Chem ch2

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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q1.
a
b

Why was the soup sample in Worked Example 2.1 heated to 110°C?
Why was it necessary to weigh the sample four times?

A1.
a
b

The soup was heated above 100°C to evaporate water from the sample.
By repeatedly heating the sample until the mass remained unchanged, the analyst
could be sure that all the water had been removed.

Q2.
Some laboratories use microwave ovens in place of conventional ovens to dry
samples. What advantage could this have?
A2.
Microwave ovens dry samples more rapidly than conventional ovens.
Q3.
Soy sauce weighing 74.6 g was heated in an oven to constant mass. The final mass
was 14.2 g. What percentage of water did the sauce contain?
A3.
Mass water = 74.6 – 14.2 = 60.4 g
100 60.4 100
mass of water
=
×
= 80.97 = 81.0%
×
%Water =
mass of sample
1
74.6
1
(three significant figures)
Q4.

Three brands of dog food were heated and dried to constant mass. The data recorded
are shown in Table 2.2.
Table 2.2 Determination of water content of dog food
Dog food tested
Phydeaux Deluxe
K9 Budget
Fresh meat – buffalo mince
a
b

Mass of dog food sample (g)
19.8
7.4
15.0

Final mass (g)
3.9
1.9
3.8

Which brand contained the highest percentage of water?
Do you consider that water content is a good guide to the relative value of
different dog foods? What other factors might be important?

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

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Worked solutions to student book questions

Chapter 2 Analysis by mass
A4.
a

b

19.8 − 3.9 100
= 80%
×
19.8
1
7.4 − 1.9 100
K9 Budget
% water =
= 74%
×
7.4
1
15.0 − 3.8 100
Fresh meat – buffalo mince % water =
= 75%
×
15.0
1
Water would not be a good guide to the nutritional value of the dog food. The
amount of protein, carbohydrates, fats, vitamins and minerals need to be
considered.
Phydeaux Deluxe

% water =

Q5.

A student determined the water content of a sample of jam. The following
measurements were obtained:
Mass of evaporating dish:
20.22 g
Mass of jam and evaporating dish
before heating:
30.95 g
after heating:
27.22 g
after more heating:
26.50 g
after more heating:
26.49 g
What was the percentage, by mass, of water in the jam?
A5.
Step 1 Find the mass of the moist jam by subtracting mass of evaporating dish.
Mass of moist jam
= 30.95 – 20.22 g
= 10.73 g
Step 2 Find the mass of water.
Mass of water = 30.95 – 26.49 g
= 4.46 g
Step 3 Find the percentage of water in moist jam.
4.46
% (H2O) =
× 100
10.73
= 41.57%
Step 4 Express the answer with the correct number of significant figures.
% (H2O) = 41.6%
Q6.

Calculate the amount (in mole) of:
a NaCl in 5.85 g of salt
b Fe atoms in 112 g of iron
c CO2 molecules in 2.2 g of carbon dioxide
d Cl– ions in 13.4 g of nickel chloride (NiCl2)
e O2– ions in 159.7 g of iron(III) oxide (Fe2O3)

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

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Worked solutions to student book questions

Chapter 2 Analysis by mass
A6.
a

Step 1

Write the formula to find amount.
mass
n=
molar mass

Step 2

Substitute values and calculate.
M(NaCl)
= 58.442 g mol–1
5.85 g
n(NaCl)
=
58.442 g mol −1

Step 3
b

c

= 0.10009 mol
Express the answer with the correct number of significant figures.
n(NaCl) = 0.100 mol

Step 1

Write the formula to find amount.
mass
n=
molar mass

Step 2

Substitute values and calculate.
M(Fe) = 55.847 g mol–1
112 g
n(Fe) =
55.847 g mol −1
= 2.005479 mol

Step 3

Express the answer with the correct number of significant figures.
n(Fe) = 2.01 mol

Step 1

Write the formula to find amount.
mass
n=
molar mass

Step 2

Substitute values and calculate.
M(CO2)
= 43.991 g mol–1
2.2 g
n(CO2)
=
43.991 g mol −1
= 0.0500 mol

Step 3

Express the answer with the correct number of significant figures.
n(CO2) = 0.050 mol

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
d

Step 1

Write the formula to find amount.
mass
n=
molar mass

Step 2

Substitute values and calculate.
M(NiCl2)
= 129.596 g mol–1
13.4 g
n(NiCl2)
=
129.596 g mol −1
= 0.10339 mol

Step 3

Each mol of NiCl2 contains 2 mol of chloride ions.
n(Cl–) = 0.10339 mol × 2
= 0.20679 mol
Express the answer with the correct number of significant figures.
n(Cl–) = 0.207 mol
= 2.07 × 10–1 mol

Step 4

e

Step 1

Write the formula to find amount.
mass
n=
molar mass

Step 2

Substitute values and calculate.
M(Fe2O3) = 159.69 g mol–1
159.7 g
n(Fe2O3)
=
159.69 g mol −1
= 1.000 mol

Step 3

Each mol of Fe2O3 contains 3 mol of oxide ions.
n(O2–) = 1.000 mol × 3
= 3.000 mol

Step 4

Express the answer with the correct number of significant figures.
n(O2–) = 3.000 mol

Q7.

Calculate the mass of:
a 3.0 mol of oxygen molecules (O2)
b 1.2 mol of aluminium chloride (AlCl3)
c 2.0 mol of nitrogen atoms
A7.
a

Step 1

Write the formula to find mass.
mass = amount × molar mass

Step 2

Substitute values and calculate.
M(O2) = 32 g mol–1
m(O2) = 3.0 mol × 32 g mol–1
= 96 g

Step 3

Express the answer with the correct number of significant figures.
m(O2) = 96 g

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
b

c

Step 1

Write the formula to find mass.
mass = amount × molar mass

Step 2

Substitute values and calculate.
M(AlCl3)
= 133.34 g mol–1
m(AlCl3)
= 1.2 mol × 133.34 g mol–1
= 160.00 g

Step 3

Express the answer with the correct number of significant figures.
m(AlCl3) = 1.6 × 102 g

Step 1

Write the formula to find mass.
mass = amount × molar mass

Step 2

Substitute values and calculate.
M(N) = 14.0 g mol–1
m(N) = 2.0 mol × 14.0 g mol–1
= 28 g

Step 3

Express the answer with the correct number of significant figures.
m(N) = 28 g

Q8.

A small oxygen cylinder carried by an ambulance has an internal volume of 1.42 L.
What mass of oxygen is present at a pressure of 15 000 kPa and temperature of
15.0°C?
A8.
Step 1 Write the formula for the mass of a gas.
pV
= nRT
pV
n
=
RT
pV
m
=
M
RT
pVM
m
=
RT

Step 2 Convert pressure, temperature and volume into the appropriate units for use in
the general gas equation.
V = 1.42 L
P = 15 000 kPa
T = 15.0°C = (15.0 + 273) K = 288 K
Step 3 Calculate the mass of O2, using M(O2) = 32 g mol–1.
15 000 × 1.42 × 32
m(O2) =
8.31 × 288
= 284.797 g
Step 4 Express the answer with the correct number of significant figures.
m(O2) = 285 g

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q9.

Calculate the mass of the following gases:
a 3.5 L of argon at SLC
b 250 mL of ammonia (NH3) at STP
A9.

