MODULE • Chemistry FORM 4
CONTENTS
THEME: FUNDAMENTALS OF CHEMISTRY
UNIT
2
2.1
2.2
2.3
2.4
THE STRUCTURE OF
ATOMS
1
3
3.1
3.2
3.3
3.4
23
Relative Atomic Mass and Relative Molecular Mass
Mole Concept
Chemical Formulae
Chemical Equations
4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
THE PERIODIC TABLE OF
ELEMENTS
SPM PRACTICE
UNIT
SPM PRACTICE
5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
CHEMICAL BONDS
Basic Formation of Compound
Ionic Bonds
Covalent Bonds
Hydrogen Bond
Dative Bond
Metallic Bond
Ionic and Covalent Compound
SPM PRACTICE
7
49
The Development of the Periodic Table of Elements
The Arrangement in the Periodic Table of Elements
Elements in Group 18
Elements in Group 1
Elements in Group 17
Elements in Period 3
Transition Elements
UNIT
95
The Role of Water in Showing Acidic and Alkaline
Properties
6.2 The Strength of Acid and Alkali
6.3 Chemical Properties of Acid and Alkali
6.4 Concentration of Aqueous Solution
6.5 Standard Solution
6.6 PH Value
6.7 Neutralisation
6.8 Salt, Crystals and Uses of Salt in Daily Life
6.9 Preparation of Salts
6.10 Action of Heat on Salt
6.11 Qualitative Analysis
SPM PRACTICE
UNIT
ACIDS, BASES AND
SALTS
6.1
SPM PRACTICE
MOLE CONCEPTS,
FORMULAE AND
CHEMICAL EQUATIONS
UNIT
6
Basic Concept of Matter
The Development of Atomic Model
The Structure of Atom
Isotope and Its Uses
UNIT
THEME: INTERACTION BETWEEN
MATTERS
7.1
7.2
7.3
7.4
RATE OF REACTION
158
Determine Rate of Reaction
Factors that Affect the Rate of Reaction
Application of Factors that Affect the Rate of
Reaction in Daily Life
The Collision Theory
SPM PRACTICE
THEME: INDUSTRIAL CHEMISTRY
UNIT
74
8
8.1
8.2
8.3
8.4
MANUFACTURED
SUBSTANCES IN INDUSTRY
196
Alloy and its Importance
Composition of Glass and its Uses
Composition of Ceramics and its Uses
Composite Materials and its Importance
SPM PRACTICE
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MODULE • Chemistry FORM 4
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MODULE • Chemistry FORM 4
UNIT
2
THE STRUCTURE OF ATOMS
Concept Map
Ion
Molecule
Particle Theory of Matter
Three types of particle
Three states
of matter
MATTER
Solid
Liquid
Kinetic Theory
of Matter
Gas
Proton
number
Nucleon
number
History of
Atomic Model
Electron
arrangement
Diffusion
[Studied in Form 2]
A
X
Z
Standard
representation
of element
Meaning
Meaning
Atom
Isotope
Calculate
RAM
Three sub atomic particles
Proton
Uses
Neutron
Electron
Note:
Differentiate between
1 State of matter
2 Type of particles
3 Sub atomic particles
1
UNIT
2
Atom
MODULE • Chemistry FORM 4
2.1
BASIC CONCEPT OF MATTER
Describe Matter
1
SK
2.1
SP 2.1.1
Particle theory of matter
What is matter?
Form 1, Unit 5
SP 2.1.1
Matter is any substance that has mass and occupies space.
UNIT
2
What is particle theory
of matter?
Matter is made up of tiny and discrete particles. Three types of tiny particles are
atoms
ions
,
and molecules .
Define element.
A substance made from only
Define compound.
A substance made from
which are bonded together.
Define atom and give
example of substance
that made up of atom.
The smallest neutral particle of an element.
Example:
(a) Pure metal:
– Copper, Cu
– Lead, Pb
one
two
type of atom.
or
more
(b) Non-metal:
– Carbon, C
– Silicon, Si
(c) Inert gases:
– Neon, Ne
– Helium, He
Carbon, C
Copper, Cu
different elements
Helium, He
Define molecule and give A neutral particle consists of similar or different non-metal atoms which are
example of substance
covalently bonded.
that made up of
Example:
molecule.
(a) Molecule of an element:
– Oxygen, O2
– Hydrogen, H2
(b) Molecule of a compound:
– Carbon dioxide, CO2
– Water, H2O
Oxygen, O2
Water, H2O
Define ion and give
example of substance
that made up of ion.
Positively or negatively charged particles, which are formed from metal atom
and non-metal atom respectively. The force of attraction between the two
oppositely charged ions forms an ionic bond.
Example:
Sodium chloride, NaCl
Sodium chloride, NaCl
– Elements can be identified as metal or non-metal by referring to the Periodic Table of Elements.
– Formation of molecule and ion will be further explained in Chapter 5, page 74.
© Nilam Publication Sdn. Bhd.
2
MODULE • Chemistry FORM 4
2
Conclusion:
Matter
Element
Atoms
Compound
Molecules
Cu, Ag, C, Ne, Ar
O2, Br2, N2, Cl2
Molecules
Ions
CO2, H2O, NO2
NaCl, KNO3, AgCl
UNIT
2
Exercise
Determine the type of particles in the following substances:
3
Substances
Type of
particle
Substances
Type of
particle
Substances
Type of
particle
Hydrogen gas
(H2)
Molecule
Sulphur dioxide
(SO2)
Molecule
Tetrachloromethane
(CCl4)
Molecule
Copper(II)
sulphate (CuSO4)
Ion
Iron (Fe)
Atom
Zinc chloride
(ZnCl2)
Ion
Argon (Ar)
Atom
Carbon (C)
Atom
Hydrogen peroxide
(H2O2)
Molecule
Kinetic theory of matter
State kinetic theory of
matter. SP 2.1.2
The tiny and discrete particles that made up matter are constantly moving.
What is the significant of
this theory?
This theory uses the particular model to explain the properties of matter. For
example, diffusion and change of physical state of matter with the energy
change (melting, boiling, freezing and condensation).
Remark:
Particles represent atoms, ions or molecules.
3
MODULE • Chemistry FORM 4
Explain the Change in Physical State of Matter
SP 2.1.2
tiny
discrete
State Kinetic Theory of
Matter.
Matter that made up of
How does the matter
exist?
Matter exists in three different states which are
gas
and
.
and
moving
always in constantly
.
Complete the following table to compare particles in solid, liquid and gas.
State of matter
Solid
particles which are
solid
,
liquid
SP 2.1.2
Liquid
Gas
UNIT
2
Draw the particles
arrangement. Each
particle (atom/ ion/
molecule) is
represented by
Particles arrangement
Particles movement
Closely
packed in
orderly
manner.
not in
Vibrate
and
Particles can vibrate ,
rotate
and
move
throughout
the liquid.
rotate
position.
at their fixed
Attractive forces
between the particles
Very
strong
Energy content of the
particles
Very
low
Define melting point.
SP 2.1.3
Closely
packed but
orderly manner .
Very widely separated
from each other.
Particles can vibrate
rotate
and
move
freely.
Strong
,
Weak
Very
Moderate
solid
The constant temperature at which a
become a liquid is called the melting point .
high
completely changes to
Remark:
For the same pure substance, melting point = freezing point. For example, pure ice melts and
liquid
freezes at 0°C (the melting point of ice). The constant temperature at which a
changes to a solid is called
Define boiling point.
SP 2.1.3
freezing point
The constant temperature at which a
become a gas.
.
liquid
completely changes to
Remark:
Pure water boils and condenses at 100°C (the boiling point of water).
When does a substance
change its physical state?
© Nilam Publication Sdn. Bhd.
A substance changes its physical state when the temperature reaches the melting
point and boiling point.
4
MODULE • Chemistry FORM 4
heat
Matter undergoes changes of state when
absorbed or
released :
absorbed
(i) When heat energy is
kinetic energy of the particles
releases
energy of the particles
by the matter (it is heated), the
increases and they vibrate faster.
kinetic
heat energy (it is cooled), the
decreases and they vibrate less vigorously.
How does the energy cause
a substance to change its
physical state?
Heat energy is absorbed to overcome the attractive forces between particles
or released when the attractive force between particles is formed.
State the change in heat
energy during physical
state change.
– During melting, boiling and sublimation, heat energy is absorbed from the
surrounding.
– During freezing and condensation heat energy is given out to the surrounding.
Sublimation
Key:
Melting
Boiling
Freezing
Condensation
Solid
Liquid
Heat is
absorbed
from the
surrounding
Gas
Heat is given
out to the
surrounding
Sublimation
What happen to the
temperature when a
substance undergoes
changes in physical states?
Explain.
– During the melting process, the temperature remains unchanged because heat
used
energy absorbed by the particles is
to overcome the forces
liquid .
between particles so that the solid changes to a
– During the freezing process, the temperature remains unchanged because the
heat released to the surrounding is balanced by the heat released
when the liquid particles attract one another to become a
Why do different substance
have different melting and
boiling points?
solid
.
– When the force of attraction between particles is stronger, more heat needed
to overcome the force. The melting and boiling points are higher.
– When the force of attraction between particles is weaker, less heat needed to
overcome the force. The melting and boiling points are lower.
Remark:
The strength of attractive force between particles is used to explain the difference in melting
point and boiling point of ionic and covalent compound in Chapter 5 (Chemical Bond)
Determine the Melting Point and Boiling Point of Naphtalene
SP 2.1.2
Describe experiment to determine the melting and freezing points of naphthalene.
Materials: Naphthalene powder, water
Apparatus: Boiling tube, conical flask, beaker, retort stand with clamp, thermometer 0 – 100°C, stopwatch, Bunsen
burner, wire gauze, tripod stand
5
2
(ii) When matter
energy is
UNIT
What causes a substance to
change its physical state?
MODULE • Chemistry FORM 4
Procedure:
I
Heating of naphthalene
(a) A boiling tube is filled 3 cm height with naphthalene powder
and a thermometer is placed into it.
(b) The boiling tube is immersed in a water bath as shown in the
diagram and make sure the water level in the water bath is
higher than naphtalene powder in the boiling tube.
(c) The water is heated and the naphthalene is stirred slowly with
thermometer.
(d) When the temperature of naphthalene reaches 60°C, the
stopwatch is started. The temperature of naphthalene is
recorded at 30 seconds intervals until the temperature of
naphthalene reaches 90°C.
Thermometer
Thermometer
/ Termometer
Boiling
tube
Boiling
tube
/ Tabung didih
Water/ Air
Water
Naphthalene
Naphthalene
/ Naftalena
Heat
Heat
Haba
UNIT
2
II Cooling of naphthalene
(a) The boiling tube and its content is removed from the water
bath and put into a conical flask as shown in the diagram.
(b) The content in the boiling tube is stirred constantly
with thermometer throughout cooling process to avoid
Naphthalene
supercooling (the temperature of cooling liquid drops below
freezing point, without the appearance of a solid).
(c) The temperature of naphthalene is recorded every 30 seconds
interval until the temperature drops to 60 °C.
(d) A graph of temperature against time is plotted for the heating and cooling process respectively.
Result:
I Heating of naphthalene
Time / s
0
30
60
90
120
150
180
210
240
0
30
60
90
120
150
180
210
240
Temperature / °C
II Cooling of naphthalene
Time / s
Temperature / °C
Temperature / °C
Temperature / °C
X
X
Time / s
Time / s
From the graph, the melting point
of naphthalene is X°C
© Nilam Publication Sdn. Bhd.
From the graph, the freezing point of
naphthalene is X°C
6
MODULE • Chemistry FORM 4
Interpretation of heating curve
It is a plot of the temperature against time to show how temperature change as
a substance is heated up.
Sketch the heating curve of
a substance with the
melting point P °C and the
boiling point Q °C from
solid to gas.
Label the part on the graph
where melting point and
boiling point take place.
Q
P
Time / s
Temperature / °C
F
D
B
E
C
A
2
Study the heating curve of a
substance.
(a) State the physical state
of the substance at the
following region:
AB, BC, CD, DE, EF
(b) Explain the changes in
physical state and
temperature of the
substance.
Temperature / °C
AB
Solid
BC
Solid and liquid
CD
Liquid
DE
Liquid and gas
EF
Gas
Time / s
AB
1 Heat energy is
naphthalene.
absorbed
by the particles in the
solid
kinetic
2 The heat energy absorbed causes
energy of the particles
increase
faster
to
and the particles vibrate
.
3 The temperature
increases .
BC
1 Heat energy is
naphthalene.
2 The heat energy
absorbed
absorbed
between particles so that the
by the particles in the
is used to overcome forces of attraction
solid
liquid .
turns to
remains constant
3 The temperature
melting point
solid
. The constant temperature is
.
CD
1 Heat energy is
naphthalene.
absorbed
by the particles in the
liquid
absorbed causes the
kinetic
2 The heat energy
energy of the
increase and the particles move
faster
particles to
.
3 The temperature
increases .
7
UNIT
What is heating curve?
SP 2.1.3
MODULE • Chemistry FORM 4
DE
absorbed
1 Heat energy is
naphthalene.
overcome the forces of attraction
freely
between particles so that the particles begin to move
to form
2 The heat
absorbed
liquid
by the particles in the
gas
a
is used to
.
3 The temperature remains constant
boiling point .
. The constant temperature is
EF
absorbed
UNIT
1 Heat energy is
2 The heat
absorbed
gas
by the particles in the
causes their
kinetic
2
faster
increase and the particles move
.
energy of the particles to
.
increases .
3 The temperature
Remark:
The diagram shows how the physical state of substance is related to its melting and boiling
points.
Melting point
Boiling point
Solid
Liquid
Gas
Solid + liquid
Interpretation of cooling curve
What is cooling curve?
Study the cooling curve of a
substance.
(a) State the physical state
of the substance at the
following region:
PQ, QR, RS
(b) Explain the changes in
physical state and
temperature of the
substance.
Liquid + gas
SP 2.1.3
It is a plot of the temperature against time to show how temperature change as
a substance is cooled.
Temperature / °C
P
Q
R
PQ
Liquid
QR
Liquid and solid
RS
Solid
S
Time / s
PQ
1 Heat is released
naphthalene.
to the surrounding by the particles in the
liquid
2 The particles in the
slower .
3 The temperature
© Nilam Publication Sdn. Bhd.
Temperature
increase
decreases .
8
lose their
heat
liquid
energy and move
MODULE • Chemistry FORM 4
QR
released
1 Heat is
naphthalene.
to the surrounding by the particles in
2 The heat released is
balanced
heat
by the
solid
as the particles attract one another to form a
3 The temperature
freezing point
remains constant
energy
liquid
released
.
. The constant temperature is
.
RS
releases
2 The heat released causes the particles vibrate
slower
.
decreases .
UNIT
3 The temperature
heat.
Exercise
1
The table below shows substances and their chemical formula.
Substance
Chemical formula
Type of particle
Silver
Ag
Atom
Potassium oxide
K2O
Ion
Ammonia
NH3
Molecule
Chlorine
Cl2
Molecule
(a) State the type of particles that made up each substance in the table.
TP2
(b) Which of the substances are element? Explain your answer.
TP2
Silver and chlorine. Silver and chlorine are made up of only one type of atom.
(c) Which of the substances are compound? Explain your answer.
TP2
Potassium oxide and ammonia. Potassium oxide and ammonia are made up of two different elements.
2
The table below shows the melting and boiling points of substances P, Q and R.
(a) (i)
TP1
Substance
Melting point / °C
Boiling point / °C
P
–36
6
Q
–18
70
R
98
230
What is meant by ‘melting point’?
The constant temperature at which a solid changes to a liquid at particular pressure.
(ii) What is meant by ‘boiling point’?
TP1
The constant temperature at which a liquid changes to a gas at particular pressure.
9
2
1 The particles in the solid naphthalene
MODULE • Chemistry FORM 4
(b) Draw the particles arrangement of substances P, Q and R at room condition.
TP2
Substance
Bahan P P
(c) (i)
TP2
Substance
Bahan Q Q
Substance
Bahan R R
State the substance/substances that exist in the form of liquid at 0 °C.
Q
(ii) Give reason to your answer in c(i).
The temperature 0 °C is above the melting point of Q and below the boiling point of Q.
UNIT
(d) (i)
TP4
Substance Q is heated from room temperature to 100 °C. Sketch a graph of temperature against time
for the heating of substance Q.
2
Temperature / °C
70
Time / s
(ii) What is the state of matter of substance Q at 70 °C?
Liquid and gas
(e) Compare the melting point of substances Q and R. Explain your answer.
TP4
The melting point of substance R is higher than substance Q. The attraction force between particles in
substance R is stronger than substance Q. More heat is needed to overcome the forces between particles
in substance R.
(f) The diagram shows a set up of apparatus to separate
TP4 a mixture of water and ethanol. During the process,
HOTS
the vapour produces at part X and pure ethanol is
collected into the conical flask.
(i) Which compound is being distilled first?
Explain your answer.
Ethanol. Boiling point of ethanol is lower than
boiling point of water.
Thermometer
Water out
X
Water +
ethanol
Porcelain Water
in
chips
(ii) Draw the arrangement of particles at part X.
Heat up
(iii) Explain how the state of matter at part X changes during the process.
Condensation process occur // vapour is condensed. Gas changes into liquid.
© Nilam Publication Sdn. Bhd.
10
Liebig
condenser
Pure ethanol
MODULE • Chemistry FORM 4
The melting point of acetamide can be determined by heating
solid acetamide until it melts as shown in the diagram below. The
temperature of acetamide is recorded every three minutes when it is
left to cool down at room temperature.
Thermometer
Boiling tube
Water
Acetamide
(a) What is the purpose of using water bath in the experiment?
TP2
To ensure even heating of acetamide. Acetamide is easily combustible.
Heat
(c) Sodium nitrate has a melting point of 310 °C. Can the melting point of sodium nitrate be determined by
TP4 using the water bath as shown in the diagram? Explain your answer.
