Focus TG4 KSSM (Chemistry) Terbitan Penerbitan Pelangi Sdn Bhd

CONTENTS

1Chapter 4.2 The Arrangement in the Periodic 63
Table of Elements 67
Introduction to Chemistry 1 4.3 Elements in Group 18 69
4.4 Elements in Group 1 76
1.1 Development in Chemistry Field and Its 4.5 Elements in Group 17 84
Importance in Daily Life 2 4.6 Elements in Period 3 90
4.7 Transition Elements 92
1.2 Scientific Investigation in Chemistry 4
SPM Practice 4
1.3 Usage, Management and Handling of
Apparatus and Materials 5

SPM Practice 1 8

2Chapter Matter and the Atomic 5Chapter Chemical Bound 94
Structure 11

2.1 Basic Concepts of Matter 12 5.1 Basics of Compound Formation 95
2.2 The Development of the Atomic 5.2 Ionic Bond 98
Model 18 5.3 Covalent Bond 103
2.3 Atomic Structure 20 5.4 Hydrogen Bond 107
2.4 Isotopes and Its Uses 26 5.5 Dative Bond 111
30 5.6 Metallic Bond 113
SPM Practice 2 5.7 Properties of Ionic Compounds
and Covalent Compounds 114
3Chapter The Mole Concept, Chemical 120
Formula and Equation 33 SPM Practice 5

3.1 Relative Atomic Mass and Relative 34 6Chapter
Molecular Mass Acid, Base and Salt 123

3.2 Mole Concept 36 6.1 The Role of Water in Showing Acidic

3.3 Chemical Formula 42 and Alkaline Properties 124

3.4 Chemical Equation 52 6.2 pH Value 130

SPM Practice 3 57 6.3 Strength of Acids and Alkalis 133

4Chapter The Periodic Table of 6.4 Chemical Properties of Acids and 135
Elements 60 Alkalis

6.5 Concentrations of Aqueous Solution 142

4.1 The Development of the Periodic Table 6.6 Standard Solution 145

of Elements 61 6.7 Neutralisation 147

iv

6.8 Salts, Crystals and Their Uses in 154 8Chapter Manufactured Substances in
Daily Life 156 Industry 223
6.9 Preparation of Salts 170
6.10 Effect of Heat on Salts 178 8.1 Alloy and Its Importance 224
6.11 Qualitative Analysis 189 8.2 Composition of Glass and Its Uses 229
8.3 Composition of Ceramics and Its
SPM Practice 6 Uses 232
8.4 Composite Materials and Its
7Chapter Rate of Reaction 193 Importance 235
239
SPM Practice 8

7.1 Determining Rate of Reaction 194

7.2 Factors Affecting Rate of Reactions 201 PRE-SPM MODEL PAPER 242

7.3 Application of Factors that Affect the 212
Rate of Reaction in Daily Life

7.4 Collision Theory 213 ANSWERS 258

SPM Practice 7 220

v



3Chapter The Mole Concept, Chemical
Formula and Equation

CHAPTER FOCUS
• Relative Atomic Mass and Relative Molecular Mass
• Mole Concept
• Chemical Formula
• Chemical Equation

In 1814, Jacob Berzelius (1779 – 1848), a Swedish chemist, had developed a system
that is used till today to represent chemicals. The only difference is that superscript
numbers are used in the system while subscript numbers are used in modern chemical
formulae. For example, he used the formula H2O for water while modern chemistry uses
the formula H2O.

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

3.1 Relative Atomic Mass and Relative Molecular Mass

Relative Atomic Mass, RAM

1. An atom has a very small mass and thus, is difficult to measure. Therefore, chemists compare the mass
of an atom with that of a standard atom.

2. The mass of an atom compared to a standard atom is called relative atomic mass. Relative atomic
mass has no unit as it is just a comparison value.


At first, hydrogen
atom which is the In 1850, oxygen In 1961, carbon-12 was
lightest atom was used atom was used as the accepted as the single standard
standard atom. atom internationally.
as the standard atom. • The mass of carbon-12 atom is
Chapter • The relative atomic
• The mass of hydrogen taken as 12 units.
masses of most • Carbon-12 was chosen due to a
atom is taken as 1 unit.
elements can be few reasons:
• Problems that arose: – Carbon-12 is the most
determined as most of
– The relative atomic abundant carbon isotope
them can combine with (almost 98%). So, the mass of
3 masses of most 12 units given to a carbon-12
elements cannot be oxygen. atom is accurate.
• Problems arose when – Carbon-12 is easy to handle in
determined as they do laboratories as it exists as a solid
three oxygen isotopes at room temperature.
not easily combine with
were found.
hydrogen.
– Chemists used natural
– Hydrogen gas is difficult
oxygen which contains
to handle.
all the three isotopes as

the standard.

– Physicists used oxygen-16 – Many elements can combine with

isotope as the standard. carbon-12.

– The mass spectrometer at that time

was already using carbon-12 as the

Figure 3.1 The history of choosing the standard atom in standard.
determining relative atomic masses

3. Based on the carbon-12 scale, the relative atomic mass of an element is defined as the average mass
1
of an atom of the element compared to 12 of the mass of a carbon-12 atom.

Relative atomic mass of an element = Average mass of one atom of the element

1 × mass of one carbon-12 atom
12

4. For example, the relative atomic mass of lithium is 7. This means that the mass of one lithium atom
1
is 7 times of 12 the mass one carbon-12 atom.

SPM Highlights

The average mass of a molybdenum atom is 96 times larger than 1 of the mass of a carbon-12 atom. What is the
relative atomic mass of molybdenum? 12

A 8 C  48

B 12 D  96

Examiner’s tip

Students need to remember the definition of ‘relative atomic mass’ (refer to the definition above). Based on the

definition, the relative mass of molybdenum is 96. Note that the relative mass of one carbon-12 atom is taken as

exactly 12 units. So, the phrase 1 of the mass of carbon-12’ equals to 1 unit.
12
Answer: D

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

5. We can solve various problems by comparing 3. One molecule is made up of a few atoms.
the relative atomic masses of elements. See Therefore, we can calculate the relative
Examples 3.1 and 3.2. molecular mass of a substance by adding
up the relative atomic masses of all the
EXAMPLE 3.1 atoms that form the molecule. To do
so, you need to know the formula of the
How many beryllium atoms have the same mass substance. See Figure 3.2 and Table 3.1.

as two silver atoms? Carbon diokside
molecule, CO2
[Relative atomic mass: Be, 9; Ag, 108] OC
O
Solution RAM for O = 16 RAM for O = 16

The number of beryllium atoms = n RAM for C = 12

The mass of n beryllium atoms = Mass of 2 RMM for CO2 = 16 + 12 + 16 = 44

silver atoms Figure 3.2 Calculating the relative molecular mass of Chapter
carbon dioxide
So, n × RAM of Be = 2 RAM of Ag

n × 9 = 2 × 108 Table 3.1 Calculation of relative molecular mass of
2 × 108 several substances
n = 9
Molecular Formula of Relative molecular mass,
= 24 atoms 3
substance substance RMM

Hydrogen H2 2(1) = 2
gas

EXAMPLE 3.2 Oxygen gas O2 2(16) = 32

How many times is the mass of a zinc atom Water H2O 2(1) + 16 = 18
larger than the mass of a helium atom?
[Relative atomic mass: He, 4; Zn, 65] Ammonia NH3 14 + 3(1) = 17

Ethanol C2H5OH 2(12) + 5(1) + 16 + 1 = 46

Solution [Relative atomic mass: H, 1; C, 12; N, 14; O, 16]

Mass of a zinc atom = RAM of Zn Relative Formula Mass, RFM
Mass of a helium atom RAM of He 1. The term ‘relative molecular mass’ is only

= 65 used for substances made up of molecules. For
4 substances made up of ions, we use the term
‘relative formula mass’.
= 16.25 times
2. In the same manner, we can calculate the
Relative Molecular Mass, RMM relative formula mass of an ionic substance by
adding up the relative atomic mass of all the
1. Based on carbon-12 scale, relative molecular atoms of elements shown in the formula. See
Table 3.2.
mass of a substance is defined as the average
Table 3.2 Calculation of relative formula mass of a few
mass of a molecule of the substance compared substances
1
to 12 of the mass of a carbon-12 atom. Ionic substance Formula of Relative formula mass,
substance RFM
Relative molecular = Average mass of one Sodium chloride
mass of a substance molecule of the substance Aluminium oxide NaCl 23 + 35.5 = 58.5
Calcium Al2O3 2(27) + 3(16) = 102
1 × mass of one hydroxide Ca (OH)2
12 carbon-12 atom Magnesium 40 + 2[16+ 1]= 74
nitrate Mg (NO3)2
2. For example, the relative molecular mass of Copper(II) 24 + 2[14 + 3(16)] = 148
sulphate hydrate CuSO4.5H2O
carbon dioxide is 44. This means that one 64 + 32 + 4(16) + 5[2(1) +
1 16] = 250
12
molecule of carbon dioxide is 44 times the [Relative atomic mass : H,1; N, 14; O, 16; Na, 23,
mass of a carbon-12 atom. Mg, 24; S, 32; Cl, 35.5; Ca, 40; Cu, 64]

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

3. Using the knowledge about relative mass, we Q2 The mass of a sulphur atom is 8 times more than
can solve various calculation problems. See the mass of a helium atom. What is the relative
Example 3.3. atomic mass of sulphur?

[Relative atomic mass: He, 4]

EXAMPLE 3.3 Q3 Calculate the relative molecular mass of the
following substances:
The relative formula mass of X2SO4 is 142. (a) Phosphorus, P4
Calculate the relative atomic mass of element X. (b) Carbon monoxide gas, CO
[Relative atomic mass: O, 16; S, 32] (c) Sucrose, C12H22O11
(d) Benzoic acid, C6H5COOH
Solution [Relative atomic mass: H, 1; C, 12; O, 16; P,
31]
RAM of X = x

Chapter Given RFM of X2SO4 = 142 Q4 Calculate the relative formula mass of the
2(x) + 32 + 4(16) = 142 following substances:
(a) Sodium oxide, Na2O
2x + 96 = 142 (b) Zinc nitrate, Zn(NO3)2
(c) Potassium thiosulphate, K2S2O3
2x = 142 – 96 = 46 (d) Hydrated sodium carbonate, Na2CO3.10H2O
46 [Relative atomic mass: H,1; C,12; N,14; O,
x = 2 = 23 16; Na, 23; S, 32; K, 39; Zn, 65]

3 RAM of X is 23.

Checkpoint 3.1 Q5 Element Z forms a chloride salt with the formula
ZCI2 and relative formula mass of
Q1 How many oxygen atoms have the same mass the relative atomic mass of element 95. What is
as three copper atoms? Z?

[Relative atomic mass: C, 12; Cu, 64] [Relative atomic mass: Cl, 35.5]

3.2 Mole Concept

1. Chemists use the ‘mole’ unit to represent the quantity of a substance.

One mole of substance is Avogadro’s constant, NA is defined
as the number of particles.
defined as the quantity of (NA = 6.02 × 1023 mol-1)

substance containing the

same number of atoms

contained in 12 g of
carbon-12, which is 6.02 ×
1023 particles



2. The number of particles per mole, which is 6.02 × 1023 is determined experimentally and is known as
Avogadro’s constant or Avogadro’s number.

3. The way of using the mole unit is exactly the same as that of using the dozen unit. See Figure 3.3.

1 dozen of pencils = 12 units of pencils 2 dozen of pencils = 24 units of pencils

36 Contains Contains
6.02 x 1023 2 x 6.02 x 1023
molecules molecules
of water of water

Figure 3.3 The mole unit is used in the same way as the dozen unit

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

4. The particles referred depends on the type of substance (Table 3.3).

Table 3.3 Type of substance and type of particles

Type of Atomic substance Molecular substance Ionic substance
substance (consists of atoms) (consists of molecules) (consists of ions)

Example of • All metal elements • Most non-metal elements such as All ionic compounds
substances • All noble gases hydrogen (H2), oxygen (O2), nitrogen such as sodium chloride
• Some non-metal elements (N2), fluorine (F2), chlorine (Cl2), and zinc oxide
bromine (Br2), iodine (I2), sulphur (S8)
such as carbon and silicon and phosphorus (P4).

• All covalent compounds

The meaning 1 mole of atomic substance 1 mole of molecular substance has 6.02 1 mole of ionic substance Chapter
of 1 mole of has 6.02 × 1023 atoms. For × 1023 molecules. For example, 1 mole has 6.02 × 1023 formula
substance example, 1 mole of zinc of oxygen gas contains 6.02 × 1023 O2 units. For example, 1
molecules. mole of sodium chloride
contains 6.02 × 1023 zinc contains 6.02 × 1023 NaCl

atoms. units.

3

5. We can calculate the number of particles in a fraction of mole of substance by using the Avogadro’s
constant, NA as follows. (NA: 6.02 × 1023 mol-1)

Number × NA Number of
of moles ÷ NA particles

EXAMPLE 3.4

Calculate the number of atoms in:
1. 0.5 mole of copper
2. 2 moles of hydrogen gas

Solution
1. Copper is an atomic substance.
So, 1 mole of copper has 6.02 × 1023 of copper atoms.
The number atoms in 0.5 mole of copper
= Number of mole × NA
= 0.5 × 6.02 × 1023 atoms
= 3.01 × 1023 atoms
2. Hydrogen gas is a molecular substance.
So, 1 mole of hydrogen gas has 6.02 × 1023 H2 molecules.
The number of H2 molecules in 2 moles of hydrogen gas
= number of mole × NA
= 2 × 6.02 × 1023 H2 molecules
Based on the formula H2, each H2 molecule is made up of 2 H atoms.
Therefore, the number of H atoms in 2 moles of hydrogen gas
= 2 × 2 × 6.02 × 1023 atoms
= 24.08 × 1023 atoms
= 2.408 × 1024 atoms

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

EXAMPLE 3.5 Table 3.3 Molar mass of a few substances

How many moles of carbon dioxide contain Substance Relative mass Molar
1.806 × 1023 CO2 molecules? mass
(g mol-1)

Solution Atomic Magnesium, Mg RAM = 24 24
The number of moles of carbon dioxide 65
substance Zinc, Zn RAM = 65

= Number of particles ÷ NA Hydrogen gas, RMM = 2(1) = 2 2
= 1.806 × 1023 16
Molecular H2 RMM = 12 + 4(1)
6.02 × 1023 substance Methane, CH4 = 16

= 0.3 mol Zinc bromide, RFM = 65 + 2(80) 225
= 225 142
Ionic ZnBr2
RFM = 2(23) + 32
substance Sodium sulphate, + 4(16) = 142

Chapter EXAMPLE 3.6 Na2SO4

0.5 mole of magnesium chloride, MgCl2 are [Relative atomic mass : H,1; C,12; O,16; Na, 23;
dissolved in a beaker of water. How many ions
Mg, 24; S, 32; Zn, 65; Br, 80]
3 are present in the beaker?
Solution Figure 3.4 Molar mass of magnesium is 24 g mol-1
Magnesium chloride, MgCl2 is an ionic
substance. So, 1 mole of magnesium chloride EXAMPLE 3.7
has 6.02 × 1023 MgCl2 units.
The number of MgCl2 units in 0.5 mole of What is the mass of 1.2 moles of sodium
magnesium chloride hydroxide, NaOH?
= number of moles × NA [Relative atomic mass: H, 1; O, 16; Na, 23]
= 0.5 × 6.02 × 1023 units
Solution
Each MgCl2 unit is made up of three ions, RFM of sodium hydroxide, NaOH
which are one magnesium ion and two chloride = 23 + 16 + 1 = 40
ions. So, the number of ions in the beaker So, the molar mass of sodium hydroxide
= 3 × number of MgCl2 = 40 g mol-1
= 3 × 0.5 × 6.02 × 1023 ions The mass of 1.2 moles of sodium hydroxide
= 9.03 × 1023 ions = number of moles × molar mass
= 1.2 × 40 = 48 g
Number of moles and mass of substance

1. Chemists measure the number of moles of EXAMPLE 3.8
substance by just weighing its mass. Mass of
substance and number of moles are linked by How many moles of molecules are there in 450
molar mass as shown below. g of water, H2O?
[Relative atomic mass: H, 1; O, 16]
Number × molar mass Mass in
of moles ÷ molar mass grams Solution

RMM of water, H2O = 2(1) + 16 = 18
2. Molar mass of a substance is the mass of one So, the molar mass of water = 18 g mol-1
mole of the substance in grams. The unit for
molar mass is gram per mole (g mol-1). The number of moles of molecules in 450 g of

water

= mass ÷ molar mass

3. The value of molar mass is the same as the = 450 = 25 mol
relative mass of the substance. 18

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

SPM Highlights SPM Tips

Which substance has a different mass with 1 mole of For molecular elements such as oxygen, make sure
glucose, C6H12O6? you read the question carefully to understand what
[Relative atomic mass: H,1; C, 12; O, 16] is required in the question. Use RAM to calculate
A 3 moles of propanol, C3H7OH the number of atoms of oxygen, O. Use RMM to
B 4 moles of propane, C3H6 calculate the number of molecules of oxygen,
C 6 moles of ethane, C2H6 O2.
D 10 moles of water, H2O
5. We can compare the number of particles in
Examiner’s tip substances by just comparing the number of
Students need to know the molar mass and the mass moles of the substances. See Example 3.10.

of each substance, including glucose.