Step 1

a

Write the formula for the mass of a gas at SLC, where the molar
volume is 24.5 L mol–1.
V
n =
24.5
V
m
=
M 24.5
V ×M
m =
24.5
Step 2
Calculate the m(Ar) where M(Ar) = 39.948 g mol–1.
3.5 L × 39.948 g mo l −1
m(Ar)
=
24.5 L mol −1
= 5.707 g
Step 3
Express the answer with the correct number of significant figures.
m(Ar) = 5.7 g

b

Step 1
n
m
M
m
Step 2

Step 3

Write the formula for the mass of a gas at STP, where the molar
volume is 22.4 L mol–1.
V
=
22.4
V
=
22.4
V ×M
=
22.4
Calculate the m(NH3) where M(NH3) = 17.0 g mol–1.
0.250 L × 17.0 g mo l −1
m(NH3)
=
22.4 L mol −1
= 0.1897 g
Express the answer with the correct number of significant figures.
m(NH3) = 0.190 g

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q10.

Determine the percentage composition of the following compounds:
a lead(IV) oxide (PbO2)
b sodium carbonate (Na2CO3)
A10.
a

Step 1

Calculate the molar mass of lead(IV) oxide.
M(PbO2)
= 207.2 g mol–1 + (16.0 g mol–1 × 2)
= 239.2 g mol–1

Step 2

Calculate the mass of lead in 1 mol (239.2 g) of PbO2.
m(Pb) = 207.2 g

Step 3

Calculate the % of lead in PbO2.
m(Pb)
% Pb =
× 100
M (PbO 2 )
207.2 g
=
× 100
239.2 g
= 86.6220%

Step 4

Express the answer with the correct number of significant figures.
% Pb = 86.6%

Step 5

Calculate the % of oxygen in PbO2.
% O = 100 – % Pb
= 100 – 86.6
= 13.4%

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
b

Step 1

Calculate the molar mass of sodium carbonate.
M(Na2CO3) = (22.9898 g mol–1 × 2) + 12.01115 g mol–1 +
(16.0 g mol–1 × 3)
= 105.99 g mol–1

Step 2

Calculate the mass of sodium in 1 mol (105.99 g) of Na2CO3.
m(Na) = (22.9898 × 2) g
= 45.9796 g

Step 3

Calculate the % of sodium in Na2CO3.
m( Na)
% Na =
× 100
M (Na 2 CO 3 )
45.9796 g
=
× 100
105.99 g
= 43.381%

Step 4

Express the answer with the correct number of significant figures.
% Na = 43.4%

Step 5

Calculate the mass of carbon in 1 mol (105.99 g) of Na2CO3.
m(C) = 12.01115 g

Step 6

Calculate the % carbon in Na2CO3.
m(C)
%C=
× 100
M (Na 2 CO 3 )
12.01115 g
× 100
=
105.99 g
= 11.332 %

Step 7

Express the answer with the correct number of significant figures.
% C = 11.3%

Step 8

Calculate the % oxygen in Na2CO3.
% O = 100 – (% Na + % C)
= 100 – (43.4 + 11.3)
= 45.3%

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q11.

A gaseous hydrocarbon that is used as a fuel for high-temperature cutting and welding
of metals contains 92.3% carbon.
a Determine its empirical formula.
b If the molar mass of the hydrocarbon is 26 g mol–1, find its molecular formula.
A11.
a

b

Step 1

As this is a hydrocarbon it will contain carbon and hydrogen.
Calculate the mass of hydrogen in a 100 g sample.
m(H) = 100 – 92.3 g
= 7.7 g

Step 2

Write the ratio by mass.
C:H
92.3 g : 7.7 g

Step 3

Calculate the ratio by amount (in moles).
92.3 g 7.7 g
:
M (C) M (H)
92.3 g
7.7 g
:
7
−1
12.01115 g mol
1.00797 g mol −1
7.6845 mol : 7.639 mol

Step 4

Divide by the smaller amount.
7.6845 7.639
:
7.639 7.639
1.0059 : 1

Step 5

Round off to whole numbers.
1:1
Therefore, the empirical formula of the compound is CH.

As the empirical formula is CH, the molecule must contain a whole number of
CH units.
The molar mass of one of these units is (12.0 + 1.01) = 13.01 g mol–1.
molar mass of the compound
The number of units in a molecule =
molar mass of one unit
26 g mol −1
=
13.10 g mol −1
=2
∴The molecular formula of the compound is C2H2. (This is ethyne, commonly
called acetylene.)

Heinemann Chemistry 2 4th edition Enhanced
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Chapter 2 Analysis by mass
Q12.

When 1.66 g of tungsten (W) is heated in excess chlorine gas, 3.58 g of tungsten
chloride is produced. Find the empirical formula of tungsten chloride.
A12.

Step 1 Calculate the mass of chlorine in this sample of tungsten chloride.
m(Cl) = 3.58 – 1.66 g
= 1.92 g
Step 2 Write the ratio by mass.
W
:
Cl
1.66 g :
1.92 g
Step 3 Calculate the ratio by amount (in moles).
1.66 g
1.92 g
:
M (W)
M (Cl)
1.66 g
1.92 g
:
−1
183.85 g mol
35.45 g mol −1
0.00903 mol
:
0.0542 mol
Step 4 Divide by the smaller amount.
0.00903 0.0542
:
0.00903 0.00903
Step 5 Round off to whole numbers
1 : 5.998
1:6
∴The empirical formula of the compound is WCl6.

Heinemann Chemistry 2 4th edition Enhanced
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Chapter 2 Analysis by mass
Q13.

A sample of blue copper(II) sulfate crystals weighing 2.55 g is heated and
decomposes to produce 1.63 g of anhydrous copper(II) sulfate. Show that the formula
of the blue crystals is CuSO4.5H2O.
A13.

Step 1 Calculate the amount of anhydrous copper sulfate.
m(CuSO 4 )
n(CuSO4) =
M (CuSO 4 )
1.63
=
159.5
= 0.0102 mol
Step 2 Calculate the amount of water in hydrated copper sulfate.
2.55 − 1.63
n(H2O) =
18.0
= 0.0511 mol
Step 3 Calculate the ratio of amount of anhydrous copper sulfate to amount of water.
n(CuSO 4 )
0.0102
=
n(H 2 O)
0.0511
=5
∴ The formula of the crystals is CuSO4.5H2O.

Heinemann Chemistry 2 4th edition Enhanced
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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q14.

Magnesium reacts with hydrochloric acid according to the equation:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
If 10.0 g of magnesium reacts completely, calculate:
a the mass of magnesium chloride that forms
b the mass of hydrogen that forms.
A14.
a

Step 1

Write a balanced equation.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step 2

Calculate the amount of magnesium consumed.
m(Mg)
n(Mg) =
M (Mg)
10 g
=
24.3 g mol −1
= 0.4115 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of
MgCl2 produced.
From the equation, 1 mol of MgCl2 is produced by 1 mol of Mg.
n(MgCl 2 )
1
=
n(Mg)
1
n(MgCl2)
= n(Mg)
= 0.4115 mol

Step 4

Calculate the mass of magnesium chloride produced.
M(MgCl2)
= 95.211 g mol–1
m(MgCl2)
= n(MgCl2) × M(MgCl2)
= 0.4115 mol × 95.211 g mol–1
= 39.179 g

Step 5

Express the answer with the correct number of significant figures.
m(MgCl2) = 39.2 g

Heinemann Chemistry 2 4th edition Enhanced
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Chapter 2 Analysis by mass
b

Step 1

Write a balanced equation.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step 2

Calculate the amount of magnesium consumed.
m(Mg)
n(Mg) =
M (Mg)
10 g
=
24.3 g mol −1
= 0.4115 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of
hydrogen gas produced.
From the equation, 1 mol of H2(g) is produced by 1 mol Mg.
n(H 2 )
1
=
n(Mg) 1
n(H2) = n(Mg)
= 0.4115 mol

Step 4

Calculate the mass of hydrogen produced.
M(H2) = 2.0016 g mol–1
m(H2) = n(H2) × M(H2)
= 0.4115 mol × 2.0016 g mol–1
= 0.08236584 g

Step 5

Express the answer with the correct number of significant figures.
m(H2) = 0.0824 g

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q15.

