Focus SPM 2022 - Chemistry

PELANGI BESTSELLER

CHEMISTRY SPM
FORM
4∙5
KSSM
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Chien Hui Siong D ual L anguage
Low Swee Neo P rogramme
Lim Eng Wah
Salida Sani NEW SPM ASSESSMENT

FORMAT 2021

CONTENTS

Form 4 4.4 Elements in Group 1 69
4.5 Elements in Group 17 76
1Chapter 4.6 Elements in Period 3 84
4.7 Transition Elements 90
Introduction to Chemistry 1 92
SPM Practice 4
Penerbitan Pelangi Sdn Bhd. All Rights Reserved1.1 Development in Chemistry Field and Its
Importance in Daily Life 2
1.2 Scientific Investigation in Chemistry 4 5Chapter
1.3 Usage, Management and Handling of
Apparatus and Materials 5 Chemical Bond 94

SPM Practice 1 8 5.1 Basics of Compound Formation 95
5.2 Ionic Bond 98
2Chapter Matter and the Atomic 5.3 Covalent Bond 103
Structure 11 5.4 Hydrogen Bond 107
5.5 Dative Bond 111
2.1 Basic Concepts of Matter 12 5.6 Metallic Bond 113
2.2 The Development of the Atomic 5.7 Properties of Ionic Compounds and
Model 18 114
2.3 Atomic Structure 20 Covalent Compounds 120
2.4 Isotopes and Their Uses 26
SPM Practice 5

SPM Practice 2 30 6Chapter

3Chapter The Mole Concept, Chemical Acid, Base and Salt 123
Formula and Equation 33
6.1 The Role of Water in Showing Acidic
3.1 Relative Atomic Mass and Relative and Alkaline Properties 124
Molecular Mass 34
36 6.2 pH Value 130
3.2 Mole Concept 42
3.3 Chemical Formula 52 6.3 Strength of Acids and Alkalis 133
3.4 Chemical Equations 57
SPM Practice 3 6.4 Chemical Properties of Acids and
Alkalis 135

6.5 Concentrations of Aqueous Solution 142

4Chapter 6.6 Standard Solution 145

The Periodic Table of 6.7 Neutralisation 147
Elements 60
6.8 Salts, Crystals and Their Uses in
4.1 The Development of the Periodic Table Daily Life 154

of Elements 61 6.9 Preparation of Salts 156

4.2 The Arrangement in the Periodic 6.10 Effect of Heat on Salts 170

Table of Elements 63 6.11 Qualitative Analysis 178

4.3 Elements in Group 18 67 SPM Practice 6 189

v

7Chapter Rate of Reaction 193 3Chapter Thermochemistry 350

7.1 Determining Rate of Reaction 194 3.1 Heat Change in Reactions 351
7.2 Factors Affecting Rate of Reactions 201 3.2 Heat of Reaction 357
7.3 Application of Factors that Affect the 212 3.3 Application of Exothermic and
213 Endothermic Reactions in Daily Life 391
Rate of Reaction in Daily Life 220 SPM Practice 3 395
7.4 Collision Theory
SPM Practice 7
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8Chapter Manufactured Substances in 4Chapter Polymer Chemistry 401
Industry 223

8.1 Alloy and Its Importance 224 4.1 Polymer 402
4.2 Natural Rubber 407
8.2 Composition of Glass and Its Uses 229 4.3 Synthetic Rubber 414
8.3 Composition of Ceramics and Its
Uses 232 SPM Practice 4 416

8.4 Composite Materials and Their
Uses 235
239 5Chapter
SPM Practice 8 Consumer and Industrial
Chemistry 419
Form 5
5.1 Oils and Fats 420
1Chapter 5.2 Cleaning Agents 423
5.3 Food Additives 432
Redox Equilibrium 242 5.4 Medicines and Cosmetics 436

1.1 Oxidation and Reduction 243 5.5 Application of Nanotechnology in
Industry 441
1.2 Standard Electrode Potential 260 5.6 Application of Green Technology in
1.3 Voltaic Cell 261
1.4 Electrolytic Cell 265 Industrial Waste Management 443
SPM Practice 5 446
1.5 Extraction of Metal from Its Ore 276
1.6 Rusting 278
SPM Practice 1 285

2Chapter Carbon Compound 289 SPM MODEL PAPER 450
ANSWERS 464

2.1 Types of Carbon Compound 290
2.2 Homologous Series 294
2.3 Chemical Properties and Interconversion
of Compounds between Homologous
Series 308
2.4 Isomers and Naming based on IUPAC
Nomenclature 329
SPM Practice 2 345

vi

3Chapter Form 4

The Mole Concept, Chemical
Formula and Equation

CHAPTER FOCUS
• Relative Atomic Mass and Relative Molecular Mass
• Mole Concept
• Chemical Formula
• Chemical Equation
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Form 4

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

33

  Chemistry SPM  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.
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At first, hydrogen
Form 4 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
as the standard atom. standard atom. atom internationally.
• The mass of hydrogen • The relative atomic • The mass of carbon-12 atom is

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

– 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 12 the mass of 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

34

Chemistry SPM  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 dioxide
molecule, CO2

OC O
RAM for O = 16

RAM for C = 12
Penerbitan Pelangi Sdn Bhd. All Rights Reserved[Relative atomic mass: Be, 9; Ag, 108]
Form 4
Solution RAM for O = 16

The number of beryllium atoms = n

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

= Mass of 2 silver atoms Figure 3.2 Calculating the relative molecular mass of
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 = 24 at9oms
= Molecular Formula of Relative molecular mass,
substance substance RMM
EXAMPLE 3.2
Hydrogen H2 2(1) = 2
How many times is the mass of a zinc atom gas
larger than the mass of a helium atom? O2 2(16) = 32
[Relative atomic mass: He, 4; Zn, 65] Oxygen gas H2O 2(1) + 16 = 18

Water

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 of [Relative atomic mass : H,1; N, 14; O, 16; Na, 23,
the mass of a carbon-12 atom. Mg, 24; S, 32; Cl, 35.5; Ca, 40; Cu, 64]

35

  Chemistry SPM  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?

EXAMPLE 3.3 [Relative atomic mass: He, 4]

The relative formula mass of X2SO4 is 142. Q3 Calculate the relative molecular mass of the
Calculate the relative atomic mass of element X. following substances:
[Relative atomic mass: O, 16; S, 32] (a) Phosphorus, P4
(b) Carbon monoxide gas, CO
(c) Sucrose, C12H22O11
(d) Benzoic acid, C6H5COOH

[Relative atomic mass: H, 1; C, 12; O, 16; P, 31]

Q4 Calculate the relative formula mass of the
following substances:
(a) Sodium oxide, Na2O
(b) Zinc nitrate, Zn(NO3)2
(c) Potassium thiosulphate, K2S2O3
(d) Hydrated sodium carbonate, Na2CO3.10H2O

[Relative atomic mass: H,1; C,12; N,14; O, 16;
Na, 23; S, 32; K, 39; Zn, 65]

Q5 Element Z forms a chloride salt with the formula
ZCI2 and relative formula mass of 95. What is
the relative atomic mass of element Z?

[Relative atomic mass: Cl, 35.5]
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Form 4 RAM of X = x

Given RFM of X2SO4 = 142

2(x) + 32 + 4(16) = 142

2x + 96 = 142

2x = 142 – 96 = 46
46
x = 2 = 23

RAM of X is 23.

Checkpoint 3.1

Q1 How many oxygen atoms have the same mass
as three copper atoms?

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

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

Chemistry SPM  Chapter 3 The Mole Concept, Chemical Formula and Equation  

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

Table 3.3 Type of substances 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
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The meaning 1 mole of atomic substance Form 41 mole of molecular substance has 6.021 mole of ionic substance
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
mole of sodium chloride
contains 6.02 × 1023 zinc molecules. contains 6.02 × 1023 NaCl

atoms. units.

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

37

  Chemistry SPM  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
Molecular H2 16
1.806 × 1023 substance RMM = 12 + 4(1)
6.02 × 1023 = 16
Methane, CH4
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Form 4 = 0.3 mol Zinc bromide, RFM = 65 + 2(80) 225
= 225 142
Ionic ZnBr2
RFM = 2(23) + 32
substance Sodium sulphate, + 4(16) = 142

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
are present in the beaker? Mg, 24; S, 32; Zn, 65; Br, 80]

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.
What is the mass of 1.2 moles of sodium
The number of MgCl2 units in 0.5 mole of hydroxide, NaOH?
magnesium chloride [Relative atomic mass: H, 1; O, 16; Na, 23]
= number of moles × NA
= 0.5 × 6.02 × 1023 units Solution
RFM of sodium hydroxide, NaOH
Each MgCl2 unit is made up of three ions, = 23 + 16 + 1 = 40
which are one magnesium ion and two chloride So, the molar mass of sodium hydroxide
ions. So, the number of ions in the beaker = 40 g mol-1
= 3 × number of MgCl2 The mass of 1.2 moles of sodium hydroxide
= 3 × 0.5 × 6.02 × 1023 ions = number of moles × molar mass
= 9.03 × 1023 ions = 1.2 × 40 = 48 g

Number of moles and mass of substance EXAMPLE 3.8

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

Number × molar mass Mass in Solution
of moles ÷ molar mass grams

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

38

Chemistry SPM  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
you read the question carefully to understand what
glucose, C6H12O6? is required in the question. Use RAM to calculate
[Relative atomic mass: H,1; C, 12; O, 16] the number of atoms of oxygen, O. Use RMM to
calculate the number of molecules of oxygen,
A 3 moles of propanol, C3H7OH O2.
B 4 moles of propane, C3H6
C 6 moles of ethane, C2H6 5. We can compare the number of particles in
D 10 moles of water, H2O substances by just comparing the number of
moles of the substances. See Example 3.10.
Penerbitan Pelangi Sdn Bhd. All Rights ReservedExaminer’s tip
Form 4Students need to know the molar mass and the mass

of each substance, including glucose.

