018 ×
6 × 1023
The answer is reasonable.
65. Practice Problem (page 242)
What is the mass, in grams, of a single atom of platinum?
What Is Required?
You must find the mass of one platinum atom.
What Is Given?
You know the number of atoms, N, of platinum, Pt(s): 1
Plan Your Strategy
Use the atomic molar mass in the periodic table to calculate the molar mass of
Pt(s).
Solution 1
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of moles, n, of the Pt(s) using the relationship n =
N
.
NA
Calculate the mass of the Pt(s) using the relationship m = n × M .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 104 Chemistry 11 Solutions
Calculate the number of atoms of Pt(s) by multiplying the mass
1 mol
, and then by the molar mass, M.
by
6.02 × 10 23 molcules
Act on Your Strategy
Molar mass, M, of Pt(s): 195.08 g/mol (from the periodic table)
Solution 1
Amount of moles, n, of the Pt(s):
N
nPt =
NA
=
1 atom
6.02 × 1023 atoms /mol
= 1.661 × 10 –24 mol
Mass, m, of the Pt(s):
mPt = n × M
= 1.661 × 10 –24 mol × 195.08 g/ mol
= 3.24 × 10 –22 g
The mass of a single platinum atom is 3.24 × 10–22 g.
Solution 2
Mass, m, of the Pt(s):
mPt = n × M
= 1 atom ×
1 mol
× 195.08 g/ mol
6.02 × 1023 atoms
= 3.24 × 10 –22 g
The mass of a single platinum atom is 3.24 × 10–22 g.
Check Your Solution
Use rounded numbers to estimate the answer:
1
× 200 = 3.3 × 10–22 g
1×
23
6 × 10
The answer is reasonable.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 105 Chemistry 11 Solutions
66. Practice Problem (page 242)
Rubbing alcohol often contains propanol, C3H7OH(ℓ). Suppose that you have
an 85.9 g sample of propanol.
a. How many carbon atoms are in the sample?
b. How many hydrogen atoms are in the sample?
c. How many oxygen atoms are in the sample?
What Is Required?
a. You need to determine the number of carbon atoms in a sample of propanol.
b. You need to determine the number of hydrogen atoms in a sample of
propanol.
c. You need to determine the number of oxygen atoms in a sample of propanol.
What Is Given?
You know the mass of the propanol: 85.9 g
You know the chemical formula for propanol: C3H7OH
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
C3H7OH(ℓ).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of C3H7OH(ℓ).
Solution 1
m
.
N
From the chemical formula for C3H7OH(ℓ), determine the amount in moles of
atoms of each element in 1 mol of molecules of C3H7OH(ℓ).
Calculate the total number of moles of atoms of each element.
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of atoms of each element in the C3H7OH(ℓ) using the
relationship N = n × N A .
Calculate the number of moles of C3H7OH(ℓ) using the relationship n =
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the number of atoms of each element by multiplying the mass
1
, by the number of moles of carbon atoms in 1 mol of molecules
by
molar mass
of C3H7OH(ℓ), and then by the Avogadro constant, NA.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 106 Chemistry 11 Solutions
Act on Your Strategy
Molar mass, M, of C3H7OH(ℓ):
M C3H7 OH = 3M C + 8M H + 1M O
= 3 (12.01 g/mol ) + 8 (1.01 g/mol ) + 1(16.00 g/mol )
= 60.11 g/mol
a. number of carbon atoms
Solution 1
Amount in moles, n, of C3H7OH(ℓ):
m
nC3H7 OH =
M
85.9 g
=
60.11 g /mol
= 1.429 mol
From the chemical formula, C3H7OH, there are 3 mol of carbon atoms per
mole of propanol molecules.
Amount in moles, n, of carbon atoms:
nC
3 mol C atoms
=
1.429 mol C3 H 7 OH 1 mol C3 H 7 OH
n × 1 mol C3 H 7 OH = 1.429 mol C3 H 7 OH ×3 mol C atoms
nC = 1.429 mol C3 H 7 OH ×
3 mol C atoms
1 mol C3 H 7 OH
= 4.287 mol C atoms
Number, N, of carbon atoms:
N = n × NA
= 4.287 mol ×
6.02 × 1023 C atoms
1 mol
= 2.58 × 1024 C atoms
There are 2.58 × 1024 carbon atoms in the sample of propanol.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 107 Chemistry 11 Solutions
Solution 2
Number, N, of carbon atoms:
1 mol
3 mol C atoms
6.02 × 1023 molecules
×
×
N = 85.9 g ×
60.11 g
1 mol
1 mol molecules
= 2.58 × 1024 C atoms
There are 2.58 × 1024 carbon atoms in the sample of propanol.
b. number of hydrogen atoms
Solution 1
Amount in moles, n, of C3H7OH(ℓ):
m
nC3H7 OH =
M
85.9 g
=
60.11 g /mol
= 1.429 mol
From the chemical formula, C3H7OH, there are 8 mol of hydrogen atoms per
mole of propanol molecules.
Amount in moles, n, of H atoms in C3H7OH(ℓ):
nH
8 mol H atoms
=
1.429 mol C3H 7 OH 1 mol C3H 7 OH
n × 1 mol C3 H 7 OH = 1.429 mol C3H 7 OH × 8 mol H atoms
nH = 1.429 mol C3 H 7 OH ×
8 mol H atoms
1 mol C3 H 7 OH
= 11.432 mol H atoms
Number, N, of hydrogen atoms:
N = n × NA
= 11.432 mol ×
6.02 × 1023 H atoms
1 mol
= 6.88 × 1024 H atoms
There are 6.88 × 1024 hydrogen atoms in the sample of propanol.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 108 Chemistry 11 Solutions
Solution 2
Number, N, of hydrogen atoms:
1 mol
8 mol H atoms
6.02 × 1023 molecules
N = 85.9 g ×
×
×
60.11 g
1 mol
1 mol molecules
= 6.88 × 1024 H atoms
There are 6.88 × 1024 hydrogen atoms in the sample of propanol.
c. number of oxygen atoms
Solution 1
Amount in moles, n, of C3H7OH(ℓ):
m
nC3H7 OH =
M
85.9 g
=
60.11 g /mol
= 1.429 mol
From the chemical formula, C3H7OH, there is 1 mol of oxygen atoms per mole
of propane molecules.
Amount in moles, n, of O atoms in C3H7OH(ℓ):
nO
1 mol O atoms
=
1.429 mol C3H 7 OH 1 mol C3H 7 OH
n × 1 mol C3 H 7 OH = 1.429 mol C3H 7 OH × 1 mol O atoms
nO = 1.429 mol C3H 7 OH ×
1 mol O atoms
1 mol C3 H 7 OH
= 1.429 mol O atoms
Number, N, of oxygen atoms:
N = n × NA
= 1.429 mol ×
6.02 × 1023 O atoms
1 mol
= 8.60 × 1023 O atoms
There are 8.60 × 1023 oxygen atoms in the sample of propanol.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 109 Chemistry 11 Solutions
Solution 2
Number, N, of oxygen atoms:
1 mol
1 mol O atoms
6.02 × 1023 molecules
N = 85.9 g ×
×
×
60.11 g
1 mol
1 mol molecules
= 8.60 × 1023 O atoms
There are 8.60 × 1023 oxygen atoms in the sample of propanol.
