21st Century Combined Science (Chemistry Part)
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
In-text activities
Checkpoint (page 75)
1
2
Substance
Number of moles of substance present
Number of particles present
Sulphur atoms
3.00 mol
1.81 x 1024 atoms
Copper(II) ions
6.50 mol
3.91 x 1024 ions
a)
Number of water molecules
= number of moles of water molecules x L
= 5.00 mol x 6.02 x 1023 mol–1
= 3.01 x 1024
b)
One water molecule contains two hydrogen atoms and one oxygen atom.
Number of atoms present
= 3 x number of water molecules
= 3 x 3.01 x 1024
= 9.03 x 1024
Checkpoint (page 78)
Substance
Hydrogen
chloride
Ethanoic acid
Aluminium
hydroxide
Magnesium
carbonate
Chemical
formula
Relative
atomic masses
Formula mass /
relative
molecular mass
Molar mass
HCl
H = 1.0
Cl = 35.5
36.5
36.5 g mol–1
CH3COOH
H = 1.0
C = 12.0
O = 16.0
60.0
60.0 g mol–1
78.0
78.0 g mol–1
84.3
84.3 g mol–1
H = 1.0
Al(OH)3
MgCO3
O = 16.0
Al = 27.0
C = 12.0
O = 16.0
Mg = 24.3
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
1
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
Checkpoint (page 82)
1
Chemical
formula
Molar mass
of substance
(g mol–1)
Mass of
substance
present (g)
Number of
moles of
substance
present (mol)
Nitrogen
dioxide
NO2
46.0
59.8
1.30
Lead(II)
oxide
PbO
223.2
Substance
Ammonium
carbonate
2
(NH4)2CO3
44.6
96.0
864
7.83 x 1023
molecules
1.20 x 1023 formula
0.200
units
5.42 x 1024
9.00
a)
Molar mass of K2S = (2 x 39.1 + 32.1) g mol–1 = 110.3 g mol–1
b)
Number of formula units of K2S
=
=
Number of
molecules /
formula units
present
formula units
number of moles of K2S x L
0.598 mol x 6.02 x 1023 mol–1
= 3.60 x 1023
One formula unit of K2S contains 2 potassium ions and 1 sulphide ion.
Number of potassium ions = 2 x 3.60 x 1023
= 7.20 x 1023
Number of sulphide ions
= 3.60 x 1023
Checkpoint (page 86)
1
Formula mass of ammonium sulphate (NH4)2SO4
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
2
= 2 x (14.0 + 4 x 1.0) + 32.1 + 4 x 16.0
= 132.1
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
2
Let m be the relative atomic mass of M.
Formula mass of MCl2 = m + 2 x 35.5
3
Formula mass of FeSO4•7H2O
=
55.8 + 32.1 + 4 x 16.0 + 7 x (2 x 1.0 + 16.0)
=
277.9
Percentage by mass of water in FeSO4•7H2O =
=
4
7 x (2 x 1.0  16.0)
x 100%
277.9
45.3%
Formula mass of Na2CO3•xH2O
= 2 x 23.0 + 12.0 + 3 x 16.0 + x(2 x 1.0 + 16.0)
= 106.0 + 18x
1 mole of Na2CO3•xH2O contains x moles of H2O.
i.e. (106.0 + 18x) g of Na2CO3•xH2O contain 18x g of H2O.
14.3 g of Na2CO3•xH2O contain 9.00 g of H2O.
18x
9.00 g
=
106.0  18x
14.3 g
x
=
10
Checkpoint (page 90)
1
a) Mass of oxide = 68.5 g
Mass of oxygen = (68.5 – 62.1) g = 6.4 g
Lead
Oxygen
Mass of element in the
oxide
62.1 g
6.4 g
Relative atomic mass
207.2
16.0
0.300 mol
=1
0.300 mol
0.400 mol
= 1.33
0.300 mol
1x3=3
1.33 x 3 = 4
Number of moles of atoms
that combine
Mole ratio of atoms
Simplest whole number
ratio of atoms
∴
the empirical formula of the oxide is Pb3O4.
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
3
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
b)
Suppose there are x mole(s) of PbO and y mole(s) of PbO2 in the lead oxide Pb3O4.