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

39

  Chemistry SPM  Chapter 3 The Mole Concept, Chemical Formula and Equation

• Room conditions refer to the temperature EXAMPLE 3.11
of 25 ºC 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
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Form 4 Number × molar volume Volume EXAMPLE 3.12
of moles ÷ molar volume of gas
100 cm3 of carbon dioxide gas is collected in a
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)

40

Chemistry SPM  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?
[Relative atomic mass: Mg, 24. Avogadro’s
constant, NA: 6.02 × 1023 mol–1] [Molar volume: 22.4 dm3 mol–1 at STP]

Solution A 56 g mol–1 C 180 g mol–1
Number of moles of Mg = number of particles ÷ NA
=36.6.0122 ×× 11002234 B 127 g mol–1 D 254 g mol–1
= 6 mol
Therefore, the mass of magnesium powder Examiner’s tip
= number of moles × molar mass
= 6 × 24 First, analyse the question. The question gives the
= 144 g
Penerbitan Pelangi Sdn Bhd. All Rights Reserved mass and volume of the gas. To determine the molar
Form 4mass of gas X, students need to calculate the mass
of 1 mole of gas X or 22 400 cm3 at STP.
Number of moles of gas X that occupies 5600 cm3

= volume of gas
molar volume

= 5 600
22 400

= 0.25 mol

EXAMPLE 3.14 So, the mass of 0.25 mole of gas X is 63.5 g.
This means that the mass of 1 mole of gas X

What is the volume of 100 g of ammonia gas, = 63.5
NH3 at STP? 0.25
[Relative atomic mass: H, 1; H, 14. Molar
volume: 22.4 dm3 mol-1 at STP] = 254 g

So, the molar mass of gas X is 254 g mol–1.

Answer: D

Solution

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 condition.]
Q1 A sample consists of 2.5 moles of carbon dioxide
= 5.88 × 22.4
gas, CO2.
= 131.71 dm3 (a) How many molecules are there in the sample?
(b) Hence, calculate the number of atoms in
EXAMPLE 3.15
the sample.
How many hydrogen molecules, H2 are there in (c) What is the volume of the sample at room
6 dm3 of the gas at room conditions?
conditions?
[Molar volume: 24 dm3 mol–1 at room Q2 Calculate the number of moles of each of the
conditions. Avogadro’s constant, NA: 6.02 ×
1023 mol–1] following.
(a) 13.8 g of sodium, Na
Solution (b) 40.4 g of potassium nitrate, KNO3
Q3 An experiment requires 0.5 mole of copper(II)
Number of moles of H2 = volume ÷ molar volume sulphate, CuSO4.
(a) Calculate the molar mass of copper(II)
= 6 = 0.25 mol
24 sulphate.
So, the number of H2 molecules (b) What is the mass of copper(II) sulphate

= number of moles × Avogadro’s constant required in the experiment?
Q4 What is the mass of carbon that has twice the
= 0.25 × 6.02 × 1023
amount of atoms found in 11.2 g of iron?
= 1.505 × 1023 molecules Q5 Arrange the following samples of gases in an

ascending order of volume.
X: 5 g of neon gas, Ne
Y: 10 g of ammonia gas, NH3
Z: 16 g of bromine gas, Br2

41

  Chemistry SPM  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
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Chemical Chemical formula Information

Form 4 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

Hydrogen gas H2 Made up of diatomic molecules • Elements
Ozone gas
Sulphur O3 Made up of triatomic molecules • Type of particles: molecules
The formula shows the number

S8 Made up of molecules with 8 atoms of atoms in each molecule

Water H2O Contains hydrogen dan oxygen with the • Compounds
NH3 ratio H:O = 2:1 • Show the elements that made
Ammonia Zn(OH)2
MgSO4 Contains nitrogen dan hydrogen with the up the chemicals
Zinc hydroxide ratio N:H = 1:3 • Show the ratio of number of
Magnesium
sulphate Contains zinc, oxygen dan hidrogen with atoms of each element in
the ratio Zn:O:H = 1:2:2
chemicals
Contains magnesium, sulphur dan oxygen • Type of particles: May be
with the ratio Mg:S:O = 1:1:4
molecules or ions

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 C:H=1:4
Vitamin C C2H4 CH2
H:H=1:1
C6H6 CH
C:H:O=3:4:3
C6H8O6 C3H4O3

Similarities

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

42

Chemistry SPM  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

Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 4Figure 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
compound. What is the empirical formula of the oxide?
[Relative atomic mass: O, 16; Al, 27]

Solution

Element Aluminium, Al Oxygen, O

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.

43

  Chemistry SPM  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)
Number of moles of atoms

Ratio of number of moles of
atoms
Simplest ratio of number of
moles of atoms
Penerbitan Pelangi Sdn Bhd. All Rights Reserved3.06 31.63 65.31

Form 4 3.06 = 3.06 31.63 = 1.02 65.31 = 4.08
1 31 16

3.06 =3 1.02 =1 4.08 =4
1.02 1.02 1.02

3 1 4

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.

44

Chemistry SPM  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
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Form 4Calculation:
Mass of magnesium used = (y – x) g
Mass of oxygen that combines with it = (z – y) g

Table 3.7

Element Magnesium, Mg Oxygen, O

Mass (g) y–x z–y

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.

45

  Chemistry SPM  Chapter 3 The Mole Concept, Chemical Formula and Equation

ACTIVITY 3.2

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
Form 4 Penerbitan Pelangi Sdn Bhd. All Rights Reserved

Thistle Hydrogen Copper(II) oxide
funnel gas
Burning of excess
Anhydrous hydrogen gas
calcium
Dilute sulphuric chloride Heat Combustion tube
acid + Porcelain dish
Retort
Copper(II) stand
sulphate solution

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.

1 0. 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

46

Chemistry SPM  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

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Form 4
[Relative atomic mass : O, 16; Cu, 64]
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.

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

47

  Chemistry SPM  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

Form 4 Penerbitan Pelangi Sdn Bhd. All Rights Reserved 2. The determination of a molecular formulaEXAMPLE 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
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
= 0.15 g = 27 g

Therefore, the relative atomic mass of W is 27.

48

Chemistry SPM  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
Number of moles of CO2
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Form 4Mass132 132
= Molar mass = 12+2(16) = 44 = 3 mol

So, the percentage by mass of carbon Based on the formula, 1 mole of CO2 contains 2

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

= 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 Br –
H+ –2 Iodide ion I–
+1 Potassium ion NH4+ –3 Hydroxide ion OH–
Hydrogen ion Ca2+ Nitrate ion NO3–
Mg2+ Hydride ion H–
Ammonium ion Zn2+ Chlorate(I) ion ClO–
Fe2+ Chlorate(V) ion ClO3–
Calcium ion Pb2+ Manganate(VII) ion MnO4–
Sn2+ Oxide ion O2-
Magnesium ion Cu2+ Sulphate ion SO42-
Al3+ Sulphide ion S2-
Zinc ion Fe3+ Thiosulphate ion S2O32-
Cr3+ Chromate(VI) ion CrO42-
+2 Iron(II) ion Pb4+ Dichromate(VI) ion Cr2O72-
Sn4+ Phosphate ion PO43-
Lead(II) ion

Tin(II) ion

Copper(II) ion

Aluminium ion

+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.

49

  Chemistry SPM  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
anions by balancing the positive Total positive charges Total negative charges
charges and negative charges. = 1(+3) = 3(–1)
= +3 = –3
③ Write the number of cations and anions
as subscript numbers.
Form 4 Penerbitan Pelangi Sdn Bhd. All Rights Reserved Formula : FeCI3

Figure 3.12 Steps in constructing the formula of an ionic compound

EXAMPLE 3.23 SPM Tips

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

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

Formula : Na2SO4 The brackets The brackets The brackets
show there are show there are show there are

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

50

Chemistry SPM  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
than one type of ion, Roman
numerals such as I and II are used to
differentiate the ions.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved Lead(II) oxide 'II’ shows Pb2+ ion
Form 4Lead(IV) oxide'IV’ shows Pb4+ ion

Figure 3.14 Naming of ionic compounds

① The name of less Hydrogen sulphide ② The name of the more
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
SO3 Carbon monoxide Meaning of Greek prefixes:
CCl4
PCl5 Sulphur trioxide Mono = 1 ; Di = 2; Tri = 3 ;
Cl2O7
Carbon tetrachloride Tetra = 4 ; Penta = 5 ; Hexa = 6;

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
Q3 Sodium thiosulphate has the formula of Na2S2O3.
(a) What is the mass of sulphur in 0.2 mole of (c) NH4Cl (f) ZnS
sodium thiosulphate?
(b) How many moles of oxygen atoms are present Q6 A mixture contains carbon powder, magnesium
in 79 g of sodium thiosulphate? hydroxide and iron(II) bromide. Write the chemical

formulae of the substances in the mixture.