Iron metal is extracted in a blast furnace by a reaction between iron(III) oxide and
carbon monoxide:
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)
To produce 1000 kg of iron, calculate:
a the mass of iron(III) oxide required
b the volume of carbon dioxide produced at SLC (25°C and 101.3 kPa)
A15.
a

Step 1

Write a balanced equation.
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

Step 2

Calculate the amount of iron produced.
m(Fe)
n(Fe) =
M (Fe)
(1000 kg × 1000) g
=
55.8 g mol −1
= 17 921.15 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of
iron(III) oxide required.
From the equation, 1 mol Fe2O3 produces 2 mol Fe.
n(Fe 2 O 3 )
1
=
2
n(Fe)
1
= × n(Fe)
n(Fe2O3)
2
1
= × 17 921.15 mol
2
= 8960.57 mol

Step 4

Calculate the mass of iron(III) oxide required.
M(Fe2O3)
= 159.694 g mol–1
m(Fe2O3)
= n(Fe2O3) × M(Fe2O3)
= 8960.57 mol × 159.694 g mol–1
= 1 430 949.8 g

Step 5

Convert to kg.
1 430 949.8 g
mass in kg =
1000
= 1430.9 kg

Step 6

Express the answer with the correct number of significant figures.
m(Fe2O3) = 1430 kg

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Worked solutions to student book questions

Chapter 2 Analysis by mass
b

Step 1

Write a balanced equation.
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

Step 2

Calculate the amount of iron produced.
m(Me)
n(Fe) =
M (Fe)
(1000 kg × 1000) g
=
55.8 g mol −1
= 17 921.15 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of
carbon dioxide produced.
From the equation, 3 mol CO2 and 2 mol Fe are produced together.
n(CO 2 )
3
=
n(Fe)
2
3
n(CO2)
= × n(Fe)
2
3
= × 17 921.15 mol
2
= 26 881.7 mol
Calculate the volume of carbon dioxide given that the molar volume of
a gas at SLC is 24.5 Lmol–1.
V
n(CO2)
=
Vm
V
26 881.7 =
24.5
V = 26 881.7 × 24.5
= 658 602.15 L
= 659 000 L (three significant figures)

Step 4

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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q16.

A solution containing 10.0 g of silver nitrate is mixed with a solution containing
10.0 g of barium chloride. What mass of silver chloride precipitate is likely to be
produced?
2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)
A16.

Step 1

Write a balanced equation.
2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)

Step 2

To determine which reactant is in excess, calculate amount of each reactant
divided by their respective coefficient. The smallest amount is the limiting
reactant and the one from which to calculate the amount of product formed.
The other is the excess reactant.
Note: These calculations can only be used to determine the excess reactant.
Continue the calculation, using original data.
n(AgNO 3 )
10.0 g/169.88 g mol −1
=
coefficient(AgNO 3 )
2
= 0.02943 mol
n(BaCl 2 )
10.0 g/ 208.24 g mol −1
=
coefficient(BaCl 2 )
1
= 0.0480 mol
Hence AgNO3 is the limiting reactant.

Step 3

From the equation, 1 mol AgCl is produced from 1 mol AgNO3.
1
n(AgCl)
=
n(AgNO 3 ) 1
n(AgCl)
= n(AgNO3)
10.0 g
=
169.88 g mol −1
= 0.05886 mol

Step 4

Calculate the mass.
m(AgCl) = 0.05887 mol × 143 032 g mol–1
= 8.4372 g

Step 5

Express the answer with the correct number of significant figures.
m(AgCl) = 8.44 g

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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q17.

A chemist determined the salt content of a sausage roll by precipitating chloride ions
as silver chloride. If an 8.45 g sample of sausage roll yielded 0.636 g of precipitate,
calculate the percentage of salt in the food. Assume that all the chloride is present as
sodium chloride.
A17.

Step 1
Step 2

Step 3

Step 4
Step 5

Step 6

Write a balanced equation.
Ag+(aq) + Cl–(aq) → AgCl(s)
Calculate amount of AgCl.
0.636 g
n(AgCl)
=
143.32 g mol −1
= 0.004438 mol
From the equation, 1 mol of NaCl produces 1 mol of AgCl.
1
n(NaCl)
=
n(AgCl)
1
n(NaCl)
= 0.004438 mol
Calculate the mass of salt.
m(NaCl) = 0.004438 × 58.44
= 0.2594 g
Convert to percentage.
0.2594
% NaCl
=
× 100
8.45
= 3.0698%
Express the answer with the correct number of significant figures.
% NaCl = 3.07%

Heinemann Chemistry 2 4th edition Enhanced
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Chapter 2 Analysis by mass
Q18.

An impure sample of iron(II) sulfate, weighing 1.545 g, was treated to produce a
precipitate of Fe2O3. If the mass of the dried precipitate was 0.315 g, calculate the
percentage of iron in the sample.
A18.

Step 1

Write an equation that is balanced for the appropriate element, Fe.
2FeSO4(s) + other reactants → Fe2O3(s) + other products

Step 2

Calculate amount of Fe2O3 precipitated.
0.315 g
n(Fe2O3) =
159.694 g mol −1
= 0.001973 mol

Step 3

From the equation, 2 mol of FeSO4 is precipitated as 1 mol of Fe2O3.
n(FeSO 4 )
2
=
n(Fe 2 O 3 )
1
n(FeSO4) = 2 × 0.001973 mol
= 0.003945 mol

Step 4

From the formula, 1 mol of Fe is present in 1 mol of FeSO4.
1
n(Fe)
=
n(FeSO 4 )
1
n(Fe)
= 0.003945 mol

Step 5

Calculate the mass of Fe in the sample.
m(Fe)
= 0.003945 mol × 55.847 g mol–1
= 0.2203 g

Step 6

Convert to percentage.
0.2203
% Fe
=
× 100
1.545
= 14.260%

Step 7

Express the answer with the correct number of significant figures.
% Fe = 14.3%

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Chapter 2 Analysis by mass
Chapter review
Q19.

Find the number of mol of:
a Ca atoms in 60.0 g of calcium
b NH3 molecules in 22 g of ammonia
c H2O molecules in 20.0 g of CuSO4.5H2O
d Cl– ions in 34 g of FeCl3
A19.
a

b

c

Step 1

Write formula for amount in moles.
m
n=
M

Step 2

Calculate the amount of Ca.
60.0 g
n(Ca) =
40.08 g mol −1
= 1.4970 mol

Step 3

Express the answer with correct number of significant figures.
n(Ca) = 1.50 mol

Step 1

Calculate the amount, using appropriate formula.
22 g
n(NH3) =
17 g mol −1
= 1.29 mol

Step 2

Express the answer with the correct number of significant figures.
n(NH3) = 1.3 mol

Step 1

Find molar mass of CuSO4.5H2O.
M(CuSO4.5H2O)
= 63.54 + 32.06 + (4 × 16.00) +
5((2 × 1.008) + 16.00)
= 249.68 g mol–1

Step 2

Calculate the amount of CuSO4.5H2O.
20.0 g
=
n(CuSO4.5H2O)
249.68 g mol −1
= 0.0801 mol

Step 3

Since 1 mol of CuSO4.5H2O contains 5 mol of water molecules, find
the amount of H2O.
n(H2O)
= 5 × 0.0801
= 0.4005 mol

Step 4

Express answer with correct number of significant figures.
n(H2O) = 0.401 mol

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Worked solutions to student book questions

Chapter 2 Analysis by mass
d

Step 1

Calculate the amount of FeCl3.
34 g
=
n(FeCl3)
162.297 g mol −1
= 0.2095 mol

Step 2

As each mol of FeCl3 contains 3 mol Cl– ions, calculate the amount of
Cl– ions.
n(Cl–) = 3 × 0.2095 mol
= 0.628 mol

Step 3

Express the answer with the correct number of significant figures.
n(Cl–) = 0.63 mol

Q20.

