Ndimes = 7.800 × 103 g ×
7 dimes
12.30 g
Ndimes = 4439 dimes
There are 4439 dimes in a bag that contains 7.800 kg of dimes.
9. mHBr = 38.40 g HBr (key value)
2 mol HBr = 159.8 g HBr (conversion factor equation)
nHBr = 38.40 g HBr ×
2 mol HBr
159.8 g HBr
−1
nHBr = 4.81× 10 mol HBr
There are 4.81 ¯ 10–1 mol of hydrogen bromide molecules in the cylinder.
10. (a) Molar mass is the mass in grams per mole of one mole (6.02 u 1023) of chemical entities. The symbol for molar
mass is M and its SI unit is g/mol.
(b) MC = 12.01 g/mol
(c) The mass of 6.02 ¯ 1023 atoms of zinc is 65.38 g. This value is equal to the molar mass of zinc (65.38 g/mol).
(d) The element could be gold.
Applied Inquiry Skills
11. (a) 1. Measure the total mass of the coin mixture (in grams).
2. Separate one penny, one nickel, one dime, and one quarter from the mixture.
3. Measure the mass of one penny, one nickel, one dime, and one quarter together (in grams).
4. To estimate the number of coins in the mixture, divide the mass of the mixture measured in step 1 by the
mass of the four coins measured in step 3, and multiply the answer by 4.
(b) To estimate the number of dimes, divide the answer in part 4 by 4.
Making Connections
12. (a) Chemists measure and mix chemicals to produce products with properties that differ from the properties of the
reactants. Similarly, chefs measure and mix food ingredients to produce food products with properties that differ
from the properties of the ingredients. Chemists balance the reactants and products in a chemical equation.
Similarly, accountants prepare balance sheets to balance the income and expenditures of a person or a business.
(b) Chemists work with entities and cannot see the particles. Chefs work with ingredients they can see and even
count individually. Chemists balance reactants and products in a chemical equation to determine the quantities
of reactants needed and the products formed in a chemical reaction. The balance between reactants and products
is always equal. Accountants balance the income and expenditures for clients to determine if a profit or loss has
occurred. The relationship between the income and expenditures do not always balance (i.e., one can be greater
or less than the other).
13. (a) Student answers will vary. The scanning tunnelling microscope (STM) is able to display the protrusions and
depressions caused by atoms on the surface of substances like metals. The shapes of atoms are detected by
changes in electric current that occur when the fine tip of a piezoelectric tube moves over the material being
observed.
(b) The STM will most likely not replace our need to calculate quantities in chemistry because atoms and molecules
are so small that we will always have to deal with huge numbers of a chemical entity than to count individual
entities.
2.2 CALCULATIONS INVOLVING THE MOLE
PRACTICE
(Page 94)
Understanding Concepts
1. nAl = 1.60 mol Al
MAl = 26.98 mol/Al
1 mol Al = 26.98 g Al
44
Unit 2 Student Book Solutions
NEL
mAl = 1.60 mol Al ×
26.98 g Al
1 mol Al
mAl = 43.2 g Al
The mass of 1.60 mol aluminum atoms is 43.2 g.
2. nS = 0.25 mol S
MS = 32.06 g/mol S
1 mol S = 32.06 g S
32.06 g S
mS = 0.25 mol S ×
1 mol S
mS = 8.0 g S
The mass of 0.25 mol sulfur is 8.0 g.
PRACTICE
(Page 95)
Understanding Concepts
3. mFe = 3.30 g Fe
MFe = 55.85 g/mol Fe
1 mol Fe = 55.85 g Fe
1 mol Fe
nFe = 3.30 g Fe ×
55.85 g Fe
nFe = 0.059 mol Fe
There is 0.059 mol iron in a 3.30-g iron nail.
4. mAg = 23.6 g Ag
MAg = 107.87 g/mol Ag
1 mol Ag = 107.87 g/mol Ag
nAg = 23.6 gAg ×
1 mol Ag
107.87 gAg
nAg = 0.219 mol Ag
There is 0.219 mol silver in the silver coin.