Number of moles of lead in the oxide = x + y = 3
Number of moles of oxygen in the oxide = x + 2y = 4
Solving the two equations gives x = 2 and y = 1.
∴ the mole ratio of PbO to PbO2 in Pb3O4 is 2 : 1.
2
Suppose we have 100 g of the compound, so there are 2.40 g of hydrogen, 39.0 g of sulphur
and 58.6 g of oxygen.
Hydrogen
Sulphur
Oxygen
2.40 g
39.0 g
58.6 g
1.0
32.1
16.0
2.40 mol
= 1.98
1.21 mol
1.21 mol
=1
1.21 mol
3.66 mol
= 3.02
1.21 mol
Mass of
element in the
compound
Relative atomic
mass
Number of
moles of atoms
that combine
Mole ratio of
atoms
∴
the empirical formula of the compound is H2SO3.
Checkpoint (page 100)
1
Method 1
2NaN3(s)
?g
2Na(s) + 3N2(g)
84.0 g
According to the equation, 2 moles of NaN3 are required to produce 3 moles of N2.
∴ number of moles of NaN3 required = 2.00 mol
Molar mass of NaN3
=
=
Mass of NaN3 required =
=
=
(23.0 + 3 x 14.0) g mol–1
65.0 g mol–1
number of moles of NaN3 x molar mass of NaN3
2.00 mol x 65.0 g mol–1
130 g
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
4
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
Method 2
2NaN3(s)
2Na(s) + 3N2(g)
?g
Molar mass of NaN3
84.0 g
= (23.0 + 3 x 14.0) g mol–1
= 65.0 g mol–1
Molar mass of N2 = 28.0 g mol–1
According to the equation, 2 moles of NaN3 are required to produce 3 moles of N2.
∴ 2 x 65.0 g of NaN3 are required to produce 3 x 28.0 g of N2.
2NaN3(s)
2 x 65.0 g
2Na(s) + 3N2(g)
3 x 28.0 g
130 g
84.0 g
Mass of NaN3 required = 130 g
2
a)
Method 1
2LiOH(s) + CO2(g)
50.0 g
?g
Molar mass of LiOH
Li2CO3(s) + H2O(l)
=
=
(6.9 + 16.0 + 1.0) g mol–1
23.9 g mol–1
According to the equation, 2 moles of LiOH can absorb 1 mole of CO2.
∴ number of moles of CO2 absorbed
Molar mass of CO2
=
=
Mass of CO2 absorbed =
=
=
Method 2
2LiOH(s) + CO2(g)
50.0 g
?g
Molar mass of LiOH
Molar mass of CO2
=
2.09
mol
2
= 1.045 mol
(12.0 + 2 x 16.0) g mol–1
44.0 g mol–1
number of moles of CO2 x molar mass of CO2
1.045 mol x 44.0 g mol–1
46.0 g
Li2CO3(s) + H2O(l)
=
=
=
=
(6.9 + 16.0 + 1.0) g mol–1
23.9 g mol–1
(12.0 + 2 x 16.0) g mol–1
44.0 g mol–1
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
5
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
According to the equation, 2 moles of LiOH can absorb 1 mole of CO2.
∴ 2 x 23.9 g of LiOH can absorb 44.0 g of CO2.
2LiOH(s) + CO2(g)
2 x 23.9 g 44.0 g
50.0 g
?g
Li2CO3(s) + H2O(l)
Mass of CO2 absorbed =
=
b)
Method 1
2LiOH(s) + CO2(g)
50.0 g x
44.0 g
2 x 23.9 g
46.0 g
Li2CO3(s) + H2O(l)
100.0 g
?g
According to the equation, 1 mole of Li2CO3 is produced when 1 mole of CO2 is
absorbed.
∴ number of moles of Li2CO3 produced
Molar mass of Li2CO3
=
=
Mass of Li2CO3 produced
Method 2
2LiOH(s) + CO2(g)
100.0 g
Molar mass of CO2
Molar mass of Li2CO3
=
=
=
2.27 mol
(2 x 6.9 + 12.0 + 3 x 16.0) g mol–1
73.8 g mol–1
= number of moles of Li2CO3 x molar mass of Li2CO3
= 2.27 mol x 73.8 g mol–1
= 168 g
Li2CO3(s) + H2O(l)
?g
44.0 g mol–1
(2 x 6.9 + 12.0 + 3 x 16.0) g mol–1
= 73.8 g mol–1
According to the equation, 1 mole of Li2CO3 is produced when 1 mole of CO2 is
absorbed.