Substance Mass(g) = number of moles × EXAMPLE 3.10 Chapter
molar mass
1 mole C6H12O6 How many times the number of atoms in 7 g of
3 moles 6(12) + 12(1) + 6(16) = 180 nitrogen gas is larger than that in 7 g of iron?
C3H7OH [Relative atomic mass: N, 14; Fe, 56]
4 moles C3H6 3 × [3(12) + 7(1) + 16 + 1] = 3 ×
6 moles C2H6 60 = 180 Solution 3
10 moles H2O We only need to compare the number of moles
4 × [3(12) + 6(1)] = 4 × 42 = 168 of both samples.

6 × [2(12) + 6(1)] = 6 × 30 = 180

10 × [2(1) + 16] = 10 × 18 = 180

Answer: B The number of moles of nitrogen atoms
= mass ÷ molar mass
4. Substances with the same number of moles = 7 ÷ 14
have equal number of particles even though = 0.5 mol
the particles are different in size and mass. The number of moles of iron atoms
= mass ÷ molar mass
12 g 28 g = 7 ÷ 56
0.5 mole of magnesium 0.5 mole of iron = 0.125 mol

Number of moles of nitrogen atoms = 0.5 =4
Number of moles of iron atoms 0.125
Figure 3.5 Both blocks of magnesium and iron have 0.5 x
6.022 × 1023 atoms Therefore, 7 g of nitrogen gas has 4 times more
[Relative atomic mass: Mg, 24; Fe, 56] of atoms than that in 7 g of iron.

EXAMPLE 3.9 Number of moles and volume of gases

What is the mass of gold that has the same 1. The number of moles of a gas can also be
number of atoms as 4 g of oxygen? determined through its volume. This is not true
[Relative atomic mass: O, 16; Au, 197] of a solid or liquid substance.

Solution 2. All gases with the same volume under the
The number of oxygen atoms, O same pressure and temperature, have the same
= mass ÷ molar mass number of particles.
= 4 ÷ 16 = 0.25 mol
0.25 mole of gold has the same number of 3. The volume of one mole of gas is known as
atoms as 0.25 mole of oxygen. So, the mass of molar volume.
gold = number of moles × molar mass • Molar volume of any gas is 22.4 dm3
= 0.25 × 197 mol-1 at STP or 24 dm3 mol-1 at room
= 49.25 g conditions.
• STP refers to the standard temperature of
0ºC and the pressure of 1 atm.

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

• Room conditions refer to the temperature EXAMPLE 3.11
of 25ºC and the pressure of 1 atm.
What is the volume of 1.2 moles of neon gas at
1 mole of H2 1 mole of O2 1 mole of N2 STP?
[Molar volume: 22.4 dm3 mol-1 at STP]
Figure 3.6 Each balloon has 6.02 x 1023 molecules and has
the volume of 22.4 dm3 at STP Solution
The volume of neon gas
4. Number of moles and volume of gas are linked = number of moles × molar volume
by molar volume as shown below. = 1.2 × 22.4 dm3
= 26.88 dm3

Chapter Number × molar volume Volume EXAMPLE 3.12
of moles ÷ molar volume of gas
100 cm3 of carbon dioxide gas is collected in a
3 SPM Tips reaction at room conditions. How many moles
of carbon dioxide is produced? [Molar volume:
In calculations, make sure that the volume of gas and 24 dm3 mol-1 at room conditions]
molar volume are of the same unit, which means that
both are in cm3 or both are in dm3. The changing of Solution
unit is as follows .
The volume of carbon dioxide gas = 100 cm3
1 dm3 = 1000 cm3
= 100 dm3 = 0.1 dm3
1000

The number of moles of carbon dioxide

produced

= Volume of gas ÷ molar volume

= 0.1 = 0.0042 mol
24

Number moles, number of particles, mass and volume
1. The following shows the relationships between all quantities of chemicals.

Number of ÷ NA × Molar mass Mass
particles × NA BNilaunmgbaenrmofol
÷ Molar mass
moles

× Molar volume ÷ Molar volume Number

of moles
calculation
Volume of gas
VIDEO

2. Number of moles is the intermediate unit to convert other quantities of chemicals such as number of
particles, mass and volume. See the examples below.

Table 3.4 Examples of conversion of quantity of chemicals

Conversion Calculation steps
From number of particles to mass Number of particles → number of moles → mass (see Example 3.13)
From mass to volume of gas Mass → number of moles → volume of gas (see Example 3.14)
From volume of gas to number of particles Volume of gas → number of moles → number of particles (see Example 3.15)

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

EXAMPLE 3.13 SPM Highlights

What is the mass of magnesium powder that 63.5 g of gas X occupies 5600 cm3 at STP. What is
has 3.612 × 1024 magnesium atoms? the molar mass of gas X?
[Molar volume: 22.4 dm3 mol-1 at STP]
[Relative atomic mass: Mg, 24. Avogadro’s
constant, NA: 6.02 × 1023 mol-1] A 56
B 127
Solution C 180
D 254
Number of moles of Mg == nu36m.6.0b12e2r××of11p00a22r34ticles ÷ NA

Examiner’s tip

First, analyse the question. The question gives the

= 6 mol mass and volume of the gas. To determine the molar
mass of gas X, students need to calculate the mass
Therefore, the mass of magnesium powder of 1 mole of gas X or 22 400 cm3 at STP.
Number of moles of gas X that occupies 5600 cm3
= number of moles × molar mass Chapter

= 6 × 24 = volume of gas
molar volume
= 144 g

= 5 600
22 400
EXAMPLE 3.14 3
= 0.25 mol
What is the volume of 100 g of ammonia gas, So, the mass of 0.25 mole of gas X is 63.5 g.
NH3 at STP? This means that the mass of 1 mole of gas X
[Relative atomic mass: H, 1; H, 14. Molar
volume: 22.4 dm3 mol-1 at STP] = 63.5
0.25

= 254 g
So, the molar mass of gas X is 254.

Solution Answer: D

Number of moles of NH3 = mass ÷ molar mass Checkpoint 3.2
=141+003(1)
[Relative atomic mass: H,1; N, 14; O,16; Ne, 20; Na,
=11070 = 5.88 mol
23; S, 32; K, 39; Fe, 56; Cu, 64; Br, 80. Avogadro’s
So, volume of NH3 constant, NA: 6.02 × 1023 mol-1. Molar volume: 24 dm3
= number of moles × molar volume mol-1 at room conditions.]

= 5.88 × 22.4 Q1 A sample consists of 2.5 moles of carbon dioxide

= 131.71 dm3 gas, CO2.
(a) How many molecules are there in the sample?

(b) Hence, calculate the number of atoms in

the sample.

EXAMPLE 3.15 (c) What is the volume of the sample at room

conditions?

How many hydrogen molecules, H2 are there in Q2 Calculate the number of moles of each of the
6 dm3 of the gas at room conditions?
following.
[Molar volume: 24 dm3 mol-1 at room
conditions. Avogadro’s constant, NA: 6.02 × (a) 13.8 g of sodium, Na
1023 mol-1]
(b) 40.4 g of potassium nitrate, KNO3
Q3 An experiment requires 0.5 mole of copper(II)

Solution sulphate, CuSO4.
(a) Calculate the molar mass of copper(II)

Number of moles of H2 = volume ÷ molar volume sulphate.

(b) What is the mass of copper(II) sulphate

= 6 = 0.25 mol required in the experiment?
24
Q4 What is the mass of carbon that has twice the

So, the number of H2 molecules amount of atoms found in 11.2 g of iron?

= number of moles × Avogadro’s constant Q5 Arrange the following samples of gases in an

= 0.25 × 6.02 × 1023 ascending order of volume.
X: 5 g of neon gas, Ne
Y:
= 1.505 × 1023 molecules Z: 10 g of ammonia gas, NH3
16 g of bromine gas, Br2

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

3.3 Chemical Formula

What is chemical formula?

1. A chemical formula is a representation of a chemical substance using letters for atoms and subscript
numbers to show the number of each type of atoms that are present in the substance.

2. A chemical formula tells us the composition of a chemical. See Table 3.5.

Table 3.5 Information from chemical formulae

Chemical Chemical formula Information

Carbon C Made up of carbon atoms • Elements
Argon Ar Made up of argon atoms • Type of particles : atoms
Zinc Zn Made up of zinc atoms
Chapter
Hydrogen gas H2 Made up of diatomic molecules • Elements
Ozone gas O3 • Type of particles: molecules
S8 Made up of triatomic molecules
3 Sulphur H2O The formula shows the number
Water Made up of molecules with 8 atoms of atoms in each molecule
NH3
Ammonia Contains hydrogen dan oxygen with the • Compounds
Zn(OH)2 ratio H:O = 2:1 • Show the elements that made
Zinc hydroxide
Magnesium MgSO4 Contains nitrogen dan hydrogen with the up the chemicals
sulphate ratio N:H = 1:3 • Show the ratio of number of

Contains zinc, oxygen dan hidrogen with atoms of each element in
the ratio Zn:O:H = 1:2:2 chemicals
• Type of particles: May be
Contains magnesium, sulphur dan oxygen molecules or ions
with the ratio Mg:S:O = 1:1:4

3. For compounds, there are two types of chemical formulae, namely molecular formula and empirical
formula.

Molecular formula Empirical formula

Definition Definition

Formula that shows the actual number Formula that shows the simplest whole
of atoms of each element present in one number ratio of atoms of each element

molecule of the compound present in the compound

Example Example

Compound Molecular Simplest whole number ratio of Empirical
formula atoms of elements formula
Water H:O=1:1
Ethane H2O H2O
Benzene C2H4 C:H=1:4 CH2
Vitamin C C6H6 CH
C6H8O6 H:H=1:1
C3H4O3
C:H:O=3:4:3

Similarities

Both show the type of elements in the compound
Figure 3.7 Comparing molecular formula and empirical formula

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

Determining empirical formula

1. The empirical formula of a compound is determined by investigating the simplest ratio of the number
of moles of the elements in the compound in a chemical laboratory.

Determining the mass Converting the mass of Determining the
of each element in the each element into the
number of moles of simplest ratio of
compound number of moles of
atoms atoms of each element


Figure 3.8 Steps in determining the empirical formula of a compound

EXAMPLE 3.16

1.08 g of aluminium powder combines completely with 0.96 g of oxygen to produce an oxide Chapter
compound. What is the empirical formula of the oxide?
[Relative atomic mass: O, 16; Al, 27]

Solution

Element Aluminium, Al Oxygen, O 3

Mass (g) 1.08 0.96

Number of moles of atoms 1.08 = 0.04 0.96 = 0.06
27 16
Ratio of number moles of
atoms 0.04 =1 0.06 = 1.5
Simplest ratio of number of 0.04 0.04
moles of atoms
2 3

2 moles of aluminium atoms combine with 3 moles of oxygen atoms. So, the empirical formula of
aluminium oxide is Al2O3.

EXAMPLE 3.17

23.05 g of lead(II) iodide contains 12.7 g of iodine. What is the empirical formula of lead(II) iodide?
[Relative atomic mass: I, 127; Pb, 207]

Solution

Element Lead, Pb Iodine, I
Mass (g)
23.05 – 12.7 = 10.35 12.7
Number of moles of atoms
10.35 = 0.05 12.7 = 0.1
Ratio of number of moles of 207 127
atoms
Simplest ratio of number of 0.05 =1 0.1 =2
moles of atoms 0.05 0.05

1 2

1 mole of lead atoms combine with 2 moles of iodine atoms. So, the empirical formula of lead(II)
iodide is PbI2.

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

EXAMPLE 3.18

Phosphoric acid contains 3.06% of hydrogen, 31.63% of phosphorus and 65.31% of oxygen. Calculate
the empirical formula of the acid. [Relative atomic mass: H,1; O, 16; P, 31]

Solution
Based on its percentage composition, each 100 g of phosphoric acid contains 3.06 g of hydrogen, 31.63
g of phosphorus and 65.31g of oxygen. The calculation based on the 100 g of the acid is as follows.

Elements Hydrogen, H Phosphorus, P Oxygen, O

Mass (g) 3.06 31.63 65.31

Number of moles of atoms 3.06 = 3.06 31.63 = 1.02 65.31 = 4.08
1 31 16

Chapter Ratio of number of moles of 3.06 =3 1.02 =1 4.08 =4
atoms 1.02 1.02 1.02

Simplest ratio of number of 3 1 4
moles of atoms

3 3 moles of hydrogen atoms combine with 1 mole of phosphorus atoms and 4 moles of oxygen atoms.
So, the empirical formula of phosphorus acid is H3PO4.

ACTIVITY 3.1

Aim: To determine the empirical formula of magnesium oxide.

Materials: Magnesium ribbon and sand paper.

Apparatus: Crucible with lid, tongs, Bunsen burner, tripod stand, pipe-clay triangle and electronic
balance.

Procedure:

1. A crucible with its lid is weighed.

2. 10 cm of magnesium ribbon is cleaned with sand paper to remove its oxide layer.

3. The magnesium ribbon is coiled loosely and placed in the crucible. The crucible, its lid and
content are weighed.

4. The apparatus is set up as shown in Figure 3.9. Crucible Lid
5. The crucible is heated strongly without its lid.
6. When the magnesium ribbon starts to burn, the crucible Magnesium
coil

is covered with its lid. Heat Pipe-clay
7. Using a pair of tongs, the lid is raised a little at intervals. triangle

8. When the burning is complete, the lid is removed and

the crucible is heated strongly for 1 to 2 minutes.

9. The crucible is covered with its lid and allowed to Figure 3.9 Determining the empirical
cool down to room temperature. formula of magnesium oxide

10. The crucible, its lid and content are weighed again.

1 1. The processes of heating, cooling and weighing are repeated a number of times until a constant
mass is obtained. The constant mass is recorded.

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

Results: Table 3.6

Items weighed Mass (g)

Crucible + lid x

Crucible + lid + magnesium ribbon y

Crucible + lid + magnesium oxide z

Calculation:
Mass of magnesium used = (y – x) g
Mass of oxygen that combines with it = (z – y) g

Table 3.7 Chapter

Element Magnesium, Mg Oxygen, O

Mass (g) y–x z–y 3

Number of moles of atoms y–x z–y
24 16

Simplest ratio of number of moles of atoms p q

[Relative atomic mass: O, 16; Mg, 24]
Based on the calculation, p mol of magnesium atoms combine with q mol of oxygen atoms. So, the
empirical formula of magnesium oxide is MgpOq.

Discussion:
1. Magnesium reacts with oxygen in the air to produce a white powder of magnesium oxide.

Magnesium + oxygen → magnesium oxide

2. The following are precautions taken during the activity.

(a) The crucible is opened at intervals to allow oxygen to enter and react with the magnesium.

(b) The crucible is then quickly closed to prevent the white powder of magnesium oxide from
escaping into the air. The loss of the white powder will affect the accuracy of mass obtained.

(c) Heating, cooling and weighing are repeated a number of times until a constant mass is obtained
to ensure that the magnesium reacts completely with oxygen.

3. This method is also used to determine the empirical formulae of oxides of reactive metals such as
calcium oxide, aluminium oxide and zinc oxide.

Conclusion:
The empirical formula of magnesium oxide is MgO whereby p = 1 and q = 1.

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

ACTIVITY 3.2

Chapter Aim: To determine the empirical formula of copper(II) oxide

Materials: 2 mol dm-3 sulphuric acid, 1 mol dm-3 copper(II) sulphate solution, zinc granules, copper(II)
oxide, anhydrous calcium chloride and wooden splinter.

Apparatus: Combustion tube with one hole at the end, Bunsen burner, flat-bottomed flask, thistle
funnel, stoppers, glass tube, retort stand and clamp, balance, U-tube, spatula and porcelain
dish.

Procedure:
1. The mass of the combustion tube with the porcelain dish in it is weighed.
2. One spatulaful of copper(II) oxide is added to the porcelain dish. The tube is weighed again.
3. The apparatus is set up as shown in Figure 3.10

Hydrogen Copper(II) oxide

Thistle gas Burning of excess

funnel hydrogen gas

3 Dilute sulphuric Anhydrous Heat Combustion tube
acid + calcium Porcelain dish
Copper(II) chloride Retort
sulphate solution stand

Zinc

Figure 3.10 Determining the empirical formula of copper(II) oxide

4. Hydrogen gas is allowed to flow through the set of apparatus for 5 to 10 minutes to remove all air
from the tube.

5. To determine whether all air has been removed, the gas that comes out from the small hole is
collected in a test tube. Then, the gas is tested with a lighted wooden splinter. If the gas burns
without any ‘pop’ sound, then all the air has been totally removed from the combustion tube.