5. mCu = 7.65 g Cu
MCu = 63.55 g/mol Cu
1 mol Cu = 63.55 g Cu
1 mol Cu
nCu = 7.65 g Cu ×
63.55 g Cu
nCu = 0.120 mol Cu
There is 0.120 mol copper in the bracelet.
PRACTICE
(Page 98)
Understanding Concepts
6. mC = 3.30 g C
MC = 12.01 g/mol C
1 mol C = 12.01 g C
1 mol C = 6.02 ¯ 1023 atoms C
NEL
Section 2.2 Student Book Solutions
45
1 mol C
NC
3.30 gC u
NC
1.65 u 10 atoms C
12.01 g C
u
6.02 u 1023 atoms C
1 mol C
23
There are 1.65 ¯ 1023 atoms of carbon in a 3.30-g diamond.
7. mNe = 6.80 g Ne
MNe = 20.18 g/mol Ne
1 mol Ne = 20.18 g Ne
1 mol Ne = 6.02 ¯ 1023 atoms Ne
1 mol Ne
NNe
6.80 g Ne u
NNe
2.03 u 10 atoms Ne
20.18 g Ne
u
6.02 u 1023 atoms Ne
1 mol Ne
23
There are 2.03 ¯ 1023 atoms of neon in a sign containing 6.80 g of neon.
8. mHg = 78.2 g Hg
MHg = 200.59 g/mol Hg
1 mol Hg = 200.59 g Hg
1 mol Hg = 6.02 ¯ 1023 atoms Hg
1 mol Hg
NHg
78.2 g Hg u
NHg
2.35 u 10 atoms Hg
200.59 g Hg
u
6.02 u 1023 atoms Hg
1 mol Hg
23
There are 2.35 ¯ 1023 atoms of mercury in the thermometer.
PRACTICE
(Page 99)
Understanding Concepts
9. MC H = 8 (MC ) + 18 (MH )
8 18
MC8H18
= 8 (12.01 g/mol) + 18 (1.01 g/mol)
= 96.08 g/mol + 18.18 g/mol
= 114.26 g/mol
The molar mass of octane is 114.26 g/mol.
10. MC9H8O4 = 9 (MC ) + 8 (MH ) + 4 (MO )
= 9 (12.01 g/mol) + 8 (1.01 g/mol) + 4 (16.00 g/mol)
MC9H8O4
= 108.09 g/mol + 8.08 g/mol + 64.00 g/mol
= 180.17 g/mol
The molar mass of acetylsalicylic acid is 180.17 g/mol.
11. MCaSO = 1(MCa ) + 1(MS ) + 4 (MO )
4
= 1( 40.08 g/mol) + 1(32.06 g/mol) + 4 (16.00 g/mol)
= 40.08 g/mol + 32.06 g/mol + 64.00 g/mol
MCaSO4 = 136.14 g/mol
The molar mass of calcium sulfate is 136.14 g/mol.
46
Unit 2 Student Book Solutions
NEL
PRACTICE
(Page 100)
Understanding Concepts
12. nNH = 0.900 mol NH3
3
MNH3 = 1(14.01 g/mol) + 3(1.01 g/mol)
MNH3 = 17.04 g/mol NH3
1 mol NH3 = 17.04 g NH3
mNH3 = 0.900 mol NH3 ×
17.04 g NH3
1 mol NH3
mNH3 = 15.3 g NH3
The mass of the ammonia is 15.3 g.
13. nCCl2F2 = 3.60 mol CCl2F2
MCCl2F2 = 1(12.01 g/mol) + 2(35.45 g/mol) + 2(19.00 g/mol)
MCCl2F2 = 120.91 g/mol CCl2F2
1 mol CCl2F2 = 120.91 g CCl2F2
mCCl2F2 = 3.60 mol CCl2F2 ×
120.91 g CCl2F2
1 mol CCl2F2
mCCl2F2 = 435 g CCl2F2
The mass of 3.60 mol freon-12 is 435 g.
PRACTICE
(Page 102)
Understanding Concepts
14. mMg(OH) = 204.0 g Mg(OH)2
2
MMg(OH)2 = 1(24.31 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol)
MMg(OH)2 = 58.33 g/mol Mg(OH)2
1 mol Mg(OH)2 = 58.33 g Mg(OH)2
NMg(OH)2 = 2.040 g Mg(OH)2 ×
1 mol Mg(OH)2
58.33 g Mg(OH)2
NMg(OH)2 = 3.497 mol Mg(OH)2
There is 3.497 mol magnesium hydroxide in 204.0 g of magnesium hydroxide.