Find the mass of:
a 0.30 mol of zinc atoms
b 0.16 mol of iron(III) oxide (Fe2O3)
c 1.5 mol of ammonium phosphate ((NH4)3PO4)
A20.
a

b

Step 1

Use the formula for finding mass of substance.
m=n×M

Step 2

Calculate the mass of Zn.
m(Zn) = 0.30 mol × 65.38 g mol–1
= 19.61 g

Step 3

Express the answer with the correct number of significant figures.
m(Zn) = 20 g

Step 1

Calculate the mass, using the appropriate formula.
m(Fe2O3)
= 0.16 mol × 159.70 g mol–1
= 25.55 g
Express the answer with the correct number of significant figures.
m(Fe2O3) = 26 g

Step 2
c

Step 1

Calculate the mass of (NH4)3PO4.
m((NH4)3PO4) = 1.5 mol × 149.096 g mol–1
= 223.644 g

Step 2

Express the answer with the correct number of significant figures.
m((NH4)3PO4) = 220 g
= 2.2 × 102 g

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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q21.

A brand of toothpaste contains 0.22% by mass sodium fluoride (NaF). Calculate the
mass of fluoride ions in a tube containing 120 g of the paste.
A21.

Step 1 Calculate the mass of NaF in the toothpaste tube.
0.22% by mass means 0.22 g NaF per 100 g toothpaste
So in 120 g of toothpaste:
120
m(NaF) = 0.22 ×
100
= 0.264 g
Step 2 Write formula for amount in moles.
m
n=
M
Step 3 Calculate molar mass.
M(NaF) = 41.9 g mol–1
Step 4 Calculate the amount of NaF.
0.264 g
n(NaF) =
41.9 g mol −1
= 0.006287 mol
Step 5 As 1 mol NaF contains 1 mol of F– ions, calculate the amount of F–.
n(F–) = 0.006287 mol
Step 6 Calculate the mass of F– ions.
m(F–) = 0.006287 mol × 19.00 g mol–1
= 0.1195 g
Step 7 Express the answer with the correct number of significant figures.
m(F–) = 0.12 g

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Chapter 2 Analysis by mass
Q22.

6.00 g of helium gas was blown into a fairground balloon. On the day, the temperature
was 28.0°C and the pressure inside the balloon was 103.4 kPa. Assuming it is
infinitely elastic, to what volume did the balloon inflate?
A22.

Step 1 Write the formula for the volume of a gas.
pV = nRT
nRT
V =
p
Step 2 Convert pressure and temperature into the appropriate units for use in the
general gas equation.
P = 103.4 kPa
T = 28.0°C = (28.0 + 273) K = 301 K
Step 3 Calculate the amount of He, using M(He) = 4.0 g mol–1.
6.00 g
n(He) =
4.0 g mol −1
= 1.5 mol
Step 4 Calculate the volume of He.
1.5 × 8.31 × 301
V(He) =
103.4
= 36.286 L
Step 5 Express the answer with the correct number of significant figures.
V(He) = 36.3 L

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Chapter 2 Analysis by mass
Q23.

Calculate the volume of the following gases:
a 1.50 mol of oxygen at STP
b 28.0 g of nitrogen at STP
c 17 g of sulfur dioxide at SLC
d 1.2 × 1022 atoms of helium at SLC
A23.
a

b

Step 1

Write the formula for the volume of a gas at STP, where the molar
volume is 22.4 L mol–1.
V
n=
22.4
V = n × 22.4

Step 2

Calculate the V(O2).
V(O2) = 1.50 mol × 22.4 L mol–1
= 33.6 L

Step 3

Express the answer with the correct number of significant figures.
V(O2) = 33.6 L

Step 1

Write the formula for the volume of a gas at STP, where the molar
volume is 22.4 L mol–1.
V
n
=
22.4
V
m
=
M
22.4
m × 22.4
V
=
M

Step 2

Calculate the V(N2) where M(N2) = 28 g mol–1.
28.0 g × 22.4 L mo l −1
V(N2) =
28 g mol −1
= 22.4 L

Step 3

Express the answer with the correct number of significant figures.
V(N2) = 22.4 L

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Worked solutions to student book questions

Chapter 2 Analysis by mass
c

d

Step 1

Write the formula for the volume of a gas at SLC, where the molar
volume is 24.5 L mol–1.
V
n
=
24.5
V
m
=
M
24.5
m × 24.5
V
=
M

Step 2

Calculate the V(SO2) where M(SO2) = 64 g mol–1.
17.0 g × 24.5 L mo l −1
V(SO2) =
64 g mol −1
= 6.508 L

Step 3

Express the answer with the correct number of significant figures.
V(SO2) = 6.5 L

Step 1

Write the formula for the volume of a gas at SLC, where the molar
volume is 24.5 L mol–1.
V
n=
24.5
number of particles
V
=
NA
24.5
24.5
V = number of particles ×
NA

Step 2

Calculate the V(He).
1.2 × 10 22 atoms × 24.5 L mo l −1
V(He) =
6.023 × 10 23
= 0.488 L

Step 3

Express the answer with the correct number of significant figures.
V(He) = 0.49 L

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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q24.

A 1.22 g sample of pure gas extracted from the gases from a car exhaust occupied
991 mL at 24.0°C and 1.00 atmosphere pressure.
a Calculate the amount of gas, in mol, present in the sample.
b What is the molar mass of the gas?
c Suggest the identity of the gas.
A24.
a

Step 1

Write the formula for the amount of a gas.
PV = nRT
PV
n =
RT

Step 2

Convert pressure, temperature and volume into the appropriate units
for use in the general gas equation.
P = 1.00 atm = (1.00 × 101.325) kPa
= 101.325 kPa
V = 991 mL = 0.991 L
T = 24.0°C = (24.0 + 273) K = 297 K
Calculate the amount of gas.
101.325 × 0.991
n =
8.31 × 297
= 0.0407 mol
Express the answer with the correct number of significant figures.
n = 0.0407 mol

Step 3

Step 4
b

c

Step 1

Write the formula for the molar mass of the gas.
m
M=
n

Step 2

Calculate the M of the gas.
1.22 g
M =
0.0407 mol
= 29.987 g mol–1

Step 3

Express the answer with the correct number of significant figures.
M = 30.0 g mol–1

Using the molar mass, the gas is NO.

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Chapter 2 Analysis by mass
Q25.

Solutions of silver nitrate and potassium chromate react to produce a red precipitate of
silver chromate:
2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)
If 0.778 g of precipitate is formed in a reaction, find:
a the mass of potassium chromate that reacted
b the mass of silver nitrate that reacted.
A25.
a

Step 1

Write a balanced equation.
2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)

Step 2

Calculate amount of precipitate, Ag2CrO4.
0.778 g
n(Ag2CrO4) =
331.74 g mol −1
= 0.002345 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of
K2CrO4 required.
From the equation, 1 mol K2CrO4 produces 1 mol Ag2CrO4.
n(K 2 CrO 4 )
1
=
n(Ag 2 CrO 4 )
1
= 0.002345 mol
n(K2CrO4)

Step 4

Calculate the mass of K2CrO4.
m(K2CrO4) = 0.002345 mol × 194.204 g mol–1
= 0.4554 g

Step 5

Express the answer with the correct number of significant figures.
m(K2CrO4) = 0.455 g

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Chapter 2 Analysis by mass
b

Step 1

Write a balanced equation.
2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)

Step 2

Calculate amount of precipitate, Ag2CrO4.
n(Ag2CrO4) = m × M
0.778 g
=
331.74 g mol −1
= 0.002345 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of
AgNO3 required.
From the equation, 2 mol AgNO3 produces 1 mol Ag2CrO4.
n(AgNO 3 )
2
=
n(Ag 2 CrO 4 )
1
n(AgNO3)
= 2 × 0.002345 mol
= 0.00469 mol

Step 4

Calculate the mass.
m(AgNO3)
= 0.00469 mol × 169.88 g mol–1
= 0.7967 g

Step 5

Express the answer with the correct number of significant figures.
m(AgNO3) = 0.797 g

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Chapter 2 Analysis by mass
Q26.