6. The excess hydrogen gas that flows out of the small hole of the combustion tube is burnt and the
copper(II) oxide is heated strongly.

7. The Bunsen flame is turned off when the copper(II) oxide turns completely brown.

8. The flow of hydrogen gas is continued until the set of apparatus cools down to room temperature.

9. The mass of the combustion tube with its content is weighed again.

10. The heating, cooling and weighing are repeated until a constant mass is obtained. The constant
mass is recorded.

Results: Table 3.8

Items weighed Mass (g)
x
Combustion tube + porcelain dish y
z
Combustion tube + porcelain dish + copper(II) oxide

Combustion tube + porcelain dish + copper

Calculation:
Mass of copper = (z – x) g
Mass of oxygen = (y – z) g

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

Element Table 3.9 Oxygen, O

Copper, Cu y–z

Mass (g) z–x y–x
16
Number of moles of atoms z–x
64 s

Simplest ratio of number of moles of atoms r

[Relative atomic mass : O, 16; Cu, 64] Chapter
Based on the calculation, r mol of copper atoms combine with s mol of oxygen atoms. So, the empirical
formula of copper(II) oxide is CurOs. 3

Discussion:

1. The function of anhydrous calcium chloride is to dry the hydrogen gas.

2. The reaction that occurs:

Copper(II) oxide  +  hydrogen gas  →  copper  +  water
(black powder) (brown solid) (released into the air)

3. The precautions and safety measures taken:
• All air from the combustion tube must be totally removed before step 6 is carried out. If not,
the mixture of hydrogen and air will explode when lighted.
• The flow of hydrogen gas is continuous throughout the activity so that no air enters the
combustion tube. If not,
– explosion may occur
– the hot copper produced will react with oxygen from the air to form copper(II) oxide again.
• Heating, cooling and weighing are repeated until a constant mass is obtained to ensure that all
copper(II) oxide has changed into copper.

4. This method can also be used to determine the empirical formula of oxides of less reactive metals
such as tin(II) oxide and lead(II) oxide.

5. Hydrogen gas cannot be replaced by reactive metals such as magnesium to determine the empirical
formula of copper(II) oxide because the products of the reaction are all in solid state. Therefore, the
mass of copper cannot be determined.

Copper(II) oxide + magnesium → copper + magnesium oxide

This mixture of products cannot be separated

Conclusion:
The empirical formula of copper(II) oxide is CuO, whereby r = 1 and s = 1.

Determining molecular formula
1. Actually, molecular formula is a multiple of empirical formula, where n is an integer.

Molecular formula = (Empirical formula)n

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

Table 3.10 Molecular formula are multiples of empirical formula

Compound Empirical formula n Molecular formula

Water H2O 1 (H2O)1 = H2O
Ethane CH3 2 (CH3)2 = C2H6
Propane CH2 3 (CH2)3 = C3H6
Glucose CH2O 6 (CH2O)6 = C6H12O6
Ethanoic acid CH2O 2 (CH2O)2 = C2H4O2 or CH3COOH

Chapter 2. The determination of a molecular formula EXAMPLE 3.20
requires the following information:
• Its empirical formula What is the mass of zinc that is needed to
• Its relative molecular mass or its molar combine with 0.5 mole of chlorine atoms to
mass form zinc chloride, ZnCl2?
[Relative atomic mass: Zn, 65]
EXAMPLE 3.19
Solution
3 A compound has an empirical formula of CH2 Based on the formula ZnCl2, 2 moles of
and a relative molecular mass of 70. What is the chlorine atoms will combine with 1 mole of
compound’s molecular formula? zinc atoms. So, 0.5 mole of chlorine atoms will
[Relative atomic mass: H, 1; C, 12] combine with 0.25 mole of zinc atoms. The
mass of zinc needed
Solution = number of moles × molar mass
Let the molecular formula of the compound to = 0.25 × 65
be (CH2)n. = 16.25 g
Based on the formula, its relative molecular
mass should be EXAMPLE 3.21
= n[12 + 2(1)] = 14n

It is given that its relative molecular mass = 70 4.05 g of metal W reacts with bromine to form
40.05 g of a compound with the empirical
So, 14n = 70 formula of WBr3. Determine the relative atomic
70 mass of element W.
n = 14 =5 [Relative atomic mass: Br, 80]

Therefore, the molecular formula of the

compound is (CH2)5, which is C5H10. Solution

It is given that the mass of metal W = 4.05 g.

Calculation involving empirical and So, the mass of bromine in the compound
molecular formula
= (40.05 – 4.05) g = 36 g
1. Empirical formula and molecular formula can
help us to solve calculation problems related The number of moles of bromine atoms
to the composition of compounds. 36
= mass ÷ molar mass = 80
2. The percentage composition based on mass can
be calculated as follows: = 0.45 mol

Based on the empirical formula of WBr3, 3
moles of Br atoms will combine with 1 mole

of W atoms. So, 0.45 mole of Br atoms will

Mass of the combine with 0.45 ÷ 3 or 0.15 mole of W atoms.
element in 1 mole
Percentage of an of the compound If the mass of 0.15 mole of W atoms is 4.05 g,

element by mass = Mass of 1 mole of × 100% therefore the mass of 1 mole of W atoms
in a compound the compound
= 4.05 g = 27 g
0.15

Therefore, the relative atomic mass of W is 27.

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

EXAMPLE 3.22 SPM Highlights

Calculate the percentage by mass of carbon in What is the mass of oxygen in 132 g of carbon
octane, C8H18 [Relative atomic mass: H, 1; C, 12]
dioxide, CO2?
Solution [Relative atomic mass: C,12; O, 16]
The mass of 1 mole of octane, C8H18
= 8(12) + 18(1) A 16 g C 48 g
= 114 g
Based on its formula, 1 mole of C8H18 has 8 B 32 g D 64 g
moles of carbon atoms.
The mass of carbon in 1mole of C8H18= 8 × 12 = 96 g Examiner’s tip

So, the percentage by mass of carbon Number of moles of CO2

Mass 132 132
= Molar mass = 12+2(16) = 44 = 3 mol

= Mass of carbon in 1 mole of C6H14 × 100% Based on the formula, 1 mole of CO2 contains 2 Chapter
Mass of 1 mole of C6H14 moles of O atoms.
96 × 100% So, 3 mol of CO2 contains 3 × 2, which is 6 moles of
= 114 O atoms.
Mass of oxygen = number of moles × molar mass
= 84.21% = 6 × 16 3

= 96 g

Answer: C

Chemical formula of ionic compounds

1. The formula of an ionic compound combines the formula of its cation (positive ion) and the formula
of its anion (negative ion).

Formula of ionic compound: Formula of cation Formula of anion

Examples Examples

Charge Name of cation Formula of cation Charge Name of anion Formula of anion
Na+ -1 Chloride ion Cl–
Sodium ion K+ Bromide ion
H+ -2 Iodide ion Br –
+1 Potassium ion NH4+ -3 Hydroxide ion
Hydrogen ion Ca2+ Nitrate ion I–
Mg2+ Hydride ion
Ammonium ion Zn2+ Chlorate(I) ion OH–
Fe2+ Chlorate(V) ion
Calcium ion Pb2+ Manganate(VII) ion NO3–
Sn2+ Oxide ion H–
Magnesium ion Cu2+ Sulphate ion
Al3+ Sulphide ion ClO–
Zinc ion Fe3+ Thiosulphate ion
Cr3+ Chromate(VI) ion ClO3–
+2 Iron(II) ion Pb4+ Dichromate(VI) ion MnO4–
Sn4+ Phosphate ion O2-
Lead(II) ion
SO42-
Tin(II) ion S2-

Copper(II) ion S2O32-
CrO42-
Aluminium ion Cr2O72-
PO43-
+3 Iron(III) ion

Chromium(III) ion

+4 Lead(IV) ion
Tin(IV) ion

Figure 3.11 Formula of ionic compounds combine the formula of cations and anions

2. However, the formula of an ionic compound is neutral because the total positive charges equal the
total negative charges.

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

① Based on its name, write the formula of Ionic compound: Iron(III) chloride
the cation followed by the formula of
the anion. Fe3+ CI-

② Determine the number of cations and 1 Fe3+ ion 3 CI- ions
Total positive charges Total negative charges
anions by balancing the positive = 1(+3) = 3(-1)
charges and negative charges. = +3 = -3

③ Write the number of cations and anions Formula : FeCI3
as subscript numbers.

Figure 3.12 Steps in constructing the formula of an ionic compound

Chapter EXAMPLE 3.23 SPM Tips

3 Write the formula of sodium sulphate. Students need to memorise the formulae of cations
Solution and anions. If not, the chemical formulae cannot be
Ionic compound: Sodium sulphate constructed correctly.

Na+ SO42– 3. When polyatomic ions such as OH–, NO3–,
SO42- and NH4+ are involved, brackets are used
2 Na+ ions: 1 SO42– ions:
Total positive charges Total negative charges to show the number of ions in the formula. See
= 2(+1) = 1(-2) Figure 3.12.
= +2 = -2
Mg(OH)2 Fe(NO3)3 (NH4)2SO4

The brackets The brackets The brackets

show there are show there are show there are

Formula : Na2SO4 2 groups of 3 groups of 2 groups of

OH– NO3– NH4+

Figure 3.13 Brackets are used to show the number
of polyatomic ions

SPM Highlights

The formula of calcium chloride and potassium bromate(V) are CaCl2 and KBrO3. What is the formula of calcium
bromate((V)?

Examiner’s tip

From the formula of KBrO3, 1 ion K+ is needed to balance 1 bromate((V) ion. So, it can be deduced that the formula
of bromate(V) ion is BrO3-.
Ionic compound: Calcium bromate(V)

Ca2+ BrO3 –

1 ion Ca2+: 2 ion BrO3–:
Total positive charges Total negative charges
= 1(+2) = 2(-1)
= +2 = -2

Formula : Ca(BrO3)2

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

Naming of chemical compounds

1. Chemical compounds are named systematically according to the guidelines given by the International
Union of Pure and Applied Chemistry (IUPAC).

Name of cation Name of anion ① The name of cation is
written first followed by
Magnesium sulphate the name of the anion.
Aluminium chloride

② For elements that can form more Lead(II) oxide 'II’ shows Pb2+ ion
than one type of ion, Roman Lead(IV) oxide 'IV’ shows Pb4+ ion
numerals such as I and II are used to
differentiate the ions.

Figure 3.14 Naming of ionic compounds Chapter

① The name of less Hydrogen sulphide ② The name of the more 3
electronegative element electronegative element is
is written first. Its name is added with the ending ~ide.
maintained as it is.

Shows that there is one carbon Carbon dioxide
atom and two oxygen atoms in one
molecule of CO2. ③ Greek prefixes such as ‘di’
are used to show the number
of atoms of each element in
the molecule.

Figure 3.15 Naming of simple molecular compounds

Table 3.11 Example of names of simple molecular compounds

Chemical formula Name Note

CO Carbon monoxide Meaning of Greek prefixes:

SO3 Sulphur trioxide Mono = 1 ; Di = 2; Tri = 3 ;
CCl4
PCl5 Carbon tetrachloride Tetra = 4 ; Penta = 5 ; Hexa = 6;
Cl2O7
Phosphorus pentachloride Hepta = 7 ; Octa = 8 ; Nona = 9;

Dichlorin heptoxide Deca = 10

Checkpoint 3.3

[Relative atomic mass: N, 14; O, 16; Mg, 24, S, 32] Q4 Write the formula of the following ionic compounds.
Q1 A compound contains 23.3% of magnesium, 30.7%
(a) Potassium sulphate (d) Ammonium carbonate
of sulphur and 46.0% of oxygen. What is the
empirical formula of the compound? (b) Zinc chloride (e) Magnesium nitrate

Q2 A compound has a mass of 92 g mol-1. A sample of (c) Tin(II) oxide (f) Sodium phosphate
the compound is found to contain 0.606 g of nitrogen
and 1.39 g of oxygen. Q5 Name each of the following ionic compound.
(a) What is the empirical formula of the compound?
(b) Hence, determine the molecular formula and (a) Al(OH)3 (d) Ca(NO3)2
name of the compound. (b) FeSO4 (e) K2CO3
(c) NH4Cl (f) ZnS
Q3 Sodium thiosulphate has the formula of Na2S2O3.
(a) What is the mass of sulphur in 0.2 mole of Q6 A mixture contains carbon powder, magnesium
sodium thiosulphate? hydroxide and iron(II) bromide. Write the chemical
(b) How many moles of oxygen atoms are present
in 79 g of sodium thiosulphate? formulae of the substances in the mixture.

Q7 Name each of the following molecular substances.
NO ; B2O3 ; SF6

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

3.4 Chemical Equations

1. A chemical equation is a precise description of a chemical reaction.
2. A chemical equation can be written in words or chemical formulae. Using chemical formulae is easier,

faster and accurate.

Starting substances are called 2H2(g) + O2(g) → 2H2O(l) New substances formed are
reactants and are written called products and are
on the left-hand side of an written on the right-hand side
equation. of an equation.

Chapter Figure 3.16 An example of chemical equation

3. According to the law of conservation of mass, a matter can neither be created nor destroyed. So, the
numbers of atoms before and after a chemical reaction are the same. Therefore, a chemical equation
must be balanced.

3 Example: Hydrogen + Oxygen → Water Chemical equation
before being

Reactants Products balanced Balancing of
chemical
Chemical equation equation
after being
VIDEO

balanced

2H2 (g) + O2 (g) → 2H2O(l)

Figure 3.17 Chemical equation must be balanced

4. Table 3.12 shows the steps in constructing a chemical equation.

Table 3.12 Constructing chemical equation

Reaction:  Iron filings react with copper(III) chloride solution to form iron(III) oxide and copper metal.

Step Example and explanation

① Identify the reactants, products and their Reactants : Iron, Fe dan copper(II) chloride, CuCl2
formulae. Products: Iron(III) chloride, FeCl3 dan copper, Cu

② Write the main part of the equation. Fe  +  CuCl2  →  FeCl3  +  Cu

③ Check the number of atoms of each Left-hand side: Atom Fe = 1 Right-hand side: Atom Fe = 1
element on both sides of the equation.
Atom Cu = 1 Atom Cu = 1

Atom Cl = 2 Atom Cl = 3

The numbers of atoms are not balanced.

④ Balance the equation by adjusting the Balancing the number of Cl atoms: Fe  + 3CuCl2  →  2FeCl3  + Cu
coefficients in front of the formulae. Balancing the number of Fe atoms: 2Fe  + 3CuCl2  →  2FeCl3  +  Cu
Balancing the number of Cu atoms:  2Fe  +  3CuCl2  →  2FeCl3  +  3Cu
⑤ Check whether equation is now
balanced. 2Fe + 3CuCl2 → 2FeCl3 + 3Cu
Left-hand side: Atom Fe = 2 Right-hand side: Atom Fe = 2

Atom Cu = 3 Atom Cu = 3

Atom Cl = 6 Atom Cl = 6

The numbers of atoms are balanced.

⑥ Write the state symbol of each 2Fe(s)  +  3CuCl2(aq)  →  2FeCl3(aq)  + 3Cu(s)
substance.

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

5. The state symbols (s), (l) and (g) represent the solid, liquid and gaseous states respectively. The symbol
(aq) represents the aqueous solution.

SPM Tips

Many students are not able to balance equations. You only need to adjust the coefficients in front of the formulae.
Never adjust the subscript numbers in the formulae because this will change the chemical formulae of the
substances. Balance all types of atoms, one at a time. Practice balancing equations regularly so that you can
master it.

ACTIVITY 3.3

Aim : To construct balanced chemical equations.

Materials : Copper(II) carbonate powder, limewater, concentrated hydrochloric acid, concentrated Chapter
ammonia solution, lead(II) nitrate solution and potassium iodide solution.

Apparatus : Test tubes, stoppers, delivery tubes with rubber bung, test tube holder, Bunsen burner and
glass tube.

Procedure: 3

A   Heating of copper(II) carbonate Copper(II) carbonate
powder
1. One spatulaful of copper(II) carbonate is placed in a test
tube.

2. The apparatus is set up as shown in Figure 3.18.

3. Copper(II) carbonate is heated and the gas produced is Heat
passed through lime water.

4. The changes of copper(II) carbonate and limewater are Limewater
observed.

5. When the reaction stops, the delivery tube is withdrawn Figure 3.18 Heating of copper(II) carbonate
from the limewater and the Bunsen burner is turned off.

B   Formation of ammonium chloride Hydrogen Ammonia
chloride gas
1. Using a glass tube, three to four drops of concentrated gas
hydrochloric acid are dropped into a test tube. The test tube
is stoppered and left aside for a minute. Figure 3.19 Formation of ammonium
chloride
2. Using a clean glass tube, step 1 is repeated using concentrated
ammonia solution.