1000 g C12H22O11
15. mC H O = 1.00 kg C12H22O11 ×
12 22 11
1 kg C12H22O11
mC12H22O11 = 1.00 × 103 g C12H22O11
MC12H22O11
12 12.01 g/mol 22 1.01 g/mol 1116.00 g/mol
MC12H22O11
342.34 g/mol C12H22O11
1 mol C12H22O11
nC12H22O11
342.34 g C12H22O11
1 mol C12H22O11
= 1.00 × 103 gC12H22O11 ×
342.34 g C12H22O11
nC12H22O11 = 2.92 mol C12H22O11
There is 2.92 mol sucrose in a bag that contains 1.00 kg of sucrose.
NEL
Section 2.2 Student Book Solutions
47
PRACTICE
(Page 103)
Understanding Concepts
16. mH O = 250.0 g H2O
2
MH2O = 2(1.01 g/mol) + 1(16.00 g/mol)
MH2O = 18.02 g/mol H2O
1 mol H2O = 18.02 g/mol H2O
NH2O = 250.0 g H2O ×
NH2O = 8.35 × 10
24
1 mol H2O
18.02 g H2O
×
6.02 × 1023 molecules H2O
1 mol H2O
moleculesH2O
There are 8.35 ¯ 1024 molecules of water in a bottle that contains 250.0 g of water.
17. mCo2 (Cr2O7 )3 = 3.30 kg Co2 (Cr2O7 )3 ×
103 g Co2 (Cr2O7 )3
1 kg Co2 (Cr2O7 )3
mCo (Cr O ) = 3.30 × 10 g Co2 (Cr2O7 )3
3
2
2
7 3
MCo2 (Cr2O7 )3 = 2 (58.93 g/mol) + 6 (52.00 g/mol) + 21(16.00 g/mol)
MCo2 (Cr2O7 )3 = 765.86 g/mol Co2 (Cr2O7 )3
1 mol Co2 (Cr2O7 )3 = 765.86 g Co2 (Cr2O7 )3
1 mol Co2 (Cr2O7 )3
NCo
2
(Cr2 O7 )3
= 3.30 × 10 g Co2 (Cr2 O7 )3 ×
NCo
2
(Cr2 O7 )3
= 2.59 × 10 formula units Co2 (Cr2 O7 )3
3
6.02 × 10 formulaunitsCo2 (Cr2 O7 )3
23
765.86 g Co2 (Cr2O7 )3
×
1 mol Co2 (Cr2O7 )3
24
There are 2.59 ¯ 10 formula units of cobalt(III) dichromate in a 3.30-kg sample.
24
PRACTICE
(Page 105)
Understanding Concepts
18. m = 4.4 g F
F2
2
MF2 = 2(19.00 g/mol)
MF2 = 38.00 g/mol F2
1 mol F2 = 38.00 g F2
NF2 = 4.4 g F2 ×
1 mol F2
38.00 g F2
×
6.02 × 1023 molecules F2
1 mol F2
×
2 atoms F
1 molecule F2
NF2 = 1.4 × 1023 atoms F
There are 1.4 ¯ 1023 atoms of fluorine in 4.4 g of fluorine gas.
1000 g N2
19. mN2 = 1.26 kg N2 ×
1 kg N2
mN2 = 1.26 × 103 g N2
48
Unit 2 Student Book Solutions
NEL
MN2
2 14.01 g/mol
MN2
28.02 g/mol N2
1mol N2
28.02 g N2
1 mol N2
NN = 1.26 × 103 g N2 ×
28.02 g N2
×
6.02 × 1023 molecules N2
1 mol N2
×
2 atoms N
1 molecule N2
NN = 5.41× 10 atoms N
25
There are 5.41 ¯ 1025 atoms of nitrogen in 1.26 kg of nitrogen gas.