No, because the boiling point of water is 100 °C which is lower than the melting point of sodium nitrate.
(d) Why do we need to stir the acetamide in the boiling tube in above experiment?
TP3
To make sure the heat is distributed evenly.
(e) The graph of temperature against time for the cooling of liquid acetamide is shown below.
TP3
Temperature / °C
T3
T2
Q
R
T1
Time / s
(i)
What is the freezing point of acetamide?
T2 °C
(ii) The temperature between Q and R is constant. Explain.
The heat lost to the surrounding is balanced by the heat released by the liquid when the liquid
acetamide particles rearrange themselves to become solid.
(f) Acetamide exists as molecules. State the name of another compound that is made up of molecules.
TP2
Water / naphthalene
(g) What is the melting point of acetamide?
TP4 T °C
2
(h) A housewife bought a mothball as insect repellent at her house. She put the mothballs
TP5
pieces into a cupboard. After a month, she found the mothballs becoming smaller.
HOTS
Explain why the mothballs becoming smaller.
Mothballs are made of naphthalene.
Naphthalene is a substance that undergoes sublimation.
11
2
(b) State the name of another substance which its melting point can also be determined by using water bath as
TP2
shown in the above diagram.
Naphthalene
UNIT
3
MODULE • Chemistry FORM 4
2.2
THE DEVELOPMENT OF ATOMIC MODEL
SK
2.2
Series of Atomic Model based on Atomic Model by Dalton, Thomson,
Rutherford, Bohr and Chadwick
Scientist
SP 2.2.3
Atomic Model
Discovery
atom .
Matter is made up of particles called
created , destroyed or
(ii) Atoms cannot be
divided .
(i)
Dalton
(iii) Atoms from the same element are
UNIT
(i)
2
Positively
charged sphere
Thomson
Negatively charged
electron
(i)
moves
outside the nucleus
Rutherford
Nucleus that contains
proton
(ii)
Proton
(iii)
Electron
(i)
Nucleus that contains
proton
(ii)
is a part of the nucleus.
moves outside the nucleus.
Discovered the existence of electron
shells
.
Electrons
nucleus
Electron
move in the shells around the
.
neutron .
(i) Discovered the existence of
(ii) Nucleus of an atom contains neutral particles
neutron and positively charged
called
Shell
Nucleus that contains
proton
and
particles called
neutron
proton
neutron
(iii) The mass of a
almost the same.
Electron
© Nilam Publication Sdn. Bhd.
nucleus as the centre of an
Discovered the
atom and positively charged .
(iv) Most of the mass of the atom found in the
nucleus .
Shell
James
Chadwick
electrons , the first subatomic
positive charge which is
(ii) Atom is sphere of
embedded with negatively charged particles
called electrons .
Electron
Neils Bohr
Discovered the
particle.
identical .
12
.
and
proton
is
MODULE • Chemistry FORM 4
What are the sub atomic
particles in an atom based
on the development of
atomic model?
What are the
characteristics of the
subatomic particles?
2.3
Form 1, Unit 6: Periodic Table
The sub atomic particles are
proton
,
neutron
and electron .
Subatomic
particles
Symbol
Charge
Relative mass
Position
Electron
e
– (negative)
1
1 840 ≈ 0
In the shells
Proton
p
+ (positive)
1
In the nucleus
Neutron
n
neutral
1
In the nucleus
THE STRUCTURE OF ATOM
SK
2.3
The Diagram of Atomic Structure and the Electron Arrangement
Describe atomic structure
based on history of the
atomic structure.
SP 2.3.4
Shell
Nucleus that contains protons and neutrons
Electron
Atom
nucleus
(a) An atom has a central
shells
around the nucleus.
(b) The
nucleus
and electrons that move in the
contains protons and neutrons.
(c) The relative mass of a neutron and a proton which are in the nucleus is 1.
proton
(d) The mass of an atom is obtained mainly from the number of
and
Explain how the electrons
are filled in specific shells.
neutron
.
Every shell can be filled only with a certain number of electrons. For the
elements with atomic numbers 1 – 20:
2
– First shell can be filled with a maximum of
electrons.
8
– Second shell can be filled with a maximum of
electrons.
– Third shell can be filled with a maximum of
8
electrons.
First shell is filled with 2 electrons (duplet)
Second shell is filled with 8 electrons (octet)
Third shell is filled with 8 electrons (octet)
What are valence electrons?
Valence electrons are the electrons in the outermost shell of an atom.
13
2
SP 2.2.2
UNIT
The Sub Atomic Particles
MODULE • Chemistry FORM 4
Atom or element?
Example:
Na
Na
Na
Sodium
element
Sodium
element
Na
Na
Na
Na
Na
Na
Na
Na
Na
Na
Na
Sodium
Sodium
element
atom
Remark:
1 Atom is the smallest neutral particle of an element.
one
2 An element is made from only
type of atom.
Define proton number.
UNIT
SP 2.3.1
element
Proton number of an
atom
of its
.
is the number of proton in the
nucleus
2
Remark:
Every element has its own proton number (Refer to Periodic Table of Elements).
Define nucleon number.
SP 2.3.1
Nucleon number of an element is the total number of protons and neutrons in
atom
the nucleus of its
.
Remark:
– Nucleon number is also known as mass number. (Refer to Periodic Table of Elements)
– Nucleon number = number of protons + number of neutrons.
Why atoms are neutral?
+1
– Each proton has charge of
. Each electron has an electrical
–1
charge
neutral ).
charge of
. The neutron has no
(it is
– An atom has the same number of protons and electrons, so the overall charge
zero
neutral .
of atom is
. Atom is
Remark:
If an atom loses or gains electrons it is called an ion – formation of ion will be studied in Chapter 4.
How to calculate the
number of protons,
neutrons and electrons in
an atom?
SP 2.3.2
Example
In an atom:
Number of protons =
Proton number
Number of electrons =
Number of proton
Number of neutrons =
Nucleon number – Proton number
atom
(i) Proton number of potassium, K is 19. Potasium
19 protons in the nucleus and
19 electrons in the shells.
atom
(ii) Proton number of oxygen, O is 8. Oxygen
8 electrons in the shells.
in the nucleus and
© Nilam Publication Sdn. Bhd.
14
has
has
8 protons
MODULE • Chemistry FORM 4
What is the symbol of an
element?
The symbol of an element is a short way of representing an element. If the
symbol has only one letter, it must be a capital letter. If it has two letters, the
first is always a capital letter, while the second is always a small letter.
Symbol
Element
Symbol
Element
Symbol
Oxygen
O
Nitrogen
N
Calcium
Ca
Magnesium
Mg
Sodium
Na
Copper
Cu
Hydrogen
H
Potassium
K
Chlorine
Cl
The first letter of each element is a capital letter to show that it is a new element.
This is helpful when writing a chemical formula. For example KCl. There are
two elements chemically bonded in KCl because there are two capital letters
represent potassium and chlorine.
How to write the standard
representation of an
element?
SP 2.3.3
Example:
27
13
Al
(i) What information can
be obtained from the
standard representation
of the element?
(ii) Construct the structure
and the electron
arrangement of
Aluminium atom.
atom
The standard representation of an
as:
Nucleon number
A
Proton number
Z
X
of an element can be written
Symbol of an element
(i) Atomic structure and electron arrangement of Aluminium atom
– The element is Aluminium.
27
– The nucleon number of Aluminium is
.
13
– The proton number of Aluminium is
.
– Aluminium atom has 13 protons , 14 neutrons and
electrons.
(ii) Atomic structure and electron arrangement of Aluminium atom
13
The structure of
Aluminium atom
The electron arrangement of
Aluminium atom
13 p
14 n
AI
15
UNIT
Element
2
Example:
MODULE • Chemistry FORM 4
Exercise
1
TP2
Complete the following table:
Standard
Number Number Number
Electron
Number
Proton Nucleon
representation
of
of
of
arrangement of valence
number number
for an atom
proton electron neutron
of atom
electron
Element
UNIT
2
2
TP3
Hydrogen
1
1
H
1
1
0
1
1
1
1
Helium
4
2
He
2
2
2
2
4
2
2
Boron
11
5 B
5
5
6
5
11
2.3
3
Carbon
12
C
6
6
6
6
6
12
2.4
4
Nitrogen
14
7 N
7
7
7
7
14
2.5
5
Neon
20
Ne
10
10
10
10
10
20
2.8
8
Sodium
23
Na
11
11
11
12
11
23
2.8.1
1
Magnesium
24
12
Mg
12
12
12
12
24
2.8.2
2
Calcium
40
Ca
20
20
20
20
20
40
2.8.8.2
2
Draw the structure of sodium atom and electron arrangement of sodium atom.
23
11 Na
3
TP2
The structure of sodium atom
The electron arrangement of sodium atom
11 p
12 n
Na
The diagram below shows the symbol of atoms P, R and S.
35
17
P
12
6
R
37
17
S
(a) What is meant by nucleon number?
Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom.
(b) What is the nucleon number of P?
35
(c) State the number of neutron in atom P.
18
(d) State number of proton in atom P.
17
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16
MODULE • Chemistry FORM 4
2.4
ISOTOPE AND ITS USES
Define isotope.
SP 2.4.1
SK
2.4
Isotopes are atoms of the same element with same number of protons but
different number of neutrons.
Or
proton
Isotopes are atoms of the same element with same
nucleon number.
different
– Number of neutron
Example:
17 p
18 n
17 p
20 n
Chlorine-35
Chlorine-37
Remark:
In any isotopes;
different
– Number of neutron is different. Hence, nucleon number is
.
– Number of proton is the same. Hence, proton number is the same.
same
– Number of electron is the
. Atoms of isotopes have same
UNIT
2
What causes atoms of the
same element to be
isotopes?
number but
electron
arrangement . Hence, isotopes have same chemical properties.
– Different physical properties (such as mass, density, melting and boiling points) due to
different number of neutron or nucleon number.
What is natural abundance
of isotopes?
Natural abundance is the percentage of isotopes present in a natural sample of
the element.
Example:
Element
Bromine
Chlorine
How to calculate the
relative atomic mass of the
isotope?
SP 2.4.2
Name of isotope
Natural abundance
Isotope
Bromine-79
50%
79
Br
35
Bromine-81
50%
81
Br
35
Chlorine-35
75%
35
Cl
17
Chlorine-37
25%
37
Cl
17
∑(% isotope × Relative mass of isotope)
100
(50% × 79) + (50% × 81)
Relative atomic mass of Bromine =
= 80
100
(75% × 35) + (25% × 37)
Relative atomic mass of Chlorine =
= 35.5
100
Relative atomic mass of isotope =
Remark:
Relative mass of isotope is equal to nucleon number.
17
MODULE • Chemistry FORM 4
Example
20
1 Neon is naturally found in three isotopes: 90.9% 10 Ne, 0.3%
22
8.8% 10 Ne. Calculate the relative atomic mass of Neon.
Relative atomic mass of Neon
(20 × 90.9%) + (21 × 0.3%) + (21 × 8.8%)
=
100
= 20.18
21
10
Ne and
63
UNIT
2 Relative atomic mass of Copper is 63.62. Isotopes of Copper: 69% 29 Cu and
q
q
31% 29 Cu. Calculate the nucleon number of isotope 29 Cu.
(69% × 63) + (31% × q)
63.62 =
100
q = 65
q
Nucleon number of isotope 29 Cu is 65.
2
Give examples of the usage
of isotopes.
(i)
SP 2.4.3
Medical field
– To detect brain cancer.
– To detect thrombosis (blockage in blood vessel).
– Iodine-131 is used to measure the rate of iodine absorption by thyroid
gland.
– Cobalt-60 is used to destroy cancer cells.
– Cobalt-60 is used to kill microorganism in the sterilising process.
(ii) In the industrial field
– To detect wearing out in machines.
– To detect any blockage in water, gas or oil pipes.
– Sodium-24 detect leakage of pipes underground.
– To detect defects/cracks in the body of an aeroplane.
(iii) In the agriculture field
– Phosphorus-32 is used to detect the rate of absorption of phosphate
fertilizer in plants.
– To sterile insect pests for plants.
(iv) In the archeology field
– Carbon-14 can be used to estimate the age of artifacts.
What are the disadvantages
of the usage of isotopes?
(i)
Some isotopes can stay in the human body for a long time and cause
harmful effects by their radiation. The tissues will be damaged and cause
burns, nausea, diseases such as leukemia and cancer.
(ii) The disposal of isotope residues not according to the procedure will result
in radioactive radiation to humans.
(iii) The production of isotopes requires a nuclear reactor where the cost is
very high.
© Nilam Publication Sdn. Bhd.
18
MODULE • Chemistry FORM 4
Exercise
1
TP2
Draw the atomic structure and electron arrangement for the atom of each element:
Standard Representation
of an Element
H
23
11
Na
16
8
O
Electron arrangement
of an atom
1p
0n
H
11 p
12 n
Na
8p
8n
O
UNIT
2
1
1
The structure of an atom
2Choose the correct statement for the symbol of element X.
TP2
23
11 X
Statement
Tick
(✔/✘)
Statement
Tick
(✔/✘)
Element X has 11 proton number.
7
Nucleon number of atom X is 23.
3
The proton number of element X is 11.
3
Number of nucleon of element X is 23.
7
The proton number of atom X is 11.
3
Atom X has 23 nucleon number.
7
The number of proton of element X is 11.
7
Neutron number of atom X is 12.
7
The number of proton of atom X is 11.
3
Number of neutron of atom X is 12.
3
Nucleon number of element X is 23.
3
Number of neutron of element X is 12.
7
19
MODULE • Chemistry FORM 4
SPM PRACTICE
Subjective Questions
1
The diagram shows the symbol of atom X, Y and Z.
12
Y
6
35
X
17
(a) (i)
TP1
37
Z
17
What is meant by isotope?
Isotopes are atoms of the same element with same number of protons but different number of
UNIT
neutrons
2
(ii) State a pair of isotope in the diagram shown.
TP2
X and Z
(iii) Give reason for your answer in (e)(ii).
TP2
Atoms X and Z have same proton number but different nucleon number//number of neutrons
(b) An isotope of Y has 8 neutrons. Write the symbol for the isotope Y.
TP3
14
6
Y
(c) The diagram shows an object discovered by an archeologist.
TP4
An isotope is used by this archeologist in his research.
(i)
Name the isotope used.
Carbon-14
(ii) State one advantage and one disadvantage of using the isotope.
Advantage: Estimate the age of artifact
Disadvantage: The isotope is expensive
2
TP4
The graph below shows the natural abundance of Germanium, Ge.
10
0
70
7.6%
20
7.7%
30
20.6%
40
27.4%
36.7%
Natural abundance (%)
75
(a) State the name of heaviest isotope of Germanium.
Germanium-76
© Nilam Publication Sdn. Bhd.
20
80
Nucleon number
MODULE • Chemistry FORM 4
(b) Use the natural abundance of each isotope to calculate the relative atomic mass of Germanium.
Relative atomic mass
=
(70 × 20.6%) + (72 × 27.4%) + (73 × 7.7%) + (74 × 36.7%) + (76 × 7.6%)
100
= 72.7
Element
Number of protons
Number of neutrons
P
1
0
Q
1
1
2
The table below shows the number of proton and neutron of atoms of elements P, Q and R.
R
6
6
UNIT
3
TP4
(a) Which of the atoms in the above table are isotope? Explain your answer.
P and Q. Atoms P and Q have same number of protons but different number of neutrons // nucleon
number.
(b) (i)
Write the standard representation of element Q.
TP2
2
1
Q
(ii) State three information that can be deduced from your answer in (b)(i).
The proton number of element Q is 1.
Nucleon number of element Q is 2.
Number of neutron = 2 – 1 = 1
Nucleus of atom Q contains 1 proton and 1 neutron
(c) (i)
Draw atomic structure for atom of element R.
TP3
6p
6n
(ii) Describe the atomic structure in (c)(i).
The atom consists of 2 parts: the centre part called nucleus and the outer part called electron shell.
The nucleus consists of 6 protons which are positively charged and 6 neutrons which are neutral.
The electrons are in two shells, the first shell consists of two electrons and the second shell consists
of four electrons. Electrons move around nucleus in the shells.
21
MODULE • Chemistry FORM 4
(d) Element P reacts with oxygen and produces liquid Z at room temperature. The graph below shows the
TP4 sketch of the graph when liquid Z at room temperature, 27 °C is cooled to –5 °C.
Temperature / °C
27
0
t1
t2
Time / s
−5
(i)
UNIT
What is the state of matter of liquid Z from t1 to t2? Explain why is the temperature remain unchanged
from t1 to t2.
Liquid and solid. Heat lost to the surrounding is balanced by the heat released when the liquid
particles rearrange themselves to become solid.
2
(ii) Draw the arrangement of particles Z at 20 °C.
(iii) Describe the change in the particles movement when Z is cooled from room temperature to –5 °C.
The particles move slower
4
TP6
You are given several materials such as string, button, ping-pong ball and thread. Create an atomic model
discovered by Neils Bohr using the materials given. Describe the model created.
String
Ping-pong ball
Button
The string represents shell of electron. The button represents the electron and the ping-pong ball represent
nucleus. The ping-pong ball is hung between two lines of string using thread.
Objective
Questions
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22
MODULE • Chemistry FORM 4
UNIT
3
MOLE CONCEPTS, FORMULAE
AND CHEMICAL EQUATIONS
Concept Map
RELATIVE MASS
– Relative Atomic Mass
– Relative Molecular Mass
– Relative Formula Mass
÷ Molar Mass (g mol–1)
Mass
(g)
× Avogadro constant
NUMBER
OF MOLE
× Molar Mass (g mol–1)
÷ Molar Volume (dm3 mol–1)
× Molar Volume (dm3 mol–1)
3
÷ Avogadro constant
Volume of
gas (dm3)
Numerical problem involving
chemical equation (Interpret
quantitatively)
Chemical
formula
Empirical
formula
Formula of
ionic
compound
Chemical
equation
Interpret qualitatively
– Reactants
– Products
– Physical state
Molecular
formula
Calculate percentage by mass of
element in a compound
Remark:
– Molar mass is mass of one mol of a substance. The unit is g mol–1.