3. Both stoppers are removed and the mouths of the test tubes
are brought together as shown in Figure 3.19.

C   Precipitation of lead(II) iodide Lead(II) Potassium
nitrate iodide solution
1. 2 cm3 of potassium iodide solution is added to 2 cm3 of solution
lead(II) nitrate solution as shown in Figure 3.20.
Figure 3.20 Precipitation of lead(II)
2. The mixture is shaken and observed. iodide

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  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

Observations:

Section Observations Inferences

A • Copper(II) carbonate changes colour • Copper(II) carbonate decomposes into copper(II)

from green to black. oxide that is black in colour.

• Limewater turns cloudy. • Carbon dioxide is released.

B • Thick white fumes are produced at the • Ammonium chloride powder is produced.
mouths of the test tubes.

C • Yellow precipitate is produced. • Lead(II) iodide precipitate is produced.

Discussion:

Chapter 1. When copper(II) carbonate is heated, it decomposes into copper(II) oxide and carbon dioxide. The
presence of carbon dioxide is confirmed by the cloudiness of limewater.

Chemical equation:

CoppCeru(ICI)Oca3r(bso)n at e  C opCpeur(OII)(osx) ide  +  CaCrbOon2(dgio)xide

2. Concentrated hydrochloric acid and concentrated ammonia solution are left aside for a minute to

3 produce hydrogen chloride gas and ammonia gas respectively.
3. The hydrogen chloride gas combines with the ammonia gas to form fine white solid of ammonium
chloride which are seen as thick white fumes.
Chemical equation:

HydrogeHn Cchl(logr)i de gas+   AmNmHon3(iga)g as   Am mNonHiu4mCcl(hsl)oride

4. When the colourless solution of potassium iodide is added to the lead(II) nitrate solution which
is also colourless, a yellow precipitate of lead(II) iodide is produced. At the same time, potassium
nitrate solution which is colourless is formed.

Chemical equation:

Pota2ssKiuIm(aiqo)d ide +  PLbea(Nd(IOI) 3n)i2tr(aatqe )    L eadP(bII)I2i(osd)ide +   Po2taKsNsiuOm3(naitqra)te

Qualitative and quantitative aspects of chemical equations

1. Qualitative information from chemical equations:
• Reactants and products
• Physical state of each reactant and product

2. Quantitatively, the coefficients in a balanced equation show the ratio of amounts of reactants and
products. See Figure 3.21.

Qualitative information: The reactants are The product is
potassium solid potassium bromide
and bromine gas
solid

Equation: 2K (s)  +  Br2(g) → 2KBr (s)
Quantitative information:
2 mol 1 mol 2 mol
or 2 atoms 1 molecule 2 formula units

Figure 3.21 Qualitative and quantitative aspects of a chemical equation

3. From the equation in Figure 3.21, we know that 2 moles of potassium will react with 1 mole of bromine
gas to produce 2 moles of potassium bromide.

4. At the microscopic level, we know that every 2 K atoms will react with 1 Br2 molecule to produce 2
units of KBr.

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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

SPM Highlights EXAMPLE 3.24 Chapter

The following is a chemical equation. 2Na(s)  +  Cl2(g)  →  2NaCl(s) 3
What is the mass of sodium needed to react
K2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2KCl(aq) with 0.5 mole of chlorine?
Which of the following statements is true? [Relative atomic mass: Na, 23]
A The reactants are potassium carbonate solid and
Solution
calcium chloride solution. 2Na(s)  +  Cl2(g)  →  2NaCl(s)
B Two moles of potassium carbonate react with one (2 mol) (1 mol)
? g 0.5 mol ①
mole of calcium chloride
C The products are calcium carbonate precipitate Based on the equation, 1 mole of Cl2 reacts
with 2 moles of Na. ②
and potassium chloride solution So, 0.5 mole of Cl2 will react with (0.5 × 2)
D Two moles of calcium carbonate and one mole of mole of Na or 1 mole of Na. ③
The mass of Na needed = number of moles
potassium chloride are produced × molar mass ④

Examiner’s tip = 1 × 23
Students need to examine each option carefully. Option = 23 g
A is false because potassium carbonate used is in
aqueous state and not in a solid state. Option B and D EXAMPLE 3.25
are false because the coefficients in the equation show
that 1 mole of potassium carbonate reacts with 1 mole of 2C3H6(g)  +  9O2(g)  →  6CO2(g)  + 6H2O(l)
calcium chloride to produce 1 mole of calcium carbonate What is the volume of carbon dioxide gas
and 2 moles of potassium chloride. produced at STP if 63 g of propene, C3H6 is
Answer: C burnt completely?
[Relative atomic mass: H, 1; C, 12. Molar volume
Solving numerical stoichiometry : 22.4 dm3 mol-1 at STP]
problems
1. Stoichiometry is a study of quantitative Solution

composition of substances involved in chemical 2C3H6(g)  +  9O2(g)  →  6CO2(g)  +  6H2O(l)
reactions. (2 mol) (6 mol)
2. Based on a balanced chemical equation, we
can solve various numerical problems. General 63 g ? dm3 ①
steps in solving problem:
Number of moles of C3H6 = Mass ÷ molar mass
① Gather information from the question. =3(12)6+3 6(1)
Convert the given unit into the number =6432
of moles. = 1.5 mol ①

② Compare the ratio of moles of the Based on the equation, 2 moles of C3H6
related substances. produce 6 moles of CO2.②
This means that 1 mole of C3H6 produces 3
③ Calculate the answer in proportion of moles of CO2.
moles. So, 1.5 moles of C3H6 produces (1.5 × 3) moles
which are 4.5 moles of CO2. ③
④ Convert the answer from moles to the The volume of CO2 produced at STP
required units. = Number of moles × molar volume ④
= 4.5 × 22.4 dm3

= 100.8 dm3

55

03 SPM CHEMISTRY F4.indd 55 27/02/2020 11:23 AM

  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

Chapter EXAMPLE 3.26 Checkpoint 3.4

2Al(s)  +  3CuO(s)  →  Al2O3(s)  + 3Cu(s) Q1 Write balanced chemical equations for the
If 2 g of excess aluminium is reacted with 0.06 following reactions.
mole of copper(II) oxide, calculate the mass of (a) Sulphur trioxide + water → sulphuric acid
aluminium left behind. (b) Magnesium + carbon dioxide gas →
[Relative atomic mass: Al, 27] magnesium oxide powder + carbon powder
(c) Hydrogen gas reacts with nitrogen gas to
Solution produce ammonia gas.
2Al(s)  +  3CuO(s)  →  Al2O3(s)  + 3Cu(s) (d) Silver nitrate solution reacts with metallic
(2 mol) (3 mol) copper to produce silver precipitate and
copper(II) nitrate solution.
?g of 2 g 0.06 mol ①
Q2 The following equation shows the decomposition
Based on the equation, 3 moles of CuO react of potassium chlorate(V) by heat.
with 2 moles of Al. ②
So, 0.06 mole of CuO reacts with KClO3(s) → KCl(s) + O2(g)
(0.06 × 2) mol of Al or 0.004 mole of Al. ③ (a) Balance the equation.
(b) Calculate the volume of oxygen produced at
3
room conditions from the decomposition of
3 The mass of Al that reacted 24.5 g of potassium chlorate(V).
= number of moles × molar mass ④ [Relative atomic mass: O, 16; Cl, 35.5; K,
= 0.04 × 27 39. Molar volume : 24 dm3 mol-1 at room
= 1.08 g conditions.]
So, the mass of Al left behind = 2 - 1.08 g
= 0.92 g Q3 Lead is extracted according to the following
equation.
SPM Highlights
2PbO(s) + C(s) → 2Pb(s) + CO2(g)
Equation for the reaction between potassium and (a) How many moles of lead is extracted from

oxygen: 0.5 mole of lead(II) oxide?
(b) How many lead atoms are produced if 44.6 g
4K(s) + O2(g) → 2K2O(s)
What is the maximum mass of potassium oxide that is of lead(II) oxide is heated with excess carbon
powder?
formed when 19.5 g potassium is burned completely [Relative atomic mass: O, 16; Pb, 207.
Avogadro’s constant: 6.02 × 1023 mol-1.]
in excess oxygen? [Relative atomic mass: O, 16; K,
Q4 Sodium reacts with water as follows.
39]
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
A 23.5 g C  70.5 g (a) What are the products of the reaction?
(b) How many moles of sodium will react with 5
B 47.0 g D  94.0 g
moles of water?
Examiner’s tip (c) Calculate the mass of sodium needed to be

The amount of product depends only on the amount used to produce 3.01 × 1023 H2 molecules?
[Relative atomic mass: Na, 23. Avogadro’s
of potassium as the oxygen is in excess.
constant: 6.02 × 1023 mol-1.]
4K(s) +  O2(g) → 2K2O(s)
4 mol 2 mol

19.5 g ? g ①

The number of moles of K involved

= mass ÷ molar mass

19.5
= 39 = 0.5 mol ②

Based on the equation, 4 moles of K produce 2

moles of K2O. This means that 2 moles of K produce
1 mole of K2O. So, 0.5 mol of K produces (0.5 ÷ 2)
mole of K2O or 0.25 mole of K2O.
The mass of K2O = number of moles × molar mass

= 0.25 × [2(39) + 16] g

= 23.5 g

Answer: A

56

03 SPM CHEMISTRY F4.indd 56 27/02/2020 11:23 AM

Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

CONCEPT MAP

Chemical equation

Number of Stoichiometry Chemical formulae
particles Number of moles
Chapter
× NA
÷ NA

Empirical formulae Molecular formulae 3

× molar ÷ molar × molar ÷ molar
volume volume mass mass

Volume of gas

Mass (g)

SPM Practice 3

Objective Questions [Relative atomic mass: O,16; 4. Figure 3.22 shows a cooking
Mg, 24] SPM
Choose the correct answer. 2017 gas cylinder.
A 3
3.1 Relative Atomic Mass and B 4 1 mole of propane
Relative Molecular Mass C 5 gas, C3H8
D 6

1. Six atoms of element Y have
SPM
2015 the same mass with three 3.2 Mole Concept
tellurium atoms, Te. What is

the relative atomic mass of 3. How many nitrate ions, Figure 3.22
Y? SPM
2016 NO3– are there in 2 mol of How many hydrogen atoms
[Relative atomic mass: Te, 128] aluminium nitrate, Al(NO3)3? are there in the gas cylinder?

A 8 C 32 [Avogadro’s constant: 6.02 × [Avogadro’s constant: 6.02 ×
1023 mol-1]
B 16 D 64 1023 mol-1] A 1 × 6.02 × 1023
B 3 × 6.02 × 1023
2. How many oxygen atoms A 1.204 × 1024 mol-1 C 8 × 6.02 × 1023
have the equal mass with two B 1.806 × 1024 mol-1 D 11 × 6.02 × 1023
C 3.010 × 1024 mol-1
magnesium atoms? D 3.612 × 1024 mol-1

57

03 SPM CHEMISTRY F4.indd 57 27/02/2020 11:23 AM

  Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation

5. Which of the following 8. What is the percentage C The number of atoms in
SPM
substances has the highest 2018 composition by mass of the ammonia gas formed
number of atoms? HOTS carbon per molecule of is twice the number of

A 0.1 mole of carbon hexane, C6H14? [Relative nitrogen atoms used
B 0.2 mole of zinc atomic mass: H,1; C, 12] D 1 mole of ammonia gas is
C 0.3 mole of chlorine gas A 76.23% C 83.72%
D 0.4 mole of neon gas B 81.46% D 84.13% formed when 0.5 mole of
nitrogen gas reacts with
1.5 moles of hydrogen

6. 200 cm3 of a gas has a mass 9. 0.60 g of metal Z reacts with gas.
of 0.15 g at room conditions. fluorine to produce 1.17 g Z
What is the molar mass of 12. The following equation
the gas? fluoride. What is the empirical SPM
formula of Z fluoride? 2018 represents the reaction
[Molar volume: 24 dm3 mol-1 between magnesium and
at room conditions] HOTS [Relative atomic mass: F,19;
Z, 40] hydrochloric acid.
A 18 g mol-1
Chapter B 30 g mol-1 A ZF DC ZZF2F4 Mg(s) + 2HCl(aq) →
C 50 g mol-1 B ZF2 MgCl2(aq) + H2(g)

10. What is the mass of calcium What is the volume of
hydrogen produced if 3.6 g

D 150 g mol-1 contained in 1 kg of calcium of magnesium reacts with

3 carbonate? excess hydrochloric acid at
3.3 Chemical Formula [Relative atomic mass: C,12; room conditions?
[Relative atomic mass: Mg,
O, 16; Ca, 40] HOTS
24. Molar volume : 24 dm3
7. The distinct smell of garlic A 100 g C 300 g
SPM B 200 g D 400 g mol-1 at room conditions]
2018 is due to the presence of
allicin compound. Figure 3.4 Chemical Equation A 0.12 dm3
B 1.2 dm3
3.23 shows the percentage C 3.6 dm3
D 4.8 dm3
composition by mass of

allicin.

11. Ammonia gas is produced 13. Photosynthesis in green
SPM SPM plants can be represented by
2018 according to the following 2017 the following equation.
equation:

3H2(g) + N2(g) → 2NH3(g) 6CO2 + 12H2O sunlight
C6H12O6 + 6O2 + 6H2O
Which of the following
C, 44.4%  H, 6.21% statements is true? What is the mass of glucose
O, 9.86%  S, 39.5% A 3 g of hydrogen gas produced if 7.2 dm3 carbon
reacts with 1 g of
Figure 3.23 nitrogen to produce 2 g of dioxide gas is used?
ammonia gas.
What is the empirical formula [Relative atomic mass : H,1;
of allicin? B Three molecules of
A CHOS nitrogen react with one C,12; O,16. Molar volume: 24
B C6H10OS2 molecule of hydrogen to dm3 mol-1 at room conditions]
C C12H5OS2 form two molecules of
D C12H10OS4 ammonia. A 18.00 g
B 9.00 g
C 5.40 g
D 4.91 g

Subjective Questions

Section A

1. The mass of an atom is very small. Therefore, chemists determine relative atomic mass of an element by
comparing the mass of the element with that of a standard atom. Figure 3.24 shows the comparison of the
mass of atom Y with carbon-12 atom.

Carbon-12

Y CC CC

Figure 3.24

58

03 SPM CHEMISTRY F4.indd 58 27/02/2020 11:23 AM

Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

(a) What is the meaning of relative atomic mass? [1 mark]

(b) Give one reason why carbon-12 is used as the standard atom to compare the mass of an atom?
[1 mark]

(c) Based on Figure 3.24, state the relative atomic mass of element Y. [1 mark]

(d) In an investigation, Y reacts with bromine gas to form the compound YBr4. [1 mark]
(i) Write the chemical equation for the reaction.

(ii) Interpret the chemical equation in 2(d)(i) quantitatively. [1 mark]

(iii) Using the relative atomic mass of element Y from your answer in 2(c), calculate the number of moles
of bromine gas required to react completely with 3.6 g of Y. [2 marks]

(IV) In another investigation, 9.6 g of Y reacts with 6.4 g of oxygen. Complete the table below to determine Chapter
the empirical formula of the oxide of Y.

[Relative atomic mass: O, 16]

Element Y O
Mass (g)
Number of moles of atoms 3
Simplest ratio of moles
[3 marks]

Section B
2. (a) The following is the description of P, Q and R. The letters are not the real symbol of the elements.

• One P atom has twice the mass of one Q atom.
• Ten P atoms has equal mass with seven R atoms.
• Relative atomic mass of P is 28



(i) Determine the relative atomic masses of Q and R. Then, arrange the elements in an ascending order
[3 marks]
of relative atomic mass.

