suggested answers to in-text activities and exercises

New 21st Century Chemistry

Suggested answers to in-text activities and unit-end exercises

Topic 3 Unit 12

In-text activities

Checkpoint (page 75)

1Substance 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

2 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 Chemical

formulaRelative

atomic massesFormula mass /

relative

molecular mass

Molar mass

Hydrogen chloride

HCl H = 1.0Cl = 35.5

36.5 36.5 g mol–1

Ethanoic acid CH3COOH H = 1.0C = 12.0

O = 16.0

60.0 60.0 g mol–1

Aluminium hydroxide

Al(OH)3 H = 1.0O = 16.0

Al = 27.0

78.0 78.0 g mol–1

Magnesium carbonate

MgCO3 C = 12.0O = 16.0

Mg = 24.3

84.3 84.3 g mol–1


Suggested answers to in-text activities and unit-end exercises 1 © Jing Kung. All rights reserved.Topic 3 Unit 12


1Substance Chemical

formulaMolar mass of substance

(g mol–1)

Mass of substance present (g)

Number of moles of

substance present (mol)

Number of molecules /

formula units present

Nitrogendioxide

NO2 46.0 59.8 1.30 7.83 x 1023 molecules

Lead(II)oxide

PbO 223.2 44.6 0.200 1.20 x 1023 formula units

Ammoniumcarbonate

(NH4)2CO3 96.0 864 9.00 5.42 x 1024

formula units

2 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 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


1 Formula mass of ammonium sulphate (NH4)2SO4 = 2 x (14.0 + 4 x 1.0) + 32.1 + 4 x 16.0

= 132.1

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 =277.9

16.0) 1.0 x (2 x 7 +x 100%

= 45.3%

4 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 106.0

18x

+=

g 14.3

g 9.00

x = 10


1 a) Mass of oxide = 68.5 g

Mass of oxygen = (68.5 – 62.1) g = 6.4 gLead Oxygen

Mass of element in the oxide

62.1 g 6.4 g

Relative atomic mass 207.2 16.0Number of moles of atoms that combine

Mole ratio of atoms

mol 0.300

mol 0.300 = 1

mol 0.300

mol 0.400 = 1.33

Simplest whole number ratio of atoms

1 x 3 = 3 1.33 x 3 = 4

∴ the empirical formula of the oxide is Pb3O4.

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

Mass of element in the compound

2.40 g 39.0 g 58.6 g

Relative atomic mass

1.0 32.1 16.0

Number of moles of atoms that combineMole ratio of atoms mol 1.21

mol 2.40 = 1.98

mol 1.21

mol 1.21 = 1

mol 1.21

mol 3.66 = 3.02

∴ the empirical formula of the compound is H2SO3.


1 Method 1

2NaN3(s) 2Na(s) + 3N2(g)

? 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 = (23.0 + 3 x 14.0) g mol–1

= 65.0 g mol–1

Mass of NaN3 required = number of moles of NaN3 x molar mass of NaN3

= 2.00 mol x 65.0 g mol–1

= 130 g

Method 2

2NaN3(s) 2Na(s) + 3N2(g)

? g 84.0 g

Molar mass of NaN3 = (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) 2Na(s) + 3N2(g)



2 x 65.0 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) Li2CO3(s) + H2O(l)

50.0 g ? g

Molar mass of LiOH = (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 =2

09.2mol

= 1.045 mol

Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1

= 44.0 g mol–1

Mass of CO2 absorbed = number of moles of CO2 x molar mass of CO2

= 1.045 mol x 44.0 g mol–1

= 46.0 g

Method 2


50.0 g ? g

Molar mass of LiOH = (6.9 + 16.0 + 1.0) g mol–1

= 23.9 g mol–1


= 44.0 g mol–1

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.


2 x 23.9 g 44.0 g

50.0 g ? g

Mass of CO2 absorbed = 50.0 g x g 23.9 x 2

g 44.0

= 46.0 g

b) Method 1




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 = 2.27 mol

Molar mass of Li2CO3 = (2 x 6.9 + 12.0 + 3 x 16.0) g mol–1

= 73.8 g mol–1

Mass of Li2CO3 produced = number of moles of Li2CO3 x molar mass of Li2CO3

= 2.27 mol x 73.8 g mol–1

= 168 g

Method 2


100.0 g ? g

Molar mass of CO2 = 44.0 g mol–1

Molar mass of Li2CO3 = (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.