20. mC H = 29.5 g C2H4
2 4
MC2H4 = 2 (12.01 g/mol) + 4 (1.01 g/mol)
MC2H4 = 28.06 g/mol C2H4
1 mol C2H4 = 28.06 g C2H4
NH = 29.5 g C2H4 ×
1 mol C2H4
28.06 g C2H4
×
6.02 × 1023 molecules C2H4
1 mol C2H4
×
4 atoms H
1 molecule C2H4
NH = 2.53 × 10 atoms H
24
There are 2.53 ¯ 1024 atoms of hydrogen in 29.5 g of ethane.
10−3 g Sr(OH)2
21. mSr(OH)2 = 0.170 mg Sr(OH)2 ×
1 mg Sr(OH)2
mSr(OH)2 = 1.70 × 10−4 g Sr(OH)2
MSr(OH)2 = 1(87.62 g/mol) + 2 (16.00 g/mol) + 2 (1.01 g/mol)
MSr(OH)2 = 121.64 g/mol Sr(OH)2
1 mol Sr(OH)2 = 121.64 g Sr(OH)2
NO = 1.70x10−4 g Sr(OH)2 ×
1 mol Sr(OH)2
121.64 g Sr(OH)2
×
6.02 × 1023 molecules Sr(OH)2
1 mol Sr(OH)2
×
2 atoms O
1 molecule Sr(OH)2
NO = 1.68 × 10 atoms O
18
There are 1.68 ¯ 1018 atoms of oxygen in 0.170 mg of strontium hydroxide.
TRY THIS ACTIVITY: COUNTING ATOMS, MOLECULES, AND OTHER ENTITIES
(Page 105)
1. mass of 50 drops of water
= 1.95 g
mass of 1 drop of water =
1.95 g
50
mass of 1 drop of water = 0.039 g
The mass of 1 drop of water is 0.039 g.
2. time of 1 drop of water to evaporate = 65 min, or 3.9 ¯ 103 s
NH2O = 0.039 g H2O ×
1 mol H2O
18.02 g H2O
×
6.02 × 1023 molecules H2O
1 mol H2O
NH2O = 1.30 × 1021molecules H2O
There are 1.30 ¯ 1021 molecules of water in every drop.
NEL
Section 2.2 Student Book Solutions
49
1.30 × 1021molecules H2O
3.9 × 103 s
evaporation rate = 3.34 × 1017 molecules/s
evaporation rate =
Water molecules evaporate at the average rate of 3.34 ¯ 1017 molecules/s.
3. mass of 1 penny = 2.44 g
NCu = 2.44 g Cu ×
1 mol Cu
63.55 g Cu
×
6.02 × 1023 atoms Cu
1 mol Cu
NCu = 2.31× 1022 atoms Cu
There are 2.31 ¯ 1022 copper atoms in a copper penny.
1 cent
2.31× 1022 atoms
value of a copper atom = 4.33 × 10−23 cents
value of a copper atom =
Each copper atom in a penny has a value of 4.33 ¯ 10–23 cents.
4. MC12H22O11 = 337.34 g/mol C12H22O11
337.34 g 1 mol
×
2
1 mol
mass of half a mole of sucrose = 168.67 g
mass of half a mole of sucrose =
When poured into a graduated cylinder, the volume of half a mole of sucrose is 185 mL.
5. One mole of sucrose, C12H22O11(s), contains 12 mol carbon atoms. Therefore, you would need one-sixth of a mole of
sucrose to get 2 mol carbon atoms.
337.34 g 1 mol
×
6
1 mol
required mass of sucrose = 56.22 g
required mass of sucrose =
When poured into a graduated cylinder, the volume of sucrose is 60 mL.
6. initial mass of chalk = 14.23 g
final mass of chalk = 14.15 g
change in mass of chalk = 0.08 g
MCaCO3 = 100.09 g/mol
NCaCO3 = 0.08 g CaCO3 ×
1 mol CaCO3
100.09 g CaCO3
×
6.02 × 1023 formula units CaCO3
1 mol CaCO3
NCaCO3 = 4.81× 10 formula units CaCO3
20
Since there are 5 atoms in each formula unit of calcium carbonate, we must multiply the answer by 5 atoms.
number of atoms = 4.81× 1020 formula unit ×
5 atoms
1 formula unit
number of atoms = 2.40 × 1021atoms
The number of atoms needed to write my name in chalk is 2.40 ¯ 1021 atoms.