– Molar volume of gas is volume occupied by one mol of any gas, 24 dm3 mol–1 at room conditions or 22.4 dm3 mol–1 at standard
temperature and pressure (s.t.p.)
– Avogadro constant is number of particles in one mol of any substance. The value is 6.02 × 1023.
23
UNIT
Numerically equal
Number of
particles
MODULE • Chemistry FORM 4
3.1
RELATIVE ATOMIC MASS AND RELATIVE MOLECULAR MASS
SK
3.1
What is relative mass?
A single atom is too small and light and cannot be weighed directly.
The masses of atoms are not determined directly but comparing them with
standard atom.
How is Relative Atomic
Mass calculated?
– Carbon-12 (an isotope of carbon) is chosen as a standard atom because its
mass can be determined very accurately using mass spectrometer. Carbon-12
isotope is given a mass of exactly 12.00.
SP 3.1.1
– Relative atomic mass based on the carbon-12 scale is the mass of one atom of
1
the elements compared with
of the mass of an atom of carbon-12:
12
– Relative atomic mass of an element (RAM)
=
UNIT
The average mass of one atom of the element
1
× The mass of atom of carbon-12
12
3
What can be interpreted?
Mass of an atom of carbon-12 = 12.00
1
of the mass of an atom carbon-12 = 1
12
C-12
Example:
Mg
C
C
Relative atomic mass of magnesium = 24
=
Mass of magnesium atom
1
× Mass of an atom carbon-12
12
Mass of magnesium atom = 24
What is the unit for relative
atomic mass? Give reason.
( 121 × mass of an atom carbon-12) = 24
It has no unit.
The relative atomic mass of an element can also be considered as the number
1
of times the mass of one atom of that element is heavier than
of carbon-12.
12
Example:
Relative atomic mass of helium is 4.
– 3 atoms of helium have the same mass as one carbon-12
© Nilam Publication Sdn. Bhd.
24
MODULE • Chemistry FORM 4
Define Relative Molecular
Mass (RMM).
SP 3.1.1
1
The average mass of one molecule of a substance when compared to
of the
12
mass of carbon-12.
The average mass of one molecule
Relative Molecular Mass =
1
× The mass of an atom of carbon-12
12
Remark: The Relative Molecular Mass is used for covalent molecules.
How to calculate Relative
Molecular Mass (RMM)?
RMM is obtained by adding up the RAM of all the atoms that are present in the
formula.
Exercise
Molecular substance
SP 3.1.2
Molecular formula
Relative molecular mass
O2
2 × 16 = 32
Water
H2O
2 × 1 + 16 = 18
Carbon dioxide
CO2
12 + 2 × 16 = 44
Ammonia
NH3
14 + 3 × 1 = 17
Oxygen
[Relative atomic mass: O = 16, H = 1, C = 12, N = 14]
2
TP2
Calculate Relative Formula Mass (RFM) for the following ionic substances:
Substance
Chemical formula
Relative formula mass
Sodium chloride
NaCl
23 + 35.5 = 58.5
Potassium oxide
K2O
2 × 39 + 16 = 94
CuSO4
64 + 32 + 4 × 16 = 160
(NH4)2CO3
2 [14 + 4 × 1] + 12 + 3 × 16 = 96
Aluminium nitrate
Al(NO3)3
27 + 3 [14 + 3 × 16] = 213
Calcium hydroxide
Ca(OH)2
40 + 2 [16 + 1] = 74
Lead(II) hydroxide
Pb(OH)2
207 + 2 [16 + 1] = 241
CuSO4•5H2O
64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250
Copper(II) sulphate
Ammonium carbonate
Hydrated copper(II) sulphate
[Relative atomic mass : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12,
Al = 27, Ca = 40, Pb = 207]
3
TP3
The formula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of
metal M?
M = Relative atomic mass for M
2M + 3 × 16 = 152
M = 52
25
3
Calculate Relative Molecular Mass (RMM) for the following molecular substances:
UNIT
1
TP2
MODULE • Chemistry FORM 4
4
Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x.
[Relative atomic mass: P = 31, Cl = 35.5]
31 + x × 35.5 = 208.5
35.5x = 208.5 – 31
35.5x = 177.5
x = 5 5
Relative atomic mass of calcium is 40 based on the carbon-12 scale.
(a) State the meaning of the statement above.
1
Mass of one calcium atom is 40 times greater than
mass of one carbon-12 atom.
12
TP3
TP3
(b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16]
Relative atomic mass of calcium
40
=
= 2.5 times
Relative atomic mass of oxygen
16
UNIT
(c) How many calcium atoms have the same mass as two atoms of bromine? [Relative atomic mass: Br = 80]
3
Number of calcium atom × 40 = 2 × 80
2 × 80
Number of calcium atom =
=4
40
3.2
MOLE CONCEPT
SK
3.2
Mole and the Number of Particles
What is Avogadro constant?
SP 3.2.2
A fixed quantity of 6.02 × 1023 of particles.
Remark:
The particles can be atom, ion or molecules
Why is Avogardro constant
useful?
It is a way for counting the particles (atoms, ions, or molecules). This is because
the size of particles is too small, so it is not possible to count physically.
What is mole?
1 A mole is a quantity of substance that contains as many particles as the
number of atoms in exactly 12 g of carbon-12.
2 A mole of a substance is the quantity of substance which contains a constant
number of particles (atoms, ions, molecules), which is 6.02 × 1023.
SP 3.2.1
Example:
© Nilam Publication Sdn. Bhd.
The concept of mole is the same as the concept of a dozen in our everyday life.
Dozen is used to represent a quantity:
1 dozen of pencil
1 × 12 pencils
2 dozens of pencil
2 × 12 pencils
3 dozens of pencil
3 × 12 pencils
1 mol of atom
1 × 6.02 × 1023 atoms
2 mol of atom
2 × 6.02 × 1023 atoms
3 mol of atom
3 × 6.02 × 1023 atoms
26
MODULE • Chemistry FORM 4
Why is mole concept useful?
When a compound is made up of two or more atoms, for example a covalent
molecule or an ionic compound, the mole concept is useful to determine the
number of respective particles.
Example:
Methane has a formula CH4. 1 methane molecule, CH4 is made up of 1 C atom
and 4 H atoms which are covalently bonded.
H
C
H
Is made up of
H
H
1 CH4 molecule
C
H
H
H
H
1 carbon atom and 4 hydrogen atoms
× Avogadro Constant
Number of
moles
÷ Avogadro Constant
Number of
particles
SP 3.2.2
Exercise
1
TP2
Complete the following table:
Substance
Chlorine
Water
Ammonia
Sulphur
dioxide
Magnesium
chloride
Aluminium
oxide
Formula
Cl2
H2O
NH3
SO2
SP 3.2.7
Number of atom per molecule/
Number of positive and negative ion
Number of particles in
1 mol of substance
6.02 × 1023 Cl2 molecules
Cl : 2
2
× 6.02 × 1023 Cl atoms
6.02 × 1023 H2O molecules
H: 2
O: 1
2
× 6.02 × 1023 H atoms
1
× 6.02 × 1023 O atoms
6.02 × 1023 NH3 molecules
N: 1
H: 3
1
× 6.02 × 1023 N atoms
3
× 6.02 × 1023 H atoms
6.02 × 1023 SO2 molecules
S : 1
O: 2
1
× 6.02 × 1023 S atoms
2
× 6.02 × 1023 O atoms
MgCl2
Mg2+ : 1
Cl– : 2
1
× 6.02 × 1023 Mg2+ ions
2
× 6.02 × 1023 Cl– ions
Al2O3
Al3+ : 2
O2– : 3
2
× 6.02 × 1023 Al3+ ions
3
× 6.02 × 1023 O2– ions
27
UNIT
What is the relationship
between number of moles
and number of particles
(atoms/ions/molecules)?
3
– Hence, 6.02 × 1023 CH4 contains 1 × 6.02 × 1023 C atoms and 4 × 6.02 × 1023
H atoms.
– Applying the mole concept:
1 mol of CH4 molecules consists of 1 mol C atoms and 4 mol of H atoms.
MODULE • Chemistry FORM 4
2
TP1
Complete the following: [Differentiate between “mole” dan “molecule”]
6.02 × 1023
(a) 1 mol of Cl2
[Chlorine gas]
molecules of chlorine, Cl2
2 × 6.02 × 10
23
(b) 1 mol of NH3
[Ammonia gas]
(c) 1 mol of NH3
4
[Ammonia gas]
atoms of chlorine, Cl
6.02 × 1023
4
molecules of ammonia, NH3
1
mol of nitrogen atom, N
mol atoms
3
mol of hydrogen atoms, H
1
23
4 × 6.02 × 10
1
mol of atom
molecules of ammonia, NH3
1
4 or 0.25
mol of N atoms,
number of N atoms =
UNIT
3
4 or 0.75
mol of H atoms,
3
number of H atoms =
(d) 2 mol of MgCl2
[Magnesium chloride]
(e) 2 mol of SO2
[Sulphur dioxide]
0.25 × 6.02 × 1023
0.75 × 6.02 × 1023
2
mol of Mg2+ ions, number of Mg2+ ions =
4
mol of Cl– ions, number of Cl– ions =
2 × 6.02 × 1023
3×2=6
molecules of SO2
2
mol of atoms
4 × 6.02 × 1023
mol of S atoms,
number of S atoms =
4
2 × 6.02 × 1023
2 × 6.02 × 1023
mol of O atoms,
number of O atoms =
4 × 6.02 × 1023
Number of Moles and Mass of Substance
Define molar mass.
SP 3.2.3
Molar mass is the mass of one mole of any substance.
State how to obtain the
molar mass for any
substance.
Molar mass of any substance is numerically equal to its relative mass (Relative
atomic mass / relative formula mass / relative molecular mass).
What is the unit of molar
mass?
Molar mass is the relative atomic mass (RAM), relative molecular mass (RMM)
and relative formula mass (RFM) of a substance in g mol–1.
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28
MODULE • Chemistry FORM 4
Example
Substance
Carbon,
C
Water,
H2O
Relative
mass
12
2(1) + 16 = 18
27
O
AI
12
C
H
Aluminium, Sodium chloride,
Al
NaCl
H
35.5 + 23 = 58.5
Na+
Cl –
1 mol
substance
Mass for
1 mol
What is the relationship
between number of moles
and any given mass of a
substance? SP 3.2.4
Example
12.01 g
18.00 g
58.50 g
58.00 g
27.00 g
12 g
18 g
27 g
58.5 g
12 g mol–1
18 g mol–1
27 g mol–1
58.5 g mol–1
Number of
moles
UNIT
Molar
mass
27.00 g
18.00 g
× (RAM/RFM/RMM) g mol–1
÷ (RAM/RFM/RMM) g mol–1
Mass in gram
(i) Calculate mass of 2 mol of water
Relative molecular mass of H2O = 18
Molar mass of 1 mol of H2O = 18 g mol–1
Mass of 2 mol of H2O = Number of moles × Molar mass
= 2 mol × 18 g mol–1
=
36
g
(ii) Calculate number of moles of 45 g of water, H2O
Mass of H2O
Number of moles of 45 g of H2O = Molar mass
45 g
=
= 2.5 mol
18 g mol–1
Exercise
Calculate:
SP 3.2.7
TP3
1 Mass of 3 mol of sodium hydroxide, NaOH
Molar mass of NaOH
= (23 + 16 + 1) g mol–1
= 40 g mol–1
3
12.00 g
2 Number of moles in 20 g of sodium hydroxide,
NaOH
Number of moles of sodium hydroxide, NaOH
20 g
=
40 g mol–1
= 0.5 mol
Mass of 3 mol of sodium hydroxide, NaOH
= 3 mol × 40 g mol–1
= 120 g
Answer: 120 g
Answer: 0.5 mol
29
MODULE • Chemistry FORM 4
3 Mass of 2.5 mol of oxygen gas, O2
4 Mass of 0.5 mol of sodium chloride, NaCl
Molar mass of oxygen gas, O2
= (16 + 16) g mol–1 = 32 g mol–1
Mass of 2.5 mol of oxygen gas, O2
= 2.5 mol × 32 g mol–1 = 80 g
Molar mass of NaCl
= (23 + 35.5) g mol–1
= 58.5 g mol–1
Mass of 0.5 mol of sodium chloride, NaCl
= 0.5 mol × 58.5 g mol–1
= 29.25 g
Answer : 80 g
5 Number of moles in 37.8 g of zinc nitrate, Zn(NO3)2
UNIT
3
Molar mass of zinc nitrate, Zn(NO3)2
= [65 + 2 (14 + 3 × 16)] g mol–1
= 189 g mol–1
Number of moles of zinc nitrate, Zn(NO3)2
37.8 g
=
189 g mol–1
= 0.2 mol
Answer: 29.25 g
6 Mass of 3.01 × 1023 copper atoms, Cu
Number of moles of Cu
Number of copper atom
=
Avogadro constant
3.01 × 1023
=
= 0.5 mol
6.02 × 1023
Mass of Cu
= Number of moles × Molar mass
= 0.5 mol × 64 g mol–1
= 32 g
Answer: 0.2 mol
Answer: 32 g
Number of Moles and Volume of Gas
Define molar volume of gas.
SP 3.2.5
State the molar volume of
any gases at room
conditions and at standard
temperature and pressure
(s.t.p).
Volume occupied by 1 mol of a gas.
Remark:
The volume of gas is affected by temperature and pressure.
Molar volume of any gases at room conditions is 24 dm3 mol–1.
Molar volume of any gases at standard temperature and pressure (s.t.p) is 22.4
dm3 mol–1.
Example:
The diagram shows the molar volume of three gases at room conditions.
Oxygen gas,
O2
Volume of gas
Number of mol
Mass
Number of
particles
24 dm3
1 mol
32 g
6.02 × 1023 O2
molecules
Remark:
1 dm3 = 1 000 cm3
© Nilam Publication Sdn. Bhd.
30
Ammonia gas,
NH3
24 dm3
1 mol
17 g
6.02 × 1023 NH3
molecules
Carbon dioxide gas,
CO2
24 dm3
1 mol
44 g
6.02 × 1023 CO2
molecules
MODULE • Chemistry FORM 4
Relationship between
number of moles and any
given volume of gas. SP 3.2.6
Example
× 24 dm3 mol–1 / 22.4 dm3 mol–1
Number of
moles of gas
÷ 24 dm mol / 22.4 dm mol
3
–1
3
–1
Volume of
gas in dm3
44.8
(i) 2 mol of carbon dioxide gas occupies
dm3 at STP.
0.5
(ii) 16 g of oxygen gas =
mol of oxygen gas. Therefore, 16 g of
12
oxygen gas occupies a volume of
[Relative atomic mass: O =16]
Formula for conversion of unit:
dm3 at room conditions.
SP 3.2.6
Volume of gas in dm3
× 24 dm3 mol–1/ 22.4 dm3 mol–1
Mass in
gram (g)
Exercise
1
TP3
÷ (RAM/RFM/RMM) g mol–1
Number of
moles
× (RAM/RFM/RMM) g mol–1
÷ (6.02 × 1023)
× (6.02 × 1023)
SP 3.2.7
A sampel of chlorine gas weighs 14.2 g. Calculate
[Relative atomic mass: Cl = 35.5]
(a) Number of moles of chlorine atoms.
Number of moles of chlorine atoms, Cl =
14.2 g
= 0.4 mol
35.5 g mol–1
(b) Number of moles of chlorine molecules (Cl2).
Number of moles of chlorine molecules, Cl2 =
14.2 g
= 0.2 mol
71 g mol–1
(c) Volume of chlorine gas at room conditions.
[Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure]
Volume of chlorine gas = 0.2 mol × 24 dm3 mol–1
= 4.8 dm3
31
Number of
particles
UNIT
3
÷ 24 dm3 mol–1/ 22.4 dm3 mol–1
MODULE • Chemistry FORM 4
2
TP3
(a) Calculate the number of atoms in the following substances:
[Relative atomic mass: N = 14; Zn = 65; Avogadro Constant = 6.02 × 1023]
(i) 13 g of zinc
13 g
= 0.2 mol
65 g mol–1
Number of zinc atom = 0.2 × 6.02 × 1023
= 1.204 × 1023
Number of mol of zinc atom =
(ii) 5.6 g of nitrogen gas
6.5 g
= 0.4 mol
14 g mol–1
Number of N atom = 0.4 × 6.02 × 1023
= 2.408 × 1023
Number of moles of N atom =
UNIT
(b) Calculate the number of molecules in the following substances:
[Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant = 6.02 × 1023]
(i) 8.5 g of ammonia gas, NH3
Molar mass of ammonia gas, NH3 = (14 + 3) g mol–1
= 17 g mol–1
3
8.5 g
17 g mol–1
= 0.5 mol
Number of moles of ammonia gas, NH3 =
Number of molecules in ammonia gas, NH3 = 0.5 mol × 6.02 × 1023
= 3.01 × 1023
(ii) 14.2 g of chlorine gas, Cl2
Molar mass of chlorine gas, Cl2, = 35.5 × 2 g mol–1 = 71 g mol–1
Number of moles of chlorine gas, Cl2 =
=
Mass of chlorine
Molar mass
14.2 g
= 0.2 mol
17 g mol–1
Number of chlorine molecules = 0.2 mol × 6.02 × 1023
= 1.204 × 1023
3
TP3
A gas jar contains 240 cm3 of carbon dioxide gas. Calculate:
[Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions]
(a) Number of moles of carbon dioxide gas:
240 cm3
Number of moles of CO2 =
= 0.01 mol
24 000 cm3 mol–1
(b) Number of molecules of carbon dioxide gas:
Number of molecules of CO2 = 0.01 × 6.02 × 1023
= 6.02 × 1021
© Nilam Publication Sdn. Bhd.