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

51

  Chemistry SPM  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 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.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved2H2(g) + O2(g) → 2H2O(l)

Form 4 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.

Example: Hydrogen + Oxygen → Water Chemical equation
before being
Reactants Products balanced Balancing of
chemical
Chemical equation equation
after being
balanced VIDEO

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.

52

Chemistry SPM  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
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Form 4Aim : To construct balanced chemical equations.

Materials : Copper(II) carbonate powder, limewater, concentrated hydrochloric acid, concentrated
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:

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

53

  Chemistry SPM  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.
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Discussion:

Form 4 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:

CuCO3(s)     CuO(s)  +  CO2(g)
Copper(II) oxide Carbon dioxide
Copper(II) carbonate

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

HCl(g)  +  NH3(g)     NH4Cl(s)
Hydrogen chloride gas Ammonia gas Ammonium chloride

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:

2KI(aq)  +  Pb(NO3)2(aq)     PbI2(s) +   2KNO3(aq)
Potassium iodide Lead(II) nitrate
Lead(II) iodide Potassium nitrate

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
2 formula units
or 2 atoms 1 molecule

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.

54

Chemistry SPM  Chapter 3 The Mole Concept, Chemical Formula and Equation  

SPM Highlights EXAMPLE 3.24

The following is a chemical equation. 2Na(s)  +  Cl2(g)  →  2NaCl(s)
What is the mass of sodium needed to react
K2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2KCl(aq) with 0.5 mole of chlorine?
[Relative atomic mass: Na, 23]
Which of the following statements is true?
A The reactants are potassium carbonate solid and Solution
2Na(s)  +  Cl2(g)  →  2NaCl(s)
calcium chloride solution. (2 mol) (1 mol)
B Two moles of potassium carbonate react with one ? g 0.5 mol ①

mole of calcium chloride Based on the equation, 1 mole of Cl2 reacts
C The products are calcium carbonate precipitate with 2 moles of Na. ②
So, 0.5 mole of Cl2 will react with (0.5 × 2)
and potassium chloride solution mole of Na or 1 mole of Na. ③
D Two moles of calcium carbonate and one mole of The mass of Na needed = number of moles
× molar mass ④
potassium chloride are produced
= 1 × 23
Examiner’s tip = 23 g
Students need to examine each option carefully. OptionPenerbitan Pelangi Sdn Bhd. All Rights Reserved
A is false because potassium carbonate used is in Form 4EXAMPLE 3.25
aqueous state and not in a solid state. Option B and D
are false because the coefficients in the equation show 2C3H6(g)  +  9O2(g)  →  6CO2(g)  + 6H2O(l)
that 1 mole of potassium carbonate reacts with 1 mole of What is the volume of carbon dioxide gas
calcium chloride to produce 1 mole of calcium carbonate produced at STP if 63 g of propene, C3H6 is
and 2 moles of potassium chloride. burnt completely?
Answer: C [Relative atomic mass: H, 1; C, 12. Molar volume
: 22.4 dm3 mol-1 at STP]
Solving numerical stoichiometry
problems Solution
1. Stoichiometry is a study of quantitative
2C3H6(g)  +  9O2(g)  →  6CO2(g)  +  6H2O(l)
composition of substances involved in chemical
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
of moles. =6432
= 1.5 mol ①
② Compare the ratio of moles of the
related substances. Based on the equation, 2 moles of C3H6
produce 6 moles of CO2.②
③ Calculate the answer in proportion of
moles. This means that 1 mole of C3H6 produces 3

④ Convert the answer from moles to the moles of CO2.
required units. So, 1.5 moles of C3H6 produces (1.5 × 3) moles
which are 4.5 moles of CO2. ③

The volume of CO2 produced at STP
= Number of moles × molar volume ④
= 4.5 × 22.4 dm3

= 100.8 dm3

55

  Chemistry SPM  Chapter 3 The Mole Concept, Chemical Formula and Equation

EXAMPLE 3.26 Checkpoint 3.4

2Al(s)  +  3CuO(s)  →  Al2O3(s)  + 3Cu(s) Q1 Write balanced chemical equations for the
following reactions.
If 2 g of excess aluminium is reacted with 0.06 (a) Sulphur trioxide + water → sulphuric acid
mole of copper(II) oxide, calculate the mass of (b) Magnesium + carbon dioxide gas →
aluminium left behind. magnesium oxide powder + carbon powder
[Relative atomic mass: Al, 27] (c) Hydrogen gas reacts with nitrogen gas to
produce ammonia gas.
Solution (d) Silver nitrate solution reacts with metallic
2Al(s)  +  copper to produce silver precipitate and
(2 mol) copper(II) nitrate solution.

? g of 2 g
Form 4 Penerbitan Pelangi Sdn Bhd. All Rights Reserved3CuO(s)  →  Al2O3(s)  + 3Cu(s)Q2 The following equation shows the decomposition
of potassium chlorate(V) by heat.
(3 mol)
0.06 mol ① KClO3(s) → KCl(s) + O2(g)

Based on the equation, 3 moles of CuO react (a) Balance the equation.
with 2 moles of Al. ② (b) Calculate the volume of oxygen produced at

So, 0.06 mole of CuO reacts with room conditions from the decomposition of
(0.06 × 2) mol of Al or 0.004 mole of Al. ③ 24.5 g of potassium chlorate(V).
[Relative atomic mass: O, 16; Cl, 35.5;
3 K, 39. Molar volume : 24 dm3 mol–1 at room
conditions.]
The mass of Al that reacted
= number of moles × molar mass ④ Q3 Lead is extracted according to the following
= 0.04 × 27 equation.

= 1.08 g 2PbO(s) + C(s) → 2Pb(s) + CO2(g)

So, the mass of Al left behind = 2 - 1.08 g (a) How many moles of lead is extracted from
0.5 mole of lead(II) oxide?
= 0.92 g
(b) How many lead atoms are produced if 44.6 g
SPM Highlights of lead(II) oxide is heated with excess carbon
powder?
Equation for the reaction between potassium and
[Relative atomic mass: O, 16; Pb, 207.
oxygen: Avogadro’s constant: 6.02 × 1023 mol–1.]

4K(s) + O2(g) → 2K2O(s) Q4 Sodium reacts with water as follows.

What is the maximum mass of potassium oxide that is 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

formed when 19.5 g potassium is burned completely (a) What are the products of the reaction?
(b) How many moles of sodium will react with 5
in excess oxygen? [Relative atomic mass: O, 16;
moles of water?
K, 39] (c) Calculate the mass of sodium needed to be

A 23.5 g C  70.5 g used to produce 3.01 × 1023 H2 molecules?
[Relative atomic mass: Na, 23. Avogadro’s
B 47.0 g D  94.0 g
constant: 6.02 × 1023 mol–1.]
Examiner’s tip

The amount of product depends only on the amount

of potassium as the oxygen is in excess.

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

CheCmheismtriystFryorSmPM4  Chapter 3 The Mole Concept, Chemical Formula and Equation  

CONCEPT MAP

The mole concept, chemical
formula and equation

Penerbitan Pelangi Sdn Bhd. All Rights ReservedNumber ofStoichiometry Chemical formula
Form 4particlesNumber of moles

× NA
÷ NA

Empirical formula Molecular formula

× 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 1 shows a cooking gas
Mg, 24] SPM
Choose the correct answer. 2017 cylinder.
A 3
3.1 Relative Atomic Mass and B 4
Relative Molecular Mass C 5
D 6
1 mole of propane
1. Six atoms of element Y have gas, C3H8
SPM
2015 the same mass with three 3.2 Mole Concept Figure 1
tellurium atoms, Te. What is
How many hydrogen atoms
the relative atomic mass of 3. How many nitrate ions, are there in the gas cylinder?
Y? SPM
2016 NO3– are there in 2 mol of [Avogadro’s constant: 6.02 ×
[Relative atomic mass: Te, 128] aluminium nitrate, Al(NO3)3? 1023 mol–1]
A 1 × 6.02 × 1023
A 8 C 32 [Avogadro’s constant: 6.02 × B 3 × 6.02 × 1023
C 8 × 6.02 × 1023
B 16 D 64 1023 mol–1] D 11 × 6.02 × 1023

2. How many oxygen atoms A 1.204 × 1024 mol–1
have the equal mass with two B 1.806 × 1024 mol–1
C 3.010 × 1024 mol–1
magnesium atoms? D 3.612 × 1024 mol–1

57

  Chemistry SPM  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
substances has the highest SPM the ammonia gas formed
2018 composition by mass of is twice the number of
number of atoms? HOTS carbon per molecule of nitrogen atoms used

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

A 76.23% C 83.72%

B 81.46% D 84.13%

6. 200 cm3 of a gas has a mass 9. 0.60 g of metal Z reacts with
of 0.15 g at room conditions. fluorine to produce 1.17 g Z
What is the molar mass of
the gas?

[Molar volume: 24 dm3 mol–1
at room conditions] HOTS

A 18 g mol–1
B 30 g mol–1
C 50 g mol–1
D 150 g mol–1
Penerbitan Pelangi Sdn Bhd. All Rights Reserved 12. The following equation
SPM
fluoride. What is the empirical 2018 represents the reaction
formula of Z fluoride? between magnesium and
Form 4
[Relative atomic mass: F,19; hydrochloric acid.