Check Your Solution
Use rounded numbers to estimate each answer:
a. carbon atoms
1
× 3 × 6 × 1023 = 2.7 × 1024 carbon atoms
90 ×
60
b. hydrogen atoms
1
× 8 × 6 × 1023 = 7.2 × 1024 hydrogen atoms
90 ×
60
c. oxygen atoms
1
× 1 × 6 × 1023 = 9 × 1023 oxygen atoms
90 ×
60
Each estimate is close to the calculated answer. The answers are reasonable.
67. Practice Problem (page 242)
a. How many formula units are in a 3.14 g sample of aluminum sulfide,
Al2S3(s)?
b. How many ions (aluminum and sulfur), in total, are in this sample?
a. number of formula units
What Is Required?
You must find the number of formula units of aluminum sulfide in a sample.
What Is Given?
You know the mass of the aluminum sulfide: 3.14 g
You know the chemical formula for aluminum sulfide: Al2S3
Plan Your Strategy
a. formula units
Solution 1
Use the periodic table to find the atomic molar masses of the elements in
Al2S3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 110 Chemistry 11 Solutions
Add these values to calculate the molar mass of Al2S3(s).
Calculate the number of moles of Al2S3(s) using the relationship n =
m
.
M
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units in the sample of Al2S3(s) using the
relationship N = n × N A .
Solution 2
The following steps may be completed in one line using the units to calculate
the answer.
Calculate the number of formula units by multiplying the mass
1
and then by the Avogadro constant, NA.
by
molar mass
Act on Your Strategy
Solution 1
Molar mass, M, of Al2S3(s):
M Al2S3 = 2M Al + 3M S
= 2 ( 26.98 g/mol ) + 3 ( 32.07 g/mol )
= 150.17 g/mol
Amount in moles, n, of the Al2S3(s):
m
nAl2S3 =
M
3.14 g
=
150.17 g /mol
= 2.0906 × 10 –2 mol
Number of formula units, N, of the Al2S3(s):
N = n × NA
= 2.0906 × 10 –2 mol ×
6.02 × 10 23 formula units
1 mol
= 1.25859 × 1022 formula units
= 1.26 × 1022 formula units
There are 1.26 × 1022 formula units of aluminum sulfide in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 111 Chemistry 11 Solutions
Solution 2
Number, N, of formula units of the Al2S3(s):
1 mol
6.02 × 1023 formula units
×
N = 3.14 g ×
1 mol
150.17 g
= 1.26 × 1022 formula units
There are 1.26 × 1022 formula units of aluminum sulfide in the sample.
Check Your Solution
Use rounded numbers to estimate the answer:
1
3×
× 3 × 6 × 1023 = 1.2 × 1022 formula units
150
The estimate is close to the calculated answer. The answer is reasonable.
b. total number of ions
What Is Required?
You must find the total number of ions in the sample of aluminum sulfide.
What Is Given?
You know from the answer to Part a that there are 1.26 × 1022 formula units of
Al2S3(s).
Plan Your Strategy
Determine the total number of ions in one formula unit of Al2S3(s).
Multiply the number of formula units of Al2S3(s) by the total number of ions
per formula unit.
Act on Your Strategy
From the chemical formula, Al2S3, there are 2 aluminum ions and 3 sulfide
ions for a total of 5 ions in one formula unit of aluminum sulfide.
Total number of ions, N, in the Al2S3(s):
N
5 ions
=
1.26 × 10 formula units 1 formula unit
22
N = 1.25859 × 1022 formula units ×
5 ions
1 formula units
= 6.29 × 1022 ions
There are a total of 6.29 × 1022 ions in the sample of aluminum sulfide.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 112 Chemistry 11 Solutions
Check Your Solution
The answer matches the ratio shown in the chemical formula for aluminum
sulfide. There are 5 times as many ions as there are formula units. The
chemical formula for aluminum sulfide, Al2S3(s), also shows 5 ions in one
formula unit.
68. Practice Problem (page 242)
Which of the following two substances contains the greater mass?
• 6.91 × 1022 molecules of nitrogen dioxide, NO2(g)
• 6.91 × 1022 formula units of gallium arsenide, GaAs(s)
What Is Required?
You must find which of two given samples has a greater mass.
What Is Given?
You know that one sample has 6.91 × 1022 molecules of nitrogen dioxide.
You know the other sample contains 6.91 × 1022 formula units of gallium
arsenide.
Plan Your Strategy
Use the Avogadro constant: N A = 6.02 × 10 23
N
.
NA
Use the periodic table to find the atomic molar masses of the elements in each
of the NO2(g) and the GaAs(s).
For each sample, multiply the atomic molar masses by the number of atoms of
each element in the compound.
Add these values to calculate the molar mass of each compound.
Calculate the mass of each sample using the relationship m = n × M .
Calculate the amount in moles of each sample using the relationship n =
Act on Your Strategy
Amount in moles, n, of each sample:
N
n=
NA
=
6.91 × 1022 particles
6.02 × 1023 particles /mol
= 0.11478 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 113 Chemistry 11 Solutions
Molar mass, M, of NO2(g):
M NO2 = 1M N + 2M O
= 1(14.01 g/mol ) + 2 (16.00 g/mol )
= 46.01 g/mol
Molar mass, M, of GaAs(s):
M GaAs = 1M Ga + M As
= 1( 69.72 g/mol ) + 1( 74.92 g/mol )
= 144.64 g/mol
Mass, m, of the NO2(g ):
mNO2 = n × M
= 0.11478 mol × 46.01 g /mol
= 5.28 g
The mass of the nitrogen dioxide is 5.28 g.
Mass, m, of the GaAs(s):
mGaAs = n × M
= 0.11478 mol × 144.64 g/ mol
= 16.6 g
The mass of the gallium arsenide is 16.6 g.
The sample of gallium arsenide has the greater mass.
Check Your Solution
Since the number of moles of each sample is the same, the compound having
the greater molar mass will have the greater mass. Gallium arsenide has the
greater molar mass and the greater mass.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 114 Chemistry 11 Solutions
69. Problem (page 242)
Many common dry-chemical fire extinguishers contain ammonium phosphate,
(NH4)3PO4(s), as their principal ingredient. If a sample of ammonium
phosphate contains 4.5 × 1021 atoms of nitrogen, what is the mass of the
sample?
What Is Required?
You need to determine the mass of a sample of ammonium phosphate in a drychemical fire extinguisher.
What Is Given?
