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12 Reacting masses

Chemists use the unit ‘mole’ to describe the quantity of particles

Chapter preview in a substance.

12.1 The mole and Avogadro constant

12.2 The mole and molar mass
12.3 Calculating percentage by mass of an element in a compound
12.4 Calculating the mass of an element in a compound
12.5 Calculating the relative atomic mass of an element
12.6 Determining empirical formulae from experimental data
12.7 Determining molecular formulae from experimental data
12.8 Reacting masses from chemical equations

Prior knowledge & Quick review

(For Chapter 12)
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Counting in Chemistry

It is easy to count 10 coins by hands. But it would be difficult to count 1000 coins and even 100 000
coins by hands. Apart from counting directly by hands or by using a coin-counting machine, is there
any effective way to ‘count’ a large number of coins?

How can we count a huge number of $0.5 coins effectively?

One way to do the counting indirectly is by weighing. Similarly, atoms, molecules and ions are so
tiny that it is impossible to count them one by one. Chemists count the huge number of particles in a
substance also by weighing.

Think about...
1 Given that the mass of a $0.5 coin is 4.92 g, how many coins are there in 246 kg of $0.5 coins?

2 What unit is used in counting particles, like atoms, molecules and ions, in Chemistry?

3 Can we predict the mass of the product formed from a chemical reaction if the masses of the

reactants are known?

After studying this chapter, you should be able to answer the above questions.

weighing 稱量
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12.1 The mole and Avogadro constant
In daily life, some special units are used to describe the quantity of items.
For example, eggs are often packed in dozens and papers are packed in
reams. A dozen of eggs refers to 12 pieces and a ream of paper refers to
500 sheets. See Figure 12.1.

Figure 12.1 Eggs are often packed in dozens and papers are often packed in reams.

History corner Unlike eggs and papers, particles (i.e. atoms, ions and molecules) are
too small to be seen. It is impossible to count these particles one by
Avogadro constant
The Avogadro constant is one. Chemists use a special unit called mole (abbreviation: mol) to
named after an Italian describe the quantity of particles in a substance.
scientist, Amedeo
Avogadro. He proposed
that equal volumes of all Chemists have chosen the number of atoms in exactly 12.0 g of
gases under the same carbon-12 as the reference unit for the mole. The number of atoms in
conditions contain the 23
same number of
exactly 12.0 g of carbon-12 is 6.02 × 10 . This number is called the
–1
molecules. Avogadro constant (symbol: L; unit: mol ). As a pure substance can be
represented by a formula, its simplest unit is called a formula unit. Hence,
23
we can also say that, one mole of any substance contains 6.02 × 10
formula units of that substance. See Table 12.1.

Substance Formula unit

 medeo Avogadro
A
(1776–1856) Copper Cu

Carbon C

Water H2O

Sodium chloride NaCl

Table 12.1 Formula units of some substances

Avogadro constant 亞佛加德羅常數 mole 摩爾 Amedeo Avogadro 阿莫迪歐․亞佛加德羅 ream 令 (紙張計算單位)

formula unit 式單位 dozen 打
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Figure 12.2 gives us ideas on how much one mole of some substances
are.

History corner 1 mole of sodium chloride

23
1 mole of copper 1 mole of contains 6.02 × 10 NaCl
Mole
metal contains 1 mole of carbon water contains formula units (or contains
The term ‘mole’ was 23
6.02 × 10 Cu contains 6.02 × 6.02 × 10
23 +
1 mole of Na ions and 1
introduced by a German atoms
23
10 C atoms H2O molecules
–
mole of Cl ions)
scientist Wilhelm
Ostwald in 1893. It came
from the word ‘Molekül’
meaning molecule. He
was one of the winners of
the Nobel Prize for
Chemistry in 1909.

–
Cl ion
Cu C Na ion
+

atom H2O
atom molecule

Figure 12.2 One mole of each of the four substances (copper, carbon, water and sodium
 ilhelm Ostwald
W 23
chloride). They all contain 6.02 × 10 formula units.
(1853–1932)

Chemists define mole in the following way:

Key point
One mole is the amount of a substance that contains the same
number of formula units as the number of atoms in exactly 12.0 g of
carbon-12.

Example 12.1

Calculating the number of moles from the number of formula units and vice versa
24
(a) A sample of nitrogen contains 1.024 × 10 molecules. Calculate the number of moles of nitrogen
molecules in the sample.
(b) Calculate the number of atoms in 3 mol of nitrogen.
23 –1
(Avogadro constant = 6.02 × 10 mol )

Solution
number of N2 molecules
(a) Number of moles of N2 molecules = –1
Avogadro constant (mol )
24
1.024 × 10
= 23 –1
6.02 × 10 mol
= 1.70 mol
Cont’d
Wilhelm Ostwald 威廉․奧斯托惠爾特
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(b) Number of N2 molecules
–1
= number of moles of N2 molecules (mol) × Avogadro constant (mol )
23 –1
= 3 mol × 6.02 × 10 mol
24
= 1.806 × 10
Since there are two N atoms in one N2 molecule, number of N atoms
24
= 2 × 1.806 × 10
24
= 3.612 × 10

Self-test 12.1

(a) Calculate the number of formula units in 1.5 mol of potassium chloride.
(b) Calculate the number of ions in 1.5 mol of potassium chloride.
23 –1
(Avogadro constant = 6.02 × 10 mol )

Class practice 12.1

1. A sample of oxygen gas contains 0.5 mol of oxygen molecules.
(a) Calculate the number of oxygen molecules in the sample.
(b) Hence, calculate the number of oxygen atoms in the sample.
23 –1
(Avogadro constant = 6.02 × 10 mol )
24
2. Given that a sample contains 2.408 × 10 sodium atoms, calculate the
number of moles of sodium atoms present.
23 –1
(Avogadro constant = 6.02 × 10 mol )

12.2 The mole and molar mass

In Chapter 5, we have learnt that the relative atomic mass of carbon-12
Learning tip 12
is 12.0. Scientists found that one mole of C has a mass of 12.0 g. The
The relative atomic
masses of elements can mass of one mole of a substance is called its molar mass. The unit of molar
be found from the mass is gram per mole (g mol ).
–1

Periodic Table on the

inside front cover of this • For substances consisting of atoms:
book.
–1
molar mass = relative atomic mass expressed in g mol

Think about it
Substance Symbol Relative atomic mass Molar mass
Which has a greater
mass, 1 mole of iron Carbon C 12.0 12.0 g mol
–1

atoms or 1 mole of gold

–1
atoms? Iron Fe 55.8 55.8 g mol

molar mass 摩爾質量

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• For substances consisting of molecules:
–1
molar mass = relative molecular mass expressed in g mol

Substance Formula Relative molecular mass Molar mass

–1
Oxygen O2 16.0 × 2 = 32.0 32.0 g mol
–1
Water H2O 1.0 × 2 + 16.0 = 18.0 18.0 g mol

• For substances consisting of ions:

–1
molar mass = formula mass expressed in g mol
Learning tip
Relative atomic mass,
Substance Formula Formula mass Molar mass
relative molecular mass
and formula mass carry Calcium –1
no units. CaCl2 40.1 + 35.5 × 2 = 111.1 111.1 g mol
chloride

Class practice 12.2

1. Calculate the molar mass of each of the following substances.
(a) Silver (b) Fluorine (c) Ammonia
(d) Ethanol (C2H5OH) (e) Iron(III) sulphate
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, F = 19.0,
S = 32.1, Fe = 55.8, Ag = 107.9)
2. Calculate the mass of each of the following substances.
(a) 1 mol of sodium sulphate
(b) 0.5 mol of tetrachloromethane (CCl4)
(Relative atomic masses: C = 12.0, O = 16.0, Na = 23.0, S = 32.1, Cl = 35.5)

Relationship between number of formula units,

number of moles and mass
The diagram below summarizes the relationship between ‘number of
formula units’, ‘number of moles’ and ‘mass’ of a substance.

÷ Avogadro constant × molar mass

number of number of
mass
formula units moles

× Avogadro constant ÷ molar mass

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We can calculate one item from another by using the following
formulae:

Key point
mass (g)
Number of moles (mol) = –1
molar mass (g mol )
number of formula units
Number of moles (mol) = –1
Avogadro constant (mol )

Example 12.2

Calculating the mass and number of formula units from the number of moles
A vessel contains 1.85 mol of methane (CH4).
(a) Calculate the mass of methane in the vessel.
(b) Calculate the number of methane molecules in the vessel.
23 –1
(Relative atomic masses: H = 1.0, C = 12.0; Avogadro constant = 6.02 × 10 mol )

Solution
–1 –1
(a) Molar mass of CH4 = (12.0 + 1.0 × 4) g mol = 16.0 g mol
–1
Mass of CH4 (g) = number of moles of CH4 (mol) × molar mass of CH4 (g mol )
–1
= 1.85 mol × 16.0 g mol
= 29.6 g
(b) Number of CH4 molecules
–1
= number of moles of CH4 (mol) × Avogadro constant (mol )
23 –1
= 1.85 mol × 6.02 × 10 mol
24
= 1.11 × 10

Self-test 12.2

Consider a 0.175 mol sample of magnesium hydroxide.

(a) Calculate the mass of magnesium hydroxide in the sample.
(b) Calculate the number of hydroxide ions in the sample.
23 –1
(Relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3; Avogadro constant = 6.02 × 10 mol )

Chemistry in daily life

Mole day
Mole day was proposed by a chemistry teacher in the United States in 1991.
It is an unofficial holiday celebrated among chemists or anyone who is
interested in Chemistry on every year of October 23, from 6:02 a.m. to
6:02 p.m. The date and time of this ‘holiday’ were derived from the
23
Avogadro constant (i.e. 6.02 × 10 ).

Let us mark this day down!

Mole day 摩爾日 vessel 容器

unofficial 非正式的
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Example 12.3

Calculating the mass of a particle of a substance

Calculate the mass of
(a) 1 Na atom. (b) 1 H2O molecule. (c) 1 formula unit of NaCl.
23 –1
(Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0, Cl = 35.5; Avogadro constant = 6.02 × 10 mol )

Solution
–1
molar mass (g mol )
Mass of 1 formula unit = –1
Avogadro constant (mol )
–1
23.0 g mol –23
(a) Mass of 1 Na atom = 23 –1
= 3.82 × 10 g
6.02 × 10 mol
–1
(1.0 × 2 + 16.0) g mol –23
(b) Mass of 1 H2O molecule = 23 –1
= 2.99 × 10 g
6.02 × 10 mol
–1
(23.0 + 35.5) g mol –23
(c) Mass of 1 formula unit of NaCl = 23 –1
= 9.72 × 10 g
6.02 × 10 mol
Self-test 12.3

Calculate the mass of

(a) 1 Mg atom. (b) 1 Br2 molecule. (c) 1 formula unit of CaCO3.
(Relative atomic masses: C = 12.0, O = 16.0, Mg = 24.3, Ca = 40.1, Br = 79.9; Avogadro constant =
23 –1
6.02 × 10 mol )

Class practice 12.3

1. Complete the following table by filling in the appropriate numbers.

Molar mass No. of moles

Substance –1 Mass / g No. of formula units
/ g mol / mol

(a) Sodium hydroxide 0.250

(b) Helium 0.20

24
(c) Sulphur dioxide 3.01 × 10
23
(d) Compound X 23.0 3.01 × 10

(Relative atomic masses: H = 1.0, He = 4.0, O = 16.0, Na = 23.0, S = 32.1; Avogadro constant =
23 –1
6.02 × 10 mol )
2. Which of the following gases contains the greatest number of molecules?
(Relative atomic masses: H = 1.0, N = 14.0, O = 16.0, F = 19.0, Cl = 35.5, Ar = 40.0)
A. 50.0 g of argon B. 50.0 g of fluorine
C. 50.0 g of hydrogen chloride D. 50.0 g of nitrogen monoxide
3. Calculate the mass of chlorine which contains the same number of molecules as there are in 1.20 mol of
water.

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12.3 Calculating percentage by mass of an
element in a compound
From the formula of a compound, we can work out the percentage by
mass of each element in the compound.

Key point
Percentage by mass of element A in a compound
relative atomic mass of A × number of atoms of A in the formula
= × 100%
formula mass of the compound

Example 12.4

Calculating the percentage by mass of an element in a compound

Bauxite is an ore of aluminium. It contains mainly aluminium oxide (Al2O3). Calculate the percentage
by mass of aluminium in aluminium oxide.
(Relative atomic masses: O = 16.0, Al = 27.0)

Solution
Formula mass of Al2O3
Al2 O3
= 27.0 × 2 + 16.0 × 3
= 102.0
Percentage by mass of Al in Al2O3
relative atomic mass of Al × number of atoms of Al in the formula
= × 100%
formula mass of Al2O3
27.0 × 2
= × 100%
102.0
= 52.9%

Self-test 12.4

Sodium hydroxide is an active ingredient of drain cleaners. Calculate the percentage by mass of
sodium in sodium hydroxide.
(Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0)

percentage by mass 質量百分比 drain cleaner 通渠劑

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Class practice 12.4
1. Calculate the percentage by mass of sulphur in the compound Na2S.
(Relative atomic masses: Na = 23.0, S = 32.1)
2. Calculate the percentage by mass of nitrogen in the compound
(NH4)2HPO4.
(Relative atomic masses: H = 1.0, N = 14.0, O =16.0, P = 31.0)
3. Calculate the percentage by mass of magnesium in magnesium
chloride-6-water MgCl2․6H2O.
(Relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3, Cl = 35.5)
4. Given that an oxide of iron has 69.9% of iron by mass and the formula
mass of the oxide is 159.6, how many iron atoms are there in a formula unit
of the oxide?
(Relative atomic mass: Fe = 55.8)
A. 1
B. 2
C. 3
D. 4

12.4 Calculating the mass of an element in a

compound
The mass of an element in a compound can be calculated from the mass
of the compound and the percentage by mass of the element in that
compound.