(ii) Explain why a sample of 14 g of P has equal number of atoms as that in a 20 g sample of of R.
[4 marks]

(iii) R is a reactive metal that can react with oxygen to form metal oxide. Describe briefly how you can
determine the empirical formula of metal R oxide in the laboratory. [7 marks]

(b) Hydrocarbon W consists of 92.3% carbon and 7.7% hydrogen by mass. The relative molecular mass of
hydrocarbon W is 26. Determine the empirical formula and molecular formula of W.
[6 marks]
[Relative atomic mass: H,1; C,12]

03 SPM CHEMISTRY F4.indd 59 59

27/02/2020 11:23 AM

JAWAPAN

1Chapter B – The solution/ liquid chemical Section B
should not be sucked using the
Introduction to Chemistry mouth. A pipette should be used 2. (a) • Hypothesis is a statement
to suck up the solution/ liquid
chemical into the apparatus. that links the manipulated

Q3 He should remove his laboratory variable to the responding
coat immediately. If the acid wets his
Checkpoint 1.1 clothes, he should rinse his whole variable. [1]
body under the safety shower. At the
same time, he should ask his friend • Manipulated variable is
to report the incident to the teacher.
Q1 Chemistry is a branch of science the factor that is purposely
that studies the composition,
structure, characteristics and changed in an experiment.
interaction of matter.
[1]

• Responding variable is the

Q2 Iron: It is used to make nails, screws factor that changes with the
and nuts; Chlorine: It is used to treat
tap water; Calcium carbonate: It is manipulated variable. [1]
used as construction material/ to
manufacture cement. SPM Practice • Constant variable is the
Objective Questions
1 factor that is kept constant

in an experiment. [1]

4

Q3 To carry out research and 1. D 2. B 3. C 4. B 5. A (b) • List out all the materials and
development of food products/ To 6. D 7. C 8. C
carry out analysis to check the apparatus needed to be
quality of products/ To provide Subjective Questions
technical support to the marketing used. [1]
section.
• Determine how to control

the manipulated and

Section A constant variables. [1]

1. (a) (i) Long hair that is not tied up • Determine how to measure

Checkpoint 1.2 easily catches fire. [1] the responding variable. [1]

• Determine how to collect,

Q1 Make observations, make an (ii) Closed-up shoes will be analyse and interpret data.
inference, identify the problem, make able to protect the feet from
a hypothesis, identify the variables, chemical spills or injury due [1]
control the variables, plan and carry to glass apparatus falling
out an experiment, collect data, onto the feet or floor. [1] 4
interpret data, make a conclusion,
write a report. (c) (i) Manipulated variable –

Mass of salt added to pure

(iii) If the excess solution is water. [1]

contaminated, the action Responding variable – Time

of pouring it back into taken to freeze the water [1]

Q2 (a) The more water is added to the its reagent bottle will 2
acid, the higher the pH of the
acid. contaminate all the solution (ii) The larger the mass of salt

(b) Manipulated variable: Volume of in the bottle. [1] added to pure water, the longer
water added
3 time it takes for the water to
Responding variable: pH value
of acid (b) I will report the incident to the freeze. [1]

Controlled variable: Type and teacher immediately. Then, (iii) Volume of pure water used. [1]
volume of acid used, pH paper
used using gloves, I will dispose Type of container used. [1]

Type of freezer used. [1]

of the broken test tube into (any two) 2

a specific container that is

prepared and clean up the spill (d) (i) The larger the mass of salt

Checkpoint 1.3 as instructed by the teacher. [3] added to pure water, the

(c) A fume cupboard is a higher is its density. [1]

Q1 Goggles – to protect the eyes from cupboard where the air in it (ii) • The manipulated variable
chemicals
is always being sucked out. is the mass of salt added
Safety shower – to remove
chemicals that the body comes in A fume cupboard is used to pure water. [1]
contact by showering all parts of the
body to carry out activities that • So, different mass of salt

Fire blanket – to put out the fire on a involve volatile, inflammable is added to pure water, for
victim’s body
or toxic chemicals so that the example 1 g, 2 g, 3 g, 4 g
Q2 A – Chemicals should not be sniffed
directly. Instead, the individual vapour from the chemicals is and 5 g. [1]
should use his hands to waft the
smell to his nose. disposed of immediately from • The responding variable is

the fume cupboard and does the density of water. [1]

not contaminate the air in the • So, he can measure the

laboratory. [3] initial mass and final mass

of each mixture of water

and salt.

258

10 ANSWER FOCUS KIMIA F4 SPM.indd 258 27/02/2020 11:29 AM

  Chemistry Form 4  Answers 

The density of the mixture • Barium sulphate salt is move throughout the liquid. They
is calculated as follows:
used to check our digestive have low energy content.

Density = system using X-ray. Patients Q2 (a) 0°C
(b) This is because the heat energy
Mass of mixture (g) will have a drink of the salt absorbed by the molecules of ice
is used to overcome the forces
Volume of mixture (cm3) before X-ray images are

[1] taken. [1]

• The constant variables (any 10 examples of between molecules so that ice
other chemicals are also can turn into water.
are pure water and type
accepted) (c)
of salt used. [1] 10 Temperature (°C)

• So, pure water is used

every time. Only one type (b) Hypothesis : Painting prevents

of salt is used throughout rusting of iron [1] 120
100
the experiment, for Variables:

example sodium chloride. Manipulated: Presence of a

[1] layer of paint on iron [1]

6 Responding: Rusting of iron [1] S
R
Section C Constant: Presence of water [1] 15
3. (a) •
We use various chemicals Procedure: –30 Q
• in many aspects of our P
lives such as in foods, Time (min)
• agriculture, industries and
• medicine. Water pipes Checkpoint 2.2
•
• For example, we use Painted Q1 (a) Hydrogen atom has 1 proton
• iron nail with relative charge +1 and 1
sodium chloride and electron with relative charge -1.
•
monosodium glutamate [1] Hence the net charge of H atom
• is (+1) + (-1) = 0
(MSG) as flavourings in our 1) Two iron nails are cleaned
• (b) All the protons and neutrons
dishes. [1] with sand paper. [1] of an atom are found in the
nucleus.
Sucrose or sugar is used 2) One iron nail is painted with
A proton or neutron is
in various foods such a layer of paint and left to approximately 2000 times
heavier than an electron.
as desserts, cakes and dry. [1]

delicacies. [1] 3) Both nails are placed

Acetic acid (ethanoic acid) in separate test tubes

is used as vinegar to pickle containing tap water and are Q2 An atom is a discrete particle, very
small and indivisible.
foods. [1] left aside for three days. [1]
Q3 (a) Electron
In agriculture, ammonia is Observation: (b) Thomson’s atom is a positively-

used to make fertilisers for The painted iron nail does not charged sphere embedded with
electrons.
crops. [1] rust while non-painted iron nail

Slaked lime (calcium rusts. [1]

hydroxide) is used to treat Conclusion: Q4 • Rutherford discovered the proton
• which is positively charged.
acidic soil. [1] Painting prevents rusting of iron. He proposed the atomic model

[1]

Hydrocarbons from 10 which consists of a positively-
petroleum are separated charged nucleus and negatively-
by fractional distillation and 2Chapter charged electrons orbiting
are important fuels. For
example, petrol and diesel Matter and the Atomic around the nucleus.
are fuels for vehicles. [1] Structure • He stated that the mass of an

atom is concentrated in the

We also use various Checkpoint 2.1 nucleus.
Bohr introduced the concept of
petrochemical products Q5 • shells.

such as plastics, detergents, Q1 • The particles in a gas are very
far apart from each other and in
synthetic textiles and random motion. The particles can • He stated that electrons orbit the
vibrate, rotate and move freely. nucleus in fixed shells.
synthetic rubber. [1] They have high energy content.

Metals such as copper, zinc, • When the gas is cooled, the Q6 (a) Neutron
particles of the gas lose energy
aluminium and iron are used and move slower. (b) • Proton and neutron are
found in the nucleus.
in various machines and • Eventually, the gas changes into
a liquid. The particles of the liquid
important metal equipment are packed closely together but • Neutrons contribute half the
not in an orderly arrangement. The overall mass of an atom.
in industries. [1] particles can vibrate, rotate and

Various slats are used to

treat malnutrition diseases Checkpoint 2.3

such as iron(II) sulphate is

used to treat lack of iron in Q1 (a) Proton number = 9
Nucleon number = 19
anaemic patients. [1]

259

10 ANSWER FOCUS KIMIA F4 SPM.indd 259 27/02/2020 11:29 AM

  Chemistry Form 4  Answers

(b) F F19 19 (ii) (e) (i) Particles move more slowly.
99 (ii) Particles lose its kinetic
(d) The heat energy lost to the
Q2 (a) 2.8.4 surrounding balances the heat energy.
energy given out when particles
(b) (i) 14 of X form bonds with one (f)
another to form a solid.
(ii) silicon Temperature (°C)

(c) 29 T1
T2
Q3 (a) C T3
(b) A and C
(c) B and D Time (min)
(d) B; C; D and E

Checkpoint 2.4

Q1 (a) Isotopes are atoms of the Section B
same element having the same
number of protons but different 2. (a) Temperature T1 T2
number of neutrons. Particles are closely- Particles are closely-packed
Arrangement of packed in an orderly but not in an orderly manner
(b) 82 electrons; 126 neutrons and particles manner
82 protons Particles only vibrate and Particles can vibrate, rotate
Movement of rotate about fixed positions and move from one point to
Q2 RAM Fe = [(5.85/100) × 54] + particles another point
[(91.75/100) × 56] + [(2.40/100) × Strong forces of attraction Particles are held by weak
57] = 55.91 Forces of between particles forces of attraction
attractions
Q3 Assume abundance of sulfur is S-32, between Low energy content Moderate energy content
x% dan S-34, (100 - x)% particles
[8]
RAM S = [(x/100) × 32] + [((100 - x) Energy content
/100) × 34] = 32.1 of particles

32x + 34 00 - 34x = 3210
2x = 190
x = 95 (b) • A graph of temperature
∴ S-32, 95% dan S-3, 5% Thermometer
against time is plotted for the
Q4 (a) Investigate uptake of Test tube heating of solid G.
phosphorus in plants Solid G [1]
Coconut oil
(b) Detect and treat thyroid Temperature (°C)
(c) Measure the thickness of paper

and film
(d) Detect leaks in underground

pipes

SPM Practice 2
Objective Questions
120°C

1. A 2. B 3. D 4. A 5. D [2] Time (min)
6. B 7. A 8. D 9. B 10. C
11. D 12. B 13. B 14. C 15. B Procedure: [2]
16. A 17. A 18. D 19. B 20. D 1. A boiling tube is filled • Melting point of solid G is
21. A 22. B 23. D 24. C 25. C with solid G to a depth of
26. C 120°C. [1]
3 cm and a thermometer
Subjective Questions Maximum: 12
is inserted into it. [1]

2. The boiling tube is 3. (a) (i) Chlorine [1]

Section A clamped in a beaker half- (ii) Chlorine isotopes:

1. (a) Sublimation filled with coconut oil [1] Chlorine-35 and Chlorine-37
(b) T2°C 3. The level of solid G must
(c) (i) [1]
be below the surface of

the oil in the beaker. [1] Chlorine-35 Chlorine-37

4. The coconut oil is heated

gently. [1] Proton number 17 17
5. Solid G is stirred slowly

with the thermometer. [1] Nucleon number 35 37

6. The temperature and Number of 17 17
condition of solid G protons

is recorded every half
minute until solid G is

completely melted. [2]

260

10 ANSWER FOCUS KIMIA F4 SPM.indd 260 27/02/2020 11:30 AM

  Chemistry Form 4  Answers 

Number of 17 17 (c) 60 dm3 1
electrons 18 20 Q2 (a) 0.6 mol compared to 2 of the mass of
2.8.7 2.8.7 (b) 0.4 mol
Number of Q3 (a) 160 g mol–1 a carbon-12 atom. [1]
neutrons (b) 80 g
Q4 4.8 g (b) Carbon-12 can combine with
Electron Q5 Z, X, Y
arrangement many elements/ Carbon-12

exists as solid at room

temperature and thus can

be handled easily during

Number of Checkpoint 3.3 investigations/ Carbon-12 is
velance electrons
7 7 also used as standard in mass

Q1 MgSO4 spectrometer / Carbon-12 is the

Maximum: [10] Q2 (a) NO2 most abundant carbon isotope,

12 (b) N2O4; Dinitrogen teroxide thus the mass of 12 units

(b) Chemical required: Liquid Q3 (a) 12.8 g assigned to a carbon-12 atom is
bromine [1] (b) 1.5 mol an accurate value. (any one) [1]

Procedure: Q4 (a) K2SO4 (c) 48 [1]
1. A few drops of liquid (b) ZnCl2 (d) (i) Y + 2Br2 → YBr4 [1]

bromine are dropped into a (c) SnO (ii) 1 mole of Y atoms reacts

gas jar. [1] (d) (NH4)2CO3 with 2 mole of Br2 molecules
to produce 1 mole of YBr4
2. The jar is immediately (e) Mg(NO3)2 units / 1 Y atom reacts with

closed and put aside for a (f) Na3PO4 2 Br2 molecules to produce

few minutes. [1] Q5 (a) Aluminium hydroxide 1 YBr4 unit. [1]
(b) Iron(II) sulphate
3. Another gas jar filled with (c) Ammonium chloride (iii) 0.15 mol of bromine gas [1]
(d) Calcium nitrate
air is inverted on the jar (e) Potassium carbonate (iv)
(f) Zinc sulphide
containing bromine vapour.

[1] Element Y O

4. The cover between the two

jars is removed. [1] Mass (g) 9.6 6.4

5. The apparatus is set aside Q6 C, Mg(OH)2, FeBr3

for a few minutes. [1] Q7 Nitrogen monoxide; diboron trioxide; Number of 9.6 = 0.2 6.4 = 0.4
moles of 48 16
Observation: sulphur hexafluoride atoms

Red-brown vapour spreads

rapidly to both gas jars in a few Checkpoint 3.4

minutes. [1] Simplest
ratio of
Conclusion: Q1 (a) SO3(g) + H2O(l) → H2SO4(aq) moles 1 2
(b) 2Mg(s) + CO2(g) → 2MgO(s) +
Bromine gas is made up of tiny
C(s)
discrete particles. [1] [3]
(c) N2(g) + 3H2(g) → 2NH3(g)
Note: You can also choose Section B
diffusion in liquid or solid. (d) 2AgNO3(aq) + Cu(s) → 2Ag(s) 2. (a) (i) • RAM of P = 2 × RAM of Q
+ Cu(NO3)2
8 28 = 2 × RAM of Q. So,
RAM of Q = 28 ÷ 2 = 14
3Chapter The Mole Concept, Chemical Q2 (a) 2KClO3(s) → 2KCl(s) + 3O2(s)
Formula and Equation [1]
(b) 7.2 dm3 • 10 × RAM of P = 7 ×
Q3 0.5 mol
Q4 1.204 × 1023 atom RAM of R
10 × 28 = 7 × RAM of R.
Q5 (a) Sodium hydroxide solution and

Checkpoint 3.1 hydrogen gas So, RAM of

Q1 16 (b) 5 mol R = 10 × 28 = 40 [1]
Q2 32 (c) 23 g 7

Q3 (a) 124 • Ascending order of RAM
(b) 28 = Q, P, R
SPM Practice 3 [1]

(c) 342 Objective Questions 3
(d) 122
(ii) • Number of moles of
Q4 (a) 62 1. D 2. A 3. D 4. C 5. C atoms in the sample of 14
g P Mass of P
(b) 189 6. A 7. B 8. C 9. B 10. D
= Molar mass of P
(c) 190 11. D 12. C 13. B 14 g

(d) 286 Subjective Questions = 28 g mol–1
Q5 24

Checkpoint 3.2 Section A = 0.5 mol of atoms [1]
• Number of moles of
Q1 (a) 1.505 × 1024 molecules 1. (a) Relative atomic mass of an
(b) 4.515 × 1024 atom element is the average mass atoms in the sample 20 g
of an atom of the element R

261

10 ANSWER FOCUS KIMIA F4 SPM.indd 261 27/02/2020 11:30 AM

  Chemistry Form 4  Answers

Mass of R • So, the empirical formula of (iii) Transition elements
= Molar mass of R W is CH.
• [1] (c) Period 2. Atom A, B and C have
2 shells filled with electrons.
20 g • Assume that the molecular
= 40 g mol–1 formula of W is (CH)n

= 0.5 mol atom The molecular relative mass Checkpoint 4.3
of W
• As both samples have equal Q1 (a) (i) X and Z
• = n [12 + 1] (ii) Inert gas
number of moles of atoms, (b) No, atom Z had achieved octet
= 13n electron arrangement. Atom
the samples have equal [1] Y does not donate, receive or
share electron with atom Y.
number of atoms. [1] So, 13n = 26
Q2 (a) Helium. Helium has lowest
• Number of atoms in each n = 26 ÷ 13 = 2 [1] density hence, it can float in the
air.
sample Therefore, the molecular
= number of moles × NA formula of hydrocarbon W is (b) Boiling point increases from
= 0.5 × 6.02 × 1023 atoms helium to argon.
C2H2. [1]
= 3.01 × 1023 atoms [1] Atomic size increases from
6 helium to argon. The van
4 der Waals force of attraction
between atoms become
(iii) – A crucible with its lid is stronger. More heat energy is
needed to overcome the force of
weighed and its mass is 4Chapter attraction between the atoms.

recorded. [1] The Periodic Table of
– About 2 g of R powder Elements

is placed in the crucible.

The crucible, its lid

and its content is Checkpoint 4.1

weighed and its mass is Q1 (a) • Both of them arranged the
elements in increasing order of
recorded. [1] relative atomic mass.

– The crucible is heated • Both of them grouped the
elements with similar chemical
strongly without its lid. [1] properties into the same vertical Checkpoint 4.4
– When R starts to column named Group.
Q1 (a) Group 1, all atom Li, Na dan K
glow, the crucible is (b) • Left empty spaces for has one valence electron.
undiscovered elements
covered by its lid. The (b) Li, Na, K
• He is able to predict the
lid is removed a little at properties of the undiscovered (c) Atomic radius increases,
elements based on its position reactivity increases.
intervals. [1] in the Periodic Table of
Elements.
– When the burning is
• He mutually exchanged the
complete , the crucible position of nickel with cobalt
and iodine with tellurium so that
is opened and heated elements with similar chemical Checkpoint 4.5
properties were placed under
strongly. [1] the same group.