7. mNaCl = 3.00 g NaCl
MNaCl = 58.44 g/mol NaCl
50
Unit 2 Student Book Solutions
NEL
NNaCl = 3.00 g NaCl ×
1 mol NaCl
6.02 × 1023 formula units NaCl
×
1 mol NaCl
57.44 g NaCl
NNaCl = 3.09 × 1022 formula units NaCl
Since there is one sodium ion in every formula unit of sodium chloride, the number of sodium ions is 3.09 ¯ 1022.
8. mnail = 4.16 g Fe
MFe = 55.85 g/mol Fe
NFe = 4.16 g Fe ×
1 mol Fe
6.02 × 1023 atoms Fe
×
1 mol Fe
55.85 g Fe
NFe = 4.48 × 1022 atoms
There are 4.48 ¯ 1022 atoms in the iron nail.
9. number of seconds in one year =
60 s
1 min
×
60 min 24 h 365 d
×
×
1h
1 d 1 year
number of seconds in one year = 3.15 × 107 s/year
years in one mole of seconds =
6.02 × 1023 s
1 year
×
1 mol
3.15 × 107 s
years in one mole of seconds = 1.91× 1016 year/mol
It takes 1.91 ¯ 1016 years to span one mole of seconds.
SECTION 2.2 QUESTIONS
(Page 106)
Understanding Concepts
1. M
= 1(M ) + 2 (M ) + 2 (M
Mg(OH)2
Mg
O
H
)
= 1(24.31g/mol) + 2 (16.00 g/mol) + 2 (1.01 g/mol)
MMg(OH)2 = 58.33 g/mol
The molar mass of magnesium hydroxide is 58.33 g/mol.
2. MO = 3(MO )
3
= 3(16.00 g/mol)
MO3 = 48.00 g/mol O3
The molar mass of ozone is 48.00 g/mol.
3. Student answers will vary.
Let the substance be represented by Z.
MZ = 67.2 g/mol
1.0 molZ = 67.2 g Z
8.0 molZ = 67.2 g Z × 8
mZ = 538 g Z
The mass of 8.0 mol of the substance is 538 g.
4. The mass of one mole of sucrose is equal to its molar mass.
MC12H22O11
12(MC ) 22(MH ) 11(MO )
12 12.01 g/mol 22 1.01 g/mol 1116.00 g/mol
MC12H22O11
342.34 g/mol C12H22O11
Since the molar mass of sucrose is equal to 342.34 g/mol, the mass of one mole of sucrose is 342.34 g.
NEL
Section 2.2 Student Book Solutions
51
5. (a) nAl = 0.1 mol Al
1 mol Al = 6.02 ¯ 1023 atoms Al
NAl = 0.1 mol Al ×
6.02 × 1023 atoms Al
1 mol Al
NAl = 6.02 × 10 atoms Al
22
There are 6.02 ¯ 1022 atoms of aluminum in 0.1 mol aluminum.
(b) nMgCl2 = 3.5 mol MgCl2
1 mol MgCl2 = 6.02 ¯ 1023 formula units MgCl2
NMgCl2 = 3.5 mol MgCl2 ×
6.02 × 1023 formula units MgCl2
1 mol MgCl2
NMgCl2 = 2.1× 1024 formula units MgCl2
There are 2.1 ¯ 1024 formula units of magnesium chloride in 3.5 mol magnesium chloride.
6. (a)
mC12H22O11 = 5.00 kg C12H22O11 ×
1000 g C12H22O11
1 kg C12H22O11
mC12H22O11 = 5.00 × 103 g C12H22O11
MC12H22O11 = 12 (12.01 g/mol) + 22 (1.01 g/mol) + 11(16.00 g/mol)
MC12H22O11 = 342.34 g/mol C12H22O11
1 mol C12H22O11 = 342.34 g C12H22O11
nC12H22O11 = 5.00 × 103 g C12H22O11 ×
1 mol C12H22O11
342.34 g C12H22O11
nC12H22O11 = 14.6 mol C12H22O11
There is 14.6 mol sucrose in 5.00 kg of table sugar.