Z, 40] C Z2F Mg(s) + 2HCl(aq) →
A ZF     MgCl2(aq) + H2(g)

B ZF2 D ZF4 What is the volume of

3.3 Chemical Formula 10. What is the mass of calcium hydrogen produced if 3.6 g
contained in 1 kg of calcium
7. The distinct smell of garlic is carbonate? of magnesium reacts with

[Relative atomic mass: C,12; excess hydrochloric acid at
O, 16; Ca, 40] HOTS
room conditions?
SPM due to the presence of allicin A 100 g C 300 g
2018 compound. Figure 2 shows B 200 g D 400 g [Relative atomic mass: Mg,
24. Molar volume : 24 dm3
the percentage composition mol-1 at room conditions]
A 0.12 dm3
by mass of allicin. 3.4 Chemical Equation B 1.2 dm3
C 3.6 dm3
D 4.8 dm3

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:
6CO2 + 12H2O sunlight
3H2(g) + N2(g) → 2NH3(g)    C6H12O6 + 6O2 + 6H2O

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

Carbon-12
Y

CC CC

Figure 1

58

Chemistry SPM  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 1, 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 1(d)(i) quantitatively. [1 mark]

(iii) Using the relative atomic mass of element Y from your answer in 1(c), calculate the number of moles
Penerbitan Pelangi Sdn Bhd. All Rights Reservedof bromine gas required to react completely with 3.6 g of Y. [2 marks]
Form 4
(IV) In another investigation, 9.6 g of Y reacts with 6.4 g of oxygen. Complete the table below to determine
the empirical formula of the oxide of Y.

[Relative atomic mass: O, 16]

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

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 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]

59

5Chapter Form 5

Consumer and Industrial Chemistry

Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 5
CHAPTER FOCUS Soap is the salt of fatty acids that
• Oils and Fats can be produced from animal fat and
• Cleaning Agents vegetable oils. Soaps are prepared by
• Food Additives hydrolysis of fats or oils in an alkaline
• Medicine and Cosmetics condition. Can we produce soap from
• Application of Nanotechnology in Industry palm oil?
• Application of Green Technology in
Industrial Waste Management 419

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

5.1 Oils and Fats 7. Fatty acids are long-chain carboxylic acids
containing 12 to 18 carbon atoms per molecule.
The Difference between Oils and Fats An example is myristic acid shown below.

1. Fats and oils belong to a group of naturally HHHHHHHHHHHHH O
occurring compounds called lipids.
HC C C C C C C C C C C C CC
2. Lipids are a group of organic compounds that
make up the structure of cells found in plants H H H H H H H H H H H H H OH
and animal tissues. Myristic acid

Penerbitan Pelangi Sdn Bhd. All Rights Reserved The lipids are water-insoluble compounds 8. A molecule of glycerol may combine with one,
that can be extracted from cell components two or three fatty acids to form a monoester,
by using organic solvents such as ether and diester or triester. A molecule of water is
benzene. eliminated when one fatty acid joins to the
glycerol molecule and the resulting bond
3. Fats and oils are chemically very similar, but formed is called an ester link (-COO-).
differ in their physical states.
9. Most fats and oils are triesters (triglyceride)
4. Fats found in animals like goats and cows are and formed through esterification reactions
solids at room temperature. as shown below.

Examples of animal fats are butter and tallow. O O

5. Fats from plants are liquids. They are called CH2 OH H O C R1 CH2 O C R1
oils. Examples include palm oil, coconut oil, O O
olive oil, soybean oil and corn oil.
CH OH + H O C R2 CH O C R2 + 3H2O
6. Fats and oils are mixtures of different esters O O
derived from a variety of longchain carboxylic
acids called fatty acids with the alcohol CH2 OH H O C R3 CH2 O C R3
propane-1,2,3-triol or commonly known as
glycerol. 1 mole of 3 moles of 1 mole 3 moles
glycerol fatty acid
H of fat of fats

HCOH CH2 OH 10. The differences between oils and fats are
shown in Table 5.1.

H C O H or CH OH

HCOH CH2 OH

Form 5 H Glycerol



Table 5.1 Differences between oils and fats

Aspect Oils Fats

Source Plants Animals

Saturation Unsaturated because it contains at Saturated because it contains a single

Physical state at room temperature least one double bond (C=C) bond (C–C) only
Melting point
Examples Liquid Solid

Less than 20 oC More than 20 oC

Palm oil, corn oil, olive oil, sesame oil Meat, egg, butter, margarine and ghee
and sunflower oil

420

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

Saturated and Unsaturated Fats In an unsaturated fatty acid, the carbon
chain has one or more carbon-carbon
1. The physical and chemical properties of a fat double bonds.
or oil molecule are greatly affected by the
parent fatty acid. The fatty acids differ from 2. If the fatty acid molecule has one carbon-
two aspects: carbon double bond, the fatty acid is known as
(a) Firstly, the length of the carbon chain monounsaturated. Oleic acid is an example
can vary from 12 to 18 carbon atoms. For of monounsaturated fatty acid that consists of
example: lauric acid (12 carbon atoms) one carbon-carbon double bond.
and stearic acid (18 carbon atoms) have
different numbers of carbon atoms. 3. If there is more than one carbon-carbon
(b) Secondly, the fatty acid may be saturated double bond, the fatty acid is known as
or unsaturated. A saturated fatty acid polyunsaturated. Linoleic acid has two
has all carbon atoms joined together by carbon-carbon double bonds whereas linolenic
carbon-carbon single bonds. Palmitic acid has three carbon-carbon double bonds.
acid is an example of saturated fatty acid.
4. The formulae of some fatty acids are given in
Table 5. 2.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 5Fatty acidTable 5.2 Formulae of some fatty acidsNumberType of
Formula of carbon fatty
acid
atoms

Lauric acid CH3 – (CH2)10 – COOH 12 Saturated

Palmitic acid CH3 – (CH2)14 – COOH 16 Saturated

Stearic acid CH3 – (CH2)16 – COOH 18 Saturated

Oleic acid CH3 – (CH2)7 – CH = CH – (CH)7 – COOH 18 Unsaturated

Linoleic acid CH3 – (CH2)4 – CH = CH – CH2 – CH = CH – (CH)7 – COOH 18 Unsaturated

Linolenic acid CH3 – CH2 – CH = CH – CH2 – CH = CH – CH2 – CH = CH – (CH)7 – COOH 18 Unsaturated

5. Fats which contain esters of glycerol and 7. Plant or vegetable fats contain a large
saturated fatty acids are classified as proportion of unsaturated fats. Unsaturated
saturated fats. Animal fats containing a large fats have lower melting points compared to
proportion of saturated esters tend to have saturated fats and exist as liquids at room
high melting points and exist as solids at room temperature. An example is shown below.
temperature. Two saturated fats are given
below. O

OO CH2 O C (CH2)7 CH CH (CH2)7 CH3
O
CH2 O C (CH2)16 CH3 CH2 O C (CH2)14 CH3
O O CH O C (CH2)7 CH CH (CH2)7 CH3
O
CH O C (CH2)16 CH3 CH O C (CH2)14 CH3
O O CH2 O C (CH2)7 CH CH (CH2)7 CH3
Triolein

CH2 O C (CH2)16 CH3 CH2 O C (CH2)14 CH3
Tristearin Tripalmitin

6. Fats which contain esters of glycerol and
unsaturated fatty acids are classified as
unsaturated fats.

421

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

Conversion of Unsaturated Fats to 2. Medical researchers have found that animal
Saturated Fats
fats pose a greater risk for heart diseases and

1. Unsaturated fats can be converted into saturated stroke compared to vegetable oils.

fats through hydrogenation reaction. (a) Consuming food high in saturated fats

H H will raise the level of cholesterol in
the blood. Cholesterol causes the fatty
+    H – H    ⎯Ni→    deposits or plaque on the walls of veins
– C = C –    –C–C–

D oub le Hydrogen and arteries. As the plaque builds up on
the walls of the blood vessels, blood
bond Single
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
bond circulation is restricted or blocked, and
this will raise blood pressure.
2. The hydrogenation reaction is carried out by

bubbling hydrogen gas through hot liquid oil (b) If the plaque deposits occur on the artery
in the presence of nickel powder as a catalyst. walls of the heart, hardening of artery
A temperature of about 200°C and a pressure takes place. This condition is known as
of 4 atm are used. arteriosclerosis and can block the supply

3. The relative molecular mass of the oil molecule of blood to the heart. Heart attack occurs.

increases, as more and more of the double If the plaque deposit takes place in the

bonds are hydrogenated. Intermolecular forces brain, the patient will experience stroke.

become stronger, and more energy is needed to Serious stroke and heart attacks are fatal.

overcome them. The boiling points of the oil (c) There are two types of cholesterol which
increases and the physical state changes from are low-density cholesterol (LDL) and
liquid to solid. high density cholesterol (HDL). High-

4. Margarine is a soft solid with low melting density cholesterol is known as ‘good’
point at room temperature. Margarine is made cholesterol because it can reduce the
by hydrogenating some of the carbon-carbon risk of heart diseases. High levels of
double bonds in polyunsaturated vegetable oil low-density cholesterol in the blood will
such as palm oil and sunflower oil. increase the risk of heart diseases.