Example 12.5

Calculating the mass of an element in a compound

How many grams of aluminium can be extracted from 2000 g of aluminium oxide?

Solution

From Example 12.4, we know that the percentage by mass of Al in Al2O3 is 52.9%. That means for
every gram of Al2O3, there is 0.529 g of Al in it.
 mass of Al in 2000 g of Al2O3
= 2000 g × 52.9%
= 1058 g

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Example 12.6

Calculating the mass of copper in hydrated copper(II) sulphate

Calculate the mass of copper in 15.0 g of copper(II) sulphate-5-water (CuSO4․5H2O).
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)

Solution
Learning tip
Formula mass of CuSO4․5H2O Crystals of some salts contain water
molecules as part of their structure.
= 63.5 + 32.1 + 16.0 × 4 + 5 × (1.0 × 2 + 16.0) These water molecules are called water
of crystallization e.g. CuSO4․5H2O .
= 249.6
water of crystallization
Percentage by mass of Cu in CuSO4․5H2O

relative atomic mass of Cu × number of atoms of Cu in the formula

= × 100%
formula mass of CuSO4․5H2O
63.5
= × 100%
249.6

= 25.4%
That means for every gram of CuSO4․5H2O, there is 0.254 g of Cu in it.
 mass of Cu in 15.0 g of CuSO4․5H2O
= 15.0 g × 25.4%
= 3.81 g

Self-test 12.6

Calculate the mass of potassium in 7.91 g of potassium dichromate (K2Cr2O7).

(Relative atomic masses: O = 16.0, K = 39.1, Cr = 52.0)

Class practice 12.5

1. Calculate the mass of calcium in 206 g of calcium oxide.
(Relative atomic masses: O = 16.0, Ca = 40.1)
2. Calculate the mass of oxygen in 40.0 g of carbon dioxide.
(Relative atomic masses: C = 12.0, O = 16.0)
3. A sample of sodium nitrate contains 100 g of sodium. Calculate the
mass of nitrogen present in this sample.
(Relative atomic masses: N = 14.0, O = 16.0, Na = 23.0)

water of crystallization 結晶水

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12.5 Calculating the relative atomic mass of an
element
The relative atomic mass of an element may be calculated from the
formula of the compound and percentage by mass of that element in the
compound.

Example 12.7

Calculating the relative atomic mass of an element

The chloride of metal M has the formula of MCl3 and contains 34.4% by mass of M. Calculate the
relative atomic mass of M.
(Relative atomic mass: Cl = 35.5)

Solution
Let the relative atomic mass of M be a.
Percentage by mass of M in MCl3
relative atomic mass of M × number of atoms of M in the formula
= × 100%
formula mass of MCl3
a Math tip
34.4% = × 100%
a + 35.5 × 3 To solve a,
a
0.344 = ×1
a = 55.8 a + 35.5 × 3
 the relative atomic mass of M is 55.8. 0.344a + 36.6 = a
0.656a = 36.6
a = 55.8
Self-test 12.7

The bromide of metal X has the formula of XBr2 and contains 25.6% by mass of X. Calculate the
relative atomic mass of X.
(Relative atomic mass: Br = 79.9)

Class practice 12.6

1. A metal oxide MO contains 79.87% by mass of M. Calculate the relative
atomic mass of M.
(Relative atomic mass: O = 16.0)
2. 26.88 g of a metal chloride MCl contains 5.68 g of chlorine. Calculate the
relative atomic mass of M.
(Relative atomic mass: Cl = 35.5)

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12.6 Determining empirical formulae from
experimental data

What is empirical formula?

In Chapter 8, we have seen the molecular formulae and structural
formulae of some simple molecular substances. These types of chemical
formulae are used to represent elements and covalent compounds
consisting of molecules. In this section, we are going to learn another type
of chemical formula, which is empirical formula.

The empirical formula of a compound is defined as the formula that

shows the simplest whole number ratio of the atoms or ions present in
the compound.

• For ionic compounds, the empirical formulae are the same as the
chemical formulae. For example, the empirical formulae for sodium
chloride and potassium carbonate are NaCl and K2CO3 respectively.

• For molecular substances, the empirical formulae may or may not be

the same as the chemical formulae. See Section 12.7 for more details.

Key point
The empirical formula of a compound is the formula which shows
the simplest whole number ratio of the atoms or ions present.

The empirical formula of a compound can be determined from its

composition by mass i.e. the mass of each element present in a known
mass of the compound. The composition of a compound has to be
determined by experiments.

Determining the empirical formula of an oxide of

Experiment 12.1
copper
Determining the
empirical formula of an In Chapter 7, we learnt that copper forms two oxides with oxygen, namely,
oxide of copper
copper(I) oxide (Cu2O) and copper(II) oxide (CuO). To determine the
Experiment video empirical formula of an oxide of copper, we have to find the ratio by mass
Determining the of copper and oxygen in the compound in an experiment.
empirical formula of
an oxide of copper

composition by mass 質量組成 simplest whole number ratio 最簡單整數比

empirical formula 實驗式
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A set-up for conducting the experiment is shown in Figure 12.3.

SBA note hole

oxide of copper
• At the beginning of the
excess town gas
experiment, town gas town gas
burns here
is passed into the supply
combustion tube to
expel the air inside the combustion tube
heat
tube.
• The hot copper formed
may react with oxygen
in the air again.
Therefore, it is
necessary to pass town
gas through the
excess town gas
combustion tube, even
burns here
after heating has town gas
stopped. supply
combustion tube
oxide of copper

Figure 12.3 Set-up for determining the empirical formula of an oxide of copper

Put a known mass of the oxide of copper (black) inside a combustion

Learning tip tube. Pass town gas into the tube while heating the oxide strongly (Figure
Hydrogen and carbon 12.3). Hydrogen and carbon monoxide in the town gas reduce the
monoxide reduce the
oxide of copper to copper (reddish brown). Keep heating the solid in the
oxide by removing the
oxygen from it. combustion tube until there is no further colour change in the solid. Then
find the mass of the solid (copper) remained.

The specimen results of the experiment are shown below:

Item Mass / g

Mass of combustion tube 18.100

Mass of combustion tube + oxide of copper 18.701

Mass of combustion tube + copper 18.579

Mass of copper in the oxide 18.579 – 18.100 = 0.479

Mass of oxygen in the oxide 18.701 – 18.579 = 0.122

town gas 煤氣
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From the experimental results, we can work out the empirical
formula of the oxide of copper as shown in ‘Problem-solving strategy 12.1’.

Problem-solving strategy 12.1

Determining the empirical formula of a compound

The steps showing how to determine the empirical formula of the above oxide of copper are as follows:

Cu O

1 Write the mass of each

Mass / g 0.479 0.122
element

2 Write the relative

atomic mass of each
element
Relative atomic mass 63.5 16.0
(After some practice,
you will be able to skip
this step.)

3 Calculate the number Number of moles of atoms / mol

0.479 0.122
of moles of each kind mass (g) = 0.00754 = 0.00763
(= –1
) 63.5 16.0
of atoms molar mass (g mol )

4 Calculate the simplest Simplest whole number mole ratio

whole number mole of atoms
ratio of atoms (divided by the smallest number of
moles) 0.00754 0.00763
=1 = 1.01  1
(multiplied by the smallest possible 0.00754 0.00754
whole number to turn all the
values into whole number if
necessary)

 the empirical formula of the oxide of copper is CuO.

Note:
Due to experimental error, the number(s) obtained for the simplest mole ratio when determining the
empirical formulae may have a small difference from a whole number. It is acceptable to ‘round off’
the value(s) to the nearest whole number(s). However, we must be very cautious when doing so. For
example, 1.01 can be rounded off to 1, but 1.2 is usually not rounded off to 1.

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Example 12.8

Determining the empirical formula of a compound from the percentage by mass

An oxide of lead contains 92.7% of lead by mass. Determine the empirical formula of this oxide.
(Relative atomic masses: O = 16.0, Pb = 207.2)

Solution

Pb O
Learning tip
Mass / g 92.7 7.3 Assume that there is
Relative atomic mass 207.2 16.0 100 g of X. Then there
are 92.7 g of Pb and 7.3 g
Number of moles of 92.7 7.3 of O.
= 0.447 = 0.456
atoms / mol 207.2 16.0

Simplest whole number 0.447 0.456 Flipped classroom

=1 = 1.02  1
mole ratio of atoms 0.447 0.447 Determining the
empirical formula of
a compound from
 the empirical formula of the oxide is PbO. the percentage by
mass
Self-test 12.8

Compound Y contains 26.95% sulphur, 13.44% oxygen and 59.61% chlorine by mass. Determine
the empirical formula of Y.
(Relative atomic masses: O = 16.0, S = 32.1, Cl = 35.5)

Class practice 12.7

1. A compound has the empirical formula CxHy. On analysis, 1.000 g of this
compound was found to contain 0.857 g of carbon. Calculate the values
of x and y.
(Relative atomic masses: H = 1.0, C = 12.0)
2. A 16.28 g sample of a metal oxide, MO, upon reduction, gives 13.08 g of
the metal M. Which of the following is the relative atomic mass of M?
(Relative atomic mass: O = 16.0)
A. 24.3
B. 40.1
C. 63.5
D. 65.4
3. A compound contains iron and oxygen only. In an experiment for
determining the empirical formula of this compound, 2.31 g of the
compound was heated with carbon monoxide. Upon complete reaction,
carbon dioxide and 1.67 g of iron were formed. Calculate the empirical
formula of this compound.
(HKDSE 2015 Paper 1B Q3b(i))

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12.7 Determining molecular formulae from
experimental data

What is molecular formula?

Ethene is a covalent compound. It has two carbon and four hydrogen
atoms in each molecule (Figure 12.4). Therefore, the empirical formula of
ethene is CH2.
Figure 12.4 Structural
formula of ethene The empirical formula cannot show the actual numbers of carbon
and hydrogen atoms in an ethene molecule. To represent their actual
numbers, we have to use the molecular formula.

Key point
The molecular formula of a substance shows the actual number of
each kind of atoms in one molecule of the substance.

The concept of empirical formula applies to both ionic and covalent

compounds. The concept of molecular formula only applies to
molecular substances. Table 12.2 shows the empirical formulae and
molecular formulae of some compounds.

Compound Empirical formula Molecular formula

Carbon dioxide CO2 CO2

Ethene CH2 C2H4

Propene CH2 C3H6

Ethanol C2H6O C2H6O

Glucose CH2O C6H12O6

Magnesium carbonate MgCO3 /

Silicon dioxide SiO2 /

Table 12.2 Empirical formulae and molecular formulae of some compounds

We should note that the empirical and molecular formulae of a

compound may be the same (e.g. carbon dioxide) or different (e.g.
ethene). The molecular formula is the empirical formula multiplied by
an integer (1, 2, 3, etc.).

ethene 乙烯 integer 整數
molecular formula 分子式 multiplied 乘以
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Determining the molecular formulae of covalent
compounds
Once the empirical formula and the relative molecular mass of a
compound are known, we can determine the molecular formula of the
compound. See Examples 12.9–12.11.

Example 12.9

Determining molecular formula from empirical formula and relative molecular mass
A covalent compound has an empirical formula of CH2 and a relative molecular mass of 42.0.
Determine its molecular formula.
(Relative atomic masses: H = 1.0, C = 12.0)

Solution
Let the molecular formula of the compound be (CH2)n, where n is an integer.
n × (12.0 + 1.0 × 2) = 42.0
n=3
 the molecular formula of the compound is (CH2)3, i.e. C3H6.

Example 12.10

Determining empirical formula and molecular formula from percentage by mass

Compound X was found to contain 80% carbon and 20% hydrogen by mass. Given that X has a
relative molecular mass smaller than 35.0, what are the empirical formula and molecular formula of X?
(Relative atomic masses: H = 1.0, C = 12.0)

Solution

C H Learning tip
Mass / g 80 20 Assume that there is
100 g of X. Then there
Relative atomic mass 12.0 1.0 are 80 g of C and 20 g of
80 20 H.
Number of moles of = 6.67 = 20.0
atoms / mol 12.0 1.0

Simplest whole number 6.67 20.0

=1 =3
mole ratio of atoms 6.67 6.67

 the empirical formula of X is CH3.