Magnesium in distress flares burns in air according to the equation:
2Mg(s) + O2(g) → 2MgO(s)
If 10.0 g of magnesium burns in air, calculate:
a the mass of magnesium oxide produced
b the mass of oxygen that reacts.
A26.
a

Step 1

Write a balanced equation
2Mg(s) + O2(g) → 2MgO(s)

Step 2

Calculate the amount of Mg.
10.0 g
n(Mg) =
24.31 g mol −1
= 0.4114 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of MgO
produced.
From the equation, 1 mol MgO is produced by 1 mol Mg.
n(MgO) 1
=
n(Mg )
1
n(MgO) = 0.4114 mol

Step 4

Convert to mass.
m(MgO) = n × M
= 0.4114 mol × 40.31 g mol–1
= 16.58 g

Step 5

Express answer with correct number of significant figures.
m(MgO) = 16.6 g

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Chapter 2 Analysis by mass
b

Step 1

Write a balanced equation.
2Mg(s) + O2(g) → 2MgO(s)

Step 2

Calculate the amount of Mg.
10.0 g
n(Mg) =
24.31 g mol −1
= 0.4114 mol

Step 3

Use the ratio of amounts of substances to calculate the amount of O2
used.
From the equation, 1 mol O2 reacts with 2 mol Mg.
n(O 2 )
1
=
n(Mg )
2
1 × 0.4114
=
mol
n(O2)
2
= 0.2057 mol

Step 4

Convert to mass.
m(O2) = 0.2057 mol × 32.0 g mol–1
= 6.5824 g

Step 5

Express answer with correct number of significant figures.
m(O2) = 6.58 g

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Chapter 2 Analysis by mass
Q27.

Lithium peroxide may be used as a portable oxygen source for astronauts. Calculate
the volume of oxygen gas, measured at 25°C and pressure of 101.3 kPa, that is
available from the reaction of 0.500 kg of lithium peroxide with carbon dioxide
according to the equation:
2Li2O2(s) + 2CO2(g) → 2Li2CO3(s) + O2(g)
A27.

Step 1 Write a balanced equation.
2Li2O2(s) + 2CO2(g) → 2Li2CO3(s) + O2(g)
Step 2 Calculate the amount of Li2O2.
0.500 × 10 3 g
n(Li2O2) =
45.878 g mol −1
= 10.898 mol
Step 3 From the equation, 1 mol O2 is produced by 2 mol Li2O2.
n(O 2 )
1
=
n(Li 2 O 2 )
2
1 × 10.898
n(O2)
=
mol
2
= 5.449 mol
Step 4 Convert pressure and temperature to the appropriate units for use in the
general gas equation.
P = 101.3 kPa
T = 25°C = (25 + 273) K = 298 K
Step 5 Calculate the volume at the conditions given, using the general gas equation.
pV
= nRT
nRT
V
=
p
5.449 × 8.31 × 298
V(O2) =
101.3
= 133.2 L
Step 6 Express answer with correct number of significant figures.
V(O2) = 133 L

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Chapter 2 Analysis by mass
Q28.

A power station burns coal at 45.5 tonnes per hour (1 tonne = 106 g). Assuming the
coal is pure carbon and that all the coal is oxidised completely to carbon dioxide gas
on combustion, what volume of carbon dioxide is released to the atmosphere per hour
when the atmospheric pressure is 758 mmHg and the temperature is 19.0°C?
A28.

Step 1 Write a balanced equation.
C(s) + O2(g) → CO2(g)
Step 2 Calculate the amount of C.
45.5 × 10 6 g
n(C) =
12 g mol −1
= 3.792 × 106 mol
Step 3 From the equation, 1 mol CO2 is produced from 1 mol C.
n(CO 2 )
1
=
n(C)
1
n(CO2)
= 3.792 × 106 mol
Step 4 Convert pressure and temperature to the appropriate units for use in the
general gas equation.
758
× 101.325 kPa
P
=
760
= 101.058 kPa
T = 19°C = (19 + 273) K = 292 K
Step 5 Calculate the volume at the conditions given, using the general gas equation.
pV
= nRT
nRT
V
=
p
3.792 × 10 6 × 8.31 × 292
101.058
= 9.105 × 107 L

V(CO2)=

Step 6 Express answer with correct number of significant figures.
V(CO2) = 9.10 × 107 L

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Chapter 2 Analysis by mass
Q29.

Silicon steel is an alloy of the elements iron, carbon and silicon. An alloy sample was
reacted with excess hydrochloric acid and the following reaction occurred:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
The carbon and silicon in the alloy did not react with the acid. If an alloy sample with
a mass of 0.160 g produced 62.0 mL of hydrogen gas, measured at SLC, calculate the:
a amount of hydrogen evolved in the reaction
b mass of iron that reacted to produce this amount of hydrogen
c percentage of iron in the alloy.
A29.
a

b

c

Step 1

Write a balanced equation.
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

Step 2

Calculate the amount of H2(g) given the volume at SLC.
0.062 L
n(H2) =
24.5 L mol −1
= 2.5306 × 10–3 mol

Step 3

Express the answer with the correct number of significant figures.
n(H2) = 2.53 × 10–3 mol

Step 1

Write a balanced equation.
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

Step 2

From the equation, 1 mol Fe produces 1 mol H2.
n(Fe) 1
=
n(H 2 ) 1
n(Fe) = n(H2) from part a
= 2.53 × 10–3 mol

Step 3

Calculate the mass.
m(Fe) = 2.53 × 10–3 mol × 55.8 g mol–1
= 0.14117 g

Step 4

Express the answer with the correct number of significant figures.
m(Fe) = 0.141 g

Step 1

Calculate the percentage of Fe in the alloy.
m(Fe)
× 100
% (Fe) =
m(alloy)
0.141
× 100
=
0.160
= 88.12%

Step 2

Express the answer with the correct number of significant figures.
% (Fe) = 88.1%

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Chapter 2 Analysis by mass
Q30.

Explain the meaning of the following terms:
a relative atomic mass
b relative molecular mass
c mole
d Avogadro’s number
e molar mass
f precipitate
g gravimetric analysis
h ionic equation
A30.
a
b
c
d
e
f
g
h

Relative atomic mass—the weighted mean of the relative masses of the isotopes
of an element on the 12C scale.
Relative molecular mass—the relative mass of a molecule on the 12C scale.
Mole—the amount of substance which contains the same number of specified
particles as there are atoms in exactly 12 g of 12C.
Avogadro’s number—the number of carbon atoms in exactly 12 g of 12C
(approximately 6.02 × 1023).
Molar mass—mass in grams of a mole of a substance.
Precipitate—solid which is formed during a chemical reaction that occurs in
solution.
Gravimetric analysis—analysis of a sample using measurement of mass. One step
in the procedure usually involves the formation of a precipitate.
Ionic equation—an equation for a reaction in which the ions that remain in
solution during the reaction (spectator ions) are omitted.

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Chapter 2 Analysis by mass
Q31.