Uses of Oils and Fats in Life 3. Vegetable oils do not contain cholesterol
because only animals make cholesterol. These
1. Table 5.3 shows the uses of oils and fats in our unsaturated fats do not have the damaging
life. effects to cause cardiovascular problems.
However, we are not encouraged to take
Table 5.3 Uses of oils and fats in life excessive unsaturated fats. Excessive intake
of fats, either saturated or unsaturated, will
Food • Oils and fats are used in the production cause obesity. We need to control our food
product of human and animal foods intake to ensure that the essential processes in
our body can function well to build a healthy
Form 5 Source of • Fats and oils are the most effective body.
energy and energy suppliers
nutrition • As a solvent of vitamins A, D, E and K

Biofuel • Used cooking oil can be used to Checkpoint 5.1
produce biodiesel fuel through the
Personal process of transesterification. Q1 Fats and oils are classified as saturated fats and
care unsaturated fats.
products • Biodiesel is increasingly being used (a) Explain the meaning of ‘saturated fat’ and
as a renewable energy source, for ‘unsaturated fat’.
example for heavy vehicle use (b) How can an unsaturated fat be converted
into a saturated fat?
• Soap
• Skin moisturisers such as body lotion

and face cream

422

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

Q2 Our bodies need fats and oils. Discuss the roles 4. The general formulae for two common
fats and oils play in helping us to maintain a detergents are:
healthy body.
O O
Q3 Give one effect of eating a diet high in saturated ' '
fats. R!O!S!O–Na+ R S!O–Na+
' '
Q4 Give one health benefit of consuming palm oil. O O

Sodium alkyl sulphate Sodium alkylbenzene
sulphonate

Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 55.2 Cleaning Agents where R represents a long-chain hydrocarbon.

Soap Preparation of Soap

1. Soap is sodium or potassium salts of a long- 1. Soap can be made from animal fats and
chain fatty acids. vegetable oils.

2. The general formula of a soap can be written 2. The animal fats most commonly used are fats
as RCOO–Na+ or RCOO–K+, where R is an from cows and goats.
alkyl group usually containing 12 to 18 carbon
atoms. R can be saturated or unsaturated. 3. The vegetable oils often used are palm oil,
olive oil and coconut oil.
Table 5.4 The formulae of some examples of soap
4. Soaps are prepared by hydrolysing fats or oils
Soap Formula under alkaline condition. The reaction is called
saponification.
Sodium laurate CH3(CH2)10COO–Na+
5. The saponification process involves boiling fats
Sodium palmitate CH3(CH2)14COO–Na+ or oils with concentrated sodium hydroxide
solution or concentrated potassium
Potassium stearate CH3(CH2)16COO–K+ hydroxide solution to produce glycerol and
the salts of fatty acids which are the soaps.
Potassium oleate CH3(CH2)7CH=CH(CH2)7COO–K+
The general equation for this reaction is:
3. The general formula for sodium and potassium
soaps are: O O
' '
O O CH2! O!C!R CH2OH R ! C!O–Na+
C O–Na+ C O–K+
CH3(CH2)n CH3(CH2)n O O
' '
Alkyl group Alkyl group CH ! O!C!R' + 3NaOH CHOH + R'! C!O–Na+
(a)  Sodium soap (b)  Potassium soap
O O
Detergent ' Alkali CH2OH '
CH2! O!C!R'' Glycerol R''! C!O–Na+
1. Any cleaning agent that is not soap is a Soap
detergent. Detergents are usually made from Oils or fats
synthetic resources such as petroleum fractions.
where the three alkyl groups (R, Rʹ dan Rʺ)
2. Detergent was developed during the Second can be the same or different groups.
World War in response to a shortage of animal
fats and vegetable oils. 6. The fats or oils are hydrolysed first to form
glycerol and fatty acids. The fatty acids then
3. Detergents are usually sodium salts of react with an alkali to form the corresponding
sulphonic acid. sodium or potassium salts.

423

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

7. The following equation shows how a soap, 9. Potassium soaps are softer and milder than
sodium palmitate, is prepared. sodium soaps and are usually used for bathing.

O SPM Highlights
'
CH2! O!C!(CH2)14 CH3 Which statements are correct about soap and
detergent?
O
' 3NaOH Soap Detergent
CH !O!C!(CH2)14 CH3 + Alkali
A Is a salt Is an acid
O
'
CH2! O!C!(CH2)14 CH3

Oil
Penerbitan Pelangi Sdn Bhd. All Rights Reserved B Less effective in hard Effective in hard water
water

C Made from petroleum Made from vegetable oil

CH2OH D Forms scum in soft Does not form scum in
water soft water

O
'
! C!O–Na+ Examiner’s Tip
CHOH + 3CH3 (CH2) Both soap and detergent are salts.
14 Soap is made from vegetable oil whereas detergent
is made from petroleum.
Sodium palmitate Soap and detergent do not form scum in soft water.

(soap) Answer: B

CH2OH

Glycerol

8. When concentrated potassium hydroxide
solution is used instead of concentrated
sodium hydroxide solution, a potassium soap,
potassium palmitate is formed.

Form 5 ACTIVITY 5.1

Aim: To prepare soap using saponification process.
Materials: Palm oil, 5 mol dm–3 sodium hydroxide solution, sodium chloride powder, filter paper,

distilled water
Apparatus: 250 cm3 beaker, 50 cm3 measuring cylinder, spatula, glass rod, filter funnel, wire gauze,

tripod stand, Bunsen burner, test tubes

Safety Precautions

• Wear safety goggles
• Concentrated sodium hydroxide solution is caustic. Avoid skin contact.

Procedure:
1. 10 cm3 of palm oil is poured into a beaker.
2. 50 cm3 of 5 mol dm–3 sodium hydroxide solution is added to the palm oil.
3. The mixture is heated until it boils.
4. The mixture is stirred with a glass rod.
5. The mixture is allowed to boil for 10 minutes.
6. The beaker is removed from the heat. 50 cm3 of distilled water and three spatulas of sodium

chloride are added to the mixture.

424

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

7. The mixture is boiled for another 5 minutes.
8. The mixture is allowed to cool.
9. The soap formed is filtered out. The soap is rinsed with a little distilled water.
1 0. The soap is pressed between a few pieces of filter paper to dry it.
11. The soap is felt with the fingers. A small amount of the soap is placed in a test tube. Water is added

into the test tube. The mixture is then shaken.
12. The observations are recorded.
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Form 5Observations:
The soap feels slippery. When the mixture of the soap and water is shaken, lather is formed.

Discussion:

1. When palm oil is heated with sodium hydroxide solution, a soap is formed. The word equation for
the saponification process involved in this activity is:

Palm oil + concentrated sodium hydroxide solution ⎯→ soap + glycerol

2. The soap formed can be precipitated by adding sodium chloride. This is because sodium chloride
lowers the solubility of soap in water.

3. The glycerol and excess sodium hydroxide solution are removed by rinsing the soap formed with
water.

4. Soaps have the following properties:
(a) Soaps feel slippery
(b) Soaps form lather when they are shaken with water.

Conclusion:
Soaps can be prepared by heating palm oil with concentrated sodium hydroxide solution.

Preparation of Detergents

1. During the preparation of detergents, a long-chain hydrocarbon obtained from petroleum fractions is
converted into an organic acid through a series of steps.

2. The organic acid is then neutralised with sodium hydroxide solution to produce a neutral salt, which
is a detergent.

Preparation of sodium alkyl sulphate

Step 1: Formation of an organic acid
A long-chain alcohol reacts with concentrated sulphuric acid to form alkyl sulphonic acid.

O O
' '
+ Sulphonation CH3 (CH2)n CH2!O!'S!O!H + H2O
CH3 (CH2)n CH2 OH H!O!S!O!H O
'

O

Long-chain alcohol Concentrated alkyl sulphonic acid Water
sulphuric acid

425

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

Step 2: Neutralisation
The resulting acid produced is then converted to a sodium salt by a reaction with sodium hydroxide.

O O
' '
CH3 (CH2)nCH2!O!'S!O!H + NaOH Neutralisation O! S!O–Na+ +
O CH3 (CH2)n CH2! ' H2O
Water
O

Alkyl sulphonic acid Sodium Sodium alkyl sulphate
hydroxide solution (detergent)
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Preparation of sodium alkylbenzene sulphonate

Step 1: Formation of an organic acid
A long-chain alkene reacts with benzene to form alkylbenzene.

CH3 (CH2)n CH " CH2 + Alkylation &CH3
dehydrated AICI3
CH3 (CH2)n C
&

H

Long-chain alkene Benzene Alkylbenzene



The alkylbenzene formed is then reacted with concentrated sulphuric acid to form alkylbenzene
sulphonic acid.

&CH3 O &CH3 O
' C '
+ H!O!S!O!H sulphonation & +
CH3 (CH2)n C ' CH3 (CH2)n H S!O!H H2O
& O '

H O

Alkylbenzene Concentrated sulphuric acid Alkylbenzene sulphonic acid Water



Form 5 Step 2: Neutralisation

The alkylbenzene sulphonic acid is then converted to a sodium salt by a reaction with sodium
hydroxide.

&CH3 O Nautralisation &CH3 O
' '
CH3 (CH2)n C S!O!H + NaOH CH3 (CH2)n C S!O –Na+ + H2O
& ' & '
O O
H H

Alkylbenzene Sodium Sodium alkylbenzene sulphonate Water
sulphonic acid hydroxide solution



426

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

SPM Highlights 2. This ability is due to the structure of soaps and
detergents.
A substance has the structural formula shown
below. 3. In water, a sodium soap dissolves to form
soap anions and sodium cations. For example,
O the following chemical equation shows the
' ionisation of sodium palmitate.
O!S!O–
' CH3(CH2)14COO–Na+  →  CH3(CH2)14COO– + Na+
O
Sodium palmitate Palmitate ion Sodium
What is the substance?
A Sulphuric acid
B Soap
C Detergent
D Fatty acid

Examiner’s Tips
The substance structural formula is the formula
of a sodium alkyl sulphate detergent.