Let the molecular formula of X be (CH3)n, where n is an integer.
Math tip
n × (12.0 + 1.0 × 3) < 35.0 The symbol ‘< ’ means
n < 2.33 (i.e. n = 2) ‘smaller than’.

 the molecular formula of X is C2H6.

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Example 12.11

Determining empirical formula and molecular formula from masses of combustion products
Compound Z contains only carbon, hydrogen and oxygen. 2.43 g of Z burns completely to give
3.96 g of carbon dioxide and 1.35 g of water. Determine the empirical formula of Z. If the relative
molecular mass of Z is 162.0, determine its molecular formula.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)

Solution

Since all the C in CO2 and H in H2O come from Z,

mass of C in Z
12.0
= 3.96 g ×
12.0 + 16.0 × 2
= 1.08 g
mass of H in Z
1.0 × 2
= 1.35 g ×
1.0 × 2 + 16.0
= 0.15 g
The rest of the mass of Z must come from oxygen.
 mass of O in Z
= (2.43 – 1.08 – 0.15) g
= 1.20 g
Now go on to find the empirical formula of Z as follows:

C H O

Mass / g 1.08 0.15 1.20

Relative atomic mass 12.0 1.0 16.0

Number of moles of 1.08 0.15 1.20

= 0.090 = 0.15 = 0.075
atoms / mol 12.0 1.0 16.0

Simplest whole 0.090 0.15 0.075

= 1.2 =2 =1
number mole ratio 0.075 0.075 0.075
of atoms
1.2 × 5 = 6 2 × 5 = 10 1×5=5

 the empirical formula of Z is C6H10O5.

Let the molecular formula of Z be (C6H10O5)n, where n is an integer.
n × (12.0 × 6 + 1.0 × 10 + 16.0 × 5) = 162.0
n=1
 the molecular formula of Z is C6H10O5.

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Example 12.12

Determining the number of water of crystallization in a hydrated salt

5.60 g of a hydrated copper(II) sulphate CuSO4․nH2O is heated in a
Experiment 12.2
crucible to remove the water of crystallization. The white residue is
Determining the
anhydrous copper(II) sulphate, which is found to have a mass of 3.59 g.
formula of hydrated
copper(II) sulphate

Experiment video
Determining the
formula of hydrated
copper(II) sulphate

Hydrated copper(II) sulphate Anhydrous copper(II) sulphate

(a) Deduce, from the experimental results, a reasonable value for n.

(b) Suggest a reason why the value of n deduced in (a) differs slightly from the value calculated.
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)

Solution
(a) Mass of water of crystallization = (5.60 – 3.59) g = 2.01 g

CuSO4 H2O
Mass / g 3.59 2.01
Formula mass 159.6 18.0

Number of moles of 3.59 2.01

= 0.0225 = 0.112
formula units / mol 159.6 18.0

Simplest whole number 0.0225 0.112

=1 = 4.98
mole ratio of formula units 0.0225 0.0225

Since n should be a whole number, a reasonable value of n would be 5.

(b) The experimental value of n (4.98) is smaller than 5. This might be because not all water of
crystallization has been removed from the hydrated salt during the heating process.

Class practice 12.8

1. A compound contains only carbon, hydrogen and oxygen. 0.81 g of the
compound gives 1.32 g of carbon dioxide and 0.45 g of water upon
complete combustion. Given that the relative molecular mass of the
compound is 320.0, determine its
(a) empirical formula.
(b) molecular formula.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
Cont’d

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2. In an experiment to determine the number of water of crystallization in

a hydrated salt of magnesium sulphate, 2.50 g sample of the salt was
heated gently in a crucible until there was no further change. The results
are tabulated below:

Item Mass / g

Mass of crucible 24.30

Mass of crucible and MgSO4․nH2O 26.80

Mass of crucible and salt after heating 25.52

(a) Calculate the mass of water evaporated from the sample.

(b) Calculate the mass of MgSO4.
(c) Calculate the mole ratio of MgSO4 to H2O.
(d) State the empirical formula of the hydrated MgSO4.
(Relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1)

12.8 Reacting masses from chemical equations

Relationship between mole ratio and stoichiometric

Learning tip
The quantitative study of
coefficients a chemical equation
reactants and products in
In Chapter 11, we have learnt that a balanced chemical equation provides
a reaction is called
stoichiometry. us information about the relative numbers of particles of reactant(s)
and the product(s) taking part in a reaction.

Consider the reaction between magnesium and oxygen:

2Mg(s) + O2(g) 2MgO(s)

2 magnesium atoms react 1 oxygen molecule to 2 formula units of

with form magnesium oxide (2
Particle ratio: magnesium ions and 2
oxide ions)

23 23 23
2 × 6.02 × 10 react 1 × 6.02 × 10 to 2 × 6.02 × 10 formula
magnesium atoms with oxygen molecules form units of magnesium oxide

2 moles of react 1 mole of oxygen to 2 moles of formula units

Mole ratio:
magnesium atoms with molecules form of magnesium oxide

reacting mass 反應質量

stoichiometry 化學計量學
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The stoichiometric coefficients in the equation indicate the mole
ratio of reactants and products in the reaction. In the above reaction, the
mole ratio is 2 : 1 : 2 for Mg to O2 to MgO.

Conservation of mass
Besides the mole ratio, the chemical equation also tells us information
about the reacting masses:

2Mg(s) + O2(g) 2MgO(s)

Think about it
–1 –1 –1
How do we calculate the 2 mol × 24.3 g mol react 1 mol × 32.0 g mol to 2 mol × 40.3 g mol
molar masses of Mg, O2 = 48.6 g of with = 32.0 g of oxygen form = 80.6 g of formula
and MgO respectively? magnesium atoms molecules units of magnesium
oxide

48.6 g + 32.0 g = 80.6 g 80.6 g

From the above calculation, the total mass of the products is equal to
the total mass of the reactants. This is called conservation of mass. Mass
is conserved because the numbers of the different kinds of atoms remain
unchanged before and after a reaction.

Example 12.13

Understanding conservation of mass

Lead(II) oxide reacts with carbon powder to give lead and carbon dioxide.
2PbO(s) + C(s) 2Pb(s) + CO2(g)
w grams of PbO(s) react with x grams of C(s) to give y grams of Pb(s) and z grams of CO2(g). Express y
in terms of w, x and z.

Solution
Learning tip
Total mass of the reactants at the start = (w + x) grams The stoichiometric coefficients in the
equation shows the quantitative
Total mass of the products at the end = (y + z) grams relationship (i.e. mole ratio), but not
According to the conservation of mass, w + x = y + z the mass relationship among the
species in a reaction.
y=w+x–z

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Calculating reacting masses from chemical
equations
Chemists often want to know the amount of product(s) that can be
produced from a given amount of reactant(s). See Problem-solving
strategy 12.2.

Problem-solving strategy 12.2

Calculating reacting masses from chemical equations

Try it now
Calculate the mass of magnesium oxide formed when 2.43 g of Calculate the mass of copper
magnesium burns completely in air. formed when 15.9 g of
(Relative atomic masses: O = 16.0, Mg = 24.3) copper(II) oxide reacts
completely with hydrogen.
1 Write the chemical equation for the reaction. (Relative atomic masses: O =
2Mg(s) + O2(g) 2MgO(s) 16.0, Cu = 63.5)

2 Convert the mass(es) of the given substance(s) into the

number of moles.
Number of moles of Mg (mol)
mass of Mg (g)
= –1
molar mass of Mg (g mol )
2.43 g
= –1
24.3 g mol
= 0.100 mol

3 Calculate the number of moles of the substance asked in

the question.
From the equation, mole ratio of Mg to MgO = 2 : 2 (or simply
1 : 1)
 number of moles of MgO formed = 0.100 mol

4 Convert the number of moles of that substance into mass.

Mass of MgO (g)
= number of moles of MgO molar mass of MgO
× –1
(mol) (g mol )
–1
= 0.100 mol × (24.3 + 16.0) g mol
= 4.03 g

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The flow chart below illustrates the steps for calculating the reacting
masses from a chemical equation.

Number of
Known divided by
Number of by mole ratio multiplied by Mass of the
moles of the
mass of A moles of A substance asked
molar (shown in the substance asked molar mass of
mass of A equation) that substance
in the question
in the question

(Note: A represents the chemical formula of a given substance.)

Example 12.14

Calculating the mass of a product from the mass of a reactant

Calculate the mass of iron formed when 1.60 g of iron(III) oxide reacts completely with excess carbon
monoxide.
(Relative atomic masses: O = 16.0, Fe = 55.8)

Solution

1 Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

2 Number of moles of Fe2O3 used

1.60 g
= –1
(55.8 × 2 + 16.0 × 3) g mol
= 0.0100 mol

3 From the equation, mole ratio of Fe2O3 to Fe = 1 : 2.

 number of moles of Fe
= 0.0100 × 2 mol
= 0.0200 mol

4 Mass of Fe formed
–1
= 0.0200 mol × 55.8 g mol
= 1.12 g

Self-test 12.14

Calculate the mass of lead produced when 2.23 g of lead(II) oxide reacts completely with excess carbon.
(Relative atomic masses: O = 16.0, Pb = 207.2)

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When the mass of the product of a reaction is known, we can
calculate the mass of the reactant(s) based on the balanced chemical
equation. See Example 12.15.

Example 12.15

Calculating the mass of a reactant from the mass of a product

A sample of lead ore contains 75% by mass of lead(II) sulphide and 25% impurities. It is known that
there are no other lead-containing compounds in the impurities. Calculate the mass of the ore
required to extract 20.72 g of lead.
(Relative atomic masses: S = 32.1, Pb = 207.2)

Solution
1 The extraction of lead involves two steps:
2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)
2PbO(s) + C(s) 2Pb(s) + CO2(g)
2 Number of moles of Pb to be extracted
20.72 g
= –1
207.2 g mol
= 0.1000 mol
3 From the equation, mole ratio of Pb to PbO to PbS = 1 : 1 : 1.
 number of moles of PbS required = 0.1000 mol
–1
4 Mass of PbS required = 0.1000 mol × (207.2 + 32.1) g mol = 23.93 g
As the ore contains 75% by mass of PbS, mass of the ore required
23.93 g
=
75%
= 31.91 g

Class practice 12.9

1. Upon strong heating, silver oxide decomposes to silver and oxygen:
2Ag2O(s) 4Ag(s) + O2(g)
Calculate the mass of silver obtained when 6.96 g of silver oxide is
strongly heated in air.
(Relative atomic masses: O = 16.0, Ag = 107.9)
2. Aluminium powder reacts with finely divided iron(III) oxide according to
the following equation:
2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe()
Calculate the mass of finely divided iron(III) oxide needed to produce
10.55 g of iron.
(Relative atomic masses: O = 16.0, Fe = 55.8)

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Calculations involving limiting reactant
Consider the reaction between hydrogen and oxygen to form water:

2H2(g) + O2(g) 2H2O()

The equation tells us that 1 molecule of oxygen is required to react

with 2 molecules of hydrogen for complete reaction. See Figure 12.5.

H2 molecule H2O molecule

O2 molecule

Figure 12.5 Two hydrogen

molecules require one oxygen
molecule for complete reaction.
Therefore, oxygen is in excess.
2 H2 molecules + 2 O2 molecules 2 H2O molecules + 1 O2 molecule

In this reaction, all the hydrogen has reacted. Oxygen is in excess. The

Flipped classroom amount of water produced is limited by the amount of hydrogen used.
Calculations Therefore, hydrogen is called the limiting reactant in this case. The
involving limiting limiting reactant limits the amount of the product formed in a reaction.
reactant
Besides, the reaction stops when the limiting reactant is used up.

Problem-solving strategy 12.3

Determining the limiting reactant in a reaction

Try it now
In the extraction of zinc from zinc oxide, 8.14 g of zinc oxide In an experiment, 6.54 g of
is heated with 6.50 g of carbon powder. Which reactant is the zinc is heated with 3.84 g of
limiting reactant? oxygen. Which one is the
(Relative atomic masses: C = 12.0, O = 16.0, Zn = 65.4) limiting reactant?

1 Write the chemical equation for the reaction.

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

2 Calculate the number of moles of each reactant.

Number of moles of ZnO
8.14 g
= –1
= 0.100 mol
(65.4 + 16.0) g mol
Number of moles of C
6.50 g
= –1
= 0.542 mol
12.0 g mol
Cont’d
limiting reactant 限量反應物
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3 Calculate, if all of a reactant is used up, the number of
moles of the other reactant that will be needed for
complete reaction.
2ZnO(s) + C(s) 2Zn(s) + CO2(g)
0.100 mol 0.542 mol
(from 2 ) (from 2 )

From the equation, mole ratio of ZnO to C is 2 : 1.

0.100
(a) 0.100 mol of ZnO requires mol = 0.0500 mol of C
2
for complete reaction
(b) 0.542 mol of C requires 0.542 mol × 2 = 1.084 mol of ZnO
for complete reaction

4 Identify the reactant that is in limited amount i.e. not in

excess. This reactant will first be used up in the reaction.
In 3(a), only 0.0500 mol of C is needed to react with all of the
ZnO (i.e. 0.100 mol). As there is 0.542 mol of C available, C is
in excess. Thus ZnO is the limiting reactant.