(b)
mC10H8 = 250 g C10H8
MC10H8 = 10 (12.01 g/mol) + 8 (1.01 g/mol)
MC10H8 = 128.18 g/mol C10H8
1 mol C10H8 = 128.18 g C10H8
nC10H8 = 250 g C10H8 ×
1 mol C10H8
128.18 g C10H8
nC10H8 = 1.95 mol C10H8
There is 1.95 mol naphthalene in 250 g of naphthalene moth balls.
(c)
mC3H8 = 35.0 g C3H8
MC3H8 = 3 (12.01 g/mol) + 8 (1.01 g/mol)
MC3H8 = 44.11 g/mol C3H8
1 mol C3H8 = 44.11 g C3H8
nC3H8 = 35.0 g C3H8 ×
1 mol C3H8
44.11 g C3H8
nC3H8 = 0.793 mol C3H8
There is 0.793 mol propane in the stove cylinder.
52
Unit 2 Student Book Solutions
NEL
(d) mC H O = 275 mg C9H8 O4 ×
9 8 4
10−3 g C9H8O4
1 mg C9H8 O4
mC9H8O4 = 0.275 gC9H8 O4
MC9H8O4 = 9 (12.01 g/mol) + 8 (1.01 g/mol) + 4 (16.00 g/mol)
MC9H8O4 = 180.17 g/mol C9H8 O4
1 molC9H8 O4 = 180.17 g/mol C9H8 O4
1 mol C9H8 O4
nC9H8O4 = 0.275 g C9H8 O4 ×
180.17 g C9H8O4
nC9H8O4 = 1.53 × 10−3 mol C9H8 O4
There is 1.53 ¯ 10–3 mol acetylsalicylic acid in a 275-mg headache relief tablet.
(e) mC3H8O = 240 gC3H8 O
MC3H8O = 3 (12.01 g/mol) + 8 (1.01 g/mol) + 1(16.00 g/mol)
MC3H8O = 60.11 g/mol C3H9O
1 mol C3H8O = 60.11 g C3H8 O
nC3H8O = 240 g C3H8 O ×
1 mol C3H8 O
60.11 g C3H8 O
nC3H8O = 4.0 mol C3H8 O
There is 4.0 mol 2-propanol in 240 g of rubbing alcohol.
7. (a) nNH3 = 2.67 mol NH3
MNH3 = 1(14.01 g/mol) + 3(1.01 g/mol)
MNH3 = 17.04 g/mol NH3
1 mol NH3 = 17.4 g NH3
mNH3 = 2.67 mol NH3 ×
17.04 g NH3
1 mol NH3
mNH3 = 45.5 g NH3
The mass of 2.67 mol ammonia is 45.5 g.
(b) nNaOH = 0.965 mol NaOH
MNaOH = 1(22.99 g/mol) + 1(16.00 g/mol) + 1(1.01 g/mol)
MNaOH = 40.00 g/mol NaOH
1 mol NaOH = 40.00 g NaOH
mNaOH = 0.965 mol NaOH ×
40.00 g NaOH
1 mol NaOH
mNaOH = 38.6 g NaOH
The mass of 0.965 mol sodium hydroxide is 38.6 g.
(c) nH O = 19.7 mol H2O
2
MH2O = 2(1.01 g/mol) + 1(16.00 g/mol)
MH2O = 18.02 g/mol H2O
1 mol H2O = 18.02 g H2O
NEL
Section 2.2 Student Book Solutions
53
mH2O = 19.7 mol H2O ×
18.02 g H2O
1 mol H2O
mH2O = 355 g H2O
The mass of 19.7 mol water vapour is 355 g.
(d) nKMnO = 3.85 mol KMnO4
4
MKMnO4 = 1(39.10 g/mol) + 1(54.94 g/mol) + 4(16.00 g/mol)
MKMnO4 = 158.04 g/mol KMnO4
1 mol KMnO4 = 158.04 g KMnO4
mKMnO4 = 3.85 mol KMnO4 ×
158.04 g KMnO4
1 mol KMnO4
mKMnO4 = 608 g KMnO4
The mass of 3.85 mol potassium permanganate is 608 g.