Answer: C
Penerbitan Pelangi Sdn Bhd. All Rights Reserved (soap) (soap anion) ion
Form 5
4. A soap anion consists of a long hydrocarbon
chain with a carboxylate group on one end.

5. The hydrocarbon chain, which is hydrophobic,
is soluble in oils or grease.

6. The ionic part is the carboxylate group, which
is hydrophilic, is soluble in water.

Cleansing Action of Soap and Detergent Hydrophobic part
CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2
1. The cleansing action for both soaps and
detergents results from their ability to lower CH2 CH2 CH2 CH2 CH2 CH2 CH2 COO–
the surface tension of water, to emulsify oil
or grease and to hold them in a suspension Hydrophillic part
in water.

7. In water, detergent dissolves to form detergent anions and sodium cations. For example, the following
chemical equations show the ionisation of sodium alkyl sulphate and sodium alkylbenzene sulphonate.

O H2O O
' '
O!S!O– + Na+
CH3 (CH2)nCH2 O!S!O –Na+ CH3 (CH2)nCH2 ' Sodium
' O ion
Sodium alkyl sulphate Alkyl sulphate ion
O (anion detergent)
(detergent)

O H2O O + Na+
' Hydroph'illic part
S!O –Na+ Sodium
CH3 (CH2)nCH2 ' AlkylbC(eaCHnnzH3ioe(3Cnn(eCHdesH2ut)e2nlp)rCgnhCHeonHn2ta)2t!e iOon!''O'SOS!!OO–– ion
O
Sodium alkylbenzene sulphonate
(detergent)

Hydrophobic part O
8. Like soap, the anion part of a detergent also consists of a hydrophobic part and a hydrophilic part.

Hydrophillic part Hydrophillic part

O O
' '
!S!O–
CH3 (CH2)n CH2!O ' CH3 (CH2)nCH2! !S!O–
Hydrophobic part '
Hydrophobic part O O


Hydrophillic part

CH3 (CH2)nCH2! O 427
Hydrophobic part '

!S!O–
'
O

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

9. The following explains the cleansing action of a soap or detergent on a piece of cloth with a greasy
stain.

Water Sodium ion (a) When soap or detergent is added to water,
Cloth soap or detergent reduces the surface tension
Soap or detergent of water. This phenomenon increases the
anion ability of water to wet the surface. So, the
surface of the cloth is wet thoroughly.
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Hydrophillic part (b) Hydrophobic part for soap or detergent
Hydrophobic part anion dissolves in grease. Hydrophilic part
Grease soluble in water.

(c) Scrubbing or mechanical agitation helps to pull the grease
away from the cloth and the grease is break into smaller
droplets.

Repulsion (d) These droplets do not clump together nor stick again onto the
surface of the cloth because of the repulsion between negative
charges on the surface of these droplets. The suspension of
droplets in water forms emulsion. Rinsing will remove these
droplets and leave a clean surface.

Form 5 SPM Highlights

Which statement is correct about cleansing action of soap?
A Soap molecules emulsify oil.
B Soap reacts with acid to form salt.
C Soap increases the surface tension of water.
D The hydrophobic part of soap molecules dissolves in water.

Examiner’s Tips
Soap reacts with acid to form long-chain fatty acids.
Soap decreases the surface tension of water.
The hydrophobic part of soap molecules dissolves in oil.
Answer: A

428

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

Comparison of Cleansing Action of Soap mass. This reduces the amount of soap
and Detergent available for cleaning. The effectiveness
of soap is thus reduced.
1. Although soap is a good cleaning agent, its
effectiveness is reduced when used in hard CH3(CH2)16COO–Na+(aq) + H+(aq)
water. Sodium stearate Hydrogen ion

2. Hard water contains a great amount of CH3(CH2)16COOH(s) + Na+(aq)
calcium and magnesium ions. These ions Insoluble stearic acid Sodium ion
react with the soap to form an insoluble
precipitate known as soap scum.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved 6. Detergents do not form scum with hard water.
2CH3(CH2)16COO–Na+(aq) + Ca2+(aq) Form 5(a) Detergent form soluble substances with
calcium salt or magnesium salt.

Sodium stearate Calcium ion

[CH3(CH2)16COO]2Ca(s) + 2Na+(aq) 2CH3(CH2)11OSO3–Na+(aq) + Ca2+(aq)

Insoluble calcium stearate Sodium ion Sodium dodecanyl sulphate Calcium ion
(scum)
[CH3(CH2)11OSO3–]2Ca2+(aq) + 2Na+(aq)
Soluble calcium Sodium ion
2CH3(CH2)16COO–Na+(aq) + Mg2+(aq)
Sodium stearate Magnesium ion dodecanyl sulphate

[CH3(CH2)16COO]2Mg(s) + 2Na+(aq) 2CH3(CH2)11OSO3–Na+(aq) + Mg2+(aq)

Insoluble magnesium stearate Sodium ion Sodium dodecanyl sulphate Magnesium ion
(scum)
[CH3(CH2)11OSO3–]2 Mg2+(aq) + 2Na+(aq)
3. The formation of soap scum reduces the
amount of soap available for cleaning, thus Soluble magnesium Sodium
causes wastage of soap.
dodecanyl sulphate ion
4. Soap is only suitable for use in soft water.
Soft water is water that contains little or no (b) This means a detergent can still perform
calcium and magnesium ions. Soap does not its cleansing action in hard water. Thus,
form scum with soft water. a detergent is more effective than a soap
in hard water.
5. The effectiveness of the cleansing agent of
soap is also reduced when used in acidic 7. Detergents do not form precipitates in acidic
water. water. Thus, their cleansing action is not
(a) The hydrogen ions in acidic water react affected.
with the soap to form long-chain fatty
acids. CH3(CH2)11OSO3–Na+(aq) + H+(aq)
(b) Long-chain fatty acids are insoluble in
water due to their high relative molecular Sodium dodecanyl sulphate Hydrogen ion

CH3(CH2)11OSO3–H+(aq) + Na+(aq)
Soluble dodecanyl Sodium

sulphonic acid ion

EEkxsppeerrimimeennt 52.1

Aim: To compare the effectiveness of the cleansing action of soap and detergent in hard water.
Problem statement: Is the cleansing action of a detergent more effective than a soap in hard water?
Hypothesis: The cleansing action of a detergent is more effective than a soap in hard water.

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  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

Variables:
(a) Manipulated variable: Soap and detergent solutions
(b) Responding variable: The oily stains on a cloth
(c) Fixed variable: Volume and concentration of magnesium sulphate solution, volume and

concentration of cleaning agents.

Operational definition: The ability of a cleaning agent to remove oily stains on a cloth indicates that
the cleaning agent is effective.

Materials: 5% soap solution, 5% detergent solution, 1.0 mol dm–3 magnesium sulphate solution, cloth
with oily stains.

Apparatus: 100 cm3 beaker, 50 cm3 measuring cylinder.

Procedure:
1. The soap and detergent solutions are prepared separately as shown in Figure 5.1.
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50 cm3 of 5% soap solution Beaker X Beaker Y 50 cm3 of 5% detergent solution
+ 20 cm3 hard water + 20 cm3 hard water

(magnesium sulphate solution) (magnesium sulphate solution)

Figure 5.1

2. A small piece of cloth with oily stains is dipped into each beaker.
3. Each cloth is washed with the soap or detergent in each beaker.
4. The cleansing action of the soap and detergent is observed, compared and recorded.

Observations:

Beaker Observation
X Oily stains remain
Y Oily stains disappear

Discussion:

1. The presence of magnesium ions in water increases the hardness of the water.

2. The magnesium ions react with the soap to form soap scum, an insoluble precipitate.

2RCOO–Na+(aq)  +  Mg2+(aq)  ⎯→ [RCOO]2Mg(s)  +  2Na+(aq)
Soap Magnesium ion Insoluble precipitate Sodium ion

Form 5 3. The formation of soap scum reduces the amount of soap available for cleaning. The effectiveness
of the cleansing action of soap is thus reduced.

4. Detergents do not form scum with hard water. They form soluble substances with magnesium ions.

2ROSO3–Na+(aq) + Mg2+(aq)  ⎯→ [ROSO3–]2Mg2+(aq)  +  2Na+(aq)
Detergent Magnesium ion Soluble substance Sodium ion

5. Thus, a detergent is more effective than a soap in hard water.

Conclusion:
The cleansing action of a detergent is more effective than a soap in hard water.
The hypothesis is accepted.

430

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

Additives in Detergent its cleaning efficiency and to meet the needs
of consumer.
1. Detergents generally contain a number of
additives. 3. The additives in a detergent and their respective
functions are shown in Table 5.5.
2. Additives are added to a detergent to enhance

Table 5.5 Additives in detergents and their functions

Additives Example Function

Biological enzymes Amylase, protease, cellulase To remove protein stains such as blood
and lipase
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Form 5Whitening agentSodium perborateTo convert stains into colourless substances

Optical whitener Fluorescent dyes To add brightness and whiteness to white fabrics

Builder Sodium tripolyphosphate To enhance the cleaning efficiency of a detergent by softening
the water.