44.0 g 73.8 g

100.0 g ? g

Mass of Li2CO3 produced = 100.0 g x g 44.0

g 73.8

= 168 g

3 a) 2Ag2O(s) 4Ag(s) + O2(g)

b) 2Ag2O(s) 4Ag(s) + O2(g)

? 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 =2

0.0604 mol

= 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

Percentage by mass of Ag2O in the sample =g 8.00

g 7.00 x 100%

= 87.5%


1 Hg(l) + Br2(l) HgBr2(s)

21.5 g 15.6 g ? g

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

= 19.6 g

Mass of Hg left = (21.5 – 19.6) g



= 1.9 g

2 a) 6Li(s) + N2(g) 2Li3N(s)

8.28 g 10.6 g ? 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 = 6

2x 1.20 mol

= 0.400 mol

Molar mass of Li3N = (3 x 6.9 + 14.0) g mol–1

= 34.7 g mol–1

Theoretical yield of Li3N = number of moles of Li3N x molar mass of Li3N

= 0.400 mol x 34.7 g mol–1

= 13.9 g

b) Percentage yield of Li3N =g 13.9

g 3.97x 100%

= 28.6%

Unit-end exercises (pages 108-113)

Answers for the HKCEE and HKALE questions are not provided.

1



2Substance Molar mass

of substance (g mol–1)


Number of moles of substance present (mol)

Number of molecules / formula

units presentSulphur 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 Molar mass of substance

(g mol–1)


Number of moles of substance present (mol)

Number of molecules / formula

units presentHydrated sodiumcarbonate

(Na2CO3•9H2O)

268.0 1 340 5.00 3.01 x 1024 formula units

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.

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

Percentage by mass of MnO2 in the nodule =g 15.0

g 3.48x 100%

= 23.2%

14 –

15 a) To prevent the condensed water from running back to the tube and crack the hot glass.



b) Test the liquid with dry cobalt(II) chloride paper. The liquid turns the paper from blue to

pink.

c) To prevent ‘sucking back’ of the liquid.

d) 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.

g 18x) (151.9

g18x

+ =g 30.6

g 13.9

x = 7

16 a) 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 elementin the compound

40.0 g 6.60 g 53.4 g

Relative atomicmass

12.0 1.0 16.0

Number of molesof atoms that combine

Mole ratio of atoms

mol 3.33

mol 3.33 = 1

mol 3.33

mol 6.60 = 2

mol 3.33

mol 3.33 = 1

∴ the empirical formula of glucose is CH2O.

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) Fe2O3(s) + 3H2O(g)

5.35 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 =2

3x 0.0501 mol

= 0.0752 mol

Molar mass of H2O = (2 x 1.0 + 16.0) g mol–1

= 18.0 g mol–1

Mass of H2O formed = number of moles of H2O x molar mass of H2O

= 0.0752 mol x 18.0 g mol–1

= 1.35 g

Method 2


5.35 g ? g

Molar mass of Fe(OH)3 = [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.


2 x 106.8 g 3 x 18.0 g

5.35 g ? g

Mass of H2O formed = 5.35 g x g 106.8 x 2

g 18.0 x 3

= 1.35 g

b) Suppose we have 100 g of the oxide, so there are 72.4 g of iron and 27.6 g of oxygen.Iron Oxygen

Mass of element in the oxide 72.4 g 27.6 gRelative atomic mass 55.8 16.0

Iron OxygenNumber of moles of atoms that combine

Mole ratio of atoms

mol 1.30

mol 1.30 = 1.00

mol 1.30

mol 1.73 = 1.33

Simplest whole number ratio 1 x 3 = 3 1.33 x 3 = 4



of atoms∴ the empirical formula of the oxide is Fe3O4.

18 a) 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)

2ZnO(s) + C(s) 2Zn(s) + CO2(g)

b) 2ZnO(s) + C(s) 2Zn(s) + CO2(g)

48.8 g ? g ? g

Molar mass of ZnO = (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.

∴ number of moles of Zn obtained = 0.600 mol

number of moles of C required =2

0.600 mol

= 0.300 mol

i) Mass of Zn obtained = number of moles of Zn x molar mass of Zn

= 0.600 mol x 65.4 g mol–1

= 39.2 g

ii) 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

19 a) 2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)

b) 2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)

1.61 g 2.58 g

According to the equation, 2 moles of Al react to give 3 moles of Cu.

∴ number of moles of Cu produced =2

3x 0.0596 mol



= 0.0894 mol

Theoretical yield of Cu = number of moles of Cu x molar mass of Cu

= 0.0894 mol x 63.5 g mol–1

= 5.68 g

Percentage yield of Cu =g 5.68

g 2.58 x 100%

= 45.4%

20 4KO2(s) + 2H2O(g) + 4CO2(g) 4KHCO3(s) + 3O2(g)

? g 14.0 g


= 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 = 0.318 mol

a) 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

b) Since the process is only 80% efficient, the mass of KO2 required = 22.6 g ÷ 80%

= 28.3 g

21 MnO2(s) + 4HCl(aq) MnCl2(aq) + Cl2(g) + 2H2O(l)

217 g 274 g ? g

Molar mass of MnO2 = (54.9 + 2 x 16.0) g mol–1

= 86.9 g mol–1



Molar mass of HCl = (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 =4

7.51 mol

= 1.88 mol

Molar mass of Cl2 = 2 x 35.5 g mol–1

= 71.0 g mol–1

Mass of Cl2 produced = number of moles of Cl2 x molar mass of Cl2

= 1.88 mol x 71.0 g mol–1

= 133 g

c) Number of moles of MnO2 used =4

7.51 mol

= 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


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