– The crucible, its lid and (c) He arranged the elements in Q 1 (a) F: 2.7 Cl: 2.8.7
ascending order of their relative
its content is weighed, atomic mass instead of proton (b) Group 17
number.
cooled and weighed. (c) 7

The mass reading is (d) Yes. All atoms in Figure 1
have same number of valence
recorded. [1] electrons.

– The heating, cooling

and weighing are (e) Fluorine atom has smaller
atomic size.
repeated a few times

until a constant mass is The outermost shell is closer

obtained. [1] to the nucleus. The nuclear
attraction on valence electrons
7 in fluorine atom is stronger.

(b) • Based on the given Checkpoint 4.2 Fluorine atom can attract one
percentage composition, electon into its outermost shell
each 100 g of hydrocarbon Q1 (a) (i) Nucleon number easier.
W contains 92.3 g of carbon (ii) 2.8.1
dan 7.7 g of hydrogen. (iii) Group 1, Period 3 Q2 (a) Blue litmus paper turns red and
then bleached.
Element CH (b) Element X and Y. X and Y atom
have same number of valence (b) The solution formed is acidic.
Mass (g) electrons.
92.3 7.7 (c) Cl2 + H2O → HCl + HOCl
Number (c) Element Y and Z. Atom Y and
of moles 92.3 7.7 Z have same number of shells (d) Yes. Bromine has same number
of atoms 12 = 7.7 1 = 7.7 filled with electrons. of valence electrons as chlorine.

Simplest Q2 (a) Ascending order of proton Checkpoint 4.6
ratio of number
moles 11 Q1 (a) Period 2
(b) (i) Alkali metal (b) Element X.
[2] (ii) Halogen
Atomic size of atom X is bigger
than atom Y.

262

10 ANSWER FOCUS KIMIA F4 SPM.indd 262 27/02/2020 11:30 AM

  Chemistry Form 4  Answers 

The valence electron is further The nuclear attraction on Q2 (a)
from nucleus.
electrons become stronger, HOH
Force of attraction between
nucleus and valence. electrons are attracted closer
Atom X can donates valence
electron easier. toward nucleus. [3]

5Chapter (b) Cl

H

Checkpoint 4.7 Chemical Bond

Q1 (a) Transition elements Checkpoint 5.1 (c) N
(b) – Form coloured ion or
compound Q1 Inert gases have stable duplet or HC H
– Has more than one octet electron arrangement. Atom of
oxidation number inert gas do not donate, receive or Q3 (a)
– Can act as catalyst share electrons with other atom.
– Can form complex ions HM H
(c) Use as catalyst in Haber Q2 Ionic bond and covalent bond.
process.

Checkpoint 5.2

SPM Practice 4 Q1 (a) Lithium atom donates one H
Objective Questions valence electron. Fluorine atom N
receive one electron into its
1. C 2. D 3. B 4. D 5. A outermost shell. (b)
6. D 7. A 8. D 9. C 10. C
11. D 12. B 13. B (b) Cation Li+, anion F–
(c) Li → Li+ + e– ; F + e– → F–

Subjective Questions Q2 (a) Li2O NM N
(b) LiCl
Section A (c) MgCl2
(d) MgO
1. (a) X: chlorine, 2.8.7 (e) Al2O3 N
Y : bromine, 2.8.8.7 Q3 (a) Ionic bond
[2] (b)
(b) Group 17 because atom Y has
+
7 valence electrons and period 2– + M

4 because it has 4 shells filled MOM (c) NN

with electrons. [3]
(c) (i) X is more reactive.
(ii) Atomic size of atom X is [1]

smaller. (c) – Electron arrangement of
atom M is 2.1.
The nuclear attraction on Checkpoint 5.4
– Atom M donates one
electrons is stronger. valence electron to achieve
duplet electron arrangement
Atom X can receive one and forming cation M+. Q1 Hydrogen atom becomes partial
positive charged because the
electron in the outermost – M → M+ + e bonded pair electrons are pulled
– Electron arrangement of closer to highly electronegative
shell easier. [3] atom. The highly electronegative
(iii) 2 Fe + 3 X2 → 2 FeCl3 [2] atom O is 2.6 atom in the molecule becomes
– Oxygen atom receives 2 partial negative charged.
(d) Element Y is less reactive than
element X. electrons from two M atoms Q2 Water is universal solvent because
[1] to achieve stable octet water is a polar molecule which is
electron arrangement and able to form hydrogen bonds. The
Section B forming anion O2–. partial positive charged end of water
– O + 2 e– → O2– molecule can form hydrogen bond
2. (a) (i) Group 1 because atom W – Cation M+ and anion O2– with partial positive charge end from
are attracted by strong nearby ammonia, NH3, hydrogen
has 1 valance electron, electrostatic force and chloride, HCl or ethanol, C2H5OH
Period 3 because W has 3 forming an ionic bond. molecules.
– An ionic compound, M2O
shells filled with electron. [2] formed. Q3 Water is a dipole molecule while
(ii) 2W + 2 H2O → 2WOH + H2 carbon dioxide is a non-polar
molecule. Stronger hydrogen bonds
[2] form between water molecules
while weak Van der Waals attraction
(iii) Red litmus paper turns blue. form 14 between carbon dioxide

[1]

(iv) Colourless solution /

solution WOH is alkaline. [1]
(b) (i) Y
(ii) 4Y + 3O2 → 2Y2O3 [1]

[2] Checkpoint 5.3
Q1 (a) XY2
(c) W, X, Y, Z. (b) Y2

Proton number increases from
W, X, Y to Z.

263

10 ANSWER FOCUS KIMIA F4 SPM.indd 263 27/02/2020 11:30 AM

  Chemistry Form 4  Answers

molecules. Hence, more heat is (ii) – B is a dipole molecule nitrogen atom and hydrogen
needed to overcome the hydrogen while A is non-polar
bonds during boiling. molecule. nucleus share a pair of

– Molecules B are attracted electrons. [3]
by stronger hydrogen
bonds while molecules A (c) – Valence electrons of copper
are attracted by weaker –
Checkpoint 5.5 Van der Waals attraction. atoms were delocalised into
–
Q1 Dative bond is a type of covalent – More heat energy is – sea of electrons.
bond in which two electrons derive needed to overcome the
hydrogen bonds between Electrons in the sea of
from one atom. molecule B during boiling.
electron can move freely
[1]
3 and carry electrical charges.

Q2 Lone pair electrons. Hence, copper can conduct

Q3 (a) electricity.

H H+ Metal cations are attracted
H N H + H+ HNH
by delocalised electrons

strongly.

High heat energy is needed

H Section B to overcome the strong
H+
(b) 2. (a) Substance Y has high melting metallic bonds during the
point while substance X has low
H melting. [4]

melting point.

H+ + O H + HOH Substance X is a covalent Section C

compound. 3. (a) (i) X: hydrogen [1]

Substance Y is an ionic Type of bond: Ionic bond [1]

compound. [4] 2

Checkpoint 5.6 (b) (i) NH3 + H2O → NH4+ + OH– [2] (ii) X2 + Cl2 → 2XCl
(ii) – Hydrogen atom shares (iii) – Number of mol X2 =
02.43 = 0.0125 mol
Q1 (a) Metallic bond electrons with a more

(b) Metallic bonds are formed when electronegative nitrogen
metal cations are attracted by
delocalised electrons in the sea atom to form an ammonia Number of mol Cl2 =
of electrons. 02.46 = 0.025 mol
molecule. [1]
(c) Aluminium can conduct
electricity in solid state because – Electrons are not
the delocalised electrons in the
sea of electrons can move freely distributed evenly – Based on the equation 1
and carry electrical charges. mol X2 gas reacts with 1
because the shared pair mol Cl2 gas to produce
1 mol XCl gas. Hence,
electrons are pulled closer 0.0125 mol X2 gas reacts

to nitrogen atom. [1]

– Ammonia molecule

becomes dipole molecule with 0.0125 mol Cl2 gas
will produce 0.0125 mol
like water molecule. [1] XCl gas

SPM Practice 5 – Ammonia molecule form
Objective Questions
hydrogen bond to water – Chlorine gas is in excess.

molecule. [1] Maximum volume of
substance Y
1. C 2. A 3. D 4. B 5. D 4
6. C 7. C 8. D 9. A 10. B
11. C 12. A (iii) Dative bond = 0.0125 × 24 000

H+ = 300 cm3 [4]

(iv) Channel the excess chlorine

Subjective Questions HNH gas into concentrated

sodium hydroxide solution.

Section A H [2]
1. (a) X
(b) (i) YZ2 [1] (iv) Nitrogen atom in ammonia (b) Compound PQ4
(ii) Covalent [1] molecule has a lone – P atom need 4 electrons
(iii) Molecule [1] pair of electrons that are
[1] not involved in covalent to achieve octet electron
(iv) bonding.
O arrangement. [1]
O Nitrogen atom contributes – P atom contributes 4
C its lone pair of electrons
for sharing with hydrogen electrons for sharing. [1]
nucleus. – Q atom needs 1 electron

(c) (i) A hydrogen nucleus to achieve octet electron
(hydrogen ion) from water
molecule is transferred to arrangement. [1]
ammonia molecule. – Four Q atom each
H– –H
––– A dative bond is formed in contribute 1 electron for
ammonium ion, NH4+ when
sharing. [1]
– One P atom and 4 Q atom
O
share 4 pair of electrons

– – – and form 4 single covalent
– –
bonds. [1]

264

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  Chemistry Form 4  Answers 

– A covalent compound PQ4 is alkaline properties and the red litmus Checkpoint 6.4
turns blue.
formed. [1] Q1 (a) 3H2SO4 + 2Al(OH)3 → Al2(SO4)3
+ 6H2O
Compound RQ Checkpoint 6.2
– R atom has 1 valence (b) 2HNO3 + PbO → Pb(NO3)2 +
H2O
electron. [1] Q1 (a) pH value is used to measure
– R atom donates 1 valence acidity and alkalinity of an (c) H3PO4 + 3NH3 → (NH4)3PO4
aqueous solution.
electron to achieve octet (d) 2HCl + Zn → ZnCl2 + H2O
(b) Definition: pH = -log [H+];
electron arrangement and i.e. negative logarithm of the (e) 2CH3COOH + Na2CO3 →
R+ cation is formed. // R→ concentration of H+ ions in mol 2CH3COONa + CO2 + H2O
R+ + e dm-3.
– Q atom has 7 valence [1] (f) (NH4)2SO4 + 2NaOH → Na2SO4
(c) At pH value = 7: + 2NH3 + 2H2O
electrons. [1] Concentration of H+ ions is 10-7
– Q atom receives 1 electron (g) 2NaOH + MgCl2 → Mg(OH)2 +
from atom R to achieve mol dm-3 2NaCl

octet electron arrangement pOH value = 14 – 7 = 7 Q2 (a) (i) Hydrogen
and Q- anion is formed // Q (ii) Zn + H2SO4 → ZnSO4 + H2
Hence, water is neutral because (b) (i) Copper
+e →Q- [1] the concentration of H+ ions is (ii) Copper is an unreactive
the same as the concentration
– R+ ion and Q- ion are of OH- ions metal and does not react
with dilute acids.
attracted by strong Q3 (a) (i) Alkali metal oxide: sodium
oxide
electrostatic force to form Q2 (a) pH = –log[0.50] = 0.30 (ii) Metal hydroxide: sodium
(b) pH = –log[0.20] = 0.70 hydroxide
ionic bond. [1] (c) pH = –log[0.04] = 1.40 (iii) M2O + H2O → 2MOH :
– An ionic compound RQ is Na2O + H2O → 2NaOH
(b) (i) Copper(II) kloride
formed. [1] (ii) CuO + 2HCl → CuCl2 + H2O

Maksimum 10 Q3 (a) pOH = –log[0.08] = 1.1; pH = 14
– 1.1 = 12.9
6Chapter Acid, Base and Salt
(b) pH = -log[0.001] = 3.0; pH = 14
Checkpoint 6.1 – 3.0 = 11 Checkpoint 6.5

Q1 Example of acid, hyrochloric acid (c) pH = -log[0.04 × 2] = -log [0.08] Q1 (a) 3.2/0.250 = 12.8 g dm–3
Hydroclhloric acid, HCI ionises in
= 1.40; pH = 14 - 1.1 = 12.9 (b) {3.2/(23+16+1)}/0.250 =
water to produces hydrogen ions, (3.2/40)/0.250 = 0.32 mol dm–3
H+. Q4 P: 1.5 = –log [H+]; [H+] = 10 – 1.5 =
Example of alkali: ammonia 0.0316 mol dm-3 Q2 (a) Al(NO3)3 → Al3+ + 3NO3–
Ammonia ionises in water to Molarity of Al3+ = 1 × (0.30 mol
produce hydroxide ions, OH–. Q: 3.0 = –log [H+]; [H+] = 10 – 3.0 =
0.001 mol dm-3 dm–3) = 0.30 mol dm–3
Q2 H atoms in ammonia cannot ionise Molarity of NO3- = 3 × (0.30 mol
to produce H+ ions. R: 4.7 = –log [H+]; [H+] = 10 – 4.7 =
2.0 × 10-5 mol dm-3 dm–3) = 0.90 mol dm–3
NH3 molecule dissolves in water and (b) MgCl2 → Mg2+ + 2Clˉ
ionises to produce OH- ion. S: 11.1 = –log [H+]; [H+] = 10 – 11.1
= 7.9 × 10-12 mol dm-3 Molarity of Mg2+ = 1 × (0.90 mol
NH3 + H2O → NH4+ + OH–
T: 13.8 = -log [H+]; [H+] = 10 – 13.8 dm–3) = 0.90 mol dm–3
Q3 Monoprotic acid: nitric acid and = 1.6 × 10-14 mol dm-3 Molarity of NO3- = 2 × (0.90 mol
ethanoic acid
Checkpoint 6.3 dm–3) = 1.80 mol dm–3
Diprotic acid: carbonic acid
Triprotic acid: phosphoric acid Q1 (a) HX ionises completely in water;
degree of ionisation is 100%.
Q4 In water, ascorbic acid ionises to
produce H+ ion. H+ ion reacts with (b) HBr(aq) → H+ (aq) + Brˉ(aq) (c) Na2SO4 → 2Na+ + SO42–
reactive metal zinc to form salt and Molarity of Na+ = 2 × (1.80 mol
hydrogen gas. Q2 (a) A weak acid is an acid that
ionises partially in water. dm–3) = 3.60 mol dm–3
Hydrogen gas causes the Molarity of SO42– = 1 × (1.80 mol
effervescence observed. In (b) (i) HCOOH(aq) → HCOOˉ(aq)
tetrachloromethane, ascorbic dm–3) = 1.80 mol dm–3
acid does not ionise. Without + H+(ak)
H+ ions, zinc do not react in Q3 (a) Number of moles of Pb(NO3)2 =
tetrachloromethane. (ii) HCOOH; HCOOˉ : H+ (0.50) × (0.500) = 0.25 mol

Q5 Moist litmus paper contains water. Q3 (a) 2K(p) + 2H2O(l) → 2KOH(aq) + Mass of(NO3)2 = (number of
Ammonia ionises in the water and H2(g) moles) × (molar mass) = (0.25
produces OH- ions. OH- ion exhibits
(b) KOH is a strong alkali because mol) x (331 g mol-1) = 82.75 g
it ionises completely in water.