You know the number of atoms of nitrogen in the sample: 4.5 × 1021
You know the chemical formula for ammonium phosphate: (NH4)3PO4
Plan Your Strategy
Solution 1
From the chemical formula, (NH4)3PO4, determine the number of nitrogen
atoms in one formula unit of ammonium phosphate.
Calculate the number of formula units of (NH4)3PO4(s).
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of (NH4)3PO4(s) using the relationship n =
N
.
NA
Use the periodic table to find the atomic molar masses of the elements in
(NH4)3PO4(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of (NH4)3PO4(s).
Calculate the mass of (NH4)3PO4(s) using the relationship m = n × M .
Solution 2
The following two steps can be completed in one line using the units to
calculate the answer.
Calculate the mass of (NH4)3PO4(s) by multiplying the number of atoms of
1 formula unit
, and then by
nitrogen by
number of N atoms per formula unit
1 mol
, and finally by the molar mass, M.
23
6.02 × 10 formula units
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 115 Chemistry 11 Solutions
Act on Your Strategy
From the chemical formula, (NH4)3PO4, there are 3 nitrogen atoms in one
formula unit of ammonium phosphate.
Number of formula units, N, of (NH4)3PO4(s)
1 formula unit (NH 4 )3 (PO) 4
N
=
21
4.5 × 10 N atoms
3 N atoms
N = 4.5 × 1021 N atoms ×
1 formula unit (NH 4 )3 (PO)4
3 N atoms
=1.5 × 1021 formula units (NH 4 )3 (PO) 4
Amount in moles, n, of the (NH4)3PO4(s):
N
n( NH4 ) PO4 =
3
NA
=
1.5 × 1021 formula units
6.02 × 10 23 formula units /mol
= 2.4916 × 10 –3 mol
Molar mass, M, of (NH4)3PO4(s):
M ( NH4 ) PO4 = 3M N + 12M H + 1M P + 4M O
3
= 3 (14.01 g/mol ) + 12 (1.01 g/mol ) + 1( 30.97 g/mol ) + 4 (16.00 g/mol )
= 149.12 g/mol
Mass, m, of the (NH4)3PO4(s):
m( NH4 ) PO4 = n × M
3
= 2.4916 × 10 –3 mol × 149.12 g/ mol
= 0.371547 g
= 0.37 g
The mass of the sample of ammonium phosphate is 0.37 g.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 116 Chemistry 11 Solutions
Solution 2
Mass, m, of the (NH4)3PO4(s):
m( NH4 ) PO4 = 4.5 × 1021 N atoms ×
3
1 formula unit
3 N atoms
1 mol
149.12 g
×
1 mol
6.02 × 10 formula units
= 0.371547 g
×
23
= 0.37 g
The mass of the sample of ammonium phosphate is 0.37 g.
Check Your Solution
Use rounded numbers to estimate the answer:
1
1
5 × 1021 × ×
× 150 = 0.42 g
3
6 × 1023
The estimate is close to the calculated answer. The answer is reasonable.
70. Problem (page 242)
Place the following three substances in order, from greatest to smallest number
of hydrogen atoms:
• 268 mg of sucrose, C12H22O11(s)
• 15.2 g of hydrogen cyanide, HCN(ℓ)
• 0.0889 mol of acetic acid, CH3COOH(ℓ)
What Is Required?
You need to determine the number of hydrogen atoms in three different
samples and identify which has the smallest number of hydrogen atoms.
What is Given?
You know the mass of the sucrose: 268 mg
You know the mass of the hydrogen cyanide: 15.2 g
You know the amount in moles of the acetic acid: 0.0889 mol
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
C12H22O11(s) and in HCN(ℓ).
Multiply the atomic molar masses by the number of atoms of each element in
each compound.
Add these values to calculate the molar masses of C12H22O11(s) and in HCN(ℓ).
Convert the mass of C12H22O11(s) from milligrams to grams: 1 mg = 1 × 10–3 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 117 Chemistry 11 Solutions
Calculate the amount in moles of C12H22O11(s) and of HCN(ℓ) using the
m
relationship n =
.
M
From the chemical formulas for each of the three compounds, determine the
amount in moles of hydrogen atoms in 1 mol of each compound.
Using the amount in moles of each compound, calculate the total amount in
moles of hydrogen atoms of each compound.
Identify which compound has the smallest amount in moles of hydrogen
atoms.
The number of hydrogen atoms is the same proportion as the amount in moles
of hydrogen atoms.
Act on Your Strategy
• sucrose
Molar mass, M, of C12H22O11(s):
M C12 H22O11 = 12M C + 22M H + 11M O
= 12 (12.01 g/mol ) + 22 (1.01 g/mol ) + 11(16.00 g/mol )
= 342.34 g/mol
Mass (in grams), m, of the C12H22O11(s):
mC12H22O11 = 268 mg × 1 × 10–3 g/mg
= 0.268 g
Amount in moles, n, of the C12H22O11(s):
m
nC12 H22O11 =
M
0.268 g
=
342.34 g /mol
= 7.828 × 10 –4 mol
From the chemical formula, C12H22O11, there is 22 mol of hydrogen atoms per
mole of sucrose.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 118 Chemistry 11 Solutions
Total amount in moles, n, of H atoms in the C12H22O11(s):
nH
22 mol H atoms
=
–4
7.828 × 10 mol C12 H 22 O11 1 mol C12 H 22 O11
nH = 7.828 × 10–4 mol C12 H 22 O11 ×
22 mol H atoms
1 mol C12 H 22 O11
= 0.0172 mol H atoms
There is a total of 0.0172 mol of hydrogen atoms in the sucrose sample.
• hydrogen cyanide
Molar mass, M, of HCN(ℓ):
M HCN = 1M H + 1M C + 1M N
= 1(1.01 g/mol ) + 1(12.01 g/mol ) + 1(14.01 g/mol )
= 27.03 g/mol
Amount in moles, n, of the HCN(ℓ):
m
nHCN =
M
15.2 g
=
27.03 g /mol
= 0.562 mols
From the chemical formula, HCN, there is 1 mol of hydrogen atoms per mole
of hydrogen cyanide.
Total amount in moles, n, of H atoms in the HCN(ℓ):
nH
1 mol H atoms
=
0.562 mol HCN
1 mol HCN
1 mol H atoms
nH = 0.562 mol HCN ×
1 mol HCN
= 0.562 mol H atoms
There is a total of 0.562 mol of hydrogen atoms in the hydrogen cyanide
sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 119 Chemistry 11 Solutions
• acetic acid
From the chemical formula, CH3COOH, there is 4 mol of hydrogen atoms per
mole of acetic acid.
Total amount in moles of H atoms in the CH3COOH(ℓ):
nH
4 mol H atoms
=
0.0889 mol CH 3COOH 1 mol CH 3COOH
nH = 0.0889 mol CH3COOH ×
4 mol H atoms
1 mol CH3COOH
= 0.3556 mol H atoms
There is a total of 0.3556 mol of hydrogen atoms in the acetic acid sample.