(e) n(NH4 )2SO4 = 0.47 mol (NH4 )2SO4
M(NH4 )2SO4 = 2(14.01 g/mol) + 8(1.01 g/mol) + 1(32.06 g/mol) + 4(16.00 g/mol)
M(NH4 )2SO4 = 132.16 g/mol (NH4 )2SO4
1 mol (NH4 )2SO4 = 132.16 g (NH4 )2SO4
m(NH4 )2SO4 = 0.47 mol (NH4 )2SO4 ×
132.16 g NH4 )2SO4
1 mol (NH4 )2SO4
m(NH4 )2SO4 = 62 g (NH4 )2SO4
The mass of 0.47 mol ammonium sulfate is 62 g.
8. (a) nCO2 = 2.5 mol CO2
1 mol CO2 = 6.02 × 1023 molecules CO2
NCO2 = 2.5 mol CO2 ×
6.02 × 1023 molecules CO2
1 mol CO2
NCO2 = 1.5 × 10 molecules CO2
24
There are 1.5 ¯ 1024 molecules OF solid carbon dioxide in 2.5 mol carbon dioxide.
(b) mNH = 2.5 g NH3
3
MNH3 = 1(14.01 g/mol) + 3(1.01 g/mol)
MNH3 = 17.04 g/mol NH3
1 mol NH3 = 17.04 g NH3
1 mol NH3 = 6.02 × 1023 molecules NH3
NNH3 = 2.5 g NH3 ×
1 mol NH3
17.04 g NH3
×
6.02 × 1023 molecules NH3
1 mol NH3
NNH3 = 8.8 × 10 molecules NH3
22
There are 8.8 ¯ 1022 molecules of ammonia in 2.5 g of ammonia gas.
54
Unit 2 Student Book Solutions
NEL
(c) mHCl = 2.5 g HCl
MHCl = 1(1.01 g/mol) + 1(35.45 g/mol)
MHCl = 36.46 g/mol HCl
1 mol HCl = 36.46 g HCl
1 mol HCl = 6.02 × 1023 molecules HCl
1 mol HCl
6.02 u 1023 molecules HCl
u
1 mol HCl
36.46 g HCl
NHCl
2.5 g HCl u
NHCl
4.1 u 1022 molecules HCl
There are 4.1 ¯ 1022 molecules in 2.5 g of hydrogen chloride gas.
9. (a) NCO = 0.10 mol CO2
2
MCO2 = 1(12.01 g/mol) + 2(16.00 g/mol)
MCO2 = 44.01 g/mol CO2
1 mol CO2 = 44.01 g CO2
mCO2
0.10 mol CO2 u
mCO2
4.4 g CO2
44.01 g CO2
1 mol CO2
The mass of 0.10 mol carbon dioxide is 4.4 g.
(b) nC6H12O6 = 0.10 mol C6H12O6
MC6H12O6 = 6(12.01g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
MC6H12O6 = 180.18 g/mol C6H12O6
1 mol C6H12O6 = 180.18 g C6H12O6
mC6H12O6
0.10 mol C6H12O6 u
mC6H12O6
18 g C6H12O6
180.18 g C6H12O6
1 mol C6H12O6
The mass of 0.10 mol glucose is 18 g.
(c) NO2 = 0.10 mol O2
MO2 = 2(16.00 g/mol)
MO2 = 32.00 g/mol O2
1 molO2 = 32.00 g O2
mO2
0.10 mol O2 u
mO2
3.2 g O2
32.00 g O2
1 mol O2
The mass of 0.10 mol oxygen gas is 3.2 g.
10. NO = 2.7 mol O2
2
1 molO2 = 6.02 × 1023 molecules O2
6.02 u 1023 molecules O2
NO2
2.7 mol O2 u
NO2
1.6 u 1024 molecules O2
1 mol O2
There are 1.6 ¯ 1024 oxygen molecules in 2.7 mol oxygen gas.