16.0 g of hydrogen sulfide is mixed with 20.0 g of sulfur dioxide and they react
according to the equation:
2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)
a What mass of sulfur is produced?
b What mass of reactant is left after the reaction?
A31.
a

Step 1

Write a balanced equation.
2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)

Step 2

To determine which reactant is in excess, calculate amount of each
reactant divided by their respective coefficient. The smallest amount is
the limiting reactant and the one from which to calculate the amount of
product formed. The other is the excess reactant.
Note: These calculations can only be used to determine the excess
reactant. Continue the calculation, using original data.
n(H 2 S)
16.0 g/ 34.0 g mol −1
=
coefficient(H 2 S)
2
= 0.23529 mol
n(SO 2 )
20.0 g/ 64.0 g mol −1
=
coefficient(SO 2 )
1
= 0.3125 mol
Hence H2S is the limiting reactant.

Step 3

From the equation, 3 mol of S is produced by 2 mol of H2S.
n(S)
3
=
n(H 2 S)
2
3 × n(H 2 S)
n(S)
=
2
3 × (16.0 g/34.0 g mol −1 )
=
2
= 0.70588 mol

Step 4

Calculate the mass.
m(S) = 0.70588 mol × 32.0 g mol–1
= 22.588 g

Step 5

Express the answer with the correct number of significant figures.
m(S) = 22.6 g

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Chapter 2 Analysis by mass
b

Step 1

As SO2 was in excess, calculate the amount of SO2 that reacted with all
the H2S, using the mol ratio from the equation.
n(SO 2 )
1
=
n(H 2 S)
2
n(SO2)reacted

16.0 g/ 34.0 g mol −1
2
= 0.23529 mol

=

Step 2

Calculate the amount of SO2 added initially.
20.0 g
n(SO2)initially =
64.0 g mol −1
= 0.3125 mol

Step 3

Calculate the amount of SO2 in excess by subtracting the amount of
SO2 that reacted from the amount of SO2 added initially.
n(SO2)excess = 0.3125 mol – 0.23529 mol
= 0.07721 mol

Step 4

Calculate the mass of SO2 that was in excess.
m(SO2) = 0.07721 g × 64.0 g mol–1
= 4.9412 g

Step 5

Express the answer with the correct number of significant figures.
m(SO2) = 4.94 g

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Chapter 2 Analysis by mass
Q32.

Calculate the volume of carbon dioxide gas produced, at SLC, when 5.00 g of calcium
carbonate is added to a solution containing 5.00 g of nitric acid.
A32.

Step 1 Write a balanced equation.
CaCO3(s) + 2HNO3(aq) → CO2(g) + H2O(l) + Ca(NO3)2(aq)
Step 2 To determine which reactant is in excess, calculate amount of each reactant
divided by their respective coefficient. The smallest amount is the limiting
reactant and the one from which to calculate the amount of product formed.
The other is the excess reactant.
Note: These calculations can only be used to determine the excess reactant.
Continue the calculation, using original data.
5.00 g
n(CaCO 3 )
100.08 g mol −1
=
coefficient (CaCO 3 )
1
= 0.04996 mol
5.00 g
n(HNO3 )
63 g mol −1
=
coefficient (HNO3 )
2
= 0.03968 mol
Hence HNO3 is the limiting reactant.
Step 3 From the equation, 1 mol of CO2 is produced by 2 mol of HNO3.
n(CO 2 )
1
=
n(HNO3 )
2
n(CO2)

1 × (5.00 g 63 g mol −1 )
2
= 0.03968 mol

=

Step 4 Calculate the volume of CO2 at SLC when the molar volume is 24.5 L mol–1.
V (CO 2 )
=
n(CO2)
24.5 L mol −1
V(CO2)
= n(CO2) × 24.5 L mol–1
= 0.03968 mol × 24.5 L mol–1
= 0.9722 L
Step 5 Express the answer with the correct number of significant figures.
V(CO2) = 0.972 L

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Chapter 2 Analysis by mass
Q33.

The following compounds are used in fertilisers as a source of nitrogen. Calculate the
percentage of nitrogen, by mass, in:
a ammonia (NH3)
b ammonium nitrate (NH4NO3)
c urea (CO(NH2)2).
A33.
a

b

Step 1

Calculate the molar mass of NH3.
M(NH3) = 17.0 g mol–1

Step 2

From the formula, 1 mol of N is present in 1 mol of NH3.
Calculate the mass of N in 1 mol (17.0 g) of NH3.
m(N) = 14.0 g

Step 3

Calculate the % of nitrogen in NH3.
m( N )
%N =
× 100
M ( NH) 3
14.0
× 100
=
17.0
= 82.35%

Step 4

Express the answer with the correct number of significant figures.
% N = 82.4%

Step 1

Calculate the mass of 1 mol of NH4NO3.
M(NH4NO3) = 80.2 g mol–1

Step 2

From the formula, 2 mol of N are present in 1 mol of NH4NO3.
Calculate the mass of N in 1 mol (80.2 g) of NH4NO3.
m(N) = 2 × 14.0 g
= 28.0 g

Step 3

Calculate the % of nitrogen in NH4NO3.
28.0
× 100
%N =
80.2
= 34.91%

Step 4

Express the answer with the correct number of significant figures.
% N = 35.0%

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Chapter 2 Analysis by mass
c

Step 1

Calculate the molar mass of CO(NH2)2.
M(CO(NH2)2) = 60.0 g mol–1

Step 2

From the formula, 2 mol of N are present in 1 mol of CO(NH2)2.
Calculate the mass of nitrogen in 1 mol (60.0 g) of CO(NH2)2.
m(N) = 2 × 14.0 g
= 28.0 g

Step 3

Calculate the % of nitrogen in CO(NH2)2.
28.0
× 100
%N =
60.0
= 46.67 %

Step 4

Express the answer with the correct number of significant figures.
% N = 46.7 %

Q34.

Find the empirical formula of:
a a compound that contains 65.2% scandium and 34.8% oxygen by mass
b an oxide of copper that contains 89% copper by mass
c a polymer used to make plastic drain pipes, which contains 38.4% carbon, 4.84%
hydrogen and 56.7% chlorine by mass.
A34.
a

Step 1

Write the ratio by mass.
Sc : O
65.2 g : 34.8 g

Step 2

Calculate the ratio by amount (in moles).
65.2 g
34.8 g
:
M (Sc)
M (O)
65.2 g
34.8 g
:
−1
44.956 g mol
16.0 g mol −1
1.450 mol
:
2.175 mol

Step 3

Divide by the smaller amount.
1.450
2.175
:
1.450
1.450
1
:
1.5

Step 5

Express as integers by multiplying by 2.
2:3
∴ The empirical formula of the compound is Sc2O3.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

38

Worked solutions to student book questions

Chapter 2 Analysis by mass
b

c

Step 1

As this is an oxide of copper, calculate the percentage oxygen in the
compound.
% O = (100 – 89)%
= 11%

Step 2

Write the ratio by mass.
Cu
:
O
89 g :
11 g

Step 3

Calculate the ratio by amount (in moles).
89 g
11 g
:
M (Cu)
M (O)
89 g
11g
:
−1
63.54 g mol
16.0 g mol −1
1.401 mol
:
0.6875 mol

Step 4

Divide by the smaller amount.
1.401
0.6875
:
0.6875
0.6875
2.038
:
1

Step 5

Round off to whole numbers.
2:1
∴ The empirical formula of the compound is Cu2O.

Step 1

Write the ratio by mass.
C
:
H
:
38.4 g :
4.84 g :

Step 2

Cl
56.7 g

Calculate the ratio by amount (in moles).
38.4 g
4.84 g
:
:
M (Cu)
M (H)
38.4 g
4.84 g
:
:
−1
12.0 g mol
1.00 g mol −1
3.2 mol
:
4.84 mol
:

56.7 g
M (Cl)
56.7 g
35.5 g mol −1
1.597 mol

Step 3

Divide by the smaller amount.
3.2 g
4.84
1.597
:
:
1.597
1.597
1.597
2.00 :
3.04 :
1

Step 4

Round off to whole numbers.
2:3:1
∴ The empirical formula of the compound is C2H3Cl.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

39

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q35.