KOH(aq) → K+(aq) + OHˉ(aq) (b) Number of moles of K3PO4 =
(1.0) × (1.5) = 1.5 mol
Q4 (a) A weak alkali is an alkali that
ionises partially in water. Mass of K3PO4 = (number of
moles) × (molar mass) = (1.5
(b) Ammonium ion and hydroxide mol) × (212 g mol-1) = 318 g
ion.
(c) Number of moles of Na2S2O3 =
(c) 0.5 × 200 = 1 (2.20) × (2.0) = 4.4 mol
100

265

10 ANSWER FOCUS KIMIA F4 SPM.indd 265 27/02/2020 11:30 AM

  Chemistry Form 4  Answers

Mass of Na2S2O3 = (number of Checkpoint 6.8 a fixed amount of acid until
moles) × (molar mass) = (4.4 the powder cannot dissolve
mol) × (158 g mol-1) = 695.2 g Q1 (a) Salt is an ionic compound anymore.
produced when hydrogen ion in
Checkpoint 6.6 an acid is replaced by a metal • The salt solution should not
ion or ammonium ion. be evaporated to dryness
Q1 Number of moles of Na2CO3 = (2.5) because the salt might
× (0.250) = 0.625 mol (b) (i) CaCO3 decompose by the heat.
(ii) Calcium ion and carbonate Evaporate the salt solution
Mass of Na2CO3 = (number of until it is saturated and
moles) × (molar mass) = (0.625 mol) ion then allow it to cool for
× (106 g mol-1) = 66.25 g crystallisation to occur.
(iii) Carbonic acid and calcium
hydroxide (b) Equation for reaction:
2CH3COOH(aq) + PbO(s) →
Q2 Number of moles of (COOH)2.2H2O Q2 (a) Ammonium ethanoate Pb(CH3COO)2(s) + H2O(l)
= (1.5) × (0.500) = 0.75 mol (b) Potassium phosphate
(c) Sodium bromide Number of moles of PbO
Mass of (COOH)2.2H2O = (number
of moles) × (molar mass) = (0.75 Q3 (a) Food additives and food = mass PbO
mol) × (126 g mol-1) = 94.5 g preservatives molar mass PbO

Q3 (2.0 mol dm-3) × (V) = (0.50 mol (b) Rising flour and antacid
dm_3) × (2.5 dm3)
Q4 The particles are closely packed and = 4.46 g
V = (0.50 × 2.5)/2.0 = 0.625 dm3 = orderly. (207 + 16) g mol–1
625 cm3
The strong forces of attraction = 4.46
between keep the particles in 223
Q4 (1.8 mol dm–3) × (888 cm3) = (1.0 position.
mol dm–3) × (888 + V cm3) = 0.020 mol

888 + V = (1.8 × 888)/1.0 From equation: 1 mol PbO
V = 710.4 cm3
Checkpoint 6.9 produces 1 mol Pb(CH3COO)2 →
0.020 mol PbO produces 0.020

Checkpoint 6.7 Q1 (a) (i) Sodium nitrate; lead(II) mol Pb(CH3COO)2
ethanoate; copper(II)
Q1 (a) pH = 7 chloride; ammonium Mass of Pb(CH3COO)2 produced
sulphate; zinc nitrate
(b) 5 cm3 = [number of moles
(ii) Iron(II) carbonate; silver Pb(CH3COO)2] × [molar mass
(c) (x mol dm–3) × (5 cm3) = (0.20 chloride
mol dm–3) × (25 cm3) Pb(CH3COO)2]
(b) (i) Ammonia solution and = 0.020 mol × [207 + 2 (2 × 12
x = 1.0 mol dm–3 sulphuric acid:
+ 3 × 1 + 2 × 16)] g mol-1
2NH3(aq) + H2SO4(aq) → = 0.020 × 325 = 6.5 g
(NH4)2PO4(aq)
Q2 Number of moles of HBr = 0.240/24 Q3 (a) Sodium chloride
= 0.01 mol (ii) Copper(II) carbonate and
hydrochloric acid: (b) 2HCl(aq) + Na2CO3(aq) →
HBr + NH3 → NH4Br 2NaCl(aq) + H2O(ce) + CO2(g)
1 mol 1mol CuCO3(s) + 2HCl(aq) →
CuCl2(aq) + H2O(l) + CO2(g) (c) Volume of HCl used = 27.60 –
Number of moles of NH3 = 0.01 mol
0.01 = (0.50 mol dm–3) × (V dm3) (iii) Iron(III) chloride and calcium 1.10 = 26.50 cm3
V = 0.02 dm3 = 20 cm3 carbonate:
Number of moles of HCl
2FeCl3(aq) + 3K2CO3(aq) → MV
Q3 Number of moles of K2CO3 = Fe2CO3(s) + 3H2O(l) = 1000
(2.0 mol dm-3)
13.8 = 13.8 = 0.10 mol (iv) Zinc hydroxide and nitric × 26.5 dm3
2(39) + 12 + 3(16) 138 acid: 1000

2H3PO4 + 3K2CO3 → 2K3PO4 + 3H2O Zn(OH)2(s) + 2HNO3(aq) → = 0.053 mol
Zn(NO3)2(aq) + 2H2O(l)
+ 3CO2 From equation: 2 mol HCl react

Mole ratio: K2CO3:H3PO4 = 3:2 with 1 mol Na2CO3 → 0.053 mol
HCl reacts with 0.053 × ½ =
Number of moles of H3PO4 required
0.0265 mol Na2CO3
= 0.10 × 2 = 2 mol Concentration of Na2CO3
3 30
solution
Molarity of H3PO4
Q2 (a) • The mixture of ethanoic number of moles
= 49.0 = 49.0 = 0.5 mol dm–3 acid and lead(II) oxide = volume
3 + 31 + 64 98 must be stirred to speed
up the reaction and ensure 0.0265
Volume of acid = 2/30 = 0.133 dm3 complete reaction. = 25/1000
0.5

= 133 cm3 • Adding acid to oxide will = 1.06 mol dm–3
result in excess acid added.
Q4 % N in MAP, NH4H2PO4 = The excess acid cannot be (d) From equation: 2 mol HCl
removed by filtration. The produce 1 mol NaCl
14 + 4 1(14) +4(16) × 100 salt becomes contaminated.
+ 2 + 31 Hence, lead(II) oxide → 0.053 mol HCl produce 0.053
14 powder must be added to mol NaCl
= 115 × 100
Mass of NaCl produces
= 12.2% = [number of moles of NaCl] ×

[molar mass of NaCl]

266

10 ANSWER FOCUS KIMIA F4 SPM.indd 266 27/02/2020 11:30 AM

  Chemistry Form 4  Answers 

= 0.053 mol × (23 + 35.5) g mol–1 10 mol of NaCl produce 5 mol Ca(OH)2(aq) + CO2(g) →
= 0.053 × 58.5 = 3.10 g CaCO3(s) + H2O(l)
Cl2
[2]
Q4 (a) Silver nitrate solution is added to 0.1709 mol NaCl produces
potassium chromate(VI) solution. 5
0.1709 × 10 = 0.08545 mol Cl2

2AgNO3(aq) + K2CrO4(aq) → Volume of Cl2 liberated Section B
Ag2CrO4(p) + 2KNO3(aq) = (number of mole) × (molar volume)
2. (a) A strong alkali such as sodium

The precipitate is removed by = 0.08545 mol × 24 dm3 mol–1 hydroxide ionises completely
filtration, washed with distilled
= 2.05 dm3 in water to produce a high

water and dried between layers concentration of hydroxide ions.
of filter papers.
(b) NMumVber of moles of Pb2+ = Checkpoint 6.11 NaOH(aq) → Na+(aq) + OH–(aq)

A weak alkali such as ammonia

Q1 (a) (i) Magnesium oxide ionises partially in water to
(ii) Carbon dioxide
1000 produce a low concentration of
(iii) Calcium carbonate
(1.0 mol dm–3) × 5 dm3 (iv) MgCO3(s) → MgO(s) + hydroxide ions. [2]
1000
CO2(g) NH3(aq) + H2O(l) L
= 0.005 mol MV (b) (i) X = nitric acid; NH4+(aq) + OH–(aq)
Y = hydrochloric acid
Number of moles of = 1000 (ii) Mg2+ [2]
(iii) MgO(s) + 2H+(aq) →
20 4
1000 Mg2+(aq) + H2O(l)
(0.5 mol dm-3) × dm3 (iv) Magnesium hydroxide (b) (i) Sodium hydroxide solution
(v) Mg2+(aq) + 2OH–(aq) → reacts with lead(II) nitrate
= 0.010 mol solution to form a white
Simple mole ratio Pb2+ : I– Mg(OH)2(s) precipitate. The precipitate
is lead(II) hydroxide. [1]
= 0.005:0.010 = 1:2
2NaOH(aq) + Pb(NO3)2(aq)
Ionic equation: Pb2+(aq) + 2I–(aq) → Pb(OH)2(p) + 2NaNO3(aq)
→ PbI2(s)

Checkpoint 6.10 atau

SPM Practice 6 Pb2+(aq) + 2OH–(aq) →
Objective Questions
Q1 Silver nitrate; ammonium chloride; Pb(OH)2(s)
iron(II) sulfate; calcium carbonate [1]

Q2 (a) Ammonia, oxygen, hydrogen, 1. B 2. D 3. A 4. A 5. A The white precipitate
carbon dioxide, sulphur dioxide, 6. D 7. B 8. B 9. B 10. D
hydrogen chloride 11. B 12. C 13. D 14. A 15. B dissolves in excess sodium
16. D 17. D 18. C 19. A 20. B
21. C 22. A hydroxide solution to form a

(b) Nitrogen dioxide Subjective Questions colourless solution. [1]

(c) Chlorine; carbon dioxide; (ii) Ammonia solution reacts
nitrogen dioxide; sulphur
dioxide; hydrogen chloride with lead(II) nitrate solution

to form a white precipitate.

The precipitate is lead(II)

(d) Ammonia Section A hydroxide. [1]

Q3 (a) Green solid turns black. 1. (a) X = calcium oxide; [1] Pb2+(aq) + 2OH–(aq) →

(b) Bubble the gas through Y = carbon dioxide; [1] Pb(OH)2(s)
limewater.
Z = limewater/calcium hydroxide [1]

Limewater turns milky (white solution [1] The white precipitate does
precipitate forms)
3 not dissolve in excess

Gas is carbon dioxide. ammonia solution. [1]

(c) CuCO3(s) → CuO(s) + CO2(g) (b) CaCO3(p) → CaO(p) + CO2(g) [1] (c) (i) A standard solution is a

(c) Calcium oxide is a base. solution which has a known

Q4 (a) 10NaCl + 8H2SO4 + 2KMnO4 → CaO dissolves in water and concentration. [1]
5Na2SO4 + K2SO4 + 2MnSO4 +
8H2O + 5Cl2 ionises to produce hydroxide (ii) Calculate the mass of

ions. NaOH required:

(b) (i) Chlorine Hydroxide ion causes the Number of moles of NaOH

(ii) Bring a piece of moist blue solution to be alkaline. [2] = 2.0 mol dm–3 × 1.0 dm3

litmus paper to the mouth of (d) Carbon dioxide reacts with water = 2.0 mol

the test tube. to form carbonic acid which Mass of NaOH

The colour of the litmus ionises to produce hydrogen = 2.0 mol × (23 + 16 + 1)

changes from blue to red ions. Hydrogen ion causes the   g mol–1

and then turns white. solution to be acidic. [2] = 80 g [1]

(c) Number of moles of NaCl (e) Carbon dioxide reacts with water Procedure:

mass of NaCI to form carbonic acid which • Weigh exactly 80 g of
molar mass of NaCI sodium hydroxide solid in
= ionises to produce hydrogen a weighing bottle. [1]

10 10 ions. Hydrogen ion causes the • Transfer the solid into a
(23 + 35.5) 58.5 1.0 dm3 volumetric flask.
= = solution to be acidic. [1]

= 0.1709 mol CaO(s) + H2O(l) →

Ca(OH)2(aq)

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  Chemistry Form 4  Answers

• Add a little distilled water (iii) Add half spatula of copper(II)
Concentration of H2SO4 (mol dm–3) sulphate which acts as a catalyst.
to dissolve the solid. [1]

• Add distilled water to Q2 (a) Zn + 2HCl → ZnCl2 + H2
(b) (i) Concentration of
make exactly 1 dm3 of

solution. [1] hydrochloric acid

• Stopper the flask and (ii) The rate of reaction of

shake to obtain a uniform experiment II is higher than

solution. [1] experiment I

(iii) Number of moles of NaOH (c) Number of moles of hydrochloric
MV 0
= 1000 Time (s) acid in
(iv)
= 2.0 × 0.321 Volume of CO2 (cm3) Set I

= 0.0642 mol [1] = 0.1(25)
1000
H2SO4(aq) + 2NaOH(aq) →
= 0.0025mol
Na2SO4(aq) + 2H2O(l)
Ratio of mole between
From equation:
hydrochloric acid and hydrogen
2 mol NaOH = 1 mol H2SO4
∴ 0.0642 mol NaOH gas

= 0.0642 × 1 mol H2SO4 0 HCl : H2
2 [1] Q3 2 mol : 1 mol

Time (s) Set I
0.0025 mol : 0.00125 mol
Bilangan mol H2SO4
MV Volume of H2 gas
= 1000

= 0.0321 mol Set I

0.0321 = M × (0.025) = 0.00125 × 24 000

Concentration of acid Water = 30 cm3 60
Average rate of reaction = 150
= 0.0321 Hydrogen
0.025 peroxide = 0.4 cm3 s–1

= 1.3 mol dm–3 (d)

[1] Volume of H2 gas (cm3)

3 Q4 (a) Overall rate of reaction 60 II
maximum volume of oxygen

7Chapter = gas collected

Rate of Reaction time taken of reaction to 30 III
I
complete
0 Time (s)
Checkpoint 7.1 = 46.50
180

= 0.26 cm3 s–1

Q1 Fast : combustion of natural gas, (b) Average rate of reaction in first Checkpoint 7.3
neutralisation of acid in mouth with minute
toothpaste. Q1 The temperature of hot frying oil is
volume of oxygen gas higher than boiling water. Potato
Slow : browning of newspaper, = collected in first 60 s strips absorbed more heat in hot oil.
fermentation (any suitable answer)
time taken
Q2 (i)
= 27.50 Q2 Smaller sized charcoal has larger
Mass of CaCO3 (g) 60 total surface area exposed to
oxygen in air. Allowing smaller sized
= 0.46 cm3 s–1 charcoal to burn easily.
(c) Average rate of reaction in third
Q3 Porridge is a diluted rice. The
minute number of rice per unit volume
in porridge is smaller. Amylase
volume of oxygen gas enzyme take shorter time to digest.
= collected from 120 s to 180 s

time taken

0 Time (s) = 50.00 – 41.50 Q4 Cut the chicken meats into smaller
60 cubes and cooks it in a pressure
(ii) cooker.
= 0.14 cm3 s–1
Mass of CaSO4 (g)

Checkpoint 7.2 The temperature of the chicken soup
in pressure cooker is higher.
Q1 Replace the zinc granules with zinc
powder. Smaller cubes of chicken meat have
larger total surface area. More heat
Replace the 0.5 mol dm-3 energy is absorbed in shorter time.
hydrochloric acid with 1.0 mol dm-3.
0 Time (s)

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  Chemistry Form 4  Answers 

Checkpoint 7.4 Subjective Questions

Q1 Collision between the reactant Section A
particles in correct orientation,
achieved the activation energy and 1. (a) Changes of hydrogen gas volume per unit time. // changes in mass of reactant
results products.
per unit time [1]
Q2 (a) CaCO3 + 2H+ → Ca2+ + H2O +
CO (b) Carbon dioxide [1]

Calcium carbonate , hydrogen (c) Volume of carbon dioxide gas evolved // mass of calcium carbonate [1]
ion
(d)
(b) Zn + 2H+ → Zn2+ + H2
Zinc atom , hydrogen ion Water Sulphuric
acid
(c) Mg + 2H+ → Mg2+ + H2 Sulphuric acid Calcium
magnesium atom, hydrogen ion Calcium carbonate 100 g carbonate

(d) S2O32- + 2H+ → + S(s) + SO2(g) [2]
+ H2O(l)
(e) CaCO3 + 2H+ → Ca2+ + H2O + (b) 90% Cu, 10% Sn
thiosulphate Ion, hydrogen ion (c)
CO2 [2]
(e) 2 H2O2(aq) → 2H2O(l) + O2(g)
hydrogen peroxide molecule 2. (a) (i) Substance that increase the

Q3 (a) Set I rate of decomposition of Zn
13.6
hydrogen peroxide. [1]
= 60
= 0.23 cm3 s–1 (ii) Manganese(IV) oxide [1] Cu
(iii) 2H2O2 → 2 H2O + O2 [2]
(iv) Manganese(IV) oxide acts (d) The presence of smaller size
zinc atoms disrupt the orderly
Set II as catalyst by lowering arrangement of copper atoms in
32.6 the activation energy of pure copper. The layer of metal
atoms cannot slide easily if force
= 60 decomposition of hydrogen is applied.
= 0.54 cm3 s–1 peroxide. More hydrogen
Q2 (a) Ductile and malleable
Volume of H2 gas (cm3) peroxide molecules can (b) (i) Duralumin
achieve the activation (ii) Lighter, stronger and

32.6 energy. The frequency of resistant to corrosion.
effective collision increased. (iii) Material used in aircraft

[2] construction.

13.4 (b) Antacid B [1]
0
Antacid B in powder form is

smaller in size and it has larger

Time (s) total surface area exposed to Checkpoint 8.2

acid. More H+ ion are neutralised

(c) Experiment II has higher rate of in short time. [2] Q1 (a) Transparent, hard but brittle,
reaction. chemically inert, electrical
3 insulator
The size of magnesium powder (c)
is smaller than magnesium (b) Silicon dioxide, SiO2
ribbon. Fast reaction Slow reaction
Q2 (a) Soda-lime glass. Soda-
The total surface area of the • Combustion of • Fermentation lime glass has high thermal
magnesium powder exposed to ethanol of yeast expansion coefficient and
hydrochloric acid is larger. it cannot withstand sudden
• Displacement of • Electrolysis temperature change.
Frequency of collision between metal from its salt
magnesium atom, Mg and solution (b) Borosilicate glass.
hydrogen ion, H+ is higher. Borosilicate glass has zero
Frequency of effective collision
is higher. thermal expansion coefficient
thus it is resistant to thermal
stress.