The listing of the compounds from the one with the most hydrogen to the least
hydrogen will be the same regardless of whether the units compared are
amount in moles of H atoms or number of H atoms.
Listed from the largest number to the smallest number of H atoms:
HCN(ℓ) (0.562 mol H atoms) > CH3COOH(ℓ) (0.3556 mol) > C12H22O11(s)
(0.0172 mol)
Check Your Solution
The units cancel properly and an estimate of the amount in moles of hydrogen
is consistent with the calculated values. The answer is reasonable.
Section 5.2 Mass and the Mole
Solutions for Selected Review Questions
Student Edition page 243
4. Review Question (page 243)
Determine the amount in moles of gallium oxide, Ga2O3(s), in a 45.2 g sample.
What Is Required?
You need to determine the amount in moles of a sample of gallium oxide.
What Is Given?
You know the chemical formula for gallium oxide: Ga2O3
You know the mass of the Ga2O3(s): 45.2 g
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 120 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
Ga2O3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of Ga2O3(s).
m
Calculate the amount in moles of the Ga2O3(s) using the relationship n =
.
M
Act on Your Strategy
Molar mass, M, of Ga2O3(s):
M Ga 2O3 = 2M Ga + 3M O
= 2 ( 69.72 g/mol ) + 3 (16.00 g/mol )
= 187.44 g/mol
Amount in moles, n, of Ga2O3(s):
m
nGa 2O3 =
M
45.2 g
=
187.44 g /mol
= 0.241 mol
There is 0.241 mol of gallium oxide in the sample.
Check Your Solution
The molar mass has been determined correctly. The answer seems reasonable
and shows the correct number of significant digits.
5. Review Question (page 243)
What is the mass of 3.2 × 102 mol of cerium nitrate, Ce(NO3)3(s)?
What Is Required?
You need to determine the mass of a sample of cerium nitrate.
What Is Given?
You know the amount in moles of cerium nitrate: 3.2 × 102 mol
You know the chemical formula for cerium nitrate: Ce(NO3)3
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 121 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
Ce(NO3)3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of Ce(NO3)3(s).
Calculate the mass of Ce(NO3)3(s) using the relationship m = n × M .
Act on Your Strategy
Molar mass, M, of Ce(NO3)3(s):
M Ce( NO3 ) = 1M Ce + 3M N + 9M O
3
= 1(140.12 g/mol) + 3 (14.01 g/mol ) + 9 (16.00 g/mol )
= 326.15 g/mol
Mass, m, of the Ce(NO3)3(s):
mCe( NO3 ) = n × M
3
= 3.2 × 102 mol × 326.15 g/ mol
= 1.04368 × 105 g
= 1.0 × 105 g
The mass of the cerium nitrate is 1.0 × 105 g.
Check Your Solution.
The units cancel properly. Check to see that correct atomic masses have been
used. Using rounded numbers:
3 × 102 × 330 = 9.9 × 104
The estimate is close to the calculated value. The answer is reasonable.
6. Review Question (page 243)
Calculate the amount in moles of strontium chloride, SrCl2(s), in a 28.6 kg
sample.
What Is Required?
You need to determine the amount in moles of a sample of strontium chloride.
What Is Given?
You know the chemical formula for strontium chloride: SrCl2
You know the mass of the strontium chloride: 28.6 kg
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 122 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
SrCl2(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of SrCl2(s).
Convert the mass of the SrCl2(s) from kilograms to grams: 1 kg = 1 × 103 g
m
Calculate the amount in moles of the SrCl2(s) using the relationship n =
.
M
Act on Your Strategy
Molar mass, M, of SrCl2(s):
M SrCl2 = 1M Sr + 2M Cl
= 1( 87.62 g/mol ) + 2 ( 35.45 g/mol )
= 158.52 g/mol
Mass (in grams), m, of the SrCl2(s):
mSrCl2 = 28.6 kg × 1 × 103 g/ kg
= 2.86 ×104 g
Amount in moles, n, of the SrCl2(s):
m
nSrCl2 =
M
2.86 ×104 g
=
158.52 g /mol
= 1.80 × 102 mol
There is 1.80 × 102 mol of strontium chloride in the sample.
Check Your Solution
The molar mass has been determined correctly. The answer seems reasonable
and shows the correct number of significant digits.
7. Review Question (page 243)
What is the mass of 0.68 mol of iron(III) sulfate, Fe2(SO4)3(s)?
What Is Required?
You need to determine the mass of a sample of iron(III) sulfate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 123 Chemistry 11 Solutions
What Is Given?
You know the chemical formula for iron(III) sulfate: Fe2(SO4)3
You know the amount in moles of the iron(III) sulfate: 0.68 mol
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
Fe2(SO4)3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the Fe2(SO4)3(s).
Calculate the mass of the Fe2(SO4)3(s) using the relationship m = n × M .
Act on Your Strategy
Molar mass, M, of Fe2(SO4)3(s):
M Fe2 (SO4 )3 = 2M Fe + 3M S + 12M O
= 2 ( 55.85 g/mol ) + 3 ( 32.07 g/mol ) + 12 (16.00 g/mol )
= 399.91 g/mol
Mass, m, of the Fe2(SO4)3(s):
mFe2 (SO4 ) = n × M
3
= 0.68 mol × 399.91 g/ mol
= 2.719 × 102 g
= 2.7 × 102 g
The mass of the iron(III) sulfate is 2.7 × 102 g.
Check Your Solution
The units cancel properly. Check to see that correct atomic masses have been
used. Using rounded numbers:
7 × 10–1 × 330 = 231 g
The estimate is close to the calculated value. The answer is reasonable.
8. Review Question (page243)
What is the mass of 2.9 × 1026 molecules of dinitrogen pentoxide, N2O5(g)?
What Is Required?
You need to determine the mass of a sample of dinitrogen pentoxide.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 124 Chemistry 11 Solutions
What Is Given?
You know the chemical formula for dinitrogen pentoxide: N2O5
You know the number of molecules of the dinitrogen pentoxide: 2.9 × 1026
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements
inN2O5(g).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Solution 1
Calculate the amount in moles of the N2O5(g) using the relationship n =
N
NA .
Calculate the mass of the N2O5(g) using the relationship m = n × M .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the mass of N2O5(g) by multiplying the given number of molecules
1 mol
of dinitrogen pentoxide by
and then by the molar mass,
6.02 × 10 23 molecules
M.
Act on Your Strategy
Molar mass, M, of N2O5(g):
M N2O5 = 2M N + 5M O
= 2 (14.01 g/mol ) + 5 (16.00 g/mol )
= 108.02 g/mol
Solution 1
Amount in moles, n, of the N2O5(g):
N
nN2O5 =
NA
2.9 × 1026 molecules
=
6.02 × 1023 molecules /mol
= 4.817 × 102 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 125 Chemistry 11 Solutions
Mass, m, of the N2O5(g):
mN2O5 = n × M
= 4.817 × 102 mol × 108.02 g/ mol
= 5.2033 × 104 g
= 5.2 × 104 g
The mass of the sample of dinitrogen pentoxide is 5.2 × 104 g.