32
MODULE • Chemistry FORM 4
(c) Mass of carbon dioxide gas:
Mass of CO2 = 0.01 mol × [12 + 2 × 16] g mol–1
= 0.44 g
4
TP3
What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as found in 3.6 g of water?
[Relative atomic mass: H = 1, O = 16, Cl = 35.5]
Number of moles of chlorine molecule, Cl2 = 2 × Number of moles of water, H2O
3.6 g
= 0.2 mol
18 g mol–1
Number of moles of chlorine molecule = 2 × 0.2 mol = 0.4 mol
Mass of Cl2 = 0.4 mol × 71 g mol–1
= 28.4 g
Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium.
[Relative atomic mass: C = 12, Mg = 24]
4g
1
=
mol
24 g mol–1
6
Number of moles of carbon = Number of moles of magnesium
1
=
mol
6
1
Mass of carbon =
mol × 12 g mol–1 = 2 g
6
Number of moles of magnesium =
6
Compare the number of molecules in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your
answer. [Relative atomic mass: S = 32, O = 16, N = 14]
32 g
Number of moles of molecules in 32 g SO2 =
= 0.5 mol
64 g mol–1
7g
Number of moles of molecules in 7 g N2 =
= 0.25 mol
28 g mol–1
Number of molecules in 32 g SO2 is two times more than 7 g N2.
Number of mole in sulphur dioxide molecules is two times more than number of mole of nitrogen molecules.
7
Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer.
[Relative atomic mass: O = 16, Zn = 65]
1.28 g
Number of moles of O atoms in 1.28 g O2 =
= 0.08 mol
16 g mol–1
1.30 g
Number of moles of Zn atoms in 1.3 g Zn =
= 0.02 mol
65 g mol–1
Number of oxygen atoms in 1.28 g oxygen is 4 times more than number of zinc atoms in 1.3 g zinc.
Number of mol of oxygen atom is 4 times more than zinc atom.
TP4
TP4
33
UNIT
5
TP3
3
Number of moles of H2O =
MODULE • Chemistry FORM 4
3.3
CHEMICAL FORMULAE
Define chemical formula.
SP 3.3.1
SK
3.3
A representative of a substance using letters to show the type of atoms and
subscripts numbers to show the number of atoms in that substance.
Example:
Substance
Water
Ammonia
Propane
What information can be
obtained from the chemical
formula?
Example:
Substance
UNIT
3
Ammonia
Define empirical formula.
SP 3.3.1
Define molecular formula.
SP 3.3.1
What is the relationship
between molecular formula
and empirical formula?
Chemical formula
H2O
NH3
C3H8
Chemical
formula
Information
NH3
(i) Elements present in the substance
Ammonia is made up of nitrogen and hydrogen
(ii) Number of atoms of each element in the
compound
Ammonia molecule consists of 1 nitrogen
atom and 3 hyrogen atoms
(iii) Relative formula mass
Relative molecular mass = 14 + (3 × 1) = 17
A formula that shows the simplest whole number ratio of atoms of each
element in a compound.
Molecular formula of a compound is a formula that shows the actual number
of atoms of each element that are present in a molecule of the compound.
Molecular formula = (Empirical formula)n, where n is an integer.
Determine the empirical formula and the value of n.
Molecular
formula
Empirical formula
Value of n
Water
H2O
H2O
1
Carbon
dioxide
CO2
CO2
1
H2SO4
H2SO4
1
Ethene
C2H4
CH2
2
Benzene
C6H6
CH
6
Glucose
C6H12O6
CH2O
6
Compound
Sulphuric
acid
Remark:
The molecular formula and the empirical formula of a compound will be the same if the value of
n = 1 but different if the value of n > 1.
© Nilam Publication Sdn. Bhd.
34
MODULE • Chemistry FORM 4
Example
1
The empirical formula for chlorinated hydrocarbon is CHCl2. The relative
formula mass of this compound is 168. Find the molecular formula of this
compound.
(CHCl2)n = 168
Molecular formula
(12 + 1 + [2 × 35.5])n
= 168 = (Empirical formula)n
(84)n = 168
= (CHCl2)2
n=2
= C2H2Cl4
Experiments to determine empirical formula of metal oxide:
Empirical formula of copper(II) oxide
Set-up of apparatus:
Copper(II) oxide powder
Magnesium
Hydrogen gas
Heat
Heat
2
Reaction occurs:
Magnesium is burnt in a crucible to react with
oxygen to form magnesium oxide.
Reaction occurs:
Hydrogen gas is passed through the heated
copper(II) oxide. Hydrogen reduces copper(II)
oxide to form copper and water.
Balanced equation:
2Mg + O2 → 2MgO
Balanced equation:
CuO + H2 → Cu + H2O
This method can also be used to determine the
empirical formulae of reactive metal oxide such as
aluminium oxide and zinc oxide.
This method can also be used to determine the
empirical formulae of less reactive metal oxide such
as lead oxide and tin oxide.
Experiment to Determine Empirical Formula of Magnesium Oxide SP 3.3.2
In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide:
Magnesium + Oxygen → Magnesium oxide
Material:
Magnesium ribbon, sand paper
Apparatus: Crucible with lid, tongs, Bunsen burner, pipe-clay triangle, balance and tripod stand
Set-up of apparatus:
Magnesium ribbon
Heat
35
UNIT
3
Empirical formula of magnesium oxide
Set-up of apparatus:
MODULE • Chemistry FORM 4
Procedure:
(a) A crucible and its lid are weighed and the mass recorded.
(b) 10 cm of magnesium ribbon is cleaned with sand paper.
(c) The magnesium ribbon is coiled loosely and placed in the crucible.
(d) The crucible together with the lid and magnesium ribbon are weighed again.
(e) The apparatus is set up as shown in the diagram.
(f) The crucible is heated strongly without its lid. When the magnesium starts to burn, the crucible is covered
with its lid.
(g) The lid of the crucible is lifted from time to time using a pair of tongs.
(h) When the magnesium ribbon stops burning, the lid is removed and the crucible is heated strongly for
another 2 minutes.
(i) The crucible, the lid and its content are allowed to cool down to room temperature.
(j) The crucible, lid and its content are weighed again and the mass recorded.
(k) The process of heating, cooling and weighing are repeated until a constant mass is obtained.
Precaution steps:
Step taken
Purpose
UNIT
Magnesium ribbon is cleaned with sand paper .
3
The
crucible lid
To remove the oxide layer on the surface of the
magnesium ribbon.
is lifted from time to time.
is then replaced
The crucible lid
quickly.
heating ,
cooling
The process of
and
weighing is repeated until a constant mass
is obtained.
oxygen
To allow
magnesium .
from the air to react with
To prevent fumes of
escaping.
magnesium oxide
from
To ensure magnesium reacts completely with
oxygen
to form magnesium oxide .
Observation:
Magnesium burns
brightly
to release
white fumes
and
white solid
is formed.
Inference:
Magnesium is a
reactive
Magnesium reacts with
metal.
oxygen
in the air to form
magnesium oxide
.
Result:
Description
Mass (g)
x
y
z
Mass of crucible + lid
Mass of crucible + lid + magnesium
Mass of crucible + lid + magnesium oxide
Calculation:
Element
Mg
O
Mass (g)
y–x
z–y
Number of mole of atoms
y–x
24
z–y
16
p
q
Simplest ratio of moles
Empirical formula of magnesium oxide is
© Nilam Publication Sdn. Bhd.
MgpOq
36
.
MODULE • Chemistry FORM 4
Experiment to Determine Empirical Formula of Copper(II) oxide
SP 3.3.3
Copper(II) oxide + Hydrogen → Copper + Water
Set-up of apparatus:
Copper(II) oxide powder
Glass tube
Heat
Hydrochloric acid (1.0 mol dm–3)
Water
Zinc granules
Precaution steps:
Purpose
Hydrogen gas is passed through the glass To
remove
air
all the
tube for 10 seconds.
(The mixture of hydrogen gas and
explosion when lighted).
The flow of hydrogen gas must be
continuous throughout the experiment.
heating
The process of
and weighing
constant
,
cooling
are repeated until a
in the glass tube.
air
will cause
copper
To prevent hot
from reacting with
oxygen
copper(II)
oxide again.
to form
copper(II) oxide
To ensure all
copper .
has changed to
mass is obtained.
Observation:
The
black
brown
colour of copper(II) oxide turns
.
Inference:
Copper(II) oxide reacts with hydrogen to produce the brown
copper metal
.
Result:
Description
Mass (g)
Mass of glass tube
x
Mass of glass tube + copper(II) oxide
y
Mass of glass tube + copper
z
Calculation:
Element
Cu
O
Mass (g)
z–x
y–z
Number of mole of atoms
z–x
64
y–z
16
p
q
Simplest ratio of moles
Empirical formula of copper(II) oxide is
CupOq
37
.
3
Step taken
UNIT
3
MODULE • Chemistry FORM 4
4
Explain why the set-up of apparatus to determine the empirical formula in both experiments are different.
(a) Magnesium is reactive metal (above hydrogen in reactivity series). Magnesium oxidised easily
to form
magnesium oxide
hydrogen
(b) Copper is below
removed by
5
.
in the metal reactivity series. Oxygen in copper(II) oxide can be
hydrogen gas
to form copper and water.
To calculate the empirical formula of a compound, the following table can be used as a guide:
Element
Calculation steps:
(a) Calculate the mass of each element in the compound.
(b) Convert the mass of each element to number of mole
of atom.
(c) Calculate the simplest ratio of moles of atom of the
elements.
Mass of element (g)
Number of mole of atom
Simplest ratio of moles
UNIT
Exercise
3
1
TP3
SP 3.3.4
When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical
formula of metal X oxide.
[Relative atomic mass: X = 207, O = 16]
Element
X
O
10.35
1.6
10.35
= 0.05
207
1.6
= 0.1
16
Ratio of moles
1
2
Simplest ratio of moles
1
2
Mass of element (g)
Number of moles of atoms
XO2
Empirical formula:
2
TP3
.
A certain compound contains the following composition:
Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass: O = 16, Na = 23, Br = 80]
(Assume that 100 g of substance is used)
Element
Na
Br
O
Mass of element (g)
15.23
52.98
31.79
Number of moles of atoms
0.66
0.66
1.99
Ratio of moles
1
1
3.01
Simplest ratio of moles
1
1
3
Empirical formula:
© Nilam Publication Sdn. Bhd.
NaBrO3
.
38
MODULE • Chemistry FORM 4
2.08 g of element X combines with 4.26 g of element Y to form a compound with formula XY3. Calculate the
relative atomic mass of element X. [RAM: Y = 35.5]
Element
Y
Mass of element (g)
2.08
4.26
Number of moles of atom
2.08
x
4.26 = 0.12
35.5
1
3
Simplest ratio of moles
2.08
1
x
=
3
0.12
x = 52
2.07 g of element Z reacts with bromine to form 3.67 g of a compound with empirical formula ZBr2. Calculate
the relative atomic mass of element Z. [RAM: Br = 80]
Element
Mass of element (g)
Number of moles of atoms
Simplest ratio of moles
z = relative atomic mass of Z
1
Mol Z
=
2
Mol Br
Z
2.07
2.07
z
1
Br
3.67 – 2.07 = 1.6
1.6 = 0.02
80
2
2.07
1
z
=
2
0.02
z = 207
5
The empirical formula of compound X is CH2 and relative molecular mass is 56. Determine the molecular
formula of compound X.
[Relative atomic mass: H = 1; C = 12]
(12 + 2)n = 56
TP3
56
=4
14
Molecular formula = (CH2)4 = C4H8
6
TP3
n=
2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86.
[Relative atomic mass: H = 1; C = 12]
(i) Calculate the empirical formula of the hydrocarbon.
Element
C
H
Mass of element (g)
2.16
2.58 – 2.16 = 0.42
Number of moles of atoms
0.18
0.42
Ratio of moles
1
Simplest ratio of moles
3
Empirical formula = C3H7
39
2
1
7
=
3
3
7
3
4
TP3
x = relative atomic mass of X
1
Mol X
=
3
Mol Y
X
UNIT
3
TP3
MODULE • Chemistry FORM 4
(ii) Determine the molecular formula of the hydrocarbon.
(3 × 12 + 7 × 1)n = 86
86
=2
43
Molecular formula = (C3H7)2 = C6H14
n=
Percentage Composition by Mass of an Element in a Compound
SP 3.3.5
UNIT
% composition by mass of an element
Total RAM of the element in the compound
=
× 100%
RMM/RFM of compound
Example
Calculate the percentage composition by mass of nitrogen in the following
compounds:
[Relative atomic mass: N = 14, H = 1, O = 16, S = 32, K = 39]
(i) (NH4)2SO4
(ii) KNO3
14
%N = 2 × 14 × 100%
%N =
× 100%
101
132
3
Formula
= 21.2%
= 13.9%
Chemical Formula for Ionic Compounds
1
Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is obtained by exchanging the
charges on each ion. The formula obtained will be XnYm. SP 3.3.5
2
Example:
(i) Sodium oxide
Ion
Na+
O2–
Charges
+1
–2
Exchange of charges
2
1
Simplest ratio
2
1
2 Na+
O2–
Number of combining ions
Formula
Na2O
(ii) Copper(II) nitrate
Cu2+
+2
1
(iii) Zinc oxide
NO3–
–1
2
(Simplest ratio)
➾ Cu(NO3)2
Zn2+
+2
O2–
–2
2
2
1
➾ ZnO
© Nilam Publication Sdn. Bhd.
40
1
(Simplest ratio)
41
CaO
Calcium oxide
CuO
Copper(II)
oxide
MgO
Magnesium
oxide
ZnO
Zinc oxide
PbO
Lead(II) oxide
Al2O3
Aluminium
oxide
Ca2+
Calcium ion
Cu2+
Copper(II)
ion
Mg2+
Magnesium
ion
Zn2+
Zinc ion
Pb2+
Lead(II) ion
Al 3+
Aluminium
ion
NH4 +
Ammonium
ion
Ag+
Silver ion
Ag2O
Silver oxide
Na2O
Sodium oxide
Na+
Sodium ion
H+
Hydrogen
ion
K 2O
Potassium
oxide
K+
Potassium
ion
O2–,
Oxide ion
ZnSO4
Zinc sulphate
(NH4)2SO4
Ammonium
sulphate
CaSO4
Calcium
sulphate
CuSO4
Copper(II)
sulphate
MgSO4
Magnesium
sulphate
Ag2SO4
Silver sulphate
PbSO4
PbCO3
Lead(II)
Lead(II) carbonate
sulphate
Al2(CO3)3
Al2(SO4)3
Aluminium
Aluminium
carbonate
sulphate
ZnCO3
Zinc carbonate
(NH4)2CO3
Ammonium
carbonate
CaCO3
Calcium
carbonate
CuCO3
Copper(II)
carbonate
MgCO3
Magnesium
carbonate
Ag2CO3
Silver carbonate
K2SO4
Potassium
sulphate
SO42–,
Sulphate ion
Cl–,
Chloride ion
KI
Potassium
iodide
I –,
Iodide ion
PbBr2
Lead(II)
bromide
AlBr3
Aluminium
bromide
PbCl2
Lead(II)
chloride
AlCl3
Aluminium
chloride
UNIT
ZnBr2
Zinc bromide
NH4Br
Ammonium
bromide
CaBr2
Calcium
bromide
CuBr2
Copper(II)
bromide
MgBr2
Magnesium
bromide
AgBr
Silver bromide
HBr
Hydrobromic
acid
3
TP1
AgOH
Silver hydroxide
KOH
Potassium
hydroxide
NaOH
Sodium
hydroxide
OH–,
Hydroxide ion
SP 3.3.5
Zn(OH)2
Zinc hydroxide
Pb(OH)2
PbI2
Lead(II)
Lead(II) iodide
hydroxide
AlI3
Al(OH)3
Aluminium
Aluminium
iodide
hydroxide
ZnI2
Zinc iodide
Ca(OH)2
CaI2
Calcium
Calcium iodide
hydroxide
CuI2
Cu(OH)2
Copper(II)
Copper(II)
iodide
hydroxide
MgI2
Mg(OH)2
Magnesium
Magnesium
iodide
hydroxide
NH4I
Ammonium
iodide
AgI
Silver iodide
HI
Hydroiodic
acid
NaBr
NaI
Sodium bromide Sodium iodide
KBr
Potassium
bromide
Br–,
Bromide ion
ZnCl2
Zinc chloride
NH4Cl
Ammonium
chloride
CaCl2
Calcium
chloride
CuCl2
Copper(II)
chloride
MgCl2
Magnesium
chloride
AgCl
Silver chloride
KCl
Potassium
chloride
NaCl
Na2CO3
Na2SO4
Sodium
Sodium carbonate Sodium sulphate
chloride
HCl
H2CO3
H2SO4
Hydrocloric
Carbonic acid
Sulphuric acid
acid
K2CO3
Potassium
carbonate
CO32–,
Carbonate ion
ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS
Pb(NO3)2
Lead(II)
nitrate
Al(NO3)3
Aluminium
nirate
Zn(NO3)2
Zinc nitrate
NH4NO3
Ammonium
nitrate
Ca(NO3)2
Calcium
nitrate
Cu(NO3)2
Copper(II)
nitrate
Mg(NO3)2
Magnesium
nitrate
AgNO3
Silver nitrate
HNO3
Nitric acid
NaNO3
Sodium nitrate
KNO3
Potassium
nitrate
NO3–,
Nitrate ion
MODULE • Chemistry FORM 4
UNIT
3
© Nilam Publication Sdn. Bhd.