∴ 73.8 g of Li2CO3 are produced when 44.0 g of CO2 are absorbed.
2LiOH(s) + CO2(g)
44.0 g
100.0 g
Li2CO3(s) + H2O(l)
73.8 g
?g
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Topic 3 Unit 12
6
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
Mass of Li2CO3 produced
3
a)
b)
2Ag2O(s)
2Ag2O(s)
?g
=
100.0 g x
=
168 g
73.8 g
44.0 g
4Ag(s) + O2(g)
4Ag(s) + O2(g)
6.52 g
According to the equation, 2 moles of Ag2O give 4 moles of Ag upon strong heating.
∴ number of moles of Ag2O decomposed
=
0.0604
mol
2
= 0.0302 mol
Molar mass of Ag2O
= (2 x 107.9 + 16.0) g mol–1
= 231.8 g mol–1
Mass of Ag2O decomposed = number of moles of Ag2O x molar mass of Ag2O
= 0.0302 mol x 231.8 g mol–1
= 7.00 g
7.00 g
Percentage by mass of Ag2O in the sample =
x 100%
8.00 g
=
Checkpoint (page 105)
1
Hg(l) + Br2(l)
21.5 g 15.6 g
87.5%
HgBr2(s)
?g
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
7
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
According to the equation, 1 mole of Hg reacts with 1 mole of Br2 to produce 1 mole of
HgBr2. During the reaction, 0.0976 mole of Br2 reacted with 0.0976 mole of Hg. Therefore Hg
was in excess. The amount of Br2 limited the amount of HgBr2 produced.
Number of moles of HgBr2 produced = 0.0976 mol
Molar mass of HgBr2 = (200.6 + 2 x 79.9) g mol–1
= 360.4 g mol–1
a) Mass of HgBr2 produced = number of moles of HgBr2 x molar mass of HgBr2
= 0.0976 mol x 360.4 g mol–1
= 35.2 g
b) Mass of Hg reacted
= number of moles of Hg x molar mass of Hg
= 0.0976 mol x 200.6 g mol–1
Mass of Hg left
2
a)
=
=
= 19.6 g
(21.5 – 19.6) g
1.9 g
6Li(s) + N2(g)
8.28 g 10.6 g
2Li3N(s)
?g
According to the equation, 6 moles of Li react with 1 mole of N2 to produce 2 moles
of Li3N. During the reaction, 1.20 moles of Li reacted with 0.200 mole of N2. Therefore
nitrogen was in excess. The amount of lithium limited the amount of nitride produced.
Number of moles of Li3N produced
Molar mass of Li3N
=
=
Theoretical yield of Li3N
=
2
x 1.20 mol
6
= 0.400 mol
(3 x 6.9 + 14.0) g mol–1
34.7 g mol–1
= number of moles of Li3N x molar mass of Li3N
= 0.400 mol x 34.7 g mol–1
= 13.9 g
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
8
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
b)
Percentage yield of Li3N
=
3.97 g
x 100%
13.9 g
=
28.6%
Unit-end exercises (pages 108-113)
Answers for the HKCEE and HKALE questions are not provided.
1
2
Molar mass
of substance
(g mol–1)
Mass of
substance
present (g)
Number of moles
of substance
present (mol)
Number of
molecules / formula
units present
Sulphur dioxide
(SO2)
64.1
1.28
0.0200
1.20 x 1022 molecules
Calcium sulphate
(CaSO4)
136.2
40.9
0.300
1.81 x 1023 formula units
Substance
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
9
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
Substance
Hydrated sodium
carbonate
(Na2CO3•9H2O)
Molar mass
Mass of
Number of moles
Number of
of substance
(g mol–1)
substance
present (g)
of substance
present (mol)
molecules / formula
units present
5.00
3.01 x 1024 formula
units
268.0
1 340
3
B
One mole of sodium oxide (Na2O) contains two moles of sodium ions and one mole of oxide
ions.