Solution 2
Mass, m, of the N2O5(g):
mN2O5 = 2.9 × 1026 molecules ×
1 mol
108.02 g
×
23
1 mol
6.02 × 10 molecules
= 5.2033 × 104 g
= 5.2 × 104 g
The mass of the dinitrogen pentoxide is 5.2 × 104 g.
Check Your Solution
Use rounded numbers to estimate the answer:
1
3 × 1026 ×
× 110 = 5.5 × 104 g
23
6 × 10
The estimate is close to the calculated answer. The answer is reasonable.
9. Review Question (page243)
Calculate the number of oxygen atoms in 15.2 g of trinitrotoluene,
C7H5(NO2)3(s).
What Is Required?
You need to determine the number of oxygen atoms in a sample of
trinitrotoluene.
What Is Given?
You know the mass of the trinitrotoluene: 15.2 g
You know the chemical formula for trinitrotoluene: C7H5(NO2)3
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 126 Chemistry 11 Solutions
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements
inC7H5(NO2)3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Solution 1
Calculate the amount in moles of the C7H5(NO2)3(s) using the relationship
m
n= .
N
From the chemical formula, C7H5(NO2)3, determine the amount in moles of
oxygen in 1 mol of trinitrotoluene.
Calculate the amount in moles of oxygen atoms.
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of oxygen atoms using the relationship N = n × N A .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the number of oxygen atoms by multiplying the mass
1
number of moles of oxygen atoms
, then by
, and then by the
by
molar mass
1 mol of molecules
Avogadro constant ( N A = 6.02 × 10 23 ) .
Act on Your Strategy
Molar mass, M, of C7H5(NO2)3(s):
M C7 H5 ( NO2 )3 = 7M C + 5M H + 3M N + 6M O
= 7 (12.01 g/mol ) + 5 (1.01 g/mol ) + 3 (14.01 g/mol ) + 6 (16.00 g/mol )
= 227.15 g/mol
Solution 1
Amount in moles, n, of the C7H5(NO2)3(s):
m
nC7 H5 ( NO2 ) =
3
N
15.2 g
=
227.15 g /mol
= 6.6916 × 10 –2 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 127 Chemistry 11 Solutions
From the chemical formula, C7H5(NO2)3, there are 6 mol of oxygen atoms per
mole of trinitrotoluene.
Amount in moles, n, of O atoms in C7H5(NO2)3(s):
nO
6 mol O atoms
=
–2
6.6916 × 10 mol C 7 H 5 (NO 2 )3 1 mol C 7 H 5 (NO 2 )3
nO = 6.6916 × 10 –2 mol C 7 H 5 (NO 2 )3 ×
6 mol O atoms
1 mol C7 H 5 (NO 2 )3
= 0.40149 mol O atoms
Number, N, of oxygen atoms in the C7H5(NO2)3(s):
N = n × NA
= 0.40149 mol O atoms × 6.02 × 1023 atoms/ mol
= 2.42 × 1023 O atoms
There are 2.42 × 1023 atoms of oxygen in the sample of trinitrotoluene.
Solution 2
Number of oxygen atoms, N, in the C7H5(NO2)3(s):
1 mol
6 mol O atoms
6.02 × 1023 atoms
N = 15.2 g ×
×
×
227.15 g
1 mol
1 mol
= 2.42 × 1023 atoms
There are 2.42 × 1023 atoms of oxygen in the sample of trinitrotoluene.
Check Your Solution
Use rounded numbers to estimate the answer:
1
15 ×
× 6 × 6 × 1023 = 2.3 × 1023 oxygen atoms
230
The estimate is close to the calculated answer. The answer is reasonable.
10. Review Question (page243)
Which has more sulfur atoms: 13.4 g of potassium thiocyanate, KSCN(s), or
0.067 mol of aluminum sulfate, Al2(SO4)3(s)?
What Is Required?
You need to determine the number of sulfur atoms in two samples and identify
which sample has the greater number.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 128 Chemistry 11 Solutions
What Is Given?
You know the chemical formula for potassium thiocyanate: KSCN
You know the chemical formula for aluminum sulfate: Al2(SO4)3
You know the mass of the potassium thiocyanate: 13.4 g
You know the amount in moles of the aluminum sulfate : 0.067 mol
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in each
of KSCN(s) and Al2(SO4)3(s).
For each compound, multiply the atomic molar masses by the number of atoms
of each element in the compound.
Add these values to calculate the molar mass of each compound.
Solution 1
m
.
M
From the chemical formulas, determine the amount in moles of sulfur atoms in
1 mol of each compound.
Calculate the amount in moles of sulfur atoms in KSCN(s) and Al2(SO4)3(s).
The amount in moles of sulfur atoms is in the same proportion as the number
of sulfur atoms. Compare the amounts of sulfur atoms in the two compounds
and identify which has the greater number.
Calculate the amount in moles of the KSCN(s) using the relationship n =
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
1
,
Calculate the number of sulfur atoms by multiplying the mass by
molar mass
number of moles of sulfur atoms
then by
, and then by the Avogadro constant
1 mol of molecules
( N A = 6.02 × 10 23 ) .
Act on Your Strategy
Molar mass, M, of KSCN(s):
M KSCN = 1M K + 1M S + 1M C + 1M N
= 1( 39.10 g/mol ) + 1( 32.07 g/mol ) + 1(12.01 g/mol ) + 1(14.01 g/mol )
= 97.19 g/mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 129 Chemistry 11 Solutions
Molar mass, M, of Al2(SO4)3(s):
M Al2 (SO4 )3 = 2M Al + 3M S + 12M O
= 2 ( 26.98 g/mol ) + 3 ( 32.07 g/mol ) + 12 (16.00 g/mol )
= 342.17 g/mol
Solution 1
Amount in moles, n, of the KSCN(s):
m
nKSCN =
M
13.4 g
=
97.19 g /mol
= 0.1387 mol
From the chemical formula, KSCN, there is 1 mol of sulfur atoms in 1 mol of
the compound.
Amount in moles, n, of S atoms in the KSCN(s):
nS
1 mol S atoms
=
0.138787 mol KSCN 1 mol KSCN
nS = 0.138787 mol KSCN ×
1 mol S atoms
mol KSCN
= 0.13787 mol S atoms
= 0.138 mol S atoms
There is 0.138 mol of sulfur atoms in the sample of potassium cyanide.
From the chemical formula, Al2(SO4)3, there is 3 mol of sulfur atoms in 1 mol
of aluminum sulfate.