42
MgCO3
ZnCO3
PbCO3
CuO
MgO
ZnO
PbO
Al2O3
Copper(II) ion
Magnesium ion
Zinc ion
Lead(II) ion
Aluminium ion
Al2(CO3)3
CuCO3
CaO
CaCO3
(NH4)2CO3
Calcium ion
Ammonium ion
Ag2CO3
Ag2O
Silver ion
Na2CO3
H2CO3
Na2O
Sodium ion
K2CO3
Carbonate ion
Hydrogen ion
K2O
Potassium ion
Oxide ion
Al2(SO4)3
PbSO4
ZnSO4
MgSO4
CuSO4
CaSO4
(NH4)2SO4
Ag2SO4
H2SO4
Na2SO4
K2SO4
Sulphate ion
AlCl3
PbCl2
ZnCl2
MgCl2
CuCl2
CaCl2
NH4Cl
AgCl
HCl
NaCl
KCl
Chloride ion
AlBr3
PbBr2
ZnBr2
MgBr2
CuBr2
CaBr2
NH4 Br
AgBr
HBr
NaBr
KBr
Bromide ion
AlI3
PbI2
ZnI2
MgI2
CuI2
CaI2
NH4 I
AgI
HI
NaI
KI
Iodide ion
Al(OH)3
Pb(OH)2
Zn(OH)2
Mg(OH)2
Cu(OH)2
Ca(OH)2
AgOH
NaOH
KOH
Hydroxide ion
Al(NO3)3
Pb(NO3)2
Zn(NO3)2
Mg(NO3)2
Cu(NO3)2
Ca(NO3)2
NH4 NO3
AgNO3
HNO3
NaNO3
KNO3
Nitrate ion
ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULAE OF THE FOLLOWING
COMPOUNDS SP 3.3.5 TP1
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
ACTIVITY 3: WRITE THE CHEMICAL FORMULAE AND TYPE OF PARTICLES FOR THE
FOLLOWING ELEMENT/COMPOUND SP 3.3.5 TP2
Formula
Type of particles
ZnCO3
Ion
(NH4)2CO3
Ion
Silver chloride
AgCl
Ion
Ion
Sulphuric acid
H2SO4
Ion
K2O
Ion
Copper(II) nitrate
Cu(NO3)2
Ion
Magnesium oxide
MgO
Ion
Hydrogen gas
H2
Molecule
Lead(II) carbonate
PbCO3
Ion
Carbon dioxide
gas
CO2
Molecule
3
Compound /
Element
Iron(III) sulphate
Fe2(SO4)3
Ion
Oxygen gas
O2
Molecule
UNIT
Compound /
Element
Formula
Type of particles
Na2SO4
Ion
Zinc carbonate
Ammonium
carbonate
(NH4)2CO3
Ion
Ammonium
carbonate
Magnesium nitrate
Mg(NO3)2
Ion
Hydrochloric acid
HCl
Potassium oxide
Magnesium
chloride
MgCl2
Ion
Aluminium
sulphate
Al2(SO4)3
Ion
Zinc sulphate
ZnSO4
Ion
Lead(II) chloride
PbCl2
Ion
Silver nitrate
AgNO3
Ion
Potassium iodide
KI
Ion
Ammonium
sulphate
(NH4)2SO4
Ion
Copper(II)
carbonate
CuCO3
Ion
Zinc oxide
ZnO
Ion
Potasium
carbonate
K2CO3
Ion
Nitric acid
HNO3
Ion
Sodium hydroxide
NaOH
Ion
Ammonia gas
NH3
Molecule
Aqueous ammonia
NH3(aq)
Ion and
Molecule
Magnesium
Mg
Atom
Ammonium
chloride
NH4Cl
Ion
Zinc
Zn
Atom
Nitrogen dioxide
gas
NO2
Molecule
CuSO4
Ion
Sodium chloride
NaCl
Ion
I2
Molecule
Silver
Ag
Atom
Cl2
Molecule
Bromine
Br2
Molecule
Sodium sulphate
Copper(II)
sulphate
Iodine
Chlorine
43
MODULE • Chemistry FORM 4
3.4
1
SK
3.4
CHEMICAL EQUATIONS
A chemical equation summerises what happen during chemical reaction.
Reactants
(Substances take part in the reaction)
Products
(Substances that are produced)
Produce
Example:
Reaction between zinc powder and hydrochloric acid produce zinc chloride aqueous and hydrogen gas.
Identify the reactant and products.
Reactants: Zinc and hydrochloric acid
Products: Zinc chloride and hydrogen gas
Write the word equation.
Zinc + hyrochloric acid ➝ Zinc chloride + hydrogen gas
SP 3.4.1
UNIT
Write the chemical formula of the reactants
and products. List down the number of atoms
of each element on both sides of the equation.
Zn
+ HCl
Left
1
1
1
➝
ZnCl2 + H2
Right
1 (Balanced)
2 (Not balanced)
2 (Not balanced)
Zn
+ 2HCl
Left
1
2
2
➝
ZnCl2 + H2
Right
1 (Balanced)
2 (Balanced)
2 (Balanced)
3
Zn atom
H atom
Cl atom
Balance the number of atoms of each element
on both sides of the equation by adjusting the
coefficients in front of the formulae.
Zn atom
H atom
Cl atom
SP 3.4.1
Put the state symbols for every reactant and
products:
Remark:
State symbols
Zn (s) + 2HCl (aq) ➝ ZnCl2 (aq) + H2 (g)
(s)
(I)
(g)
(aq)
Solid
Liquid
Gas
Aqueous
Qualitative interpretation of the chemical
equation (the reactants and the products)
SP 3.4.2
Quantitative interpretation of the chemical
equation (the coefficient of each formula
shows the number of moles of reactants react
and products formed)
SP 3.4.2
© Nilam Publication Sdn. Bhd.
(i) The reactants are solid zinc and hydrochloric acid.
(ii) The products are zinc chloride solution and hydrogen
gas.
Coefficient
Zn + 2HCl ➝ ZnCl2 + H2
1 2
1 1
1 mol of zinc reacts with 2 mol of hydrochloric acid to
produce 1 mol of zinc chloride and 1 mol of hydrogen.
44
MODULE • Chemistry FORM 4
Exercise
Write a balanced chemical equation for each of the following reactions:
TP3
1
Copper(II) carbonate
CuCO3 CuO + CO2
2
Ammonia + Hydrogen chloride
NH3 HCl + NH4Cl
3
Lead(II) nitrate + Potassium iodide
Pb(NO3)2 + 2KI PbI2 + 2KNO3
Lead(II) iodide + Potassium nitrate
4
Sulphuric acid + Sodium hydroxide
H2SO4 + 2NaOH Na2SO4 + 2H2O
Sodium sulphate + Water
5
Copper(II) oxide + Hydrochloric acid
CuO + 2HCl CuCl2 + H2O
6
Sodium + Water Sodium hydroxide + Hydrogen
2Na + 2H2O 2NaOH + H2
7
Potassium oxide + Water
K2O + H2O 2KOH
8
Zinc oxide + Nitric Acid Zinc nitrate + Water
ZnO + 2HNO3 Zn(NO3)2 + H2O
9
Lead(II) nitrate Lead(II) oxide + Nitrogen dioxide + Oxygen
2Pb(NO3)2 2PbO + 4NO2 + O2
Copper(II) oxide + Carbon dioxide
Ammonium chloride
UNIT
3
Copper(II) chloride + Water
Potassium hydroxide
10 Aluminium nitrate Aluminium oxide + Nitrogen dioxide + Oxygen
4Al(NO3)3 2Al2O3 + 12NO2 + 3O2
Numerical Problems involving Chemical Equations
SP 3.4.3
Example:
The equation shows the reaction between zinc and hydrochloric acid.
Zn + 2HCl
ZnCl2 + H2
Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at
room conditions. [Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions]
Write a balanced equation
Write the information from the question above the equation
Convert the given quantitity into moles by using the method
shown in the chart below.
Use the mole ratio of the substances involved to find the
number of moles of the other substance.
Remark:
The coefficient of each formula shows the number of moles of reactants react
and products formed.
45
?g
excess
6 dm3
Zn(s) + 2HCl ➝ ZnCl2 + H2
6 dm3
Number of moles of H2 = 24 dm3 mol–1
= 0.25 mol
From the equation,
1 mol H2 : 1 mol Zn
0.25 mol H2 : 0.25 mol Zn
MODULE • Chemistry FORM 4
Convert the mole into the quantity required in the question
by using the method shown in the chart below.
Mass
(g)
÷ (RAM/RFM/RMM) g mol–1
× (RAM/RFM/RMM) g mol–1
Mass of Zn
= 0.25 mol × 65 g mol–1
= 16.25 g
÷ 24 dm3 mol–1
22.4 dm3 mol–1
No. of moles
(n)
× 24 dm3 mol–1
22.4 dm3 mol–1
Volume of gas
(dm3)
SPM PRACTICE
Subjective Questions
1
UNIT
TP3
The equation shows the reaction between potassium and oxygen.
4K + O2
2K2O
3
Calculate the mass of potassium required to produce 23.5 g of potassium oxide.
[Relative atomic mass: K = 39, O = 16]
Number of moles of K2O =
23.5 g
23.5
=
= 0.25 mol
–1
(2 × 39 + 16) g mol
94
From the equation,
2 mol K2O : 4 mol K
0.25 mol K2O : 0.5 mol K
2
TP3
Mass of K
= 0.5 mol × 39 g mol–1
= 19.5 g
8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II)
nitrate produced. [Relative atomic mass: N = 14, O = 16, Cu = 64]
CuO + 2HNO3 Cu(NO3)2 + H2O
Number of moles of CuO
8g
=
(64 + 16) g mol–1
= 0.1 mol
From the equation,
1 mol CuO : 1 mol Cu(NO3)2
0.1 CuO : 0.1 mol Cu(NO3)2
© Nilam Publication Sdn. Bhd.
Mass of Cu(NO3)22
= 0.1 mol × 188 g mol–1
= 18.8 g
46
MODULE • Chemistry FORM 4
Zn + H2SO4
ZnSO4 + H2
Number of moles of Zn
1.3 g
=
= 0.02 mol
65 g mol–1
From the equation,
1 mol Zn : 1 mol H2
0.02 mol Zn : 0.02 mol H2
4
TP3
Volume of H2
= 0.02 mol × 22.4 dm3 mol–1
= 0.448 dm3
= 448 cm3
0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate
the volume of chlorine gas that has reacted.
[Relative atomic mass: Na = 23, Molar volume of gas = 24 dm3 mol–1 at room conditions]
2Na + Cl2
2NaCl
Number of moles of Na
0.46 g
=
= 0.02 mol
23 g mol–1
From the equation,
2 mol Na : 1 mol Cl2
0.02 mol Na : 0.01 mol Cl2
5
TP3
Volume of Cl2
= 0.01 mol × 24 dm3 mol–1
= 0.24 dm3
= 240 cm3
The equation shows the combustion of propane gas.
C3H8 + 5O2
3CO2 + 4H2O
720 cm3 of propane gas (C3H8) at room conditions burns in excess oxygen. Calculate the mass of carbon
dioxide formed.
[Relative atomic mass: C = 12, O = 16, Molar volume of gas = 24 dm3 mol–1 at room conditions]
C3H8 + 5O2 3CO2 + 4H2O
Number of moles of C3H8
=
720 cm3
= 0.03 mol
24 000 cm3 mol–1
Mass of CO2
= 0.09 mol × 44 g mol–1
= 3.96 g
From the equation,
1 mol C3H8 : 3 mol CO2
0.03 mol C3H8 : 0.09 mol CO2
47
3
HOTS
1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate
the volume of hidrogen gas released at STP.
[Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1 at STP]
UNIT
3
TP3
MODULE • Chemistry FORM 4
6
TP3
The diagram below shows a car fitted with air bag which will inflate in an accident.
HOTS
The air bag contains solid sodium azide, NaN3 which will decompose rapidly to form sodium and nitrogen
gas. The nitrogen gas formed fills the air bag. [Relative atomic mass: N = 14; H = 1; Na = 23 and 1 mol of gas
occupies the volume of 24 dm3 at room temperature and pressure]
(a) Construct equation for the decomposition of sodium azide.
2NaN3 → 2Na + 3N2
(b) In an accident, an air bag fills with 72 dm3 of nitrogen at room temperature and pressure. What is the mass
of sodium azide needed to provide 72 dm3 of nitrogen?
Number of moles of nitrogen =
UNIT
72 dm3
= 3 mol
24 dm3 mol–1
3
Number of moles of NaN3 = 2 mol
Mass of NaN3 = 2 mol × [23 + 3(14)] g mol–1 = 130 g
(c) Sodium azide, NaN3, reacts with dilute hydrochloric acid to produce sodium chloride and compound A.
Compound A contains 2.33% of hydrogen and 97.7% of nitrogen by mass.
(i) What is the empirical formula for compound A?
Element
H
N
Mass (g)
2.33
2.33
–—–
1 = 2.33
2.33
–—–
2.33 = 1
97.7
97.7
–—–
14 = 6.98
6.98
–—–
2.33 ≈ 3
Number of mole
Simplest ratio
Empirical formula: HN3
(ii) Construct the equation for the reaction between sodium azide and dilute hydrochloric acid.
NaN3 + HCl → NaCl + HN3
Objective
Questions
© Nilam Publication Sdn. Bhd.
48
MODULE • Chemistry FORM 4
UNIT
4
THE PERIODIC TABLE OF ELEMENTS
Concept Map
Historical Development – Contribution of Scientists
Elements are arranged in order
of increasing proton number
PERIODIC TABLE
Electron arrangement
in an atom
Group 18
(Noble
gases)
Atoms of
elements have
one valence
electron
Group 1
(Alkali
metal)
(a) Monoatomic
and inert
(b) Uses in daily
life
(a) Similar chemical
properties. (React
with H2O, O2 &
Cl2)
(b) Reactivity
increases down
the group.
Atoms of
elements have
seven valence
electrons
(a) Similar chemical
properties. (React
with H2O, NaOH
& Fe)
(b) Reactivity
decreases down
the group.
Atoms of
elements have
three shells
occupied with
electrons
Period 3
Group 17
(Halogen)
Physical
properties &
changes in
physical
properties down
the group
Period
Across Period 3 from
left to right:
(a) Change in atomic
size
(b) Change in
electronegativity
(c) Change in metallic
properties (metal ➝
semi metal ➝ non
metal)
(d) Change in oxide
properties (Basic
oxide ➝ amphoteric
oxide ➝ acidic
oxide)
49
Transition
element
4
Group
Number of shells occupied
with electrons in an atom
Located
between
Group 2 to
Group 13
(a) Metallic properties
(shiny, conducts
electricity, malleable,
high tensile strength,
high melting point &
density)
(b) Special
characteristics:
(i) Most elements
form coloured
compound.
(ii) Most elements
have more than
one oxidation
number.
(iii) Many transition
elements can
form complex
ion
(iv) Many elements
can act as a
catalyst
UNIT
Number of
valence electrons
in an atom
Atoms have
achieved stable
duplet/octet
electron
arrangement
Studied in Form 1,
Unit 6: Periodic Table
MODULE • Chemistry FORM 4
4.1
THE DEVELOPMENT OF THE PERIODIC TABLE OF ELEMENTS
Advantages of Classifying the Elements in the Periodic Table
SK
4.1
SP 4.1.2
What is the Periodic Table?
It is an arrangement of the elements in the orders of increasing proton number.
What is the advantages of
arranging elements in the
Periodic Table?
Periodic Table enables:
(a) Chemists to study, understand and remember the chemical and physical
properties of all the elements and compounds in an orderly manner.
(b) Properties of elements and their compounds to be predicted based on the
position of elements in the Periodic Table.
(c) Relationship between elements from different groups to be known.
Contribution of Scientist to the Historical Development of the Periodic Table
Scientists
SP 4.1.1
Discoveries
UNIT
Substances were classified into 4 groups with similar chemical properties.
J.W Dobereiner
Substances were arranged into groups of 3 elements with similar chemical properties.
Groups of element with similar chemical properties were called Triads.
Triad system was confined to some elements only.
John Newlands
Elements were arranged in ascending atomic mass.
Law of Octaves because similar chemical properties were repeated at every eighth element.
This system was inaccurate because there were some elements with wrong mass numbers.
Lothar Meyer
(g)
The atomic volume = Mass of 1 mol –3
Density (g cm )
4
Antoine Lavoisier
Plotted graph for the atomic volume against atomic mass.
Found that elements with similar chemical properties were positioned at equivalent
places along the curve.
Mendeleev
Elements were arranged in ascending order of increasing atomic mass.
Elements with similar chemical properties were in the same group.
Empty spaces were allocated for elements yet to be discovered.
Contributor to the formation of the modern Periodic table.
Henry Moseley
Classified element based on concepts of proton number and arranged elements in order of
increasing proton number.
Contributor to the formation of the modern Periodic Table.
© Nilam Publication Sdn. Bhd.
50
MODULE • Chemistry FORM 4
4.2
SK
4.2
THE ARRANGEMENT IN THE PERIODIC TABLE OF ELEMENTS
Write the electron arrangement for atom of each element in the Periodic Table below.
Nucleon number
Proton number
A
Z
X
SP 4.2.1
Symbol of an element
GROUP
1
1
P
E
R
I
O
D
2
3
18
4
2
H*
1
7
3
Li
2
8
4
Na
2.2
24
12
2.8.1
39
19
K
11
5
Be
2.1
23
11
13
Mg
2.8.2
40
20
14
12
6
B
27
13
2.4
28
14
Al
3 4 5 6 7 8 9 10 11 12 2.8.3
14
7
C
2.3
TRANSITION METALS
15
16
16
8
N
O
2.5
31
15
Si
2.8.4
32
16
2.8.5
S
19
9
35
17
Ne
2.8
40
18
Cl
2.8.7
80
35
2.8.8.1 2.8.8.2
20
10
F
2.7
2.8.6
Ca
2
17
2.6
P
He
Ar
2.8.8
Br
What is the basic principle
of arranging the element in
the Periodic Table?