Anti suspension agent Carboxylmethyl cellulose To prevent the dirt particles removed from redepositing onto
cleaned fabrics

Filler Sodium sulphate, sodium To increase the quantity of detergent and enable it to be poured
silicate easily

Foam control agent Silicones To control foaming in a detergent

Fragrance Rose, lavender To add fragrance to both the detergent and the fabrics

Drying agent Sodium sulphate and To ensure that the detergent powder is always
sodium silicate dry in its container

SPM Tips

The comparison between a soap and a detergent is given in Table 5.6.

Table 5.6

Cleaning agent Soap Detergent

Effectiveness Soap is a cleaning agent that is effective Detergent is a cleaning agent that is effective

in soft water in both hard and soft water

Formation of scum Soap forms scum in hard water Detergent does not form scum in hard water

Sources Soap is made from natural resources Detergent is made from synthetic resources

(animal fats or vegetable oils) like petroleum fractions

Formation of Soap forms precipitate in acidic water Detergent does not form precipitate in acidic
precipitate water

Effect on the Soap is biodegradable and does not cause Non-biodegradable detergent give thick foams
environment
any pollution that kills aquatic lives

Checkpoint 5.2

Q1 (a) What is soap? Q2 Consider the following cleaning agents.
(b) Figure 5.2 shows the structure of a soap P: RR--COOSOO–3K–K+ +
Cleaning agent Q:
particle. Cleaning agent

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COO–Na+ Which of the cleaning agent is more suitable as
a cleaning agent to remove oily stains in hard
X Y water? Explain your answer.

Figure 5.2 Q3 State two examples of additives in detergents and
their respective functions.
Which part of the structure is hydrophilic and
which part is hydrophobic? Give reasons for
your answer.

(c) Explain the cleansing action of soap on
greasy stains.

431

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

5.3 Food Additives (c) Some preservatives can cause side effects
on health.
1. A food additive is a natural or synthetic
substance which is added to food to prevent (i) Sodium nitrite is added to meat to
spoilage or to improve its appearance, taste preserve and to stabilise its red colour.
or texture. However, sodium nitrite can cause
stomach cancer. During cooking, the
Types of Food Additives and Their nitrites produce nitrosamine, which
Functions are carcinogenic.
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1. Food additives that are commonly used are: (ii) Eating too much food preserved with
salt such as salted fish increases the
Preservatives Emulsifiers risk of high blood pressure, heart
attack and stroke.
Antioxidants Thickeners
(iii) Eating too much food preserved with
Flavourings Dyes sugar can cause obesity, tooth decay
and diabetes.
Stabilisers
(iv) Sulphur dioxide may cause asthma
2. Preservatives and allergies in certain people who
are sensitive to this additive.
(a) Preservatives are substances added to food
to slow down or to prevent the growth 3. Antioxidants 
of microorganisms so that food can be
kept for longer periods of time. (a) Food containing fats and oils can turn
rancid, that is, fats and oils are oxidised
(b) Table 5.7 shows some examples of to become unpleasant-smelling acids, on
preservatives and how they work. exposure to the air.

Table 5.7 Some examples of preservatives (b) All food stuffs are easily oxidised. For
example, apples and potatoes turned
Preservative Example How it works brown when exposed to air.

Table salt Salted fish Table salt or sugar (c) Antioxidants are added to food to prevent
Sugar Jam draws the water oxidation that causes fats to be rancid and
out of the cells of fruits to turn brown.
microorganisms and
retards the growth (d) Table 5.8 shows several examples of
of microorganisms antioxidant.

Vinegar Pickled mango Vinegar provides an SPM Tips
acidic condition that
Form 5 inhibits the growth Definition of antioxidant ought to consist of two main
of microorganisms concepts:
 chemical substance that is added to oily food,
Sodium nitrite Burger, These preservatives  to avoid oils in food from being oxidised by oxygen
or sodium
nitrate sausage and slow down in air.

Benzoic acid canned meat the growth of
and sodium
benzoate Oyster sauce, microorganisms

tomato sauce,

chilli sauce and

fruit juice

Sulphur dioxide Fruit juice

432

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

Table 5.8 Several examples of antioxidants (f) Vitamins C and E are among the safest
antioxidants known.
Antioxidant Example Function
(i) Vitamin C inhibits the formation
Butyl hydroxyanisole Margarine To prevent oil from of nitrosamine that is carcinogenic,
(BHA) and becoming rancid stimulates the immune system and
Butyl hydroxytoluene protect the chromosome against
(BHT) damage.

Ascorbic acid Fruit juice To preserve the (ii) Vitamin E neutralises free radicals
(Vitamin C) colour of fruit compounds before they can destroy
juices the cell membrane and helps to reduce
the risk of heart diseases and cancer.

4. Flavourings
(a) Flavourings are used to improve the taste
of food and to restore taste loss due to
food processing.
(b) Examples of flavouring are sugar, table
salt, vinegar, monosodium glutamate
(MSG), aspartame and esters like pentyl
ethanoate.
Penerbitan Pelangi Sdn Bhd. All Rights ReservedAlpha tocopherolVegetable To prevent oil from
Form 5(Vitamin E)oils becoming rancid

Sodium citrate Preserved To prevent fats
cooked from becoming
meat rancid

(e) The use of BHA and BHT has been
controversial because they have produced
bad reactions in dogs. Thus, there are
restrictions on the quantity used on these
antioxidants.

Flavouring Table 5.9 Several examples of flavourings Function
Examples

Monosodium glutamate (MSG) Frozen meat, spice mixes, canned To bring out the flavour in many
soup, salad dressings, meat or fish- types of food
based products

Aspartame Diet drinks, frozen desserts with low To sweeten food

• It is a non-sugar sweetener calories and soft drinks.

• It is approximately 200 times sweeter than

sugar

• It has less calories than sugar

• It is stable when dry or frozen but it breaks

down and loses its sweetness over time

when kept stored in liquids at temperatures

above 30°C.

Ester Pentyl ethanoate (banana flavour), To produce artificial flavours which

• It contains compounds belonging to the ethyl butanoate (pineapple flavour), resemble natural flavours

homologous series of esters methyl butanoate (apple flavour)

• It is cheaper to use these artificial flavours and octyl ethanoate (orange flavour)

than to use real fruits.

5. Stabilisers (d) Examples of stabilisers are starch, agar,
(a) Stabilisers are added to maintain its pectin and gelatin.
consistency and texture of food so that it
is more solid. 6. Emulsifiers
(b) Stabilisers are not an emulsifier although (a) Emulsifiers are a mixture of oil droplets
it can stabilise an emulsion. in water.
(c) Stabilisers are used in chocolate milk, ice (b) Emulsifiers are used to prevent the
cream and jam. separation of oil droplets from the mixture.

433

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

(c) Emulsifiers are used to make bread, orange drinks, custard powder, sweets
chocolate cake and ice cream. and apricot jam. Tartrazine is believed to
cause hyperactivity in children.
(d) Examples of natural emulsifiers are egg
yolk and milk. (i) Bright blue FCF, a blue triphenyl dye, is
found in beverages, jellies, confections and
(e) Synthetic emulsifier includes lecithin and syrups. This food dyes can be combined
pectin. with tartrazine to produce various shades
of green.
7. Thickeners
(j) Chromatography enables us to identify
(a) Thickeners are used to thicken food. whether there are food dyes that are
dangerous in certain food.
(b) Acacia gum can act as a thickener as well
as a stabiliser. (k) Chromatography is a technic used to
separate the different components in a
(c) Table 5.10 shows some examples of mixture.
thickeners.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved (l) Chromatography technic comprised of
Table 5.10 Some examples of thickeners two main components; the stationary
phase and the mobile phase.
Thickener Example
(m) The stationary phase is the strip of
Starch Instant soups and puddings chromatography paper whereas the mobile
phase is a suitable solvent.
Pectin Jam
(n) This technique separates the different
Acacia gum Chewing gum, jelly and wine components in food dyes based on their
relative solubility in specific solvents.
Gelatin Curd (cheese)
(o) These components will move along the
Xanthan gum Sauce, salad dressing stationary phase with different speed,
and as a result got separated in the
8. Dyes chromatography paper.

(a) Food colourings are dyes. (p) By comparing the Rf value of each
component with the Rf value of a known
(b) Food processing often caused loss of substance, the components in the mixture
colour. can be identified.

(c) Food dyes are used to add or restore (q) Rf value for each component is calculated
the colour of food to enhance their using the following formula:
appearance and to fulfil consumers’
Form 5 expectations. Rf = Distance travelled by the component
Distance travelled by solvent
(d) Food dyes can be classified into natural
and artificial food dyes. Distance End
travelled line
(e) Artifical food dyes are usually used by solvent
because they are more uniform, less Distance travelled by
expensive and have brighter colours than different colours
natural food dyes.
Starting
(f) Many food dyes are azo compounds or line
triphenyl compounds.
Figure 5.3 Paper chromatography with different distance
(g) Azo dyes have colours such as red, orange travelled by components in a food dye
and yellow, whereas triphenyl dyes have
colours such as blue and green.

(h) Tartrazine, a yellow azo dye, is used in

434

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

ACTIVITY 5.2

Aim: To separate the different components in food dyes using chromatography method.