Gypsum is hydrated calcium sulfate (CaSO4•xH2O). A residue of 5.65 g of anhydrous
calcium sulfate is obtained by heating 7.15 g of gypsum. Determine the empirical
formula of gypsum.
A35.

Step 1 Calculate the mass of water present in the sample.
m(H2O)
= m(CaSO4•xH2O) – m(CaSO4)
= 7.15 g – 5.65 g
= 1.50 g
Step 2 Write the ratio by mass.
CaSO4 : H2O
5.65 g : 1.50 g
Step 3 Calculate the ratio by amount (in moles).
5.65 g
1.50 g
:
−1
136.08 g mol
18.0 g mol −1
0.0415 mol
: 0.08333 mol
Step 4 Divide by the smaller amount.
0.0415 0.08333
:
0.0415 0.0415
1
: 2.008
Step 5 Round off to whole numbers.
1:2
∴ The empirical formula of the compound is CaSO4.2H2O.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

40

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q36.

A 2.203 g sample of an organic compound, CxHyOz, was extracted from a plant. When
it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of
water and the carbon was oxidised to 3.23 g of carbon dioxide.
a Find the empirical formula of the compound.
b Another sample was analysed in a mass spectrometer. The mass spectrum
produced showed that the molar mass of the compound was 60.0 g mol–1. What is
its molecular formula?
A36.

1.32
= 0.15 mol
18
m(H) = 0.15 × 1 = 0.15 g
3.23
= 0.073 mol
n(C) = n(CO2) =
44
m(C) = 0.073 × 12 = 0.88g
m(O) = 2.203 – m(C) – m(H) = 2.203 – 0.88 – 0.15 = 1.173g
C
:
H
:
O
Mass
0.88 :
0.15 :
1.17
0.88
0.15
1.17
No mole
:
:
12
1
16
=
0.073 :
0.15 :
0.073
=
1
:
2
:
1
Empirical formula is CH2O

a

n(H) = 2n(H2O) =

b

mass of empirical formula = 12+1 ×2 + 16 = 30
molecular mass is 60, hence molecular formula must be double empirical
formula, i.e. C2H4O2

Q37.

What mass of barium chloride (BaCl2) will remain after a 15.0 g sample of the
hydrated salt BaCl2•2H2O is heated to drive off all of the water?
A37.

Step 1 Find the percentage of BaCl2 in the sample.
M (BaCl2 )
× 100
=
% BaCl2
M (BaCl 2 ⋅ 2H 2 O)
208.34
=
× 100
244.34
= 85.266%
Step 2 Calculate the mass of the sample which is BaCl2.
m(BaCl2)
= 85.266% of 15.0 g
= 12.79 g
Step 3 Express the answer with correct number of significant figures.
m(BaCl2) = 12.8 g

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

41

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q38.

A student is given solutions of lead(II) nitrate, copper(II) chloride and barium
hydroxide.
a Using Table 2.5 on page 23 of the student book, name the precipitates that could
be made by mixing together pairs of solutions.
b Write full and ionic equations for each of the reactions.
A38.
a
b

lead(II) chloride, lead(II) hydroxide, copper(II) hydroxide
Pb(NO3)2(aq) + CuCl2(aq) → PbCl2(s) + Cu(NO3)2(aq)
Pb2+(aq) + 2Cl–(aq) → PbCl2(s)
Pb(NO3)2(aq) + Ba(OH)2(aq) → Pb(OH)2(s) + Ba(NO3)2(aq)
Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s)
CuCl2(aq) + Ba(OH)2(aq) → Cu(OH)2(s) + BaCl2(aq)
Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)

Q39.

Design a flowchart to show how the salt content of a savoury spread could be
determined by gravimetric analysis.
A39.

1
2
3
4
5
6
7
8

Weigh a sample of the savoury spread.

Mix the sample with water to dissolve Cl– ions.

Filter the mixture.

Add excess silver nitrate solution to precipitate silver chloride.

Filter the precipitate and wash with water.

Dry the precipitate in an oven at 110°C.

Weigh the precipitate.

Repeat steps 6 and 7 until a constant mass is obtained.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

42

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q40.

The iodide ions in a solution containing 0.300 g of sodium iodide were precipitated as
silver iodide. What mass of silver iodide was formed?
A40.

Step 1 Write a balanced equation.
Ag+(aq) + I–(aq) → AgI(s)
Step 2 Calculate amount of NaI.
0.300 g
n(NaI)
=
149.89 g mol −1
= 0.002001 mol
Step 3 Use the ratio of amounts of substances to calculate the amount of AgBr
produced.
From the equation, 1 mol AgI is produced by 1 mol NaI.
1
n(AgI)
=
n(NaI)
1
n(AgI)
= 0.002001 mol
Step 4 Calculate the mass.
m(AgI)
= 0.002001 g × 234.77 g mol–1
= 0.4698 g
Step 5 Express the answer with the correct number of significant figures.
m(AgI) = 0.470 g
Q41.

A precipitate of Fe2O3, of mass 1.43 g, was obtained by treating a 1.5 L sample of
bore water. What was the concentration of iron, in mol L–1, in the water?
A41.

Step 1 Calculate amount of Fe2O3.
1.43 g
=
n(Fe2O3)
159.7 g mol −1
= 0.008954 mol
Step 2 From the formula, 2 mol Fe are obtained from 1 mol Fe2O3.
n(Fe) = 2 × 0.008954
= 0.017908 mol
Step 3 Calculate the concentration in mol L–1.
0.017908 mol
c(Fe) =
1.5 L
= 0.01193 mol L–1
Step 4 Express the answer with the correct number of significant figures.
c(Fe) = 0.012 mol L–1

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

43

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q42.

The chlorine in a 0.63 g sample of a chlorinated pesticide, DDT (C14H9Cl5), is
precipitated as silver chloride. What mass of silver chloride is formed?
A42.

Step 1 Write an appropriately balanced equation.
C14H9Cl5(aq) + 5Ag+(aq) → 5AgCl(s) + other products
Step 2 Calculate the amount of DDT (C14H9Cl5).
0.63 g
n(C14H9Cl5) =
354.462 g mol −1
= 0.001777 mol
Step 3 From the equation, 5 mol of AgCl is precipitated from 1 mol of C14H9Cl5.
n(AgCl)
= 5 × 0.001777 mol
= 0.008886 mol
Step 4 Calculate the mass of AgCl.
m(AgCl)
= 0.008886 × 143.32
= 1.2698 g
Step 5 Express the answer with the correct number of significant figures.
m(AgCl) = 1.3 g

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

44

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q43.

A 0.693 g sample of a silver alloy used to make cutlery is dissolved completely in
nitric acid. Excess sodium chloride solution is added to produce a precipitate of silver
chloride. The precipitate is filtered, dried and found to weigh 0.169 g.
a Find the percentage of silver in the alloy.
b If the precipitate was not completely dry when weighed, what effect would this
have on the answer for part a?
A43.
a

b

Step 1

Write a balanced equation.
Ag+(aq) + Cl–(aq) → AgCl(s)

Step 2

Calculate amount of AgCl precipitated.
0.169 g
n(AgCl)
=
143.32 g mol −1
= 0.001179 mol

Step 3

From the equation, 1 mol of Ag+ from the alloy is precipitated as 1 mol
of AgCl.
1
n(Ag + )
=
n(AgCl)
1
+
n(Ag )
= 0.001179 mol

Step 4

Calculate the mass of silver in the alloy.
m(Ag) = 0.001179 mol × 107.87 g mol–1
= 0.1272 g

Step 5

Convert to percentage.
0.1272
× 100
% Ag =
0.693
= 18.355%

Step 6

Express the answer with the correct number of significant figures.
% Ag = 18.4%

The result would be too high.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

45

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q44.