Elements in the Periodic Table are arranged horizontally in increasing order of
proton number . SP 4.2.2
State two main components
of the Periodic Table:
(a) Group
(b) Period
Group
What is a Group?
SP 4.2.2
How is the group number
related to the number of
valence electrons?
They
vertical
column of elements in the Periodic Table arranged
according to the number of valence electron in they outermost shell
of atoms is called groups.
There are 18 vertical columns, called Group 1, Group 2, and Group 3 until
Group 18.
Number of
valence electrons
1
2
3
4
5
6
7
8
(except Helium)
Group
1
2
13
14
15
16
17
18
For atoms of elements with 3 to 8 valence
electrons, the group number is:
10 + number of valence electrons.
Certain groups have a
special name. What are the
names for these certain
groups?
Group
1
2
3 – 12
17
18
Special name
Alkali metals
Alkali-earth metals
Transition elements
Halogens
Noble gases
51
UNIT
4
4
1
1
MODULE • Chemistry FORM 4
How is the group number
related to types of
substance?
Group 1, 2 and 13
Group 3 – 12
Group 14 – 18
Metals
Metals (Transition element)
Non-metal
Period
What is a Period?
The
SP 4.2.2
horizontal
same number of
called period.
How is the period number
related to the number of
shells?
row of elements in the Periodic Table consists of the
shells
atom
occupied with electrons in an
Number of shells
Period
(a) Period 1 has
1
1
2
2
2
elements
(b) Period 2 and 3 have 8 elements #
UNIT
(c) Period 4 and 5 have 18 elements
(d) Period 6 has 32 elements
(e) Period 7 has 23 elements
3
3
4
4
5
5
6
6
7
7
Short periods, # Period 3
will be studied in detail
with respect to physical and
chemical properties
Long periods
4
Exercise
1
TP2
Complete the table below.
Element
Proton
number
Electron
arrangement
Number of valence
electrons
Group
Number of shell
Period
H
1
1
1
1
1
1
He
2
2
2
18
1
1
Li
3
2.1
1
1
2
2
Be
4
2.2
2
2
2
2
B
5
2.3
3
13
2
2
C
6
2.4
4
14
2
2
N
7
2.5
5
15
2
2
O
8
2.6
6
16
2
2
F
9
2.7
7
17
2
2
Ne
10
2.8
8
18
2
2
Na
11
2.8.1
1
1
3
3
Mg
12
2.8.2
2
2
3
3
Al
13
2.8.3
3
13
3
3
© Nilam Publication Sdn. Bhd.
52
MODULE • Chemistry FORM 4
The diagram below shows the chemical symbols which represent elements X, Y and Z.
23
11 X
12
6
Y
39
19
Z
(a) Explain how to
determine the position
of element X in the
Periodic Table.
11
The proton number of element X is
and the number of protons in
atom
11
X is
. The number of electrons in atom X is
11
atom
2.8.1
. The electron arrangement of
X is
.
(b) (i) State the position of
element Y in the
Periodic Table.
(ii) Explain how to
determine the
position of element
Y in the Periodic
Table.
(i) Element Y is located in Group 14 and Period 2.
1
atom
Element X is located in Group
because
X has
one valence electron . Element X is in Period
3
because
atom
three
X has
shells occupied with electrons .
(ii) – The proton number of element Y is 6 and the number of proton in atom
Y is 6.
– The electron arrangement of atom Y is 2.4.
– Element Y is located in Group 14 because atom Y has 4 valence
– Element Y is in Period 2 because atom Y has 2 shells occupied/filled
with electrons.
(c) Which of the above
elements show the
similar chemical
properties? Explain
your answer.
4.3
– Element X and element Z.
– Electron arrangement of atom X is 2.8.1 and electron arrangement of
atom Z is 2.8.8.1.
– Atoms X and Z have the same number of valence electron.
ELEMENTS IN GROUP 18
State the special name for
Group 18 elements.
List the elements in Group
18 of the Periodic Table and
write the electron
arrangement of the atoms
of elements.
SK
4.3
Noble gases
Elements
Helium (He)
Electron arrangement
2
Neon (Ne)
2.8
Argon (Ar)
2.8.8
Krypton (Kr)
2.8.18.8
Xenon (Xe)
–
Radon (Rn)
–
53
4
electrons.
UNIT
2
TP3
MODULE • Chemistry FORM 4
Group 18 are monoatomic
gases. Explain what is
meant by monoatomic.
These gases exist as single uncombined atoms.
SP 4.3.1
Explain why the noble gases
are monoatomic and
chemically inert.
SP 4.3.1
State the uses of noble
gases.
SP 4.3.3
– The atom has achieved duplet electron arrangement for helium and octet
electron arrangement for others.
– Noble gases do not react with other elements (the atom does not lose, gain or
share electrons).
Noble gases
Uses
Helium
To fill weather balloons and airship
Neon
To fill neon light (for advertisement board)
Argon
To fill electrical bulb
Krypton
To fill photographic flash lamp
Radon
UNIT
State the physical
properties and the changes
going down Group 18.
4
SP 4.3.2
Explain why argon does not
react with hot tungsten
filament in term of electron
arrangement.
© Nilam Publication Sdn. Bhd.
To treat cancer
1 All noble gases are insoluble in water and cannot conduct electricity in all
conditions.
2 Going down Group 18:
shells
(a) The atomic size is increasing because the number of
increases.
(b) The melting point and boiling points are very low because atoms of
weak
noble gases atoms are attracted by
Van der Waals forces,
less energy is required to overcome these forces. However, the melting
and boiling points increase going down the group because atomic size
more
increases, causing the Van der Waal forces to increase and
energy is required to overcome these forces.
(c) The density is low and increases gradually because the mass increases
greatly compared to the volume when going down the group.
– Argon atom has achieved stable
– Argon atom does not need to
electrons with other elements.
54
octet
gain
electron arrangement.
lose
share
,
or
MODULE • Chemistry FORM 4
ELEMENTS IN GROUP 1
List the elements in Group
1 of the Periodic Table and
write the electron
arrangements and number
of shells of the atoms of
elements.
Alkali metals
Elements Symbol
Proton
number
Electron
arrangement
Number of
shells
Lithium
Li
3
2.1
2
Sodium
Na
11
2.8.1
3
Potassium
K
19
2.8.8.1
4
State the physical
properties of Group 1
elements. SP 4.4.1
(a) Grey solid with shiny surface.
(b) Softer and the density is lower compared to other metals.
(c) Lower melting and boiling points compared to other metals.
Explain the changes in
physical properties going
down Group 1 elements.
(a) Atomic size increases because the number of shells increases.
(b) Density increases because mass increases faster than the increase in radius.
(c) Melting and boiling points decrease because when the atomic size increases,
the metal bonds get weaker.
SP 4.4.1
Explain the similarities in
chemical properties of the
Group 1 elements.
SP 4.4.4
(a) All
atoms
of elements in Group 1 have
1
valence
electron and achieve a stable duplet/octet electron arrangement by releasing
one
+1
electron to form
charged ions.
Example:
(i) Lithium atom releases one electron to achieve stable duplet electron
arrangement:
Li
Li+
Electron arrangement: 2.1
Number of protons = 3,
total charge: +3
Number of electrons = 3,
total charge: –3
neutral
Lithium atom is
.
+
e–
Electron arrangement: 2
Number of protons = 3,
total charge: +3
Number of electrons = 2,
total charge: –2
Positively charges lithium
ion, Li+ is formed.
(ii) Sodium atom releases one electron to achieve stable octet electron
arrangement:
Na
Na+
Electron arrangement: 2.8.1
Number of protons = 11,
total charge: +11
Number of electrons = 11,
total charge: –11
neutral .
Sodium atom is
55
+
e–
Electron arrangement: 2.8
Number of protons = 11,
total charge: +11
Number of electrons = 10,
total charge: –10
Positively charges sodium
ion, Na+ is formed.
4
State the special name for
Group 1 elements.
SK
4.4
UNIT
4.4
MODULE • Chemistry FORM 4
(b) All elements in Group 1 have similar chemical properties because all
atoms
in Group 1 have one valence electron and achieve the stable
electron
releasing
duplet/octet
arrangement by
its valence
electron to form a
SP 4.4.3
charged ions.
– Atoms of Group 1 metals achieve a stable duplet/
octet
– The valence electron is loosely held and it is
easier
for the electron to be released.
UNIT
4
Compare and explain the
reactivity of elements X and
Y.
Element
Proton number
X
11
Y
17
How are the Group 1
elements stored? Explain.
down Group 1
valence electron in the outermost shell gets
further away from the nucleus.
– The strength of attraction from the proton in the
nucleus to the valence electron gets weaker .
increases
electron arrangement by releasing
one
valence electron to form +1
charged ion.
shells
– Going down Group 1, the number of
increases, the atomic size increases and the
Reactivity
Explain the increase in the
reactivity of the elements
going down the Group 1.
positively
Na
K
– Element Y is more reactive than element X.
– Electron arrangement of X atom is 2.8.1 and Y atom is 2.8.8.1.
– The number of shells occupied with electrons of atom Y is more than atom
X.
– The size of atom Y is larger than atom X.
– Force of attraction between the nucleus and the valence electron for atom
Y is weaker than atom X.
– Therefore, it is easier for atom Y to release the valance electron compared
to X atom.
The elements are stored under the paraffin oil.
To prevent them from reacting with water vapour and air.
Experiment for the Chemical Properties of Group 1 Elements:
SP 4.4.2
(a) Metal Group 1 reacts with water to produce alkali and hydrogen gas.
2X + 2H2O
2XOH + H2 , X is the metal of Group 1
Lithium
Water
Procedure:
(i) Pour water into a basin until half full.
(ii) Cut a small piece of lithium using a knife and forceps.
(iii) Dry the oil on the surface of the lithium with filter paper.
(iv) Place the lithium slowly onto the water surface in a water trough.
(v) When the reaction stop, test the solution produced with red litmus paper.
(vi) Record the observation.
(vii) Repeat steps (i) – (vi) using sodium and potassium to replace lithium one by one.
© Nilam Publication Sdn. Bhd.
Li
56
MODULE • Chemistry FORM 4
Observation:
slowly
Lithium moves
Inference
on the
Sodium moves
quickly
on the
K
Potassium moves
very quickly
on the water surface and produce
yellow
flame. The colourless
solution formed turns red litmus
blue
paper to
.
reacts with water to produce
alkaline solution, lithium
hydroxide:
Balanced chemical equation:
2Li + 2H2O 2LiOH + H2
Sodium is
reactive
metal reacts
with water to produce alkaline
solution, sodium hydroxide.
Balanced chemical equation:
2Na + 2H2O 2NaOH + H2
Potassium is
the most reactive
metal reacts with water to produce
alkaline solution, potassium
hydroxide.
down Group 1
water surface and produces
yellow
flame. The colourless
solution formed turns red litmus
blue
paper to
.
reactive metal
increases
Na
the least
Reactivity
water surface and produces
red
flame. The colourless
solution formed turns red litmus
blue
paper to
.
Lithium is
Reactivity
Balanced chemical equation:
2K + 2H2O 2KOH + H2
(b) Metal Group 1 reacts with oxygen to form metal oxide. The metal oxide dissolves in water to produce alkaline
solution.
X2O + H2O
4X + O2 2X2O
2XOH, X is a metal element of Group 1 (Li, Na and K)
Combustion spoon
Gas jar
Oxygen gas
Burning lithium
Procedure:
(i)
Cut a small piece of lithium using a knife and forceps.
(ii) Dry the oil on the surface of the lithium with filter paper.
(iii) Place the lithium in a combustion spoon and heat lithium until it start to burn.
(iv) Put the burning lithium into a gas jar of oxygen.
(v) When the reaction stop, add water to dissolve the compound formed.
(vi) Add a few drops of universal indicator to the solution formed.
(vii) Record the observation.
(viii) Repeat steps (i) – (vii) using sodium and potassium to replace lithium one by one.
57
4
Li
Observation
UNIT
Element
MODULE • Chemistry FORM 4
Observation:
Element
Li
Observation
Inference
slowly
– Lithium burns
red
with a
flame
to produce white solid .
– The white solid dissolves
in water to form
colourless solution.
– The solution turns
green
universal
indicator to
.
produce
– Lithium oxide
form alkaline
hydroxide.
produce
4
purple
.
.
– The solution turns
indicator to
© Nilam Publication Sdn. Bhd.
– Sodium
reactive
metal towards
reacts
with oxygen to
sodium oxide .
Balanced chemical equation:
4Na + O2 2Na2O
reacts
– Sodium oxide
with water to
form alkaline solution, sodium
hydroxide.
Balanced chemical equation:
Na2O + H2O 2NaOH
– Potassium
dissolves in water to form
colourless solution.
green
solution, lithium
– Potassium is the most reactive
towards oxygen.
– Potassium burns
very brightly with a
purple
flame to
produce white solid
– The white solid
with water to
universal
purple .
produce
metal
reacts
with oxygen to
potassium oxide .
Balanced chemical equation:
4K + O2 2K2O
reacts
– Potassium oxide
with water
to form alkaline solution, potassium
hydroxide.
Balanced chemical equation:
K2O + H2O 2KOH
58
down Group 1
indicator to
reacts
increases
UNIT
to produce white solid .
– The white solid
– The solution turns
green
universal
.
Balanced chemical equation:
Li2O + H2O 2LiOH
– Sodium is
oxygen.
colourless solution.
lithium oxide
Balanced chemical equation:
4Li + O2 2Li2O
– Sodium burns brightly
yellow
with a
flame
dissolves in water to form
K
– Lithium is the least reactive metal
towards oxygen.
reacts
– Lithium
with oxygen to
Reactivity
Na
purple
Reactivity
MODULE • Chemistry FORM 4
(c) Metal Group 1 reacts with chlorine to produce metal chloride.
2X + Cl2
2XCl, X is a metal element of Group 1 (Li, Na and K)
Combustion spoon
Gas jar
Chlorine gas
Burning of metal Group 1
Observation:
slowly
– Lithium burns
red
with a
to produce
solid.
Inference
flame
white
Reactivity
– Lithium is the least reactive metal
towards chlorine.
reacts
– Lithium
with chlorine to
produce
lithium chloride
.
Na
to produce
solid.
white
– Sodium is
chlorine.
– Sodium
produce
reactive
metal towards
reacts
with chlorine to
sodium chloride .
K
– Potassium burns
very brightly with a
purple
flame to
produce
white
solid.
most reactive
– Potassium is the
metal
towards chlorine.
reacts
– Potassium
with chlorine to
produce
potassium chloride .
Balanced chemical equation:
2K + Cl2 2KCl
59
down Group 1
Balanced chemical equation:
2Na + Cl2 2NaCl
increases
– Sodium burns brightly
yellow
with a
flame
Reactivity
Balanced chemical equation:
2Li + Cl2 2LiCl
4
Li
Observation
UNIT
Element
MODULE • Chemistry FORM 4
Complete the following:
1
SP 4.4.4
2
Metal Group 1 react with water.
Metal Group 1 reacts with chlorine.
2X + 2H2O → 2XOH + H2
2X + Cl2 → 2XCl
(a) 2 Li + 2H2O →
2LiOH + H2
(a) 2 Li + Cl2 →
2LiCl
(b) 2 Na + 2H2O →
2NaOH + H2
(b) 2 Na + Cl2 →
2NaCl
2 K + 2H2O →
2KOH + H2
(c)
2 K + Cl2 →
2KCl
(c)
Group 1 Metal
Li, Na, K
X
3 Metal Group 1 reacts with oxygen or air to form metal oxide.
The metal oxide reacts with water.
UNIT
4
4.5
4X + O2 → 2X2O
X2O + H2O → 2XOH
(a)
4
Li + O2 →
Li2O
+ H2O →
2Li2O
2LiOH
(b)
4
Na + O2 →
Na2O
+ H2O →
2Na2O
2NaOH
(c)
4
K + O2
→
K2O
+ H2O →
2K2O
2KOH
ELEMENTS IN GROUP 17
State the special name for
Group 17 elements.
List the elements in Group
17 of the Periodic Table and
write the electron
arrangements and number
of shells of the atoms of
elements.
Halogens
Elements
Symbol
Proton
number
Electron
arrangement
Number of
shells
Fluorine
F2
9
2.7
2
Chlorine
Cl2
17
2.8.7
3
Bromine
Br2
35
2.8.18.7
4
I2
53
2.8.18.18.7
5
Iodine
State the physical
properties of Group 17
elements. SP 4.5.1
© Nilam Publication Sdn. Bhd.
SK
4.5
Halogens cannot conduct heat and electricity in all states.
60
MODULE • Chemistry FORM 4
Explain the changes in
physical properties going
down Group 17 elements.
SP 4.5.1
(a) The melting and boiling points are low because the molecules are attracted
by weak Van der Waals forces, and small amount of energy is required to
overcome these forces. However, the melting and boiling points increase
going down the group.
– The atomic size
increases
increasing in number of
going down the Group 17 because of
shell
, the size of molecules get larger.
– The inter molecular forces of attraction (Van der Waals forces) between
molecules become stronger.
– More energy is needed to overcome the stronger attractive forces
between molecules during melting or boiling.
(b) Physical properties change from gas (fluorine and chlorine) to liquid
(bromine) and to solid (iodine) at room temperature due to increase in the
strength of inter molecular forces from fluorine to iodine.
(c) The density is low and increases going down the group.
SP 4.5.2
(a) All
atoms
of elements in Group 17 have
seven
valence
electrons and achieve a stable octet electron arrangement by accepting
one
electron to form negatively charged ions.
Example:
atoms
(i) Fluorine
receives one electron to achieve stable octet
electron arrangement:
F
+
e–
Electron arrangement: 2.7
Number of protons = 9,
total charge: +9
Number of electrons = 9,
total charge: –9
neutral
Fluorine atom is
F–
.