Example 12.16

Calculating reacting masses involving limiting reactant

Calculate the mass of water formed when 5.4 g of hydrogen is heated with 3.2 g of oxygen.
(Relative atomic masses: H = 1.0, O = 16.0)

Solution
1 2H2(g) + O2(g) 2H2O()
5.4 g
2 Number of moles of H2 = –1
= 2.7 mol
(1.0 × 2) g mol
3.2 g
Number of moles of O2 = –1
= 0.10 mol
(16.0 × 2) g mol
3 From the equation, mole ratio of H2 to O2 = 2 : 1.
0.10 mol of O2 requires only 0.10 mol × 2 = 0.20 mol of H2 for complete reaction.
 H2 is in excess while O2 is the limiting reactant.
From the equation, mole ratio of O2 to H2O = 1 : 2.
Number of moles of H2O formed = 0.10 mol × 2 = 0.20 mol
–1
Mass of H2O formed = 0.20 mol × (1.0 × 2 + 16.0) g mol = 3.6 g

Self-test 12.16

Calculate the mass of nitrogen dioxide formed when 26.58 g of nitrogen monoxide reacts with
8.06 g of oxygen according to the following equation:
2NO(g) + O2(g) 2NO2(g)
(Relative atomic masses: N = 14.0, O = 16.0)

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Concept check
The reactant which is in the least number of moles must be the
limiting reactant.
A limiting reactant is NOT necessarily the one which is in the
least number of moles. It is the one that is not in excess and is
first used up in a reaction.

Theoretical yield, actual yield and percentage yield

Theoretical yield is the maximum amount of product expected if the
reaction proceeds exactly as shown in the chemical equation. Actual yield
is the amount of product actually obtained at the end of the reaction.
The actual yield of a reaction is often smaller than the theoretical yield
because:
• the reaction may not complete.
• impurities are present in the reactants i.e. the actual mass of the
reactant(s) used is smaller than expected.
• there are side reactions which give unwanted side products.
• some product is lost in some of the experimental processes, such as
purification.

The efficiency of a chemical reaction can be expressed by the

percentage yield. The higher the percentage yield, the more efficient is
the chemical reaction.

Key point
actual yield
Percentage yield = × 100%
theoretical yield

Example 12.17

Calculating the theoretical yield and actual yield of a product

In an experiment, 31.8 g of copper(II) oxide was heated with 6.0 g of carbon powder.
(a) Calculate the theoretical yield of copper.
(b) Given that the percentage yield of copper is 82%, calculate the actual yield of copper.
(Relative atomic masses: C = 12.0, O = 16.0, Cu = 63.5)

Solution
(a) 1 2CuO(s) + C(s) 2Cu(s) + CO2(g)
31.8 g
2 Number of moles of CuO = –1
= 0.400 mol
(63.5 + 16.0) g mol
6.0 g
Number of moles of C = –1
= 0.50 mol
12.0 g mol
Cont’d
actual yield 實際產量 side product 副產品
92 percentage yield 百分產率 theoretical yield 理論產量
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3 From the equation, mole ratio of CuO to C = 2 : 1.
0.400
0.400 mol of CuO only requires mol = 0.200 mol of C for complete reaction.
2
 C is in excess while CuO is the limiting reactant.
From the equation, mole ratio of CuO to Cu = 1 : 1.
 number of moles of Cu = 0.400 mol
–1
Theoretical yield of Cu = 0.400 mol × 63.5 g mol = 25.4 g
(b) Actual yield of Cu (g) = theoretical yield of Cu (g) × percentage yield of Cu (%)
= 25.4 g × 82% = 20.8 g

Example 12.18
Integrated
Calculating the total mass of ions in a cube of sodium chloride crystal
Sodium chloride is one of the most abundant minerals on the
Earth. It can be extracted from sea water or some ores in the
Earth’s crust. Crystals of sodium chloride have a giant ionic
structure.
(a) Outline how to obtain crystals of sodium chloride from a
muddy sea water sample in the school laboratory.
(b) Suggest a method for extracting sodium from sodium
chloride. Sodium chloride crystals
(c) The diagram below shows a part of the structure of sodium chloride crystal with some ions
missing.

+
Na ion
–
Cl ion

Complete the diagram by filling in the missing ions.

+ –
(d) From an experiment, it is found that there are 4 Na ions and 4 Cl ions in a cube of sodium
–8
chloride crystal, and the length of each side of the cube is 5.65 × 10 cm.
+ –
(i) Express the total mass of 4 Na ions and 4 Cl ions in terms of the Avogadro constant, L.
(Relative atomic masses: Na = 23.0, Cl = 35.5)
(ii) Calculate the volume of the cube of sodium chloride crystal.
(iii) Hence, determine the value of the Avogadro constant, L, given that the density of sodium
–3
chloride is 2.17 g cm .
(e) Refer to the following chemical reaction:
2Na(s) + Cl2(g) 2NaCl(s)
n moles of Na(s) are allowed to react with 2n moles of Cl2(g) under suitable conditions until the
reaction stops. Calculate the number of moles of NaCl(s) formed in terms of n.
Cont’d
sodium chloride crystal 氯化鈉晶體
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Solution
(a) Filter the sea water sample to get rid of the mud. Then heat the filtered sea water until it
becomes saturated. After that, allow the hot saturated sea water to cool slowly to room
temperature. Crystals of sodium chloride will form gradually. Finally, separate the crystals of
sodium chloride from the remaining sea water by filtration.
(b) Electrolysis of the molten sodium chloride Learning tip
(c) In part (a),the crystals
obtained are not
completely pure. They
also contain other salts
e.g. CaCl2, MgSO4, etc.

–1 –1
+ 23.0 g mol – 35.5 g mol
(d) (i) Mass of 1 Na ion = –1
; mass of 1 Cl ion = –1
L mol L mol
+ –
Total mass of 4 Na ions and 4 Cl ions
–1 –1
23.0 g mol 35.5 g mol 234
=4× –1
+4× g –1
=
L mol L mol L
(ii) Volume of the cube of sodium chloride crystal
–8 3 –22 3
= (5.65 × 10 cm) = 1.80 × 10 cm
(iii) Mass of the cube of NaCl crystal
–3 –22 3 –22
= 2.17 g cm × 1.80 × 10 cm = 3.906 × 10 g
Mass of the cube of NaCl crystal
+ –
= total mass of 4 Na ions and 4 Cl ions in the cube of crystal
234 –22
 g = 3.906 × 10 g
L
23
L = 5.99 × 10
(e) From the given equation, mole ratio of Na to Cl2 to NaCl = 2 : 1 : 2.
n
 n moles of Na requires only moles of Cl2 for complete reaction. Hence, Na(s) is the limiting
reactant. 2

 number of moles of NaCl formed = n moles

Class practice 12.10

1. Iron reacts with dilute hydrochloric acid according to the following
equation:
Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g)
0.5 mol of iron is added to an aqueous solution containing 0.8 mol of
hydrochloric acid. Which of the following statements is correct?
A. 0.5 mol of hydrogen forms.
B. 0.8 mol of iron(II) chloride forms.
C. Iron is in excess by 0.1 mol.
D. Hydrochloric acid is in excess by 0.3 mol.
Cont’d

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2. Under certain conditions, 1.0 g of SiO2 is allowed to react with 1.0 g of Mg.
The equation for the reaction is shown below:
SiO2 + 2Mg 2MgO + Si
Calculate the theoretical mass of Si that can be formed.
(Relative atomic masses: O = 16.0, Mg = 24.3, Si = 28.1)
(HKDSE 2021 Paper 1B Q3c(ii))

3. A student performed an experiment to prepare calcium hydroxide. 1.50 g

of calcium granules was added to a sufficient amount of water. The calcium
hydroxide precipitate formed was then filtered off, washed and dried.
(a) Write the chemical equation for the reaction of calcium with water.
(b) Calculate the theoretical yield of calcium hydroxide.
(c) The actual yield of calcium hydroxide was much smaller than the
theoretical yield. Suggest a reason why there was such a difference.
(Relative atomic masses: H = 1.0, O = 16.0, Ca = 40.1)

Activity 12.1
Determining the chemical reaction occurring in the thermal decomposition of baking soda
Sodium hydrogencarbonate (commonly called baking soda) is a main ingredient of baking powder. Baking
powder is used in bread-making to ensure that the bread can ‘rise’ during baking. When the temperature of
the bread mixture reaches about 50°C, the sodium hydrogencarbonate (NaHCO3) in the powder decomposes
and carbon dioxide evolves.
Sodium hydrogencarbonate is proposed to have undergone the following three decomposition reactions
during the baking process. In fact, only one of these reactions actually occurs.
• sodium hydrogencarbonate sodium hydroxide + carbon dioxide
• sodium hydrogencarbonate sodium oxide + carbon dioxide + water
• sodium hydrogencarbonate sodium carbonate + carbon dioxide + water

Experiment 12.3
Studying the thermal
decomposition of
sodium
hydrogencarbonate

Experiment video
Studying the thermal
Baking powder decomposition of
sodium
Tasks hydrogencarbonate
Form groups of two to answer the following questions.
1. Transcribe the above word equations into chemical equations. (Hint: except carbon dioxide and water, all
substances are solids in these reactions.)
2. To identify the chemical reaction that has taken place during the baking process, 2.53 g of sodium
hydrogencarbonate was heated until there was no further changes. It was found that 1.59 g of a solid
remained at the end of the experiment. Which of the above reactions matches with the result?
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0)

baking soda 梳打粉 baking powder 發粉

sodium hydrogencarbonate 碳酸氫鈉
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Chapter review
Progress check
Chapter review

After studying this chapter, you should be able to:

perform calculations related to moles, Avogadro constant and molar masses
calculate the percentage by mass of an element or the mass of an element in a compound using
appropriate information
calculate the relative atomic mass of an element using appropriate information
determine empirical formulae and molecular formulae from composition by mass and molar
masses
understand the quantitative relationship of the reactants and products in a reaction as revealed by
a chemical equation
calculate masses of reactants and products in a reaction from the relevant equation and state the
interrelationship between them
solve problems involving limiting reagents

e-Dictionary
Key terms (For Chapter 12)
Page Page
1. actual yield 實際產量 92 7. mole 摩爾 67
2. Avogadro constant 亞佛加德羅常數 67 8. molecular formula 分子式 81
3. composition by mass 質量組成 77 9. percentage by mass 質量百分比 73
4. empirical formula 實驗式 77 10. percentage yield 百分產率 92
5. limiting reactant 限量反應物 90 11. theoretical yield 理論產量 92
6. molar mass 摩爾質量 69

Key equations
mass (g)
1. Number of moles (mol) = –1
molar mass (g mol ) (Section 12.2, p.71)
number of formula units
2. Number of moles (mol) = –1
Avogadro constant (mol ) (Section 12.2, p.71)
3. Percentage by mass of element A in a compound
relative atomic mass of A × number of atoms of A in the formula
= × 100%
formula mass of the compound (Section 12.3, p.73)

4. Percentage yield is a measure of the efficiency of a chemical reaction.

actual yield
Percentage yield = × 100%
theoretical yield (Section 12.8, p.92)

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Key concepts
12.1 The mole and Avogadro constant

Chapter review
1. Chemists use mole (abbreviation: mol) to describe the quantity of particles in a substance.

2. The Avogadro constant (symbol: L) is the number of atoms in exactly 12.0 g of carbon-12, i.e.
23 23 –1
6.02 × 10 . Avogadro constant is equal to 6.02 × 10 mol .

12.2 The mole and molar mass

3. The molar mass of a substance is the mass of one mole of the substance. The unit of molar
–1
mass is g mol .

12.3 Calculating percentage by mass of an element in a compound

4. Percentage by mass of element A in a compound

relative atomic mass of A × number of atoms of A in the formula
= × 100%
formula mass of the compound

12.4 Calculating the mass of an element in a compound

5. The mass of an element in a compound can be calculated from the mass of the compound and
the percentage by mass of the element in that compound.

12.5 Calculating the relative atomic mass of an element

6. The relative atomic mass of an element can be calculated from the formula of the compound
and percentage by mass of that element in the compound.

12.6 Determining empirical formulae from experimental data

7. The empirical formula of a compound is the formula which shows the simplest whole
number ratio of the atoms or ions present.

8. The empirical formula of a compound can be determined from its composition by mass. The
composition of a compound has to be determined by experiments.

12.7 Determining molecular formulae from experimental data

9. The molecular formula of a substance shows the actual number of each kind of atoms in one
molecule of the substance.

10. The molecular formula of a compound can be determined from its empirical formula and
relative molecular mass.

12.8 Reacting masses from chemical equations

11. The total mass of the products is equal to the total mass of the reactants. This is the
conservation of mass.

12. The limiting reactant is the reactant that is completely used up in a reaction. It limits the
amount of the product(s) formed in a reaction.