Materials: Distilled water, food dyes (red, yellow and blue)

Apparatus: 250 ml beaker, filter paper, rules, stick

Procedure:

End Stick
line
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Form 5
Filter Filter paper Beaker
paper
Distilled
Starting water
line

(a) (b)

Figure 5.4 Set up of apparatus to carry out chromatography

1. A piece of filter paper measuring 5 cm × 12 cm is prepared.

2. A straight line, 1.5 cm from the edge of paper is drawn with pencil. This is the starting line.

3. Another straight line, 0.5 cm from the opposite edge is drawn. This is the end line.

4. Three concentrated dots of food dyes of different colours are placed onto the starting line.

5. The filter paper is hung in the beaker using a stick making sure the bottom part of the paper is
soaked in the distilled water without the distilled water touching the three colouring dots.

6. The apparatus is left to allow the distilled water to be absorbed from the bottom to the top of the
paper. Once the distilled water reaches the end line, the filter paper is removed from the beaker.

7. The position of the colour spots (one or several spots) for each food dyes on the chromatography
paper is identified and marked.

Observation:

Food dye Observation
Red There are two spots of different colours
There is only one spot of yellow only
Yellow There are three spots of different colours
Blue

Discussion:

1. Chromatography method can identify original or mixed food dye. If the food dye is a mixture and
not original, other components in the food dye will dissolves in water and move together with
water at a different rate. At the end, these components will form several spots of different colours
on the paper chromatography. If the food dye is original, at the end there will only be one spot on
the chromatography paper.

2. For yellow dye, there is only one yellow spot on the chromatography paper. This means the dye is
original and not a mixture.

3. For red and blue dyes, there are more than one different spot on the chromatography paper. This
means the dye is not original; it is mixture of component colourings.

4. If the food dye is not soluble in water, the colouring dot is stationary and will not move.

Conclusion:
Chromatography method can be used to separate the different components in the food dye.

435

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

Justification for Using Food Additives 5.4 Medicines and Cosmetics

1. The advantages of using food additives are as 1. A medicine is used to prevent or cure a
follows. disease or to relieve pain.

(a) They make the food stay fresh longer, 2. Medicines can be classified into traditional
look nicer and taste better. medicines and modern medicines.

(b) They make seasonal crops and fruits Traditional Medicines
available throughout the year. Penerbitan Pelangi Sdn Bhd. All Rights Reserved
1. Traditional medicines are derived from plants
2. The bad effects of food additives when taken or animals.
excessively are as follows.
2. Table 5.11 and Table 5.12 show the functions
(a) Some of the food additives are associated of some common medicinal plants and animals.
with diseases such as cancer, asthma,
allergies and hyperactivity. Table 5.11 Some common medicinal plants and their functions

(b) Some food additives make the food less Medicinal plant Function
nutritious.
Aloe vera To treat skin wounds
3. Many countries have enacted laws to control (lidah buaya)
the use of food additives so as to safeguard the
health of their citizens. In Malaysia, the public Centella asiatica To treat depression and for longevity
are protected against health hazard in the use (pegaga)
of food additive by Food Act 1983 and Food
Regulations 1985. Eurycoma To increase the male libido
longifolia
4. Before a food additive is used, it must be (tongkat Ali)
approved by the government.
Orthosiphon To treat gout, diabetes and rheumatism
5. Food additives are never given permanent aristatus
approval, but are continually reviewed and (misai kucing)
modified or withdrawn when necessary.
Andrographis To treat diarrhoea, fever and diabetes
paniculata
(hempedu bumi)

Ocimum To treat coughs, colds and bronchitis
basilicum
Checkpoint 5.3 (selasih)

Q1 State seven types of chemicals used in food Table 5.12 Animal that are claimed to have medicinal properties
additives and their respective functions.
Medicinal Function
animal
Q2 Name one type of food that can be preserved
Form 5 by pickling in vinegar. Explain how this method Sea cucumber To cure wounds
works. (gamat)

Q3 Figure 5.5 shows part of the label of a tin of Centipede To treat tetanus and epilepsy
sardines.
Ants To treat hepatitis B

Day Day Sardines 3. Traditional medicines are usually not processed.
Ingredients:
4. Any medicinal plants containing alkaloid is
Sardines, tomato sauce, salt, xanthan gum potentially toxic to the liver.

Figure 5.5 5. Traditional medicines can cause serious side
effects.
(a) List out all the food additives used in the
sardines. 6. Most research on the toxic and adverse effects
of traditional medicine should be carried out
(b) Which type of food additives does each of to ensure the safety of these medicines to the
the ingredients mentioned in (a) belong? public.

436

Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry  

Modern MedicinesPenerbitan Pelangi Sdn Bhd. All Rights Reserved stomach wall, aspirin is swallowed
Form 5 with plenty of water and is taken
1. Modern medicines are made by scientists after food.
in laboratories and are based on substances
found in nature. (ii) It is believed to cause brain and liver
damage if given to children with flu
2. The active ingredients in the substances are or chicken pox.
identified, extracted and purified.
5. Paracetamol can be taken to relieve mild to
3. The medicine is then tested repeatedly using moderate pain such as headaches, muscle
different methods before it is marketed. This and joint pains, backaches and period pains.
allows scientists to ensure that the medicine is
safe and to identify its side effects. (a) Unlike aspirin, paracetamol can be given
to children as it does not irritate the
4. Modern medicine is made in different forms stomach or cause it to bleed.
such as liquids, powders, capsules and tablets.
(b) The side effects of paracetamol are rare
5. Modern medicine consists of several types when it is taken at the recommended dose.
including analgesics, antimicrobials, psychotic
drugs, anti allergies and corticosteroids. (c) However, it causes skin rashes, blood
disorders and acute inflammation of the
Analgesics pancreas when it is taken on a regular
basis for a long time.
1. Analgesics are medicine used to relieve pain
without causing numbness or affecting (d) An overdose of paracetamol can cause
consciousness. liver damage.

2. These medicines do not treat the cause of the 6. Codeine is an analgesic used in headache
pain but merely provide enough relief to the tablets and in cough medicines.
patients to allow them to carry out their daily
routines. (a) Most codeine is synthesised from
morphine.
3. Examples of analgesics are aspirin, paracetamol
and codeine. (b) Codeine may cause drowsiness when
consumed.
4. Aspirin is used to relief pain, especially when
there is infection (swollen and painful), such (c) Abuse of this medicine may lead to
as arthritic pain and toothache. addiction, depression and nausea.
(a) The active ingredient in aspirin is
acetylsalicylic acid. Antimicrobials
(b) Aspirin has the following structural
formula. 1. Antimicrobials are used to kill or to slow down
the growth of microorganism such as bacteria,
COOH O virus and fungus.
'
2. Antimicrobials consist of antibiotics, antifungal,
O C CH3 antivirus and antiparasitic.

3. Antibiotics are obtained from microorganisms
(c) Aspirin is acidic in nature. such as bacteria and fungus. Different
(d) The side effects of aspirin are as follows. antibiotics can fight against different kinds of
(i) It cause internal bleeding and microorganisms.

ulceration. To reduce irritation of the 4. Examples of antibiotics are penicillin and
streptomycin.

437

  Chemistry SPM  Chapter 5 Consumer and Industrial Chemistry

5. Penicillin 10. This is to make sure that all the microorganisms
are killed. Otherwise, the patient may become
(a) Penicillin is one of the earliest and most sick again and the microorganisms may become
widely used antibiotics. more resistant to the antimicrobials. When this
happens, antimicrobial such as the antibiotic is
(b) Penicillin is extracted from the fungus, no longer effective. The doctor will then have
Penicillium notatum. to prescribe a different and stronger antibiotic
to fight the same infection.
(c) This antibiotic is used to cure diseases
caused by bacterial infection such as Psychotic drugs
tuberculosis (TB) and pneumonia.
1. People with mental illness are not in control
6. Streptomycin of their thoughts, feelings and behavior.

(a) Streptomycin is an antibiotic produced by 2. Psychotic drugs are used to alter abnormal
soil bacteria of the genus Streptomyces. thinking, feelings or behavior.

(b) Streptomycin is used to treat tuberculosis, 3. These medicines do not cure mental illness.
whooping cough and pneumonia. However, these medicines can reduce many
of the symptoms of mental illnesses and help
7. Both penicillin and streptomycin can be broken a person to get on with life.
down by the acid in the stomach. Therefore,
these antibiotics are usually given by injection 4. Psychotic drugs are classified into stimulants,
and not taken orally. antidepressant and antipsychotic.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
8. Antifungal and antiparasitic medicines are
usually applied externally to cure skin diseases
such as shingles or tinea.

9. The patient should take the full course of the
antimicrobials that the doctor prescribes to him
even if he feels better.

Table 5.13 The functions and side effects of some psychotic drugs

Psychotic drug Function Example Side effects

Stimulant To reduce fatigue and Methamphetamine, A high dose or an excessive use of stimulants
elevate mood
dextroamphetamine, over long periods of time can lead to anxiety,

amphetamine hallucination, severe depression or physical and

psychological dependence

Form 5 Antidepressant To reduce tension and Tranquillisers Tranquillisers cause drowsiness, poor
anxiety coordination or dizzy. An overdose of these
drugs can lead to respiratory difficulties,
insomnia, coma and even death.

Barbiturates Barbiturates cause addiction. People who rely on
barbiturates to fight insomnia can sometimes kill
themselves accidentally by taking an overdose.

Antipsychotic To treat psychiatric illness Chlorpromazine, Antipsychotic medicines cause drowsiness,
palpitations and dizziness.
such as schizophrenia haloperidol and

clozapine

438