Amount in moles, n, of S atoms in the Al2(SO4)3(s):
nS
3 mol S atoms
=
0.067 mol Al 2 (SO 4 )3 1 mol Al 2 (SO 4 )3
nS = 0.067 mol Al 2 (SO 4 )3 ×
3 mol S atoms
1 mol Al2 (SO 4 )3
= 0.201 mol S atoms
There is 0.201 mol of sulfur atoms in the sample of aluminum sulfate.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 130 Chemistry 11 Solutions
Solution 2
Amount in moles, n, of S atoms in the KSCN(s):
1 mol
1 mol S atoms
nS = 13.4 g ×
×
97.19 g
1 mol KSCN(s)
= 0.138 mol S atoms
Amount in moles, n, of S atoms in 0.067 mol of the Al2(SO4)3
3 mol S atoms
nS = 0.067 mol ×
mol
= 0.201 mol S atoms
The amount in moles of sulfur atoms is greater in the sample of aluminum
sulfate.
Check Your Solution
Use rounded numbers to estimate the amount in moles of S atoms in the
sample of KSCN(s):
1
× 1 = 0.13 mol of sulfur atoms
13 ×
100
Use rounded numbers to estimate the amount in moles of S atoms in the
sample of Al2(SO4)3:
0.07 × 3 = 0.21 mol of sulfur atoms
Each estimate is close to the calculated answer. The answer is reasonable.
11. Review Question (page 243)
Which has a greater amount in moles: a sample of sulfur trioxide containing
4.9 × 1022 atoms of oxygen or a 4.9 g sample of carbon dioxide?
What Is Required?
You need to determine the amount in moles of two samples and identify which
sample has the greater amount.
What Is Given?
You know the chemical formula for sulfur trioxide: SO3
You know the chemical formula for carbon dioxide: CO2
You know the number of oxygen atoms in the sample of sulfur trioxide: 4.9 ×
1022 atoms
You know the mass of the carbon dioxide: 4.9 g
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Plan Your Strategy
Solution 1
For the sulfur trioxide:
From the chemical formula, SO3, determine the number of oxygen atoms per
molecule of sulfur trioxide.
Calculate the number of SO3(g) molecules.
Use the Avogadro constant: N A = 6.02 × 10 23 .
N
.
Calculate the amount in moles of the SO3(g) using the relationship n =
NA
For the carbon dioxide:
Use the periodic table to find the atomic molar masses of the elements in
CO2(g).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
m
.
Calculate the amount in moles of the CO2(g) using the relationship n =
M
Compare the amounts in moles of the two compounds and identify which has
the greater amount.
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
For the sulfur trioxide:
Calculate the amount in moles of SO3(g) by multiplying the number of atoms
1 molecule SO3
of oxygen by
, and then by
number of O atoms per molecule
1 mole
.
6.02 × 10 23 molecules
For the carbon dioxide:
Calculate the amount in moles by multiplying the mass of CO2(g)
1
by
.
molar mass
Act on Your Strategy
Solution 1
For the sulfur trioxide:
From the chemical formula, SO3, there are 3 atoms of oxygen in one molecule
of sulfur trioxide.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 132 Chemistry 11 Solutions
Number of molecules, N, of SO3(g):
1 molecule SO3
N
=
22
4.9 × 10 O atoms
3 O atoms
N = 4.9 × 1022 O atoms ×
1 molecule SO3
3 O atoms
= 1.6333 × 1022 molecules SO3
Amount in moles, n, of SO3(g):
N
nSO3 =
NA
1.6333 × 1022 molecules
6.02 × 1023 molecules /mol
= 0.027131 mol
= 0.027 mol
=
The amount in moles of the sulfur trioxide is 0.027 mol.
For the carbon dioxide:
Molar mass, M, of CO2(g):
M CO2 = 1M C + 2M O
= 1(12.01 g/mol ) + 2 (16.00 g/mol )
= 44.01 g/mol
Amount in moles, n, of the CO2(g):
m
nCO2 =
M
4.9 g
=
44.01g /mol
= 0.11 mol
The amount in moles of the carbon dioxide is 0.11 mol.
The amount in moles of the sample of carbon dioxide is greater.
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Solution 2
For the sulfur trioxide:
Amount in moles, n, of the SO3(g)
1 molecule SO3
1 mol
×
nSO3 = 4.9 × 1022 O atoms ×
23
6.02 × 10 molecules SO3
3 O atoms
= 0.027131 mol
= 0.027 mol
The amount in moles of the sulfur trioxide is 0.027 mol.
For the carbon dioxide:
Amount in moles, n, of the CO2(g):
m
nCO2 =
M
4.9 g
=
44.01g /mol
= 0.1113 mol
= 0.11 mol
The amount in moles of the carbon dioxide is 0.11 mol.
The amount in moles of the sample of carbon dioxide is greater.
Check Your Solution
Use rounded numbers to estimate the amount in moles of the SO3(g):
1
1
5 × 1022 × ×
= 2.8 × 10–2 mol
23
3 6 × 10
Use rounded numbers to estimate the amount in moles of the CO2(g):
5
= 0.125 mol
40
In each case, the estimate is close to the calculated answer. The answer is
reasonable.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 134 Chemistry 11 Solutions
12. Review Question (page 243)
Imagine that you are an environmental chemist who is testing drinking water
for water hardness. A sample of water you test has 3.5 × 10–2 mg of dissolved
calcium carbonate, CaCO3(aq).
a. How many formula units of calcium carbonate are in the sample?
b. How many oxygen atoms are in the calcium carbonate in the sample?
What Is Required?
a. You have to find the number of formula units of calcium carbonate in the
sample.
b. You have to find the number of oxygen atoms in the calcium carbonate in
the sample.
What Is Given?
You know the chemical formula for calcium carbonate: CaCO3
You know the mass of the calcium carbonate: 3.5 × 10–2 mg
a. formula units of calcium carbonate
Plan Your Strategy
Use the periodic table to find the atomic molar masses of the elements in
CaCO3(aq).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Convert the mass of the CaCO3(aq) from milligrams to grams:
1 mg = 1 × 10−3 g
Solution 1
Calculate the number of moles of CaCO3(aq) using the relationship n =
m
.
M
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the number of formula units in the sample of CaCO3(aq) using the
relationship N = n × N A .
Solution 2
The following steps may be completed in one line using the units to calculate
the answer.
Calculate the number of formula units by multiplying the mass
1
and then by the Avogadro constant, NA.
by
molar mass
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Act on Your Strategy
Molar mass, M, of CaCO3(aq):
M CaCO3 = 1M Ca + 1M C + 3M O
= 1( 40.08 g/mol ) + 1(12.01 g/mol ) + 3 (16.00 g/mol )
= 100.09 g/mol
Mass (in grams), m, of the CaCO3(aq):
mCaCO3 = 3.5 × 10 –2 mg × 1 × 10−3 g/ mg
= 3.5 × 10−5 g
Solution 1
Amount in moles, n, of the CaCO3(aq):
m
nCaCO3 =
M
3.5 × 10 –5 g
=
100.09 g /mol
= 3.4968 × 10 –7 mol
Number of formula units, N, of the CaCO3(aq):
N = n × NA
= 3.4968 × 10 –7 mol ×
6.02 × 1023 formula units
1 mol
= 2.1 × 1017 formula units
There are 2.1 × 1017 formula units of calcium carbonate in the sample.