97
 III Metals
All answers TE
13. Theoretical yield is the maximum amount of product expected if the reaction proceeds exactly
as shown in the chemical equation.

14. Actual yield is the amount of product actually obtained at the end of the reaction.
Chapter review

Concept map
Complete the concept map by filling in the items listed below:

Avogadro constant, molar mass, molecular formula, molecule, relative atomic masses, relative
molecular mass

Number of
Mass formula units
equals equals
Number
(a) of moles (b)
equals
6.02 × 1023 mol ¯1

without unit,
equals

(d)
Formula mass

equals equals (f)

determine
(c)
Sum of Sum of relative
atomic masses of
of all atoms/ions in all atoms in a
(e)
a formula unit of a
substance

Empirical
formula

98
 Reacting masses 12
TE All answers
Chapter exercise
A. Fill in the blanks
Section 12.1 Section 12.2
1. One mole of any substance contains 2. The of a substance is the mass
formula units. This number is in grams of one mole of the substance.
called the .

Chapter exercise
Section 12.3
3. Percentage by mass of element A in a compound
of A × of atoms of A in the formula
= × 100%
Formula mass of the compound

Section 12.8 5. Percentage yield = × 100%

4. is the reactant that is not
in excess in a reaction. The reaction stops when it is
used up.

B. Practice questions
Section 12.2
6. Calculate the number of atoms in each of the following substances.
(a) 2 mol of sodium
(b) 1.5 mol of carbon monoxide
(c) 2.0 g of oxygen
23 –1
(Relative atomic mass: O = 16.0; Avogadro constant = 6.02 × 10 mol )

7. Calculate the number of ions in each of the following substances.

(a) 2 mol of ammonium chloride
(b) 0.5 mol of sodium carbonate
(c) 24.6 g of calcium nitrate
23 –1
(Relative atomic masses: N = 14.0, O = 16.0, Ca = 40.1; Avogadro constant = 6.02 × 10 mol )

Section 12.3
8. Calculate the percentage by mass of
(a) carbon in methane, CH4.
(b) sulphur in sodium sulphate, Na2SO4.
(c) oxygen in iron(II) sulphate-7-water, FeSO4․7H2O.
(d) water in sodium carbonate-10-water, Na2CO3․10H2O.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0, S = 32.1, Fe = 55.8)

99
 III Metals
All answers
Section 12.4
9. Calculate the mass of
(a) hydrogen in 10.0 g of methane, CH4.
(b) magnesium in 28.5 g of magnesium ethanoate, (CH3COO)2Mg.
(c) chlorine in 2 mol of iron(III) chloride-6-water, FeCl3․6H2O.
(d) water in 1.25 mol of calcium chloride-6-water, CaCl2․6H2O.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Mg = 24.3, Cl = 35.5, Ca = 40.1, Fe = 55.8)

Section 12.6
Chapter exercise

10. Determine the empirical formulae of compounds having the following composition by mass:
(a) 75% carbon, 25% hydrogen
(b) 86.6% lead, 13.4% oxygen
(c) 36.5% sodium, 25.4% sulphur, 38.1% oxygen
(d) 40.67% carbon, 23.73% nitrogen, 27.13% oxygen, 8.47% hydrogen
(e) A hydrated salt containing 37.11% copper, 41.68% chlorine (the rest being water of crystallization)
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, S = 32.1, Cl = 35.5, Cu = 63.5, Pb = 207.2)

C. Multiple-choice questions
Section 12.1 Section 12.6
11. Which of the following substances contains the least Direction: Questions 13 and 14 are about nandrolone.
number of ions? 13. Nandrolone is a drug banned by the International
A. 1 mol of silver sulphide, Ag2S Olympic Committee. It can improve the performance
B. 1 mol of copper(II) dichromate, CuCr2O7 of athletes by increasing the oxygen-carrying ability
C. 1 mol of aluminium sulphate, Al2(SO4)3 of their blood. Nandrolone has the molecular
D. 1 mol of magnesium nitrate, Mg(NO3)2 formula C18H26O2.

Section 12.2
12. One mole of oxygen and one mole of nitrogen have
the same
(1) mass.
(2) number of atoms.
(3) number of molecules.
Which of the following is the empirical formula of
A. (1) only
nandrolone?
B. (2) only
C. (1) and (3) only A. C9H13O
D. (2) and (3) only B. C18H26O2
C. C9H26O2
D. C18H13O

14. Which of the following is the molar mass of

nandrolone?
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
–1
A. 137.0 g mol
–1
B. 166.0 g mol
–1
C. 245.0 g mol
–1
D. 274.0 g mol
100
 Reacting masses 12
All answers
15. An oxide of metal X contains 83.0% of X. What is the Section 12.8
formula of the oxide? Direction: Questions 17 and 18 are about thermal
(Relative atomic masses: O = 16.0, X = 39.1) decomposition of calcium carbonate.
A. XO2 17. Calcium carbonate decomposes on heating to give
B. X2O calcium oxide and carbon dioxide.
C. X2O3 CaCO3(s) CaO(s) + CO2(g)
D. X3O2
What would be the theoretical mass of calcium
16. In an experiment, 4.40 g of a sample of oxide produced when 10.01 g of calcium carbonate
ZnSO4․xH2O was heated and 2.47 g of anhydrous decomposes completely by heat?

Chapter exercise
ZnSO4 was obtained. What is the value of x? (Relative atomic masses: C = 12.0, O = 16.0, Ca =
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, 40.1)
Zn = 65.4) A. 4.84 g
A. 2 B. 5.61 g
B. 5 C. 7.21 g
C. 7 D. 9.62 g
D. 10
18. If 4.31 g of calcium oxide is finally obtained, what is
the percentage yield of the reaction?
A. 44.8%
B. 59.8%
C. 76.8%
D. 89.0%

D. Structured questions
Section 12.2
19. For each of the following statements, decide whether or not it is right. If it is wrong, you are required to correct it.
(a) 32.0 g of oxygen (O2) contains the same number of molecules as 32.0 g of ozone (O3). (1 mark)
(b) 1 mol of Pb(NO3)2 contains 1 mol of ions. (1 mark)

Section 12.4
20. Fluoride ions, in the form of tin(II) fluoride, are often added to toothpastes.
(a) What is the chemical formula of tin(II) fluoride? (1 mark)
(b) Calculate the formula mass of tin(II) fluoride. (1 mark)
(c) Calculate the percentage by mass of fluorine in tin(II) fluoride. (1 mark)
(d) A tube of toothpaste contains 1.50 g of tin(II) fluoride. Calculate the mass
of fluorine in this tube of toothpaste. (1 mark)
(Relative atomic masses: F = 19.0, Sn = 118.7) Fluoride has long been used
in toothpastes to prevent tooth
decay.

Learn more: Building a better answer (see p.153–154)

101
 III Metals
All answers TE
Exam practice
A. Multiple-choice questions
Section 12.2 Section 12.3
1. 0.25 moles of a gas has a mass of 7 g. Which of the 5. Which calcium compound contains the greatest
following could be the molecular formula for the percentage by mass of calcium?
gas? A. calcium carbonate
A. C2H6 B. calcium nitrate
B. C2H4 C. calcium hydroxide
C. C3H8 D. calcium sulfate
D. C3H6 (OCR A Level Chemistry A 2017 Jun H432/01 Q4)
(Copyright © Scottish Qualifications Authority
Section 12.4
National 5 2017 May Section 1 Q8)
6. How many grams of iron can be extracted from
2. Which of the following statements about 36 g of 100 g of iron ore that contains 70% by mass of
pure water is/are correct? Fe2O3? (Assume that the impurities do not contain
Exam practice

(Relative atomic masses: H = 1.0, O = 16.0, Avogadro iron.)

23 –1
constant = 6.02 × 10 mol ) (Relative atomic masses: O = 16.0, Fe = 55.8)
(1) It contains 4 mol of hydrogen atoms. A. 12.3 g
23
(2) It contains 3 × 6.02 × 10 of atoms. B. 24.5 g
23
(3) It contains 3 × 6.02 × 10 of molecules. C. 48.9 g
A. (1) only D. 97.8 g
B. (2) only
7. What is the mass of oxygen in 24.0 g of
C. (1) and (3) only
CuSO4․5H2O(s)?
D. (2) and (3) only
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1,
3. If 2 g of helium contains y molecules, how many Cu = 63.5)
molecules are present in 38 g of fluorine?
A. 6.2 g
(Relative atomic masses: He = 4.0, F = 19.0) B. 9.6 g
1 C. 13.8 g
A. y
2 D. 21.7 g
B. y
C. 2y (HKDSE 2020 Paper 1A Q3)
D. 4y Section 12.5
4. If 8.0 g of sulphur dioxide gas contains n 8. Hydrated salt X․nH2O contains 62.9% of water by
molecules, how many molecules does 2.0 g of mass. Given that the molar mass of X is 106.0 g
–1
oxygen gas contain? mol , what is the value of n?
(Relative atomic masses: O = 16.0, S = 32.0) (Relative atomic masses: H = 1.0, O = 16.0)
A. 2.0 n A. 2
B. 4.0 n B. 5
C. 0.25 n C. 6
D. 0.50 n D. 10
(HKDSE 2018 Paper 1A Q4)

102
 Reacting masses 12
All answers
9. The composition by mass of element X in the 13. Refer to the following chemical equation:
compound K2XO4 is 26.8%. Which of the following Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
statements concerning X is/are correct?
N moles of Fe2O3 are allowed to react with 2N moles
(Relative atomic masses: O = 16.0, K = 39.1) of CO under suitable conditions until the reaction
(1) X is a transition metal. stops. How many moles of Fe are formed?
(2) X is an element in Group VI of the Periodic A. N
Table. B. 2N
(3) X is an element in the fourth period of the 2
C. N
Periodic Table. 3
A. (1) only 4
D. N
3
B. (2) only
(HKDSE 2014 Paper 1A Q4)
C. (1) and (3) only
D. (2) and (3) only 14. Zinc chloride has the formula ZnCl2. Which of the
(HKDSE 2021 Paper 1A Q19) following statements about zinc chloride is correct?
(Relative atomic masses: Cl = 35.5, Zn = 65.4)
Section 12.6
A. 1 mol of zinc atoms has been combined with 2
10. The relative atomic mass of metal X is 40.1. 8.42 g
mol of chlorine molecules.

Exam practice
of an oxide of X contains 2.40 g of oxygen. What is
B. 1 g of zinc has been combined with 2 g of
the mole ratio of X to oxygen in the oxide?
chlorine.
(Relative atomic mass: O = 16.0) C. It contains a higher percentage by mass of
A. 1:1 chlorine than zinc.
–1
B. 1:2 D. Its molar mass is 130.4 g mol .
C. 2:1
15. When 0.1 mol of an element Y is heated in air, there
D. 1:3
is an increase in mass of 2.4 g. Assume that Y reacts
Section 12.8 only with the oxygen in air and the reaction is
11. Silicon can be made by heating silicon tetrachloride, complete. Which of the following elements might Y
SiCl4, with zinc. be?

SiCl4 + 2Zn Si + 2ZnCl2 (Relative atomic masses: C = 12.0, O = 16.0, Al =

27.0, K = 39.1, Ca = 40.1)
8.50 g of SiCl4 is reacted with an excess of zinc. The
percentage yield of silicon is 90%. What is the mass A. Carbon
of silicon made? B. Aluminium
C. Potassium
A. 1.26 g
D. Calcium
B. 1.31 g
C. 1.40 g
D. 1.55 g
(OCR A Level Chemistry A 2017 Jun H432/01 Paper 1
Q7)

12. 0.80 g of element X is reacted with 0.40 g of O2 to

form an oxide with the formula X2O3. What is the
identity of element X?
A. Aluminium
B. Titanium
C. Germanium
D. Molybdenum
(OCR A Level Chemistry A 2020 Nov H432/01 Q3)

103
 III Metals

B. Structured questions
Section 12.7
16. Vanillin is a flavouring commonly used in the food industry. It has a relative molecular mass of 152.0, and contains
63.1% carbon, 5.3% hydrogen and 31.6% oxygen by mass. Deduce the molecular formula of vanillin.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0) (3 marks)

17. Paracetamol is the active ingredient of some commonly used painkiller. Paracetamol has a relative molecular mass
of 151.0 and it has the following composition by mass: carbon 63.58%, hydrogen 5.96%, nitrogen 9.27% and
oxygen 21.19%. Deduce the molecular formula of paracetamol.
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0) (3 marks)

Section 12.8
18. In an experiment, steam was passed over a piece of magnesium heated in a combustion tube.

combustion tube magnesium

steam in unreacted steam and

Exam practice

hydrogen out

heat

(a) Write the chemical equation for the reaction involved. (1 mark)
(b) Calculate the theoretical mass of magnesium oxide produced if 4.86 g of magnesium was used in the
experiment. Assume that all of the magnesium had reacted. (2 marks)
(Relative atomic masses: O = 16.0, Mg = 24.3)