Electron arrangement: 2.8
Number of protons = 9,
total charge: +9
Number of electrons = 10,
total charge: –10
Negatively charged uoride
ion, F– is formed.
(ii) Chlorine atom receives one electron to achieve stable octet electron
arrangement:
+
Cl
e–
Electron arrangement: 2.8.7
Number of protons = 17,
total charge: +17
Number of electrons = 17,
total charge: –17
neutral
Chlorine atom is
61
Cl–
.
Electron arrangement: 2.8.8
Number of protons = 17,
total charge: +17
Number of electrons = 18,
total charge: –18
Negatively charged chloride
ion, Cl– is formed.
UNIT
Explain the similarities in
chemical properties of the
Group 17 elements.
4
darker
(d) The colour of the elements becomes
going down the group:
fluorine (light yellow), chlorine (greenish yellow), bromine (brown) and
iodine (purplish black).
MODULE • Chemistry FORM 4
(b) All elements in Group 17 have similar chemical properties because
seven
atoms
in Group 17 have
valence electrons and
achieve the stable octet electron arrangement by receiving one electron
to form a negatively charged ion.
SP 4.5.3
– All the atoms of Group 17 have seven valence
electrons and achieve a stable octet electron
arrangement by accepting one electron to form
negatively charged ion.
4
Compare and explain the
reactivity of elements X and
Y. SP 4.5.4
Element
Proton number
X
9
Y
17
Elements in Group 17 exist
as diatomic molecules.
Explain.
Cl
down Group 17
UNIT
– The strength of a halogen atom to attract electron
decreases
from fluorine to astatine
(electronegativity decreases).
F
decreases
– Going down Group 17, the number of
shells
size
increases, atomic
increases.
further
– Outer shell becomes
from the
nucleus.
– The strength of attraction from the proton in the
nucleus to attract one electron into the outermost
weaker .
occupied shell becomes
Reactivity
Explain the decrease in the
reactivity of the elements
going down the Group 17.
Br
– Element Y is less reactive than element X.
– Electron arrangement of X atom is 2.7 and Y atom is 2.8.7.
– The number shells occupied with electrons of atom Y is more than atom X.
– The size of atom Y is larger than atom X.
– Thus, the force of attraction the nucleus to attract one electrons on the
outermost shells of atom Y is weaker than atom X.
Two atoms of element sharing one pair of valence electrons to achieve stable
octet electron arrangement.
Example:
Two fluorine atoms share one pair of electrons to form one fluorine molecule:
F
Share
Fluorine atom
F
Fluorine atom
F
F
Fluorine molecule
Chlorine, bromine and iodine exist as diatomic molecules. (Cl2, Br2 and I2)
© Nilam Publication Sdn. Bhd.
62
MODULE • Chemistry FORM 4
Experiments for the Chemical Properties of Group 17 Elements:
SP 4.5.1
(a) Halogen reacts with water with different reactivity:
X2 + H2O
HX + HOX, X is halogen. (Cl2, Br2 and I2 )
Chlorine gas
Bromine water
Chlorine
Florin, Klorin
gas
Iodine crystals
Bromine
water
Fluorine, Chlorine
Water
water
Water
air
Procedure:
– Some iodine crystals are added
to water in a test tube.
– The test tube is shaken.
– The solution produced is tested
with blue litmus paper.
Observation:
Observation:
Observation:
Chlorine dissolves rapidly
in water to form light yellow
solution:
slowly
Bromine dissolves
in water to form brown solution:
Cl2 + H2O
Br2 + H2O
HCl + HOCl
Haba
slightly
Iodine dissolves
in
water to form brown solution:
I2 + H2O
HBr + HOBr
HI + HOI
4
Klorin atau
Bromin
Procedure:
Procedure:
– Chlorine gas is passed
– A few drops of bromine water
through water in a test tube.
are addedIodine
to water in a test tube.
Iodin
to absorb
– The solution produced isNaOH
– The test tube is shaken.Iron wool
Chlorine / bromine
Heat
testedHaba
with blue litmus paper.
NaOH untuk menyerap
– The solution producedWul
is Besi
tested
klorin / bromin
with blue litmusHeatpaper.
The solution changes
blue
litmus paper to
The solution changes
blue
litmus paper to
The solution changes
blue
litmus paper to
red
and slowly
decolourises it.
red
and quickly
decolourises it.
red
does not
. The litmus paper
decolourise .
Inference:
Chlorine, bromine and iodine react with water to form acidic solution. Apart from the acidic solution,
chloride and bromine formed bleaching agent.
Solubility decreases from chlorine to iodine.
(b) Halogens react with hot iron to form brown solid, iron(III) halide.
Iron wool
Iodine
Chlorine or
Bromine
Heat
Heat
NaOH to absorb
chlorine / bromine
63
Heat
Iron
wool
UNIT
Chlorine or
Bromine
Iodine
crystals
Water
MODULE • Chemistry FORM 4
2Fe + 3X2
2FeX3, X2 represents any halogen. (Cl2, Br2 or I2 )
Halogen
Observation
Chemical equation
Chlorine
Iron wool burns very brightly
brown solid when cooled.
Bromine
Iron wool burns brightly
solid when cooled.
Iodine
slowly
Iron wools glows
with a dull glow and
forms a brown solid when cooled.
and forms a
and forms a brown
Experiment (a), (b) and (c) show that all halogens have
reactivity decreases going down the group:
2Fe + 3Cl2
2FeCl3
2Fe + 3Br2
2FeBr3
2Fe + 3I2
similar
2FeI3
chemical properties but their
decreases
Reactvity
F2, Cl2, Br2 and l2
(c) Halogens react with sodium hydroxide solution
X2 + 2NaOH
NaX + NaOX + H2O, X2 is halogen. (Cl2, Br2 and I2 )
Complete the following:
UNIT
Cl2 + 2NaOH
NaCl + NaOCl + H2O
(ii) Br2 + 2NaOH
NaBr + NaOBr + H2O
(i)
4
NaI + NaOI + H2O
(iii) I2 + 2NaOH
4.6
1
2
Reactivity decreases
ELEMENTS IN PERIOD 3
SK
4.6
There are seven periods known as period 1, 2, 3, 4, 5, 6, 7.
The number of period of an element represents the number of shells occupy with electrons in each atom of
element.
Element
Proton number
Electron arrangement
Number of shells
Period
Li
3
2.1
2
2
Na
11
2.8.1
3
3
K
19
2.8.8.1
4
4
List the elements of Period
3 and write the electron
arrangement and number
of shells of the atom of
elements.
© Nilam Publication Sdn. Bhd.
Element
Proton
number
Electron
arrangement
Number of
shells
Radius
(nm)
Na
11
2.8.1
3
0.191
Mg
12
2.8.2
3
0.160
Al
13
2.8.3
3
0.130
Si
14
2.8.4
3
0.118
P
15
2.8.5
3
0.110
S
16
2.8.6
3
0.102
Cl
17
2.8.7
3
0.099
Ar
18
2.8.8
3
0.095
64
MODULE • Chemistry FORM 4
State the change in atomic
size across Period 3
(from left to right).
decreases from sodium to chlorine
The atomic radius of the atoms
SP 4.6.1
Na
Mg
Al
Si
P
S
Cl
16 p
Na
Mg
Al
Si
P
S
Cl
11 p 12 p 13 p 14 p 15 p 16 p 17 p
+11 +12 +13 +14 +15 +16 +17
2.8.1 2.8.2 2.8.3 2.8.4 2.8.5 2.8.6 2.8.7
– All the atoms of Period 3 elements have
with electrons .
– The proton number
3
shells
occupied
increases
by one unit from sodium to chlorine.
positive
– Increasing in proton number causes the number of
charge in
the nucleus to increase .
– The strength of attraction from the proton in the nucleus to the electrons in
the shells increases .
– The size of atom
Define electronegativity.
SP 4.6.1
Explain the change in
electronegativity of
elements across Period 3
from left to right.
SP 4.6.1
Compare and explain
electronegativity of
elements X and Y. SP 4.6.1
Element
Proton number
X
11
Y
17
decreases
across Period 3.
The strength of an atom in a molecule to attract
nucleus.
electron
towards its
– The atomic radius decreases due to the increasing of nuclei attraction
on the electrons in the shells from sodium to chlorine.
– Tendency of nucleus to attract electron to the outermost shells increases
from sodium to chlorine.
– The electronegativity increases across Period 3 from sodium to chlorine.
– Element Y is more electronegative than element X.
– Electron arrangement of X atom is 2.8.1 and Y atom is 2.8.7.
– Atoms X and Y have same number shells occupied with electrons.
– The number of protons in the nucleus of atom Y is more than atom X.
– The attraction forces between the nucleus and the electrons in the shells of
atom Y is stronger than atom X.
– The size of atom Y is smaller than atom X.
– The tendency to attract electrons of atom Y is stronger than atom X.
65
4
SP 4.6.1
Atom
Number of proton
Positive charge
Electron arrangement
UNIT
Explain the change in
atomic size of elements
across Period 3 from left to
right.
MODULE • Chemistry FORM 4
State the changes in
properties of elements
across Period 3 from left to
right.
SP 4.6.2
SP 4.6.3
(a) Physical state:
– The physical state of elements in a period changes from solid to gas
from left to right.
– Metals on the left are solid while non-metals on the right are usually
gases.
(b) Changes in metallic properties and electrical conductivity:
Element
Na Mg Al
Metallic
properties
Metal
Semi metal or metaloid
Non-metal
Electrical
conductivity
Good
conductors
of electric.
Weak conductor of
electric but it increases
with and increase in
temperature and the
presence of boron or
phosphorous.
Uses: semiconductor
Cannot conduct
Si
P
S
Cl Ar
(c) Changes in properties of oxide of elements Period 3:
Na
Mg
Al
UNIT
Basic oxide
Amphoteric oxide
4
Basic oxide + Water
Alkali
Example:
Na2O + H2O
2NaOH
Basic oxide + Acid
Salt + Water
Example:
MgO + 2HCl
MgCl2 + H2O
Define basic oxide,
amphoteric oxide and
acidic oxide.
Si
S
Cl
Acidic oxide
Amphoteric oxide + Acid Acidic oxide + Water
Salt + Water
Acid
Amphoteric oxide + Alkali
Example:
Salt + Water
SO2 + H2O H2SO3
Example:
Al2O3 + 6HNO3
2Al(NO3)3 + 3H2O
Acidic oxide + Alkali
Al2O3 + 2NaOH
Salt + Water
2NaAlO2 + H2O
Example:
SiO2 + 2NaOH
Na2SiO3 + H2O
– Basic oxide is metal oxide that can react with
salt
water
and
.
acid
to form
– Acidic oxide is non-metal oxide that can react with
salt
water
and
.
alkali
to form
– Amphoteric oxide is oxide that can react with both
alkali
salt
water
to form
and
.
© Nilam Publication Sdn. Bhd.
P
66
acid
and
MODULE • Chemistry FORM 4
Aim
To investigate the properties of the oxides of elements in Period 3.
Problem
statement
How do the properties of the oxides of elements change across Period 3?
Variables
Manipulated: Type of oxide of elements in Period 3
Responding: pH value in water, reaction of oxide with acid and alkali
Constant: Water, nitric acid and sodium hydroxide solution
Hypothesis
The oxide of elements across Period 3 change from basic oxide (sodium oxide, magnesium
oxide) to amphoteric oxide (aluminium oxide) and then to acidic oxide (phosphorus(V) oxide,
silicone oxide and dichloride heptaoxide solution).
Apparatus
Test tubes, boiling tubes, measuring cylinder, Bunsen burner, glass rod and spatula
Materials
Sodium oxide, magnesium oxide, aluminium oxide, silicon(IV) dioxide, phosphorus oxide,
sulphur dioxide gas in a covered gas jar, dichloride heptaoxide, universal indicator, distilled
water, 2 mol dm–3 of sodium hydroxide solution, 2 mol dm–3 nitric acid
Procedure
I Acidic/basic properties of the oxides of elements in Period 3
1 Half spatula of sodium oxide powder is added to 5 cm3 of distilled water in a test tube.
The mixture is stirred with a glass rod until no further change occurs.
2 Two drops of universal indicator is added and the test tube is shaken. The pH value of
the solution is recorded.
3 Steps 1 and 2 are repeated by replacing sodium oxide with magnesium oxide, aluminium
oxide, silicon dioxide, phosphorus oxide and dichloride heptaoxide.
4 For sulphur dioxide gas, step 2 is repeated to the solution formed when sulphur dioxide
gas is passing through the distilled water in a test tube.
II Reaction with nitric acid and sodium hydroxide solution
1 Half spatula of sodium oxide powder is added to 2 cm3 of dilute nitric acid in a test tube.
The mixture is heated gently and shaken until no further change occurs.
2 Step 1 is repeated by replacing sodium oxide with aluminium oxide and silicon oxide.
3 Another experiment is conducted by repeating steps 1 and 2 by using 2 mol dm–3 of
sodium hydroxide solution.
Observation
I
Oxide
Solubility in water
pH
Type of oxide
Sodium oxide, Na2O
The white solid dissolves in water.
14
Basic oxide
Magnesium oxide,
MgO
The white solid slightly dissolves in
water.
9
Basic oxide
Aluminium oxide, Al2O3
Insoluble.
–
–
Silicon(IV) oxide, SiO2
Insoluble.
–
–
Phosphorous oxide, P4O10 The white solid dissolves in water.
3
Acidic oxide
Sulphur dioxide, SO2
The colourless gas dissolves in water.
3
Acidic oxide
Dichloride heptaoxide,
Cl2O7
The liquid dissolves in water to form
colourless solution.
1
Acidic oxide
67
4
SP 4.6.2
UNIT
Experiments to Investigate the Properties of the Oxide Elements Change Across Period 3:
MODULE • Chemistry FORM 4
II
Observation
Reaction with dilute
Reaction with sodium Type of oxide
nitric acid
hydroxide solution
Oxide
Magnesium oxide,
MgO
The white solid
dissolves to form
colourless solution.
No change. The white
solid does not dissolve.
Basic oxide
Aluminium oxide,
Al2O3
The white solid
dissolves to form
colourless solution.
The white solid
dissolves to form
colourless solution.
Amphoteric
oxide
No change. The white
solid does not dissolve.
The white solid
dissolves to form
colourless solution.
Acidic oxide
Silicon(IV) oxide,
SiO2
Conclusion
3
Hypothesis is accepted.
Steps to compare and explain the change in atomic size / radius / electronegativity across Period 3, reactivity
down Group 1 and Group 17:
UNIT
Na
4
Mg
Atomic radius of the atoms
sodium to chlorine.
S
P
16 p
across Period 3 from
(ii) Compare the strength of proton in the nucleus to attract
valence electron (Group 1) // to attract electron
F
Cl
Br
outermost shells
(Group 17).
release
(iii) Compare tendency of the atom to
receive
(Group 1) //
electron (Group 17).
© Nilam Publication Sdn. Bhd.
down Group 17
(b) To Compare Reactivity Down Group 1 and Group 17:
shells
(i) Compare number of
in each atom.
decreases
(iii) Compare the strength of attraction from the nucleus to the
electrons in the shells .
(iv) Compare the atomic size / Compare the electronegativity.
to the
Cl
Reactivity
down Group 1
K
decreases
Si
(a) To Compare Atomic Size / Radius and Electronegativity
Across Period 3:
shells
(i) Compare number of
in each atom.
proton
(ii) Compare number of
in the nucleus.
increases
Na
Reactivity
Li
Al
68
electron
The Periodic
Table
MODULE • Chemistry FORM 4
4.7
TRANSITION ELEMENTS
SK
4.7
State the position of the
transition element in the
Periodic Table SP 4.7.1
Situated between Groups 2 and 13
Examples:
Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn.
What are the metallic
properties of transition
element?
1
2
3
4
5
What are the special
characteristics of transition
element?
1 Most transition elements formed coloured compounds.
Examples:
(i) Iron(III) chloride is brown.
(ii) Iron(II) chloride is green.
(iii) Copper(II) sulphate is blue.
SP 4.7.2
Shiny
Conducts heat and electricity
Malleable
High tensile strength
High melting point and density
Copper
Iron
Compound
Copper(I) chloride
Copper(II) oxide
Iron(II) chloride
Iron(III) chloride
Oxidation number
+1
+2
+2
+3
• Oxidation number of element in a compound will be studied in topic
“redox”.
3 Many of the transition elements are able to form complex ion:
Element
Iron
Copper
Nyatakan kegunaan unsur
peralihan dalam industri.
SP 4.7.3
Complex ion
Hexacyanoferrate(II)
Copper(II) tetramine
Formula
Fe(CN)64Cu(NH4)42+
Many of the transition elements can act as a catalyst in industries.
Catalyst is a substance that can change the rate of reaction.
A catalyst does not change chemically after a reaction.
Examples:
(i) Iron: Haber process in the manufacture of ammonia
(ii) Vanadium(V) oxide: Contact process in the manufacture of sulphuric acid
(iii) Platinum: Ostwald process in the manufacture of nitric acid
69
UNIT
Element
4
2 Most transition elements have more than one oxidation number in their
compounds.
Examples:
MODULE • Chemistry FORM 4
SPM PRACTICE
Subjective Questions
1
The diagram below shows the electron arrangement for atoms P and Q.
PP
Q
Q
(a) Elements P and Q are placed in the same group in Periodic Table. State the group.
TP2
Group 1
(b) How are elements P and Q kept in the laboratory? Give reason for your answer.
TP1
In paraffin oil. To prevent them from reacting with oxygen or water vapour in the atmosphere.
(c) (i)
TP3
Write chemical equation for the reaction between element P with water.
2P + 2H2O 2POH + H2
UNIT
(ii) What is the expected change of colour when a few drops of phenolphthalein are added into the
TP3
aqueous solution of the product? Explain your answer.
Colourless to purple / pink. The solution formed is alkaline.
4
(iii) Between element P and element Q, which is more reactive in the reaction with water?