SPM Practice 7 8Chapter Manufactured Substances Checkpoint 8.3
Objective Questions in Industry

1. A 2. A 3. D 4. D 5. B Checkpoint 8.1 Q1 (a) Inert chemically to make dental
6. D 7. C 8. B 9. B 10. C enamel.
11. C 12. C 13. D Q1 (a) Alloy is a mixture of two or
more elements where the main Hard and strong to make
element is a metal. construction metals.

Heat insulator to make inner
wall of oven.

(any suitable answers)

269

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  Chemistry Form 4  Answers

(b) (i) Advanced ceramic atom disrupt the orderly 1. (a) (i)

(ii) • Lithium silicon oxide arrangement of iron atoms.

is used as heat shield The presence of carbon

that can withstand high atom prevents the iron

temperature and is used atoms from, sliding when a

as heat insulator in rocket force is applied. [2]

nozzle. Section B

• Alumina Al2O3 and 2. (a) (i) Fibre glass [1] [1]
(ii) Iron atoms are closely-
silica SiO2 is used (ii) Glass fibre and plastic [2]
in manufacture of packed and orderly. There
are empty spaces between
microchips. (iii) Lighter and can be moulded iron atoms. When a force
is applied, layers of iron
(any suitable answers) into any shape [2] atoms slide easily over one
(iv) • Use of composite another.
Checkpoint 8.4 Iron atoms move to fill
material can improve theses gaps, producing a
new shapes.
Q1 Composite material is a new living standards because
substance produced from combining composite material is [2]
two or more non-homogeneous superior to its original (b) (i)
substances.
component. [1]
Q2 • Composite material

Composite produced through
materials
Uses chemical engineering can
reduce the exploitation
Fibre glass Make helmet of natural resources and

Reinforced Building materials for conserves a balanced
concrete tall building ecosystem. [1]
• Producing composite
Optical fibre Cable for computer
networking material which is more
durable and long lasting
Photochromic Optical lenses can reduce the waste [1]
glass
generation by human. [1] (ii) Carbon atoms disrupt the
orderly arrangement of iron
Superconduktor Components in 3 atoms in iron metal.
Megnetic Reson-
once Imaging (MRI) (b) (i) Advanced ceramic is Layers of iron atoms are
produced by mixing additive difficult to slide over each
Q3 Original component: silica glass fibre substances such as oxides, other.
and plastic polymer carbides and nitrides into
ceramic. [2] Hence, pure iron becomes
Silica glass fibre breaks easily but stronger and harder. [2]
fibre glass is durable. Semiconductor, heat
insulator which can
Plastic has low stretching strength [2] (c) (i) As pH value decreases, rate
but fibre glass has high stretching withstand high heat.
strength. of rusting increases.

As pH value decreases,

concentration of H+ ions

SPM Practice 8 increases.
Objective Questions
PRE-SPM MODEL PAPER H+ ions react with iron; iron

Objective Questions corrodes.

1. B 2. B 3. A 4. D 5. D 1. D 2. D 3. C 4. B 5. A The higher the concentration
6. D 7. A 8. D 9. A 10. C 6. B 7. A 8. B 9. A 10. B
11. A 12. C 13. D 14. B 15. C 11. B 12. D 13. D 14. B 15. C of H+ ions, the faster the
16. C 16. B 17. D 18. B 19. C 20. D
21. B 22. C 23. A 24. B 25. D rusting of iron. [3]
26. B 27. D 28. C 29. D 30. A
31. A 32. C 33. D 34. C 35. A (ii) The higher the temperature,
36. C 37. D 38. D 39. D 40. A
41. C 42. B 43. D 44. B 45. B the higher the rate of
4 6. D 47. A 48. D 49. C 50. D
Subjective Questions rusting. [1]
Subjective Questions
Section A 2. (a) Powder cannot dissolve / no
Section A more effervescence occurs. [1]
Paper 2
1. (a) Alloy is a mixture of two or (b) To ensure all the acid has [1]
reacted.
more elements where the main

element is metal. [2] (c) (i) Zinc nitrate
(b) W: Carbon [4]
(ii) ZnCO3(s) + 2HNO3(aq) →
X: Pewter
Y: Aluminium Zn(NO3)2(aq) + H2O(l) +
Z: Stainless steel
CO2(g) [1]

(c) (i) To increase the strength and (d) (i) To prepare a hot saturated
salt solution.
resistance to corrosion. [2]
(ii) To prevent decomposition of
(ii) Carbon atom has smaller salt by high heat.

atomic size. Carbon

270

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  Chemistry Form 4  Answers 

(e) Remove the crystals by filtration 3. (a) Sodium, Na (b) (i) Chlorine gas

and dry crystals between layers Has the least number of Chlorine gas turns moist

of filter paper. [1] protons. blue litmus paper red and

(f) Number of moles of HNO3 Nuclear attraction to valence then white (bleached). [2]

= 1.5 × 100 shell is weakest. [2] (ii) Hydrogen chloride
1000
= 0.15 mol (b) From sodium to sulphur – solid Hydrogen combines with

state at room temperature and chlorine to form hydrogen

Mole ratio: pressure. chloride.

HNO3:Zn(NO3)2 = 2:1 From chlorine to argon – gas HCl is a strong acid; ionises

Mass of Zn(NO3)2 state at room temperature and completely in water to

= 0.15 × 1 pressure. [1] produce high concentration
2
(c) Compare to sodium, chromium of H+ ions. [2]

= 0.075 mol has (iii) Ammonia gas

Mass of Zn(NO3)2 formed • higher strength and Nitrogen combines with

= 0.075 × (65 + 28 + 96) hardness hydrogen to produce
= 14.18 g
• higher melting point and ammonia.

[2] boiling point

(g) Zinc oxide or zinc hydroxide [2] • higher density [1] Ammonia is used to make

nitrogenous fertilisers

(NH4Cl). [2]

(d) (i) 6. (a) (i) Neutron and proton [1]

Period 3 Na Mg Al Si P S Cl (ii) Neutron: James Chadwick
element
Proton: Ernest Rutherford [1]
3
(b) (i) Chlorine-35: 4 × 100 = 75%

Formula of Na2O MgO Al2O3 SiO2 P4O6 SO2 Cl2O7 Chlorine-37: 1 × 100
oxide P4O10 SO3 4

= 25% [1]

Acid- bes Bes Amfoterik Asid Asid Asid Asid (ii) Relative atomic mass of
baseproperry [2] chlorine

= (35 × 75) + (37 × 25)

100

(ii) Na2O(s) + H2O(l) → Mass of NaN3 = 2.083 × = 35.5 [2]
2NaOH(aq) [23+3(14)] = 2.083 × 65 = 135.4
(c) (i)
• Na2O dissolves in water; g [3]
forming an alkaline solution.
[2] (c) Number of moles of NaN3 O
Cl Cl
4. (a) (i) 2NaNH2(s) + N2O(g) → = 75
65

NaN3(s) + NaOH(s) + = 1.154 mol

NH3(g) [1] Mole ratio: NaN3:HNO2 = 1:1 [1]

(ii) Number of moles of NaN3 (ii) O
61.2
= 100 = 1000 Number of moles of HNO2 = Cl 16
23+3(14) 65 1.154 38.8 3.825
35.5 3.825
= 15.385 mol = 0.250 [3] 1.093 = 3.5

Mole ratio: NaN3:NaNH2 = = 4.616 ≈ 4.62 mol dm–3
1:2
5. (a) (i) Lead(II) nitrate solution 1.093

Number of moles of NaNH2 Pb2+(aq) + 2Cl-(aq) → 1.093
= 2(15.385) = 30.77 mol 1.093
PbCl2(s); white precipitate
Mass of NaNH2 = 30.77 × Pb2+(aq) + 2I-(aq) → PbI2(s);
(23+14+2) = 1200 g = 1.2 kg
yellow precipitate [2]

[3] (ii) Excess sodium hydroxide 27
[2]
(b) Number of moles of N2 solution:
75
Al3+(aq) + 3OH-(aq) → Empirical formula = Cl2O7
= 24 = 3.125 mol (iii) 183 = n[(35.5×2) + (16×7)] =
Al(OH)3(s); white precipitate n[71 + 112]

Mole ratio: N2:NaN3 : = 3:2 dissolves in excess alkali.

Number of moles of NaN3 Mg2+(aq) + 2OH-(aq) = 183n
→ n = 1; Empirical formula
= 2 × 2 × 3.125 → Mg(OH)2(s); white
3 3 precipitate insoluble in = Cl2O7 [2]

= 2.083 mol excess alkali. [2]

271

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  Chemistry Form 4  Answers

Section B Atom A becomes an ion of Atomic radius of astatine ≈ 150
charge -1 and achieve a stable pm [5]
7. (a) Ions duplet electron arrangement. [1]
X is an electrical conductor (b) Boiling point drops sharply from
Y is a non-electrical conductor Strong electrostatic forces of HF to HCl; then increases from
attraction hold the oppositely- HCl to HI.
Electrical conductivity requires
charged ions together to form All HX exist as diatomic
charge carriers – mobile molecules – simple discrete
ionic bonds. [1] molecules.
electrons or ions
X could be an aqueous Compound Y is a covalent

electrolyte solution – containing compound with covalent bonds Covalent bonds are found in
each H-X molecule; bonding a H
mobile ions to conduct between atoms. [1] atom to a halogen atom, X.

electricity. Atom A has 1 valence electron; Weak intermolecular forces
Y is made up of uncharged A contributes 1 electron for

particles such as molecules. [4] sharing. [1] of attraction (van der Waals
forces) are found between HX
(b) (i) C2H5OH(l) + 3O2(g) → [2] Atom C has 6 valence electrons;
2CO2(g) + 3H2O(l) C contribute 2 electrons for molecules.

(ii) Number of moles of C2H5OH sharing [1] On descending Group 17, HX
molecule size increases from HF
10 Two A atoms share 1 pair of
= 24 + 6 + 16
electrons with one C atom for to HI. [5]
10 atom A to achieve a stable
= 46 The van der Waals forces
duplet electron arrangement and become stronger.
atom C to achieve a stable octet
= 0.2174 mol More energy is needed to

Number of moles of O2 electron arrangement. [1] overcome these forces; boiling
10 10 point increases.

= 24 = 0.4167 mol 8. (a) The boiling point of HF is
140 far higher than the other HX
Mole ratio: C2H5OH:O2
= 1:3 Atomic radius (pm) because of the presence
Number of moles of O2 to
burn 0.2174 mol of ethanol 120 of intermolecular hydrogen
completely
= 3(0.2174) bonding. [5]
= 0.6522 mol
100

80

Number of moles of O2 60
provided is 0.4167 mole; 0 20 40 60 80
Number (atom)
insufficient for complete
Atom F has higher
combustion. electronegativity

At the same temperature

and pressure, mole ratio =

volume ratio

Mole ratio: O2:CO2 HF HF HF
= 3:2

Volume ratio: O2:CO2

= 3:2 or 10: 2 × 10 Hydrogen bond
3 between molecules
= 6.67 dm3 [4] (c) Heat a small piece of sodium metal in a combustion spoon until it ignites.
Place it into a gas jar filled with chlorine gas.
(c) Compound X is an ionic Sodium continues to burn and a white solid is formed.

compound with ionic bonds Combustion
spoon
between oppositely-charged Gas jar cover

ions. [1] Gas jar
Chlorine gas
Atom A has 1 valence electron.
Sodium
Atom B has 2 valence electrons.

[1]

Each atom B donates 2
electrons to two A atoms; 1
electron to each atom A. [1]

Atom B becomes an ion of

charge +2 and achieve a stable

octet electron arrangement. [1]

[5]

272

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  Chemistry Form 4  Answers 

(d) (i) Changing Ag+ ion to Ag (ii) Dissolve a little sodium 0.480
= 24 = 0.02 dm3
atom: chloride in distilled water.

Add one electron to Ag+ ion: Dissolve a little silver nitrate Mole ratio: H2:HCl = 1:2
Number of moles of HCl = 0.02
Ag+ + e- → Ag in distilled water.
× 2 = 0.04 mol
Changing Ag atom to Ag+ Mix the two solutions and Use HCl of concentration 0.40

ion: stir. mol dm–3.
Hence, volume of acid
Remove one electron from Filter, wash and dry the
0.04
Ag atom: precipitate formed. [3] = 0.40 = 0.10 dm3

Ag → Ag+ + e- [2]

9. (a) = 100 cm3 [13]
Method
Reactant Reactant II Salt
NH3(ak) NH4CI
AgCI Paper 3
AgNO3(ak) PbCI2
Titration HCI Pb(NO3)2(ak) CuCI2 1. (a) 0; 13; 25; 38; 48; 59; 70; 79; 88;

CuCO3 96 [3]

Precipitation HCI (b) (i) Effervescence / gas bubbles

/ aluminium foil dissolves [3]

HCI (ii) The higher the concentration

of sulphuric acid, the higher

the rate of reaction. [3]

Neutralisation HCI (iii) Manipulated variable:

Concentration of sulphuric

• HCl(aq) + NH3(aq) → 10. (a) Experiment II has a higher rate acid
NH4Cl(aq)
• of reaction. Responding variable:
HCl(aq) + AgNO3(aq) →
• ∆V Rate of reaction

• Gradient of graph = ∆t Constant variable:

(b) • AgCl(s) + HNO3(aq) change in volume Temperature; volume of
= of H2 gas
• 2HCl(aq) + (NO3)2(aq) → acid; size of aluminium foil
PbCl2(s) + 2HNO3(aq) change in time
• [3]
• 2HCl(aq) + CuCO3(s) →
CuCl2(aq) + H2O(l) + CO2(g) (c)
•
= Gradient of graph for Volume of gas (cm3)
experiment II is steeper
[10] than gradient of graph for
experiment I.
HCl acid reacts with lead to 100 Experiment I
The steeper the graph, the Experiment II
form lead(II) chloride. [1] higher the rate of reaction.

2HCl(aq) + Pb(s) → Concentration factor of
hydrochloric acid:
PbCl2(s) + H2(g) [1]
• Size of iron solid is fixed –
PbCl2 is an insoluble salt. powder

The layer of PbCl2 covers • Temperature is fixed – 0 Time (s)
the lead solid; preventing room temperature 200

further reaction. [1] • No catalyst is needed [7] [3]

Hence, the salt formed is (b) Equation of reaction: (d) (i) Experiment I [3]
2HCl + Fe → FeCl2 + H2
not pure. [1] (ii) Higher concentration of
Experiment I:
Alternative procedure: sulphuric acid. [3]
• Add excess lead powder to Number of moles of H2
0.240 (e) The aluminium foil has reacted
100 cm3 of dilute nitric acid.
[1] = 24 = 0.01 mol completely.

(f) Higher rate of reaction.

• Heat the mixture while Powder has a larger total

stirring it. [1] exposed surface area.

Mole ratio: H2:HCl = 1:2 Frequency of collisions
Number of moles of HCl = 0.01
• Filter out the unreacted increases.
× 2 = 0.02 mol
powder. [1] Use HCl of concentration 0.20 Frequency of effective collisions

• Add hydrochloric acid or mol dm–3. increases; increasing rate of
Hence, volume of acid
sodium chloride solution to reaction. [3]
0.02
the filtrate. [1] = 0.20 = 0.10 dm3 (g) Measuring cyclinder is not a

• Remove precipitate formed = 100 cm3 standard apparatus – reading is

by filtration, wash precipitate Experiment II: not accurate (±0.5 cm3)
Number of moles of H2
with distilled water and dry High % error. [3]

the precipitate. [1] (h) Use a burette; more accurate

10 reading, up to ±0.1 cm3. [3]

273

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  Chemistry Form 4  Answers

2. Problem statement: Is brass harder 3. Release the weight so as to hit the stainless steel cone. [1]

than pure copper? 4. Measure the depth of the dent formed on the surface of the copper block. [1]

Variables: [1] 5. Repeat steps 1 to 4 three more times at different places on the surface of the
copper block to obtain an average value. [1]
• Manipulated:

Type of metal [1] 6. Repeat steps 1 to 5 with a brass block to replace the copper block. [1]

• Responding: 7. Record your observations in a table. [1]

Hardness of metal [1] Results:

• Controlled: Size of metal block,

mass of weight, distance weight Depth of dent (cm)

is dropped [1] Type of

• Hypothesis: Brass is harder metal

than pure copper. [1] 1 2 3 Average

• Material: [1]

Copper block, brass block

• Apparatus: 1-kg weight,

stainless steel cone, retort stand

and clamp, meter ruler and

cellophane tape. [1]

Procedure: Copper

1. Tape a stainless steel cone on

the surface of a copper block

with cellophane tape. [1]

2. Hang a 1-kg weight above the

stainless steel cone, 50 cm [1]

from the surface of the block as

shown in the figure below. [1]

Meter ruler

Thread Brass

1 kg weight [1]
Stainless steel 17
cone
Cellophane tape
Copper block

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