19. Ethyne (C2H2) reacts with bromine (Br2) to form 1,1,2,2-tetrabromoethane (C2H2Br4).
C2H2(g) + 2Br2() C2H2Br4()
In an experiment, 2.00 g of ethyne reacted with 5.20 g of bromine.
(a) Calculate the theoretical yield of C2H2Br4(). (3 marks)
(b) Given that the yield of 1,1,2,2-tetrabromoethane was 90.0%, calculate the actual yield of
1,1,2,2-tetrabromoethane. (1 mark)
(Relative atomic masses: H = 1.0, C = 12.0, Br = 79.9)

20. Iron can be extracted from iron pyrite (FeS2) in two steps:

4FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

In an extraction, 12.96 g of iron pyrite was used and iron pyrite was the limiting reactant.
(a) Calculate the theoretical yield of iron. (2 marks)
(b) The mass of iron obtained from the extraction was 5.98 g.
(i) Calculate the percentage yield of iron.
(ii) Compare the actual yield with the theoretical yield and suggest TWO possible reasons for the difference.
(3 marks)
(Relative atomic masses: C = 12.0, S = 32.1, Fe = 55.8)

104
Answers to Chapter exercise (ii) No. (This is because magnesium is
a reactive metal/more reactive than
Chapter 10 carbon.) 1
16. (a) mercury(II) sulphide + oxygen
A. Fill in the blanks (p.22)
mercury + sulphur dioxide 1
1. free elements
copper(I) sulphide + oxygen
2. (a) Heating
copper + sulphur dioxide 1
(b) carbon
(b) Mercury, copper, aluminium 1
(c) Electrolysis
(c) Mercury, copper, aluminium 1
3. extraction
The more easily a metal can be extracted
4. easier; earlier
from its ores, the earlier it was discovered. 1
5. non-renewable; conserving

B. Practice questions (p.22) Chapter 11

6. (a) zinc blende A. Fill in the blanks (p.58)
(b) zinc oxide 1. metal reactivity series
(c) oxygen; sulphur dioxide; zinc oxide; zinc; 2. electrons; positive; electrons
carbon dioxide 3. lower
7. (a) Haematite 4. displace
(b) coke; limestone 5. ionic; spectator
(c) carbon monoxide
(d) iron(III) oxide; iron(III) oxide; carbon dioxide B. Practice questions (p.58)
8. (a) bauxite; aluminium oxide 6. (a) 4Al(s) + 3O2(g) 2Al2O3(s)
(b) electrolysis; melted (b) 2Al(s) + 3F2(g) 2AlF3(s)
(c) aluminium oxide; oxygen (c) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)
(d) 2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g)
C. Multiple-choice questions (p.23) (e) 3CuO(s) + 2NH3(g)
9. C 3Cu(s) + N2(g) + 3H2O(l)
Copper pyrite is a common ore of copper. Copper (f) 4FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)
exists as copper(II) iron(II) sulphide (CuFeS2) in 7. (a) Calcium burns quite vigorously with a brick-
this ore. red flame to produce a white powder.
10. B 2Ca(s) + O2(g) 2CaO(s)
11. A (b) Sodium melts to form a silvery ball and moves
mercury(II) oxide mercury + oxygen about quickly on the water surface. In addition,
12. B it burns with a golden yellow flame.
The more easily a metal can be extracted from its 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
ores, the earlier it was discovered. (c) An intense white light is produced. A white
13. D solid forms.
Mg(s) + H2O(g) MgO(s) + H2(g)
D. Structured questions (p.23)
(d) Zinc dissolves and colourless gas bubbles are
14. (a) (Metal) ores 1
given out.
(b) In the Earth’s crust 1
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
(c) (i) Galena 1
(e) Magnesium dissolves and colourless gas
(ii) Lead(II) sulphide 1
bubbles are given out.
15. (a) Lead 1
Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g)
(b) Carbon dioxide 1 + –
8. (a) H (aq) + OH (aq) H2O(l)
(c) lead(II) oxide + carbon – +
(b) Cl (aq) + Ag (aq) AgCl(s)
lead + carbon dioxide 1 + 2+
(c) Mg(OH)2(s) + 2H (aq) Mg (aq) + 2H2O(l)
(i) Yes. 2+ –
(d) Pb (aq) + 2I (aq) PbI2(s)
copper(II) oxide + carbon +
(e) 2H (aq) + CaCO3(s)
copper + carbon dioxide 1 2+
Ca (aq) + CO2(g) + H2O(l)

T1
 C. Multiple-choice questions (p.59) 16.0 × 11
(c)
9. B 55.8 + 32.1 + 16.0 × 4 + 7.0 × (1.0 × 2 + 16.0)
Sodium reacts vigorously with water while there is × 100% = 63.3%
no reaction between silver and water. (1.0 × 2 + 16.0) × 10
(d)
10. D 23.0 × 2 + 12.0 + 16.0 × 3 + (1.0 × 2 + 16.0) × 10
11. A × 100% = 62.9%
12. A 9. (a) Percentage by mass of H in CH4
1.0 × 4
13. D = × 100% = 25%
12.0 + 1.0 × 4
14. A Mass of H in 10.0 g of CH4
15. B = 10.0 g × 25% = 2.5 g
P is the least reactive as it exists as a free element (b) Percentage by mass of Mg in (CH3COO)2Mg
in nature. Q is more reactive than R as Q can 24.3
=
displace R from the aqueous solution of a (12.0 × 2 + 1.0 × 3 + 16.0 × 2) × 2 + 24.3
compound of R. × 100% = 17.1%
16. B Mass of Mg in 28.5 g of (CH3COO)2Mg
The ionic equation of the reaction is Ag (aq) +
+
= 28.5 g × 17.1% = 4.87 g
–
Cl (aq) AgCl(s) (c) Formula mass of FeCl3․6H2O
= 55.8 + 35.5 × 3 + (1.0 × 2 + 16.0) × 6 = 270.3
D. Structured question (p.60) Percentage by mass of Cl in FeCl3․6H2O
17. (a) P: aluminium oxide; Q: hydrogen; 35.5 × 3
= × 100% = 39.4%
R: aluminium chloride; 270.3
S: aluminium sulphate 4 Mass of 2 mol of FeCl3․6H2O
(b) 2Al(s) + 3H2O(g) Al2O3(s) + 3H2(g) 1 = 2 × 270.3 g = 540.6 g
+ 3+
(c) 2Al(s) + 6H (aq) 2Al (aq) + 3H2(g) 1 Mass of Cl in 2 mol of FeCl3․6H2O
(d) Aluminium atoms lose electrons to form = 540.6 g × 39.4% = 213 g
positive ions less readily than (d) Formula mass of CaCl2․6H2O
magnesium atoms. 1 = 40.1 + 35.5 × 2 + (1.0 × 2 + 16.0) × 6 = 219.1
Percentage by mass of H2O in CaCl2․6H2O
Chapter 12 (1.0 × 2 + 16.0) × 6
= × 100% = 49.3%
219.1
A. Fill in the blanks (p.99) Mass of 1.25 mol of CaCl2․6H2O
23
1. 6.02 × 10 ; Avogadro constant = 1.25 × 219.1 g = 273.9 g
2. molar mass Mass of H2O in 1.25 mol of CaCl2․6H2O
3. relative atomic mass; number = 273.9 g × 49.3% = 135.0 g
4. Limiting reactant 10. (a) C H
5. actual yield; theoretical yield
Mass / g 75 25
B. Practice questions (p.99) Number of
23 24 75 25
6. (a) 2 × 6.02 × 10 = 1.204 × 10 moles of atoms
12.0
= 6.25
1.0
= 25
23
(b) 1.5 × 2 × 6.02 × 10 = 1.806 × 10
24 / mol
2.0 23 22
Simplest whole
(c) × 2 × 6.02 × 10 = 7.525 × 10 6.25 25
16.0 × 2 number mole =1 =4
7.
23
(a) 2 × 2 × 6.02 × 10 = 2.408 × 10
24 6.25 6.25
ratio of atoms
23 23
(b) 0.5 × 3 × 6.02 × 10 = 9.03 × 10
24.6 23
∴ the empirical formula of the compound is
(c) × 3 × 6.02 × 10 CH4.
40.1 + (14.0 + 16.0 × 3) × 2
23
= 2.71 × 10
12.0
8. (a) × 100% = 75%
12.0 + 1.0 × 4
32.1
(b) × 100% = 22.6%
23.0 × 2 + 32.1 + 16.0 × 4