Solution 2
Number of formula units, N, of the CaCO3(aq):
1
6.02 × 1023 formula units
N = 3.5 × 10 –5 g ×
×
100.09 g / mol
1 mol
= 2.1 × 1017 formula units
There are 2.1 × 1017 formula units of calcium carbonate in the sample.
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 136 Chemistry 11 Solutions
b. number of oxygen atoms
Plan Your Strategy
Multiply the number of formula units of the CaCO3(s) by the number of
oxygen atoms in one formula unit.
Act on Your Strategy
From the chemical formula, CaCO3, there are 3 oxygen atoms for every
formula unit of calcium carbonate.
Number of oxygen atoms, N, in the CaCO3(s):
N
3 O atoms
=
17
2.1 × 10 formula units CaCO3 1 formula unit CaCO3
N = 2.1 × 1017 formula units CaCO3 ×
3 O atoms
1 formula unit CaCO3
= 6.3 × 1017 O atoms
There are 6.3 × 1017 atoms of oxygen in the calcium carbonate sample.
Check Your Solution
Check to see that the correct atomic molar masses have been used.
Using rounded numbers, estimate the answers:
a. 3.5 × 10–2 × 10–3 ×
1
× 6 × 1023 = 2.1 × 1017 formula units of calcium
100
carbonate
b. 2 × 1017 × 3 = 6 × 1017 oxygen atoms
The estimates are close to the calculated answers. The answers are reasonable.
13. Review Question (page 243)
Again, imagine that you are the chemist who is testing water samples in the
previous question. This time, however, you are testing for lead content
downstream from a battery manufacturing plant. Health Canada suggests that
drinking water should have a maximum lead content of 0.010 mg/L of water. If
your test reveals that a 1 L water sample contains 3.1 × 1017 atoms of lead, is
the water you tested safe to drink? Explain.
What Is Required?
You need to determine the mass of lead in a sample of water to see if it
exceeds the limit of 0.010 mg/L of water.
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What Is Given?
You know the number of atoms of lead, Pb(s), in the sample: 3.1 × 1017 atoms
You know the maximum safe content for lead in drinking water: 0.010 mg/L
Plan Your Strategy
Solution 1
Use the periodic table to determine the atomic molar mass of Pb(s).
Calculate the amount in moles of Pb(s) using the relationship n =
N
.
NA
Calculate the mass of the sample of Pb(s) using the relationship m = n × M .
Convert the mass from grams to milligrams: 1 g = 1 × 103 mg.
Solution 2
The following steps may be completed in one line using the units to calculate
the answer.
1 mol
molar mass
Multiply the number of Pb atoms by
, then by
,
23
6.02 × 10 atoms
1 mol
1 × 103 mg
.
and then by
1g
Act on Your Strategy
Molar mass, M, of Pb(s): 207.2 g/mol (from the periodic table)
Amount in moles, n, of the Pb(s):
N
nPb =
NA
=
3.1 × 1017 atoms
6.02 × 10 23 atoms /mol
= 5.149 × 10 –7 mol
Mass, m, of the Pb(s):
mPb = n × M
= 5.149 × 10 –7 mol × 207.2 g/ mol
= 1.06692 × 10 –4 g
= 1.066 × 10 –4 g
The mass (in milligrams), m, of the lead in the sample:
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 138 Chemistry 11 Solutions
mPb = 1.066 × 10 –4 g ×
1 × 103 mg
1 g
= 1.1 × 10 –1 mg
Solution 2
Mass, m, of the Pb(s):
mPb =
207.2 g
1 mol
1 × 103 mg
×
×
6.02 × 1023 atoms
1 mol
1 g
= 1.06692 × 10 –1 g ×
1 × 103 mg
1 g
= 1.1 × 10 –1 mg
The amount of lead exceeds the safe limit of 1.0 × 10–2 mg and the water is not
safe to drink.
Check Your Solution
Use rounded numbers to estimate the answer:
1
3 × 1017 ×
× 210 × 103 = 0.10 mg of lead atoms
23
6 × 10
The estimate is close to the calculated answer. The answer is reasonable.
15. Review Question (page 243)
Methyl salicylate, C6H4(OH)COOCH3(s), is used in many consumer products,
such as mouthwash, as a flavouring. A mouthwash sample contains 1.38 × 1018
molecules of methyl salicylate. What is the mass of the methyl salicylate in the
sample?
What Is Required?
You need to determine the mass of methyl salicylate in a sample of
mouthwash.
What Is Given?
You know the chemical formula for methyl salicylate: C6H4(OH)COOCH3
You know the number of molecules of the methyl salicylate:
1.38 × 1018 molecules
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 139 Chemistry 11 Solutions
Plan Your Strategy
Solution 1
Use the periodic table to find the atomic molar masses of the elements in
C6H4(OH)COOCH3(s).
Multiply the atomic molar masses by the number of atoms of each element in
the compound.
Add these values to calculate the molar mass of the compound.
Use the Avogadro constant: N A = 6.02 × 10 23
Calculate the amount in moles of the C6H4(OH)COOCH3(s) using the
N
relationship n =
.
NA
Calculate the mass of the C6H4(OH)COOCH3(s) using the relationship
m=n × M .
Solution 2
The following steps can be completed in one line using the units to calculate
the answer.
Calculate the mass of the methyl salicylate by multiplying the given number of
1
molecules of the C6H4(OH)COOCH3(s) by
and then by
23
6.02 × 10 molecules
the molar mass, M.
Act on Your Strategy
Molar mass, M, of C6H4(OH)COOCH3(s):
M C6 H4 ( OH )COOCH3 = 8M C + 8M H + 3M O
= 8 (12.01 g/mol ) + 8 (1.01 g/mol ) + 3 (16.00 g/mol )
= 152.16 g/mol
Amount in moles, n, of the C6H4(OH)COOCH3(s):
N
nC6 H4 ( OH )COOCH3 =
NA
=
1.38 × 1018 molecules
6.02 × 1023 molecules /mol
= 2.292 × 10 –6 mol
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 140 Chemistry 11 Solutions
Mass, m, of the C6H4(OH)COOCH3(s):
mC6 H4 ( OH )COOCH3 = n × M
= 2.292 × 10 –6 mol × 152.16 g/ mol
= 3.488 × 10 –4 g
= 3.49 × 10 –4 g
The mass of the methyl salicylate in a sample of mouthwash is 3.49 × 10–4 g.
Solution 2
Mass, m, of the C6H4(OH)COOCH3(s):
mC6 H4 ( OH )COOCH3 = 1.38 × 1018 molecules ×
1 mol
152.16 g
×
23
6.02 × 10 molecules
1 mol
= 3.49 × 10 –4 g
The mass of the methyl salicylate in a sample of mouthwash is 3.49 × 10–4 g.