4
D
Molar mass of CaCO3
=
=
(40.1 + 12.0 + 3 x 16.0) g mol–1
100.1 g mol–1
Number of formula units of CaCO3
=
=
number of moles of CaCO3 x L
0.415 mol x 6.02 x 1023 mol–1
=
2.50 x 1023
5
A
6
C
7
D
8
C
9
B
10
C
11
A
12
B
(2) 2 moles of sodium ions contain 2 x 6.02 x 1023 sodium ions.
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
10
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21st Century Combined Science (Chemistry Part)
13
Molar mass of MnO2 = (54.9 + 2 x 16.0) g mol–1
= 86.9 g mol–1
Mass of MnO2 in the nodule = 0.0400 mol x 86.9 g mol–1 = 3.48 g
3.48 g
Percentage by mass of MnO2 in the nodule =
x 100%
15.0 g
=
23.2%
14
–
15
a)
b)
To prevent the condensed water from running back to the tube and crack the hot glass.
Test the liquid with dry cobalt(II) chloride paper. The liquid turns the paper from blue to
pink.
c)
d)
To prevent ‘sucking back’ of the liquid.
Formula mass of FeSO4•xH2O
= (55.8 + 32.1 + 4 x 16.0) + x(2 x 1.0 + 16.0)
= 151.9 + 18x
1 mole of FeSO4•xH2O contains x moles of H2O.
i.e. (151.9 + 18x) g of FeSO4•xH2O contain 18x g of H2O.
30.6 g of FeSO4•xH2O contain 13.9 g of H2O.
18x g
13.9 g
=
(151.9  18x) g
30.6 g
x
16
a)
=
7
Suppose we have 100 g of glucose, so there are 40.0 g of carbon, 6.60 g of hydrogen and
53.4 g of oxygen.
Carbon
Hydrogen
Oxygen
Mass of element
in the compound
40.0 g
6.60 g
53.4 g
Relative atomic
mass
12.0
1.0
16.0
3.33 mol
=1
3.33 mol
6.60 mol
=2
3.33 mol
3.33 mol
=1
3.33 mol
Number of moles
of atoms that combine
Mole ratio of atoms
∴ the empirical formula of glucose is CH2O.
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
11
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
b)
Let (CH2O)n be the molecular formula of glucose.
Relative molecular mass of glucose
= n(12.0 + 2 x 1.0 + 16.0)
=
30n
∴ 30n
=
180
n
=
6
∴ the molecular formula of glucose is (CH2O)6 or C6H12O6
17
a)
Method 1
2Fe(OH)3(s)
5.35 g
Fe2O3(s) + 3H2O(g)
?g
Molar mass of Fe(OH)3 =
=
[55.8 + 3 x (16.0 + 1.0)] g mol–1
106.8 g mol–1
According to the equation, 2 moles of Fe(OH)3 give 3 moles of H2O upon heating.
∴ number of H2O formed
Molar mass of H2O
Mass of H2O formed
Method 2
2Fe(OH)3(s)
5.35 g
Molar mass of Fe(OH)3
=
=
=
=
=
=
3
x 0.0501 mol
2
= 0.0752 mol
(2 x 1.0 + 16.0) g mol–1
18.0 g mol–1
number of moles of H2O x molar mass of H2O
0.0752 mol x 18.0 g mol–1
1.35 g
Fe2O3(s) + 3H2O(g)
?g
= [55.8 + 3 x (16.0 + 1.0)] g mol–1
= 106.8 g mol–1
Molar mass of H2O
= (2 x 1.0 + 16.0) g mol–1
= 18.0 g mol–1
According to the equation, 2 moles of Fe(OH)3 give 3 moles of H2O upon heating.
∴ 2 x 106.8 g of Fe(OH)3 give 3 x 18.0 g of H2O upon heating.
2Fe(OH)3(s)
2 x 106.8 g
5.35 g
Fe2O3(s) + 3H2O(g)
3 x 18.0 g
?g
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
12
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
Mass of H2O formed
b)
=
5.35 g x
3 x 18.0 g
2 x 106.8 g
= 1.35 g
Suppose we have 100 g of the oxide, so there are 72.4 g of iron and 27.6 g of oxygen.