T2
 (b) Pb O C. Multiple-choice questions (p.100)
Mass / g 86.6 13.4 11. B
+
For (A): the substance contains 2 mol of Ag ions
Number of 2–
86.6 13.4 and 1 mol of S ions
moles of atoms = 0.418 = 0.838
207.2 16.0 2+
For (B): the substance contains 1 mol of Cu ions
/ mol
2–
and 1 mol of Cr2O7 ions
Simplest whole 3+
0.418 0.838 For (C): the substance contains 2 mol of Al ions
number mole =1 =2
0.418 0.418 2–
and 3 mol of SO4 ions
ratio of atoms
2+
For (D): the substance contains 1 mol of Mg ions
∴ the empirical formula of the compound is –
and 2 mol of NO3 ions
PbO2.
12. D
(c) Na S O The mass of 1 mole of O2 is 32.0 g and that of 1
Mass / g 36.5 25.4 38.1 mole of N2 is 28.0 g.
Number of 13. A
36.5 25.4 38.1
moles of = 1.59 = 0.791 = 2.38 14. D
23.0 32.1 16.0
atoms / mol Molar mass of nandrolone
–1
Simplest = (12.0 × 18 + 1.0 × 26 + 16.0 × 2) g mol
–1
whole 1.59 2.38 = 274.0 g mol
0.791
number mole 0.791 =1 0.791 15. B
= 2.01 ≈ 2 0.791 = 3.01 ≈ 3
ratio of
Mole ratio of X atoms to O atoms
atoms
83.0 100 – 83.0
= :
∴ the empirical formula of the compound is 39.1 16.0
Na2SO3. = 2.123 : 1.063
(d) C N H O =2:1
Mass / g 40.67 23.73 8.47 27.13
16. C
2.47 4.40 – 2.47
Mole ratio of ZnSO4 to H2O = :
Number of 40.67 23.73 8.47 27.13 161.5 18.0
moles of 12.0 14.0 1.0 16.0 = 0.0153 : 0.107
atoms / mol = 3.39 = 1.70 = 8.47 = 1.70 = 1 : 6.99
Simplest ≈1:7
whole 3.39 1.70 8.47 1.70 17. B
number mole 1.70 1.70 1.70 1.70 Number of moles of CaCO3
ratio of = 1.99 ≈ 2 = 1 = 4.98 ≈ 5 = 1 10.01
atoms = mol = 0.100 mol
40.1 + 12.0 + 16.0 × 3
∴ the empirical formula of the compound is ∴ theoretical mass of CaO produced = 0.100 ×
C2NH5O. (40.1 + 16.0) g = 5.61 g
(e) 18. C
Cu Cl H2O 4.31 g
Percentage yield of the reaction = × 100% =
100 – 37.11 5.61 g
76.8%
Mass / g 37.11 41.68 – 41.68 =
21.21 D. Structured questions (p.101)
Number of 19. (a) False
23
moles of 37.11 41.68 21.21 32.0 g of oxygen (O2) contains 6.02 × 10
= 0.584 = 1.17 = 1.18
formula units 63.5 35.5 18.0 molecules while 32.0 g of ozone (O3)
/ mol 32.0 g 23 –1
contains –1 × 6.02 × 10 mol
Simplest 16.0 × 3 g mol
23
whole 1.18 = 4.01 × 10 molecules. 1
0.584 1.17
number mole =1 =2 0.584 (b) False
0.584 0.584
ratio of = 2.02 ≈ 2 2+
1 mol of Pb(NO3)2 contains 1 mol of Pb
formula units –
ions and 2 mol of NO3 ions. Hence, 1 mol
∴ the empirical formula of the compound is of Pb(NO3)2 contains 3 mol of ions. 1
CuCl2․2H2O.
T3
 20. (a) SnF2 1 Distilled water contains dissolved air and the
(b) 118.7 + 19.0 × 2 = 156.7 1 dissolved air (oxygen) can be removed from the
19.0 × 2 water by boiling.
(c) × 100% = 24.3% 1
156.7
(d) 1.50 g × 24.3% = 0.365 g 1 D. Structured questions (p.137)
15. (a) 4Fe(s) + 3O2(g) + 2nH2O(l)
Chapter 13 2Fe2O3․nH2O(s) 1
A. Fill in the blanks (p.135) (b) It removes water (moisture) from the air. 1
1. Corrosion; oxygen; water (c) It prevents air from dissolving in the
2. Rusting boiled distilled water. 1
3. potassium hexacyanoferrate(III); phenolphthalein (d) (i) Tube 1 and Tube 4 1
4. (a) acidic substances This is because there is no water in
(b) soluble ionic Tube 1 and there is no air (oxygen)
(c) High in Tube 4. 1
(d) less reactive (ii) Tube 2
(e) scratched; bent; sharp This is because tap water contains
5. (b) Electroplating soluble ionic compounds. The presence
(c) Sacrificial of dissolved ions greatly increases the
(d) Cathodic conductivity of tap water. This speeds
6. aluminium oxide; anodization up the rusting of iron. 1
16. (a) Galvanized iron refers to an iron that has
B. Practice question (p.135) a zinc layer coated on its surface. 1
7. (a)  (b)  (b) The zinc in galvanized iron prevents iron
(c)  (d)  from contacting oxygen and water. 1
(e)  (f)  (c) No. Iron can still be protected from
(g)  (h)  corrosion even the zinc layer of galvanized
(i)  (j)  iron fences is damaged. This is because 1
zinc is more reactive than iron, it will
C. Multiple-choice questions (p.136)
corrode (or will be oxidized) in preference
8. D
to iron. 1
When copper corrodes, green solid is found on its
(d) By painting 1
surface. When silver corrodes, it becomes
17. (a) When aluminium is exposed to air, a thin
tarnished.
layer of aluminium oxide quickly forms on
9. B
its surface. 1
Sugar is water soluble but no ions would form when
The oxide layer sticks to the surface of
it dissolves in water. This is because sugar is not an
aluminium very strongly and is impermeable
electrolyte.
to both oxygen and water. Hence, it can
10. C
protect the aluminium from further
When an iron nail is wrapped with, attached or
corrosion. 1
connected to a less reactive metal, rusting becomes
(b) (i) Positive terminal 1
faster.
(ii) Negative terminal 1
11. C
(c) The mobile phone cases can be dyed to
In option (C), the iron nail is prevented from rusting
give attractive colours. 1
by sacrificial protection.
The mobile phone cases are harder. 1
12. B
13. C
The aluminium object to be anodized is connected
to the positive terminal of the d.c. power source.
Answers to Exam practice
Dilute H2SO4 is used as an electrolyte.
14. D
Chapter 10
A. Multiple-choice questions (p.25)
1. A
T4
2. D 5. B
3. D (78%) 6. B
Upon strong heating, silver oxide decomposes to Reactant Product
silver and oxygen. The oxygen escapes from the side side
test tube and hence the mass of the contents in the Number of Ti atoms 1 y (1)
test tube decreases. As the heating proceeds, Number of O atoms 2 z (2)
more silver oxide would have decomposed. When Number of Cl atoms 2w 4y (3)
all the silver oxide is reduced, the mass of the Number of C atoms x z (4)
contents in the test tube would remain unchanged. From (1), y = 1
From (2), z = 2
B. Structured questions (p.25)
From (4), x = z; ∴ x = 2
4. (a) Gold, silver, iron, aluminium 1
From (3), 2w = 4y; ∴ w = 2
(b) The less reactive a metal, the more easily
7. B (83%)
the metal can be extracted from its ores. 1 Reactant Product
(c) Gold, silver, iron, aluminium 1 side side
(d) Gold, silver, iron, aluminium 1 Number of Ni atoms 3 3
(e) This is because aluminium is extracted Number of O atoms 6 6
from its molten ores by electrolysis. 1
Number of H atoms 6+x 12 (1)
5. (a) This is because platinum is the least
Number of Cl atoms x + 4y 12 + z (2)
reactive among the four metals and exists
Number of Au atoms y y
as a free element in nature. 1
Net charge –y –z (3)
It can be extracted easily from its ores
From (1), 6 + x = 12; ∴ x = 6
by physical separation. 1
From (3), –y = –z
(b) Platinum is very rare in the Earth’s crust. 1
From (2), x + 4y = 12 + z; ∴ y = 2; z = 2
(c) Copper ores are running out/very hard
8. D (54%)
to find. 1
The more reactive a metal, the more stable is its
The demand for copper is very high. 1
carbonate and the carbonate is more stable to
(d) (i) Aluminium 1
heat. Z may be zinc as the thermal decomposition
This is because it is corrosion
of zinc carbonate gives zinc oxide, which is yellow:
resistant and it is relatively cheaper
ZnCO3(s) ZnO(s) + CO2(g)
than copper. 1
9. D (65%)
(ii) Iron 1
This is because it is very strong and B. Structured questions (p.62)
relatively cheap. 1 10. (a) Magnesium oxide 1
(iii) Platinum 1 (b) Magnesium burns with a very bright white
This is because it is rare and very light. 1
corrosion resistant. 1 (c) 2Mg(s) + O2(g) 2MgO(s) 1
(d) When calcium burns with oxygen, a
Chapter 11 brick-red light instead of a very bright
A. Multiple-choice questions (p.61) white light is produced. The brightness of
1. B the light produced by calcium is lower.
2. C Hence, calcium cannot be used to replace
The balanced equation for the reaction in (2) is: magnesium in a flashbulb. 1
(2): 2MgCO3 + 4HCl 2MgCl2 + 2CO2 + 2H2O 11. (a) Hydrogen/H2(g) 1
3. B (b) Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g) 1
A more reactive metal will displace the less reactive (c) The solution turns blue because the
metal from the oxide of the less reactive metal. reaction produces calcium hydroxide,
4. A which is alkaline. 2
X is more reactive than Y because Y cannot (d) No observable change 1
displace X from the oxide of X upon heating. Copper does not react with cold water. 1

T5
 (e) No. Potassium reacts vigorously with Number of molecules in 2 g of He = 0.5 × L = y,
water and it burns/the reaction evolves a where L represents the Avogadro constant. This
gas very rapidly in the set-up. This poses y
means that L = .
a danger. 1 0.5
y
12. (a) Oxygen 1 Hence, number of molecules in 38 g of F2 = 1 ×
0.5
(b) Silver 1 = 2y.
(c) Hydrogen 1 4. D (60%)
(d) Potassium/sodium 1 Number of moles of SO2 in 8.0 g of SO2
8.0
(e) (i) Zn(s) + 2HCl(aq) = mol = 0.125 mol
32.0 + 16.0 × 2
ZnCl2(aq) + H2(g) 1 Number of moles of O2 in 2.0 g of O2
+
Or, Zn(s) + 2H (aq) 2.0
2+ = mol = 0.0625 mol
Zn (aq) + H2(g) (1) 16.0 × 2
2+
(ii) Zn(s) + Cu (aq)
2+
Zn (aq) + Cu(s) 1 Number of molecules in 8.0 g of SO2 = 0.125 × L =
(f) A, C, B 1 n, where L represents the Avogadro constant. This
13. (a) Lead 1 n
means that L = .
(b) 2Al(s) + 3PbO(s) Al2O3(s) + 3Pb(s) 1 0.125
Hence, number of molecules in 2.0 g of O2 = 0.0625
(c) (i) Yes. 2Al(s) + Fe2O3(s)
n
Al2O3(s) + 2Fe(s) 1 × = 0.50n.
0.125
(ii) No. This is because aluminium is less 5. C
reactive than magnesium. 1 Percentage by mass of Ca in CaCO3
14. (a) Zinc is more reactive than copper. 1 40.1
= × 100% = 40.0%
Zn(s) + CuO(s) ZnO(s) + Cu(s) 1 40.1 + 12.0 + 16.0 × 3
(b) Zinc is less reactive than aluminium. 1 Percentage by mass of Ca in Ca(NO3)2
40.1
15. (a) S, Q, R, P 1 = × 100% = 24.4%
40.1 + (14.0 + 16.0 × 3) × 2
(b) (i) P 1
Percentage by mass of Ca in Ca(OH)2
(ii) Q 1 40.1
(c) P 1 = × 100% = 54.1%
40.1 + (16.0 + 1.0) × 2
(d) The blue aqueous solution turns colourless. 1 Percentage by mass of Ca in CaSO4
Reddish brown solid/precipitate forms. 1 40.1
= × 100% = 29.4%
2+
Mg(s) + Cu (aq)
2+
Mg (aq) + Cu(s) 1 40.1 + 32.1 + 16.0 × 4
6. C
Chapter 12 Percentage by mass of Fe in Fe2O3
55.8 × 2
A. Multiple-choice questions (p.102) = × 100% = 69.9%
55.8 × 2 + 16.0 × 3
1. B Mass of Fe2O3 present in 100 g of iron ore = 100 ×
7 –1 –1
Molar mass of the gas = g mol = 28.0 g mol 70% = 70 g
0.25
The molar mass of C2H4 is (12.0 × 2 + 1.0 × 4) g ∴ mass of Fe that can be extracted = 70 g × 69.9%
–1
mol = 28.0 g mol
–1 = 48.9%
2. A 7. C (82%)
Number of moles of H2O in 36 g of pure water Mass of O in 24.0 g of CuSO4․5H2O
36 16.0 × 9
= mol = 2 mol = 24.0 g ×
1.0 × 2 + 16.0 63.5 + 32.1+ 16.0 × 4 + 5 × (1.0 × 2 + 16.0)
Hence, 36 g of pure water contains 2 × 3 × 6.02 × = 13.8 g
23
10 of atoms and contains 2 × 6.02 × 10 of
23 8. D
100 – 62.9 62.9
molecules. Mole ratio of X to H2O = :
106.0 18.0
3. C = 0.35 : 3.49
2
Number of moles of He in 2 g of He = mol = 0.5 = 1 : 9.97
4.0
mol ≈ 1 : 10
38
Number of moles of F2 in 38 g of F2 = mol
19.0 × 2
= 1 mol

T6
9. C (55%) For (D): molar mass of ZnCl2 = 65.4 + 35.5 × 2 g
–1 –1
Let the relative atomic mass of X be a. mol = 136.4 g mol
a 26.8 15. B
=
39.1 × 2 + a + 16.0 × 4 100 Number of moles of O atoms in the oxide
100a = (142.2 + a) × 26.8 2.4
a = 52.0 = mol = 0.15 mol
16.0
From the Periodic Table, X is Cr. It is a transition Number of moles of Y atoms = 0.1 mol (given)
metal and is in the fourth period of the Periodic Mole ratio of Y atoms to O atoms = 0.1 : 0.15 = 2 :
Table. 3
10. A So, the formula of the oxide is Y2O3. Y might be
Mole ratio of X atoms to O atoms aluminium because the ratio of Al atoms to O atoms
8.42 – 2.40 2.40 in the empirical formula of aluminium oxide is 2 : 3.
= :
40.1 16.0
= 0.150 : 0.150 B. Structured questions (p.104)
=1:1 16. C H O
11. A Mass / g 63.1 5.3 31.6
From the equation, mole ratio of SiCl4 to Si = 1 : 1.
Number of
8.50 63.1 5.3 31.6
Theoretical yield of Si = × 28.1 g = moles of = 5.26 = 5.30 = 1.98
28.1 + 35.5 × 4 12.0 1.0 16.0
1.40 g atoms / mol
Mass of Si made = 1.40 g × 90% = 1.26 g Simplest 5.26 5.30 1.98
12. A whole = 2.66 = 2.68 =1
1.98 1.98 1.98
Let the relative atomic mass of X be a. number mole
0.80 ratio of atoms 8 8 3
Number of moles of X = mol
a ∴ the empirical formula of vanillin is C8H8O3. 1
0.40
Number of moles of O = mol = 0.025 mol Let the molecule formula of vanillin be
16.0
(C8H8O3)n, where n is an integer.
0.80
a n × (12.0 × 8 + 1.0 × 8 + 16.0 × 3) = 152.0 1
2
Mole ratio of X : O = = n=1
0.40 3
16.0 ∴ the molecular formula of vanillin is C8H8O3. 1
a = 48.0 17. C N H O
∴ X is titanium. (The relative atomic mass of titanium Mass / g 63.58 9.27 5.96 21.19
is 47.9) Number of 63.58 9.27 5.96 21.19
13. D (62%) moles of atoms 12.0 14.0 1.0 16.0
N mol of Fe2O3 requires 3N mol of CO for complete / mol = 5.30 = 0.662 = 5.96 = 1.32
reaction. However, only 2N mol of CO is available.
Simplest whole 5.30 0.662 5.96 1.32
Hence, CO is the limiting reactant. number mole 0.662 0.662 0.662 0.662
From the equation, mole ratio of CO to Fe = 3 : 2. ratio of atoms = 8 =1 =9 =2
2N 4N
∴ number of moles of Fe formed = ×2= ∴ the empirical formula of paracetamol is
3 3
14. C C8NH9O2. 1
Zinc reacts with chlorine according to the following Let the molecule formula of paracetamol be
equation: (C8NH9O2)n, where n is an integer.
Zn(s) + Cl2(g) ZnCl2(s) n × (12.0 × 8 + 14.0 + 1.0 × 9 + 16.0 × 2)
For (A): 1 mol of Zn atoms requires 1 mol of chlorine = 151.0 1
molecules for complete reaction. n=1
1g ∴ the molecular formula of paracetamol is
For (B): 1 g of Zn requires –1 × (35.5 ×
65.4 g mol
–1 C8NH9O2. 1
2) g mol = 1.09 g of chlorine for complete reaction.
18. (a) Mg(s) + H2O(g) MgO(s) + H2(g) 1
For (C): percentage by mass of Cl in ZnCl2 =
35.5 × 2
× 100% = 52.1%
65.4 + 35.5 × 2