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Section 2.2 Student Book Solutions
55
11. mC6H8O6 = 90 mg C6H8 O6 ×
10−3 g C6H8 O6
1 mg C6H8 O6
mC6H8O6 = 9 × 10−2 g C6H8 O6
MC6H8O6 = 6(12.01 g/mol) + 8(1.01 g/mol) + 6(16.00 g/mol)
MC6H8O6 = 176.14 g/mol C6H8 O6
1 mol C6H8O6 = 176.14 g/mol
1 mol C6H8O6 = 6.02 × 1023 molecules C6H8 O6
NC6H8O6
9 u 102 g C6H8O6 u
NC6H8O6
3.1 u 10
20
1 mol C6H8O6
176.14 g C6H8 O6
u
6.02 u 1023 molecules C6H8 O6
1 mol C6H8O6
molecules C6H8 O6
The number of molecules of vitamin C taken each day would be 3.1 ¯ 1020 molecules.
12. water
mH2O 450 gH2O
MH2O
2(1.01 g/mol) 1(16.00 g/mol)
MH2O
18.02 g/mol H2O
1 mol H2O 18.02 g H2O
nH2O
450 gH2O u
nH2O
25 mol H2O
18.02 g H2O
sugar
mC12H22O11
100 g C12H22O11
MC12H22O11
12(12.01 g/mol) 22(1.01 g/mol) 11(16.00 g/mol)
MC12H22O11
342.34 g/mol C12H22O11
1 mol C12H22O11
342.34 g C12H22O11
nC12H22O11
100 g C12H22O11 u
nC12H22O11
0.3 mol C12H22O11
1 mol C12H22O11
342.34 g C12H22O11
vinegar
mHC2H3O2
2.4 g HC2H3O2
MHC2H3O2
4(1.01 g/mol) 2(12.01 g/mol) 2(16.00 g/mol)
MHC2H3O2
60.06 g/mol HC2H3O2
1 mol HC2H3O2
56
1 mol H2O
60.06 g HC2H3O2
1mol HC2H3O2
nHC2H3O2
2.4 g HC2H3O2 u
nHC2H3O2
0.040 mol HC2H3O2
Unit 2 Student Book Solutions
60.06 g HC2H3O2
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salt
mNaCl
2 g NaCl
MNaCl
1(22.99 g/mol) 1(35.45 g/mol)
MNaCl
58.44 g/mol NaCl
1 molNaCl 58.44 g NaCl
1 mol NaCl
58.44 g NaCl
nNaCl
2 g NaCl u
nNaCl
0.03 mol NaCl
The recipe calls for 25 mol water, 0.3 mol sugar, 0.040 mol acetic acid (vinegar), and 0.03 mol salt.
Applied Inquiry Skills
13. Materials
eye protection
stirring rod
lab apron
coil of copper wire
250-mL beaker
filter paper
silver nitrate
scoopula
distilled water
Warning: Silver chloride may stain skin and cause skin irritation. Avoid contact with skin. Do not taste.
Procedure
1. Prepare a solution of silver ions by dissolving 0.1 g of silver nitrate in 100 mL water in a 250-mL beaker.
2. Place a coil of copper wire into the silver nitrate solution and let sit for 24 h.
3. Measure the mass of a piece of filter paper, mfilter paper.
4. Carefully remove the copper coil from the silver nitrate solution and scrape all the silver crystals from the coil
onto the filter paper.
5. Allow the silver crystals to dry.
6. Determine the mass of the filter paper and dry silver crystals, mfilter paper + Ag.
7. Calculate the mass of the silver crystals by subtracting the mass of the filter paper (Step 4) from the mass of the
filter paper and silver crystals (step 7).
mAg = mfilter paper + Ag – mfilter paper
8. Calculate the amount of silver recovered, nAg, as follows.
1 mol
nAg( ) mAg u
s
107.87 g
Making Connections
14. Student answers will vary. The mole deserves to win this award because without the mole, chemists could not
calculate the number of chemical entities they use in chemical reactions. Knowing the number of entities used in
reactions enables chemists to maximize the amount of product formed, and minimize the amount of reactants that
remain unreacted. When chemicals are mixed haphazardly, excess reactants mix with products and must be separated
in order to purify the products. This effect increases the cost associated with the industrial preparation of chemicals,
and may lead to increased levels of chemical waste and chemical pollution.
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Section 2.2 Student Book Solutions
57