Check Your Solution
Use rounded numbers to estimate the answer:
1
1.4 × 1018 ×
× 150 = 3.5 × 10–4 g
6.02 × 1023
The estimate is close to the calculated answer. The answer is reasonable.
16. Review Question (page 243)
Determine the order of the following three substances, from smallest to
greatest number of carbon atoms: 5.6 × 1023 molecules of benzoic acid,
C6H5COOH(s); 1.3 mol of acetic acid, CH3COOH(ℓ); 0.17 kg of oxalic acid,
HOOCCOOH(s).
What Is Required?
You need to determine the number of carbon atoms in three different samples
and list the compounds from the smallest to greatest number of carbon atoms.
What Is Given?
You know the number of molecules of benzoic acid: 5.6 × 1023 molecules
You know the amount in moles of acetic acid:1.3 mol
You know the mass of the oxalic acid): 0.17 kg
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Plan Your Strategy
Solution 1
For the benzoic acid:
From the chemical formula C6H5COOH, determine the number of carbon
atoms per molecule of benzoic acid.
Calculate the number of carbon atoms in 5.6 × 1023 molecules of
C6H5COOH(s).
For the acetic acid:
From the chemical formula CH3COOH, determine the amount in moles of
carbon atoms per mole of acetic acid molecules.
Calculate the number of carbon atoms in the CH3COOH(ℓ) using the
relationship N = n × N A .
For the oxalic acid:
Convert the mass of HOOCCOOH(s) from kilograms to grams:
1 kg = 1 × 103 g
Use the periodic table to find the atomic molar masses of the elements in
HOOCCOOH(s). Multiply the atomic molar masses by the number of atoms of
each element in the compound.
Add these values to calculate the molar mass of the compound.
Calculate the amount in moles of the HOOCCOOH(s) using the relationship
m
.
n=
M
Calculate the number of molecules of the HOOCCOOH(s) using the
6.02 × 1023 molecules
.
conversion factor
1 mol
From the chemical formula, determine the number of carbon atoms in one
molecule of HOOCCOOH(s).
Calculate the total number of carbon atoms by multiplying the number of
molecules of HOOCCOOH(s) by the number of carbon atoms per molecule.
Compare the number of carbon atoms in each compound and list the
compounds from smallest to greatest number of carbon atoms.
Solution 2
Calculate the number of carbon atoms using the conversion factors
6.02 × 1023 units
1 mol
and
.
1 mol
molar mass
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Act on Your Strategy
Solution 1
For the benzoic acid:
From the chemical formula, C6H5COOH, there are 7 carbon atoms per
molecule of benzoic acid.
Number of C atoms, N, in the C6H5COOH(s):
N
7 C atoms
=
23
5.6 × 10 molecules 1 molecule
7 C atoms
= 5.6 × 10 23 molecules ×
1 molecule
= 3.9 × 1024 C atoms
There are 3.9 × 1024 carbon atoms in the sample of benzoic acid.
For the acetic acid:
From the chemical formula CH3COOH, there are 2 carbon atoms in one
molecule of acetic acid.
Number of molecules, N, of the CH3COOH(ℓ):
N = n × NA
= 1.3 mol × 6.02 × 1023 molecules/ mol
= 7.826 × 1023 molecules
Number of C atoms, N, in the CH3COOH(ℓ):
N
2 C atoms
=
23
7.826 × 10 molecules 1 molecule
N = 7.826 × 1023 molecules ×
2 C atoms
1 molecule
= 1.6 × 1024 C atoms
There are 1.6 × 1024 carbon atoms in the sample of acetic acid.
For the oxalic acid:
Mass (in grams), m, of the HOOCCOOH(s):
mHOOCCOOH = 0.17 kg × 1 × 103 g/ kg
= 1.70 × 102 g
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Molar mass, M, of HOOCCOOH(s):
M HOOCCOOH = 2M C + 2M H + 4M O
= 2 (12.01 g/mol ) + 2 (1.01 g/mol ) + 4 (16.00 g/mol )
= 90.04 g/mol
Amount in moles, n, of the HOOCCOOH(s):
m
nHOOCCOOH =
N
1.70 × 102 g
=
90.04 g /mol
= 1.888 mol
Number of molecules, N, of the HOOCCOOH(s):
N = n × NA
= 1.888 mol × 6.02 × 1023 molecules/ mol
= 1.1366 × 1024 molecules
From the chemical formula, HOOCCOOH(s), there are 2 carbon atoms in one
molecule of oxalic acid.
Number of C atoms, N, in HOOCCOOH(s):
N
2 C atoms
=
24
1.1366 × 10 molecules 1 molecule
N = 1.1366 × 1024 molecules ×
2 C atoms
1 molecule
= 2.3 × 1024 C atoms
There are 2.3 × 1024 carbon atoms in the sample of oxalic acid.
Solution 2
For the benzoic acid:
Number of C atoms, N, in 5.6 × 1023 molecules of C6H5COOH(s):
N
7 C atoms
=
23
5.6 × 10 molecules 1 molecule
7 C atoms
= 5.6 × 10 23 molecules ×
1 molecule
= 3.9 × 1024 C atoms
978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 144 Chemistry 11 Solutions
There are 3.9 × 1024 carbon atoms in the sample of benzoic acid.
For the acetic acid:
Number of C atoms, N, in 1.3 mol of CH3COOH(ℓ):
2 mol C atoms
6.02 × 1023 atoms
N = 1.3 mol ×
×
1 mol
1 mol
= 1.6 ×1024 C atoms
There are 1.6 × 1024 carbon atoms in the sample of acetic acid.
For the oxalic acid:
Mass (in grams), m, of the HOOCCOOH(s):
mHOOCCOOH = 0.17 kg × 1 × 103 g/ kg
= 1.70 × 102 g
Number of C atoms, N, in the HOOCCOOH(s):
1 mol
2 mol C atoms
6.02 × 1023 atoms
N = 1.70 × 102 g ×
×
×
1 mol
90.04 g
1 mol
= 2.3 × 1024 C atoms
There are 2.3 × 1024 carbon atoms in the sample of oxalic acid.
Listing the compounds from the smallest number of carbon atoms to the
greatest number:
CH3COOH(ℓ)(1.6 × 1024) < HOOCCOOH(s)(2.3 × 1024)
< C6H5COOH(s)(3.9 × 1024)
Check Your Solution
Use rounded numbers to estimate the number of carbon atoms:
In the sample of C6H5COOH(s):
6 × 6 × 1023 = 3.6 × 1024
In the sample of CH3COOH(ℓ):
1 × 2 × 6 ×1023 = 1.2 × 1024
In the sample of HOOCCOOH(s):
1
2 × 102 × × 2 × 6 × 1023 = 2.7 × 1024
90
In each case, the units cancel properly and an estimate of the number of atoms
of carbon is consistent with the calculated value. Each answer seems
reasonable.
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