T7
 (b) From the equation, mole ratio of Mg to 5. A
MgO = 1 : 1. In cathodic protection, a conductor such as
Theoretical mass of MgO produced graphite or platinum alloy is connected to the
4.86 positive terminal of the d.c. power supply.
= × (24.3 + 16.0) g 1
24.3 6. D (93%)
= 8.06 g 1
7. D (55%)
19. (a) Number of moles of C2H2
In option (D), the iron nail corrodes fastest. This is
2.00
= mol = 0.0769 mol because the iron nail is connected to a less reactive
12.0 × 2 + 1.0 × 2 1
Number of moles of Br2 metal and is immersed in an aqueous solution of a
5.20 soluble ionic compound. Besides, the positive pole
= mol = 0.0325 mol
79.9 × 2 of the d.c. power supply draws electrons away from
0.0325
0.0325 mol of Br2 requires only mol the iron nail in the set-up. This also promotes the
2 2+
= 0.0163 mol of C2H2 for complete reaction. formation of Fe (aq).
Hence, Br2 is the limiting reactant. 8. A (66%)
From the equation, mole ratio of Br2 to The oxide of aluminium and that of iron are both
C2H2Br4 = 2 : 1. insoluble in water.
Theoretical yield of C2H2Br4
B. Structured questions (p.140)
1
= 0.0325 × × (12.0 × 2 + 1.0 × 2 + 79.9 9. (a) Fe2O3․nH2O 1
2
× 4) g 1 (b) The rust formed is just loosely attached to
= 5.62 g 1 the iron surface. It falls off from the iron
(b) Actual yield of C2H2Br4 = 5.62 g × 90% surface easily. 1
= 5.06 g 1 When the fresh iron surface is exposed to
20. (a) From the equations, 1 mol of FeS2 air, it reacts with oxygen and water. Thus,
produces 1 mol of Fe. rusting continues until the iron object
Theoretical yield of Fe corrodes completely. 1
12.96 (c) (i) Tin layer prevents iron from contacting
= × 55.8 g 1
55.8 + 32.1 × 2
oxygen and water. 1
= 6.03 g 1
(ii) Magnesium is more reactive than iron.
(b) (i) Percentage yield of Fe
5.98 g It will lose electrons (corrode) instead
= × 100% = 99.2% 1 of iron. 1
6.03 g
(ii) Any TWO of the following: OR
• The reactions may not complete. 1 Magnesium blocks offer sacrificial
• Impurities are present in the iron protection to iron for preventing
pyrite. 1 underground water pipelines from
• Some iron is lost during transfer or rusting. (1)
weighing. (1) (iii) The car battery provides electrons to
(Accept other reasonable answers.) the car body, preventing iron from
forming iron(II) ions. 1
Chapter 13 10. (a) Tube 1: iron nail rusts because air (oxygen)
A. Multiple-choice questions (p.139) and water are present. 1
1. C Tube 2: iron nail does not rust because
Rusting refers to the corrosion of iron only. there is no water. 1
2. B (87%) Tube 3: iron nail does not rust because
The iron hook in contact with a less reactive metal there is no air (oxygen). 1
(e.g. copper) would corrode faster. Tube 4: iron nail does not rust because
3. D the paint can serve as a protective layer
Zinc offers sacrificial protection to iron for that prevents the iron nail from contacting
preventing the crash barrier from rusting. oxygen and water. 1
4. B (86%) Tube 5: iron nail does not rust because
stainless steel is resistant to corrosion. 1
T8
 (b) Magnesium is more reactive than iron. 1 2. C
It provides sacrificial protection to iron for Potassium burns with a lilac flame and hydrogen is
preventing steel from rusting. 1 produced when it is added to water. The reaction is
(c) Aluminium has a coating of aluminium exothermic.
oxide. 1 3. A
It protects the metal from further corrosion. 1 4. B
(The author is responsible for the solutions and A more reactive metal can displace any less
that (a) they have neither been provided nor reactive metal from an aqueous solution of the
approved by AQA and (b) they may not necessarily compound of the less reactive metal.
constitute the only possible solutions.) 5. A
11. (a) Steel is mainly made up of iron. The iron Number of moles of atoms in 28.0 g of N2
reacts with oxygen and water as the cables 28.0
= mol × 2 = 2 mol
are exposed to air. 1 14.0 × 2
Number of moles of atoms in 2.0 g of H2
(b) Steel is much stronger than aluminium. It
2.0
can support a heavier load (weight). 1 = mol × 2 = 2 mol
1.0 × 2
(c) (i) The paint would be scratched off Number of moles of atoms in 18.0 g of O2
when cable cars move. 1 18.0
= mol × 2 = 1.13 mol
(ii) The oil can serve as a protective layer 16.0 × 2
that prevents the iron from contacting Number of moles of atoms in 34.0 g of Cl2
oxygen and water. 1 34.0
= mol × 2 = 0.958 mol
35.5 × 2
(iii) Oil can also serve as a lubricant. 1
Number of moles of atoms in 40.0 g of Ar
(iv) Repeated applications are required. 1 40.0
OR = mol = 1 mol
40.0
Dust would stick to the oil. (1) 6. D
12. (a) Oxygen and water 2 Molar mass of the substance
(b) Zinc 1 36 g –1
= = 90.0 g mol
(c) Magnesium is higher than iron in the 0.4 mol
reactivity series. It would lose electrons 7. D
(corrode) instead of iron. 1 Mole ratio of NiSO4 to H2O
2.29 3.91 – 2.29
Copper is lower than iron in the reactivity = :
154.8 18.0
series. Placing copper blocks on the ship’s = 0.0148 : 0.09
hull would speed up the corrosion of iron. 1 = 1 : 6.08
(d) (i) Fe2O3 + 2H3PO4 2FePO4 + 3H2O 1 ≈1:6
(ii) Iron(III) phosphate acts as a barrier 8. A
which prevents iron from contacting Let the relative atomic mass of metal M be a.
oxygen and water. 1 For oxide X, mole ratio of M atoms to O atoms
(Cambridge Assessment International Education 3.760 4.707 – 3.760
= :
bears no responsibility for the example answers to a 16.0
questions taken from its past question papers =1:1
which are contained in this publication. In ∴ a = 63.5
examinations, the way marks are awarded may be For oxide Y, mole ratio of M atoms to O atoms
different.) 3.760 4.234 – 3.760
= :
63.5 16.0
= 0.0592 : 0.0296
=2:1
Answers to Part exercise ∴ the formula of oxide Y is M2O.
9. D
Chapter 13
Number of moles of Al
A. Multiple-choice questions (p.142) 1.00
= mol = 0.0370 mol
1. D 27.0
In general, the more easily the metal that can be Number of moles of Cr2O3
extracted from its ore, the earlier it was discovered. 4.00
= mol = 0.0263 mol T9
52.0 × 2 + 16.0 × 3
 0.0370 14. (a) (i) The iron wool reacted with oxygen in
As 0.0370 mol of Al requires only mol =
2
the air. 1
0.0185 mol of Cr2O7 for complete reaction. Hence,
As the oxygen inside the burette was
Cr2O3 is in excess.
consumed, the pressure inside the
Theoretical yield of Cr = 0.0370 × 52.0 g = 1.924 g
burette was reduced. The atmospheric
∴ percentage yield of the reaction
pressure forced the water in the trough
1.75 g
= × 100% = 91.0% into the burette. 1
1.924 g
10. C (ii) Fe changed to Fe2O3․nH2O in rusting. 1
Let the relative atomic mass of metal T be a. Fe2O3․nH2O has a higher formula
T + Cl2 TCl2 mass than Fe. 1
4.86 (b) (i) To study the effect of temperature on
Number of moles of T = mol
a
the rate (speed) of rusting 1
Number of moles of TCl2
19.06 (ii) Independent variable: temperature 1
= mol Dependent variable: mass of iron wool
a + 35.5 × 2
From the equation, 1 mole of T produces 1 mole of after the experiment 1
TCl2. Control variable: mass of iron wool at
4.86 19.06 the start of the experiment/volume of
∴ : =1:1
a a + 71.0 air inside the burette at the start of
a = 24.3 the experiment 1
11. D (38%) (iii) The rate (speed) of rusting increases
Oxygen is the limiting reactant as the copper or with temperatures. 1
iron used is in excess here. 15. (a) X, iron, Y 1
Helium has no reaction with the metals. Y can displace both iron and X from their
12. D respective chloride solutions. 1
2+ 3+
Fe can form Fe and Fe ions while Mg can form Iron can displace X from the chloride
2+
only Mg ions. solution of X, whereas it cannot displace
Y from the chloride solution of Y. 1
B. Structured questions (p.143)
(b) (i) Y(s) + FeCl2(aq) YCl2(aq) + Fe(s) 1
13. (a) Potassium compounds are mostly soluble in
OR
water. Thus, these compounds dissolve in 2+ 2+
Y(s) + Fe (aq) Y (aq) + Fe(s) (1)
the water body forming dissolved salts. 1
(ii) Fe(s) + XCl2(aq) FeCl2(aq) + X(s) 1
Calcium carbonate is water insoluble. Thus,
OR
it is found in sediments. 1 2+ 2+
Fe(s) + X (aq) Fe (aq) + X(s) (1)
(b) The electronic arrangements of potassium
(c) Displacement reaction 1
and calcium atoms are 2,8,8,1 and 2,8,8,2
(d) Y
respectively. 1
This is because Y is more reactive than iron
In order to attain the stable electronic
and it would lose electrons (corrode)
arrangement of argon atom, a potassium
instead of iron. 1
atom has to lose only one outermost shell
electron while a calcium atom has to lose
two. 1
(c) Both potassium and calcium are reactive
Answers to Revision test
metals. They are both extracted by
electrolysis of the molten metal ores.
Chapter 13
However, electrolysis can only be used A. Multiple-choice questions (p.146)
after the invention of electricity. 1 1. D
(d) For potassium compounds: lilac 1 2. B
For calcium compounds: brick-red 1 3. B

T10
4. D Each sodium atom has fewer protons
Reactant Product (positively charged) in the nucleus than
side side
each calcium atom. The outermost shell
Number of Cu atoms x x
electron in the sodium atom is less
Number of H atoms y 8 (1) strongly bound to the nucleus compared
Number of N atoms y 2x + 2 (2) to those in the calcium atom. (1)
Number of O atoms 3y 6x + 6 13. (a) 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) 1
From (1), y = 8 (b) 2ZnO(s) + C(s) 2Zn(s) + CO2(g) 1
From (2), y = 2x + 2 Carbon is used as a reducing agent in
i.e. 8 = 2x + 2 this step. 1
∴x=3 (c) Number of moles of ZnS
5. B 2437.5
= mol = 25.0 mol 1
6. D 65.4 + 32.1
1 mole of C2H4 contains 4L hydrogen atoms. Number of moles of C
Let the number of moles of C2H4 that contains n 480.0
= mol = 40.0 mol 1
hydrogen atoms be x. 12.0
1 4L From the equations, 25.0 mol of ZnS
By simple proportion, = 25.0
x n requires only mol = 12.5 mol of C
n 2
x=
4L for complete reaction. Hence, C is in excess.
7. C Theoretical yield of Zn
Mass of water of crystallization in 5.0 g of = 25.0 × 65.4 g = 1635 g 1
CuSO4․5H2O (d) (i) Sacrificial protection 1
(1.0 × 2 + 16.0) × 5 (ii) Zinc loses electrons more readily than
= 5.0 g ×
63.5 + 32.1 + 16.0 × 4 + (1.0 × 2 + 16.0) × 5 iron. It ‘sacrifices’ itself while the iron
= 1.8 g
does not rust. 1
8. C
(iii) Magnesium 1
Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g)
(iv) Cathodic protection 1
Mass of MgSO4 produced
3.0 14. (a) Town gas is flammable. 1
= × (24.3 + 32.1 + 16.0 × 4) g = 14.9 g (b)
24.3 Cu O
9. C 22.16 – 20.10 22.68 – 22.16
10. A Mass / g
= 2.06 = 0.52
11. D Number of
The reactivity of the elements in Group I of the 2.06 0.52
moles of atoms = 0.0324 = 0.0325
63.5 16.0
Periodic Table increases down the group. / mol

B. Structured questions (p.147) Simplest whole

0.0324 0.0325
number mole =1 =1
12. (a) Sodium burns with a golden yellow flame 0.0324 0.0324
ratio of atoms
while calcium does not burn. 1
Sodium moves about quickly on the water ∴ the empirical formula of the oxide of
surface while calcium sinks in water. 1 copper is CuO. 3
(b) (i) 2Na(s) + 2H2O(l) (2 marks for correct calculations steps)
2NaOH(aq) + H2(g) 1 (c) CuO(s) + H2(g) Cu(s) + H2O(g) 1
(ii) Ca(s) + 2H2O(l) OR
Ca(OH)2(s) + H2(g) 1 CuO(s) + CO(g) Cu(s) + CO2(g) (1)
(c) Each sodium atom has to lose one (d) This is to prevent the hot copper formed
outermost shell electron in order to attain from reacting with oxygen again. 1
a stable octet electronic arrangement but
each calcium atom has to lose two. 1
OR

T11