Iron
Oxygen
72.4 g
27.6 g
55.8
16.0
Iron
Oxygen
1.30 mol
= 1.00
1.30 mol
1.73 mol
= 1.33
1.30 mol
1x3=3
1.33 x 3 = 4
Mass of element in the oxide
Relative atomic mass
Number of moles of atoms
that combine
Mole ratio of atoms
Simplest whole number ratio
of atoms
∴ the empirical formula of the oxide is Fe3O4.
18
a)
b)
2ZnS(s) + 3O2(g)
2ZnO(s) + C(s)
2ZnO(s) + C(s)
48.8 g ? g
Molar mass of ZnO
2ZnO(s) + 2SO2(g)
2Zn(s) + CO2(g)
2Zn(s) + CO2(g)
=
=
?g
(65.4 + 16.0) g mol–1
81.4 g mol–1
According to the equation, 2 moles of ZnO require 1 mole of C for reduction to give 2
moles of Zn.
∴
i)
number of moles of Zn obtained
=
0.600 mol
number of moles of C required
=
0.600
mol
2
Mass of Zn obtained
=
=
=
= 0.300 mol
number of moles of Zn x molar mass of Zn
0.600 mol x 65.4 g mol–1
39.2 g
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Topic 3 Unit 12
13
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21st Century Combined Science (Chemistry Part)
ii)
19
a)
b)
Mass of C required
=
=
number of moles of C x molar mass of C
0.300 mol x 12.0 g mol–1
=
3.60 g
2Al(s) + 3CuSO4(aq)
2Al(s) + 3CuSO4(aq)
1.61 g
Al2(SO4)3(aq) + 3Cu(s)
Al2(SO4)3(aq) + 3Cu(s)
2.58 g
According to the equation, 2 moles of Al react to give 3 moles of Cu.
∴ number of moles of Cu produced
Theoretical yield of Cu =
=
=
Percentage yield of Cu =
=
20
=
3
x 0.0596 mol
2
= 0.0894 mol
number of moles of Cu x molar mass of Cu
0.0894 mol x 63.5 g mol–1
5.68 g
2.58 g
x 100%
5.68 g
45.4%
4KO2(s) + 2H2O(g) + 4CO2(g)
4KHCO3(s) + 3O2(g)
?g
14.0 g
Molar mass of CO2
= (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
According to the equation, 4 moles of CO2 require 4 moles of KO2 for complete reaction.
∴ number of moles of KO2 required
=
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
0.318 mol
14
© Jing Kung. All rights reserved.
21st Century Combined Science (Chemistry Part)
a)
b)
21
Molar mass of KO2
=
=
(39.1 + 2 x 16.0) g mol–1
71.1 g mol–1
Theoretical mass of KO2 required = number of moles of KO2 x molar mass of KO2
= 0.318 mol x 71.1 g mol–1
= 22.6 g
Since the process is only 80% efficient, the mass of KO2 required
= 22.6 g ÷ 80%
= 28.3 g
MnO2(s) + 4HCl(aq)
217 g
274 g
Molar mass of MnO2
Molar mass of HCl
=
MnCl2(aq) + Cl2(g) + 2H2O(l)
?g
(54.9 + 2 x 16.0) g mol–1
=
86.9 g mol–1
=
=
(1.0 + 35.5) g mol–1
36.5 g mol–1
a)
According to the equation, 1 mole of MnO2 reacts with 4 moles of HCl to produce 1 mole
of Cl2. In this case, the mole ratio of MnO2 to HCl was 1:3. Therefore all the HCl would
be used up. The limiting reagent was HCl.
b)
Number of moles of Cl2 produced =
=
Molar mass of Cl2 =
=
Mass of Cl2 produced
7.51
mol
4
1.88 mol
–1
2 x 35.5 g mol
71.0 g mol–1
= number of moles of Cl2 x molar mass of Cl2
= 1.88 mol x 71.0 g mol–1
= 133 g
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Topic 3 Unit 12
15
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21st Century Combined Science (Chemistry Part)
c)
Number of moles of MnO2 used
=
7.51
mol
4
= 1.88 mol
Mass of MnO2 used = number of moles of MnO2 x molar mass of MnO2
= 1.88 mol x 86.9 g mol–1
= 163 g
Mass of MnO2 (excess reagent) left
= (217 – 163) g
= 54 g
Suggested answers to in-text activities and unit-end exercises
Topic 3 Unit 12
16
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