A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium
sulfate (Al2(SO4)3) were dissolved in a volume of water, and excess barium chloride
was added to precipitate barium sulfate.
What was the total mass of barium sulfate produced?
A44.

The Na2SO4 and Al2(SO4)3 react independently with the BaCl2. Treat them as separate
reactions and write two balanced equations. Add the masses of BaSO4 from each to
find the total mass.
Step 1
For the Na2SO4 solution, write a balanced equation.
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
Step 2

Calculate amount of Na2SO4.
0.500 g
n(Na2SO4) =
142.04 g mol −1
= 0.003520 mol

Step 3

From the equation, 1 mol of BaSO4 is precipitated from 1 mol of Na2SO4.
n(BaSO 4 )
1
=
n(Na 2 SO 4 )
1
= 0.003520 mol
n(BaSO4)

Step 4

Calculate mass of BaSO4 from this reaction with Na2SO4.
m(BaSO4) = 0.003520 mol × 233.4 g mol–1
= 0.821568 g

Step 5

For the Al2(SO4)3 solution, write a balanced equation.
3BaCl2(aq) + Al2(SO4)3(aq) → 3BaSO4(s) + 2AlCl3(aq)

Step 6

Calculate amount of Al2(SO4)3.
0.500 g
=
n(Al2(SO4)3)
342.14 g mol −1
= 0.001461 mol

Step 7

From the equation, 3 mol of BaSO4 is precipitated from 1 mol of Al2(SO4)3.
n(BaSO 4 )
3
=
n(Al 2 (SO 4 ) 3 )
1
n(BaSO4)
= 3 × 0.001461 mol
= 0.004383 mol

Step 8

Calculate mass of BaSO4 from this reaction with Al2(SO4)3.
m(BaSO4) = 0.004383 mol × 233.4 g mol–1
= 1.0229 g

Step 9

Calculate the total mass of BaSO4 precipitated from both reactions.
m(BaSO4) = (0.821568 + 1.0229) g
= 1.84456 g

Step 10

Express the answer with the correct number of significant figures.
m(BaSO4) = 1.85 g

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

46

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q45.

Water pollution can result from the phosphates added to washing powders to improve
the stability of their suds. The phosphorus in a 2.0 g sample of washing powder is
precipitated as Mg2P2O7. The precipitate weighs 0.085 g.
a What is the percentage, by mass, of phosphorus in the washing powder?
b Suppose you were in charge of an advertising campaign to promote the washing
powder. Would you advertise the percentage of phosphorus or phosphate in the
product? Explain.
A45.
a

Step 1

Find the percentage of P in Mg2P2O7.
2 × 30.97
%P =
× 100
222.56
= 27.83%

Step 2

Using the % P in Mg2P2O7, calculate the mass of P in precipitate.
m(P) = 27.83% of 0.085 g
27.83
× 0.085 g
=
100
= 0.02366 g

Step 3

Calculate the percentage P in 2.0 g of washing powder.
0.02366 g
%P =
× 100
2.0 g
= 1.183%

Step 4

b

Express the answer with the correct number of significant figures.
% P = 1.2%
You would need to consider the fact that the percentage of phosphate in the
washing powder is greater than the percentage of the phosphorus.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

47

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q46.

A 2.10 g sample of a commercial antacid powder is treated with excess hydrochloric
acid. The volume of carbon dioxide evolved is 430 mL, measured at 21.0°C and
109.6 kPa pressure. If magnesium carbonate is the active ingredient in the antacid,
calculate the percentage of magnesium carbonate in the sample.
A46.

Step 1 Write a balanced equation.
MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l)
Step 2 Convert pressure, volume and temperature to appropriate units for use in the
general gas equation.
P
= 109.6 kPa
V
= 430 mL
= 0.430 L
T
= 21°C = (21 + 273) K = 294 K
Step 3 Calculate the amount of CO2(g).
pV
n(CO2) =
RT
109.6 × 0.430
=
8.31 × 294
= 0.01928 mol
Step 4 From the equation, 1 mol MgCO3 produces 1 mol CO2.
n(MgCO 3 )
1
=
n(CO 2 )
1
= 0.01928 mol
n(MgCO3)
Step 5 Calculate the mass.
m(MgCO3) = 0.01928 mol × 84.3 g mol–1
= 1.6261 g
Step 6 Calculate the % MgCO3.
m(MgCO 3 )
× 100
=
% MgCO3
m(sample)
1.6261
× 100
=
2.10
= 77.435%
Step 7 Express the answer with the correct number of significant figures.
% MgCO3 = 77.4%

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

48

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q47.

When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of
phosphorus is produced. The molar mass of the oxide is 284 g mol–1.
a Determine the empirical formula of the phosphorus oxide.
b Determine the molecular formula of the phosphorus oxide.
A47.
a

Step 1

Write the ratio by mass.
P
:
O
0.100 g :
(0.228 – 0.100) g
0.100 g :
0.128 g

Step 2

Calculate the ratio by amount (in moles).
0.100 g
0.128 g
:
−1
30.974 g mol
16.0 g mol −1
0.003229 mol
:
0.008 mol

Step 3

Divide by the smaller amount.
0.003229
0.008
:
0.008
0.008
0.4036

Step 4

b

:

1

Express as integers by multiplying by 5.
2:5
∴ The empirical formula of the compound is P2O5.

As the empirical formula is P2O5, the molecule must contain a whole number of
P2O5 units.
The molar mass of one of these units is ((2 × 30.974) + (5 × 16)) =
141.948 g mol–1.
The number of units in a molecule = molar mass of the compound/molar mass of
one unit
284 g mol −1
=
141.948 g mol −1
=2
∴ The molecular formula of the compound is P4O10.

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

49

Worked solutions to student book questions

Chapter 2 Analysis by mass
Q48.

Excessive salt intake in the diet can cause high blood pressure and heart disease. The
salt content of a 14.96 g sample of powdered chicken soup was determined by
dissolving it in water to make a volume of 250.0 mL. A 20.00 mL volume of this
stock solution was pipetted into a conical flask and excess silver nitrate was added.
The silver chloride precipitate that formed was then filtered, washed and dried. Its
mass was 0.246 g.
a Write an ionic equation for the formation of the silver chloride precipitate.
b Calculate the amount, in mol, of silver chloride that was produced.
Assume all the chloride in the powdered soup came from sodium chloride
(common salt).
c Determine the amount of sodium chloride in the 20.00 mL volume of stock
solution in the conical flask.
d Calculate the amount of sodium chloride in 250.0 mL of the stock solution.
e What mass of sodium chloride was in the sample?
A48.
a
b

Ag+(aq) + Cl–(aq) → AgCl(s)
Step 1
Write a balanced equation.
Ag+(aq) + Cl–(aq) → AgCl(s)
Step 2

Calculate amount of AgCl produced.
Note: Round off to the appropriate number of significant figures for
this part of the answer, but keep all the digits running in your calculator
for the next parts of the question. Do this for all parts.
0.246 g
n(AgCl)
=
143.32 g mol −1
= 0.001716 mol
= 0.00172 mol

c

From the equation for the reaction in the 20.00 mL of stock solution, 1 mol of Cl–
ions produces 1 mol of AgCl.
1
n(Cl − )
=
n(AgCl)
1

= 0.001716 mol
n(Cl ) in 20.00 mL
= 0.00172 mol

d

Calculate the amount of salt in the 250.0 mL of stock solution.
250.0
mol
= 0.001716 ×
n(Cl–) in 250.00 mL
20.00
= 0.021455 mol
= 0.0215 mol

e

Calculate the mass of salt in 250.0 mL.
m(NaCl) in 250.00 mL = 0.021455 mol × 58.44 g mol–1
= 1.25386 g
= 1.25 g

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Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

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