TP4
Element Q is more reactive than P.
(iv) Explain your answer in (c)(iii).
TP4
The size of atom Q is larger than atom P. The valence electron of atom Q is further away from the nucleus
compared to atom P. The attractive forces between proton in the nucleus to the valence electron of atom Q is
weaker than atom P. It is easier for atom Q to release the valence electron compared to atom P.
(d) Name one element that has the same chemical properties as P and Q.
TP2
Potassium
2
The diagram below shows the information regarding elements W and X which are from the same group in the
Periodic Table.
19
9W
(a) (i)
TP2
35
17
X
Write the electron arrangement of atom of elements W and X.
2.7
Atom W:
Atom X:
(ii) State the position of elements W and X in the Periodic Table.
TP2
Element W: Group 17, Period 2
Element X: Group 17, Period 3
© Nilam Publication Sdn. Bhd.
70
2.8.7
MODULE • Chemistry FORM 4
(iii) Do elements W and X show the similar chemical property? Explain your answer.
TP2
Elements W and X have the similar chemical property. Atoms W and X have the same number of
valence electrons.
(b) State the type of particles in W and X.
TP2
Molecule
(c) Compare the boiling point of elements W and X. Explain the difference.
TP4 • The boiling point of element X is higher than element W.
• The size of molecule X2 is bigger than molecule W2.
• The forces of attraction between molecules X2 is stronger than molecule W2.
• More heat energy is needed to overcome the stronger forces between molecules.
(d) (i)
TP3
X can react with sodium to form a compound. Write the chemical equation for the reaction.
X2 + 2Na → 2NaX
(ii) How does the reactivity of element W and element X differ? Explain your answer.
TP4
• Element W is more reactive than element X. The size of atom W is smaller than atom X.
UNIT
• The strength of the nucleus of atom W to attract electron to the outermost shell is stronger than
atom X.
3
The table below shows the number of neutron and relative atomic mass of eight elements represented as P, Q,
R, S, T, U, V and W.
Atom
P
Q
R
S
T
U
V
W
Number of neutron
12
12
14
14
16
16
18
22
Relative atomic mass
23
24
27
28
31
32
35
40
Number of proton
11
12
13
14
15
16
17
18
2.8.1
2.8.2
2.8.3
2.8.4
2.8.5
2.8.6
2.8.7
2.8.8
Electron arrangement
(a) Complete the above table by writing the number of proton and electron arrangement for the atom of each
TP1 element.
TP2
(b) (i)
TP2
4
• The outermost occupied shell of atom W is nearer to the nucleus compare to atom X.
State the period of elements P – W in the Periodic Table. Explain your answer.
Period 3 because P – W atoms have three shells occupied with electrons.
(ii) State the proton number of another element that is in the same group as P.
TP2
3/19
(c) Write the standard representation for element Q.
TP3 24
12 Q
71
MODULE • Chemistry FORM 4
(d) Which element exists as
TP2
monoatomic gas? W
diatomic gas? T/ U/ V
(e) (i)
Which element can react vigorously with water to produce hydrogen gas?
P
TP2
(ii) Write the balanced equation for the reaction in (e)(i).
TP3
2P + 2H2O → 2POH + H2
(f) State the arrangement of elements T, U and V in the order of increasing atomic radius. Explain your
answer.
V, U and T. Atoms of T, U, and V have three shells occupied with electrons. The proton number //
TP2
positive charges in the nucleus increases from T to V. The forces of attraction between proton in the
nucleus and the electrons in the shells increase from T to V. The shells filled with electrons are pulled
nearer to the nucleus from T to V.
4
The diagram below shows part of the Periodic Table of Elements. X, Y, Z, A, B, D, E, F and G do not represent
the actual symbols.
UNIT
X
4
A
F
(a) (i)
TP2
Y
Z
B
D
E
G
State the position of element B in the Periodic Table.
Period 3, Group 13
(ii) Explain your answer in (a)(i).
TP2
Electron arrangement atom B is 2.8.3. Atom B has three valence electrons, element B is in Group
13. Atom B has three shells occupied with electrons, element B is in Period 3.
(b) (i)
TP2
Which element is monatomic gas?
Element Y/Z
(ii) Explain your answer in (b)(i).
TP2
Atom Y has achieved stable duplet electron arrangement // has electron arrangement 2. OR
Atom Z has achieved stable octet electron arrangement // has electron arrangement 2.8.
(c) Element X is hydrogen gas and element Y is helium gas. The diagram below shows a meteorological
HOTS balloon filled with helium gas.
Helium gas
(i)
TP2
Explain why helium gas is used to fill the meteorological balloon.
Helium gas is light and inert.
© Nilam Publication Sdn. Bhd.
72
MODULE • Chemistry FORM 4
(ii) Can hydrogen gas replace helium gas in the balloon? Give reason for your answer.
TP5
Cannot. Hydrogen gas is flammable, it will explode with the presence of oxygen gas at high
temperature.
(d) Choose an element that:
TP2
(i)
exists in the form of molecule
(ii)
forms acidic oxide
(iii)
has atoms that have no neutron
(iv)
is an alkali metal
(v)
forms amphoteric oxide
B
(vi)
has a proton number of 15
D
X/D/E
D/E
X
A/F
(vii) is most electropositive
F
(viii) forms basic oxide
forms coloured compound
G
4
(ix)
A/F
UNIT
(e) Arrange Y, A, B, D and E according to the order of increasing atomic size.
TP2
Y, E, D, B, A
(f) (i)
Write the electron arrangement for an atom of element:
2.8.5
D:
E:
TP2
2.8.7
(ii) Compare electronegativity of elements D and E.
TP4
Element E is more electronegative than element D.
(iii) Explain your answer in (f)(ii).
TP4
Atoms E and D have the same number of shells occupied with electrons. The number of proton in
the nucleus of atom E is more than atom D. The strength of proton in nucleus to attract electrons to
the outermost shells in atom E is stronger than of atom D.
Objective
Questions
73
MODULE • Chemistry FORM 4
UNIT
CHEMICAL BONDS
5
Concept Map
CHEMICAL BOND
To achieve stable duplet/octet
electron arrangement
Metallic Bond
Ionic Bond
Involves
Involves
Transfer of electron from
METAL ATOM to NONMETAL ATOM
Metal atom
donates
electron
Sharing one pair/two
pairs/three pairs of
electrons between
NON-METAL ATOMS
Non-metal
atom receives
electron
UNIT
Positive ion
Negative ion
5
Ionic compound
Example:
Electrons are
contributed by
both atoms
Molecule of covalent
compound
Simple
molecule
Sodium chloride
Electrons are
contributed
by one atom
only
Dative
Bond
Example:
H
N
Example:
H
H
Ammonium ion
Between hydrogen
atom and nitrogen/
oxygen/fluorine atom
Diamond
Weak Van
der Waals
forces
between
molecules
Strong covalent bond between atoms in the molecules
Carbon dioxide
74
Hydrogen Bond
Example:
H δ+
Example:
© Nilam Publication Sdn. Bhd.
Electrostatic
force between
sea of electrons
and metal ion
Dative bond
2.8.8
Strong electrostatic forces between
positive and negative ions
Sea of electrons
from the valence
electrons of
METAL ATOM
and positive
METAL ION
H
Giant
molecular
structure
Example:
2.8
Involves
Covalent Bond
H
δ+
O
δ–
H δ–
H
δ+
O
δ–
Hydrogen bond
Water
MODULE • Chemistry FORM 4
BASIC FORMATION OF COMPOUND
Compound is formed when two or more elements are combined.
SP 5.1.1
Why Group 18 elements are inert
gases? SP 5.1.1
The atoms have achieved duplet electron arrangement for helium and
octet electron arrangement for others.
What are chemical bonds?
Chemical bonds are formed when two or more atoms of elements bonded
together. There are two types of chemical bond, Ionic Bond and
Covalent Bond.
SP 5.1.1
Why do certain atoms form
chemicals bond with other atoms?
5.2
Atoms form chemical bonds to achieve a stable duplet or octet electron
arrangement.
SK
5.2
IONIC BONDS
What types of elements formed
ionic bonds?
Ionic bond is formed between atoms of metal elements that release
electrons to atoms of non-metal elements.
How ionic bond is formed?
– Atoms of elements that release electrons form positive ions and atoms
that receive electrons form negative ions to achieve a stable octet or
duplet electron arrangement.
– Ionic bond is usually formed between atoms from Groups 1, 2 and 13
(metal) with atoms from Groups 15, 16 and 17 (non-metal).
SP 5.2.1
Complete the following table:
Changes
SP 5.2.1
Na
Na+ + e–
Ca
Electron
arrangement
2.8.1
2.8
2.8.2
2.8
Total of positive
charges (From
number of proton)
+11
+11
+12
Total of negative
charges (From
number of electron)
–11
–10
0
+1
Total charges
Type of particles
Define ionic bond.
SP 5.2.1
Sodium
atom
O2–
Cl + e–
Cl–
2.6
2.8
2.8.7
2.8.8
+12
+8
+8
+17
+17
–12
–10
–8
–10
–17
–18
0
+2
0
–2
0
–1
Calcium
ion
Oxygen
atom
Oxide
ion
Chlorine
atom
Chloride
ion
Sodium Calcium
ion
atom
Ca2+ + 2e– O + 2e–
The electrostatic force between the positive and negative ions forms
ionic bond.
75
5
What is compound?
SK
5.1
UNIT
5.1
MODULE • Chemistry FORM 4
Example 1:
Explain the formation of sodium
chloride:
Predict formula:
Proton
number
11
17
Electron
arrangement
2.8.1
2.8.7
Na
Na+ + e– Na+
1
Cl + e–
Cl–
Cl–
1 ⇒ NaCl
Element
Na
Cl
Strong electrostatic force between ions
Transfer
Cl
Na
Sodium atom,
Na
Cl
Na
Chlorine atom,
Cl
Sodium ion,
Na+
Chloride ion,
Cl–
2.8.1
(a) Electron arrangement of sodium atom is
. Sodium
one
atom has
valence electron. Therefore sodium atom
stable
one
is not
. Sodium atom releases
electron
to achieve a stable octet electron arrangement to form
sodium ion , Na+ with electron arrangement
2.8
.
2.8.7
(b) Electron arrangement of chlorine atom is
. Chlorine
seven
atom has
valence electrons. Chlorine atom receives
one
electron to achieve stable octet electron arrangement
chloride
ion, Cl– with an octet arrangement of
to form
UNIT
2.8.8
electron
.
(c) Sodium ions , Na+ and chloride ions , Cl– ions are attracted
with strong electrostatic force. The bond formed is called ionic
bond.
5
Example 2:
Explain the formation of magnesium
oxide:
Element
Mg
O
Proton
number
12
8
Mg
Mg2+ + 2e–
O + 2e–
O2–
Draw the electron arrangement of the compound formed.
Transfer
Electron
arrangement
2.8.2
2.6
Mg2+
2
Mg
O
Magnesium atom, Mg
Oxygen atom, O
2–
2+
O2–
2 ⇒ MgO
Mg
Magnesium ion, Mg2+
O
Oxide ion, O2–
2.8.2
(a) The electron arrangement of magnesium atom is
.
2
Magnesium atom has
electrons in the outer shell.
stable
Therefore, magnesium atom is not
. Magnesium
2
atom releases
valence electrons to achieve a stable
octet electron arrangement to form magnesium ion, Mg2+ with
2.8
electron arrangement
.
AR
© Nilam Publication Sdn. Bhd.
76
MODULE • Chemistry FORM 4
2.6
(b) The electron arrangement of oxygen atom is
Oxygen atom is also unstable. Oxygen atom
receives
.
two
electrons to achieve a stable octet electron arrangement to form
oxide ion , O2– with electron arrangement
2.8
.
(c) Strong electrostatic force is formed between magnesium ion ,
Mg2+ and oxide ion , O2– to form ionic bond.
Steps to Explain the Formation of Ionic Bond for Ionic Compound
Predict formula
5.3
(d) State the electron arrangement
of non-metal atom (number of
valence electron is 5/6/7).
(e) State the number of electron
received by the atom.
(f) State the name of negative
ion formed and the electron
arrangement achieved
(stable duplet / octet electron
arrangement).
COVALENT BONDS
• The number of positive ions and
negative ions in the compound is
based on the formula.
• Draw all the electrons in the shells
of positive ion and negative ion.
• Write the charge of each ion.
(a) State the electron arrangement of
metal atom (number of valence
electron is 1/2/3).
(b) State the number of electron
released by the atom.
(c) State the name of positive
ion formed and the electron
arrangement achieved
(stable duplet / octet electron
arrangement).
SK
5.3
What types of elements
formed covalent bonds?
Covalent bond is formed when similar or different non-metal atoms bond
together. [Atoms from Groups 14, 15, 16 and 17]
How covalent bond is formed?
– This bond is formed when two or more similar or different atoms share
their valence electrons to achieve stable duplet or octet electron
arrangement.
– The product of the sharing of electrons between atoms is called molecule.
– The molecules are neutral as there is no electron transfer involved. During
the formation of covalent bond, each atom contributes same number of
electrons for sharing. The number of electrons shared can be one pair, two
pairs or three pairs.
– The forces that exist between molecules are Van der Waals forces that
are weak. These forces become stronger when the molecule size increases.
Define covalent bond.
Covalent bond is the bond formed by the shared of electrons between atoms.
SP 5.3.1
77
5
(g) State the
electrostatic
force between
the positive
and negative
ion formed.
• By referring to the number of
valence electron for each atom,
determine the charge of ion
formed from each atom.
• Cross the coefficient of charge
to get the formula of the ionic
compound formed.
Explanation
Write the electron
arrangement of
metal atom and
non-metal atom.
Draw the electron arrangement for the
compound formed.
UNIT
Determine the formula of the ionic compound
formed.
MODULE • Chemistry FORM 4
Example 1:
Explain the formation of
hydrogen molecule, H2
SP 5.3.1
Covalent bond between atoms
Share
H
or:
H
H + H
H H
H
or
H
H–H
Lewis Structure
(a) Hydrogen atom has one electron in the first shell, with an electron
arrangement of 1, needs one electron to achieve a stable duplet electron
arrangement.
(b) Two hydrogen atoms share a pair of electrons to form a hydrogen
molecule.
(c) Both hydrogen atoms achieve a stable duplet arrangement of electron.
one
(d) The number of electron pairs shared is
pair. Single covalent
bond is formed.
Example 2:
Explain the formation of
oxygen molecule, O2
Share
O
O
O
O
SP 5.3.1
Oxygen atom, O
or:
O
Oxygen atom, O
O
O
Oxygen molecule, O2
O
or
O=O
UNIT
(a) Oxygen atom with an electron arrangement 2.6 needs two electrons to
octet
achieve a stable
electron arrangement.
5
two
(b) Two oxygen atoms share
pairs of electrons to achieve a
stable octet arrangement of electron, form an oxygen molecule. Each
octet
oxygen atom achieves stable
electron arrangement.
(c) The number of electron pairs shared is
covalent bond is formed.
Example 3:
Explain the formation of
nitrogen molecule, N2
Share
N
Nitrogen atom, N
or:
N
N
N
Nitrogen atom, N
N
2
N
pairs. Double
N
Nitrogen molecule, N2
N
or
(a) Nitrogen atom with an electron arrangement 2.5 needs
octet
electrons to achieve stable
electron arrangement.
3
3
(b) Two nitrogen atoms share
pairs of electrons to achieve a
octet
stable
electron arrangement, form a nitrogen molecule.
octet
Each nitrogen atom achieves stable
electron arrangement.
(c) The number of electron pairs shared is
covalent bond is formed.
AR
© Nilam Publication Sdn. Bhd.
78
3
pairs. Triple
MODULE • Chemistry FORM 4
Example 4:
Explain the formation of
Hydrogen chloride, HCl
Hydrogen atom, H
or:
H
Predict formula:
Cross the
needs 1 electron number of
H
electrons
each atom
Cl needs 1 electron
needs ⇒
HCl
Chlorine atom, Cl
Cl
One
One
hydrogen atom.
atom
achieves stable
(f) Chlorine
atom
and hydrogen
arrangement.
Example 5:
Explain the formation of
carbon dioxide, CO2
C
needs 4 electrons
O
2 electrons
needs
one
one
electron and
hydrogen atom contributes one electron for sharing.
one
chlorine atom forms
single covalent bond with
one
Cross the
number of
electrons
each atom
needs ⇒
CO2
H – Cl
chlorine atom contributes
One
(e)
Predict formula:
or
chlorine atom shares
one
C
O
Cl
pair of electrons with
one
hydrogen atom to form hydrogen chloride molecule
HCl
with the formula
.
(d)
Proton
Electron
number arrangement
6
2.4
8
2.6
Hydrogen chloride molecule, HCl
atom
1
(a) Hydrogen
with an electron arrangement
needs
one
duplet
electron to achieve a stable
electron
arrangement.
atom
(b) Chlorine
with an electron arrangement 2.8.7 needs
one
octet
electron to achieve stable
electron arrangement.
(c)
Element
H
Cl
Share
achieves stable
electron arrangement
duplet
electron
Share
O
C
Oxygen atom, O
octet
Carbon atom, C
O
O
Oxygen atom, O
C
O
O
C
O
C
or
O=C=O
atom
2.4
(a) Carbon
with an electron arrangement
four
octet
electrons to achieve a stable
arrangement.
atom
2.6
(b) Oxygen
with an electron arrangement
two
octet
electrons to achieve stable
arrangement.
(c)
Two
one
formula
oxygen atoms shares
O
Carbon dioxide
molecule, CO2
or:
O
5
H
Cl
Proton
Electron
number arrangement
1
1
17
2.8.7
H
Cl
UNIT
Element
Share
H
needs
electron
needs
electron
four
pairs of electrons with
carbon atom form carbon dioxide molecule with the
CO2 .
79
AR
MODULE • Chemistry FORM 4
(d)
One
four
carbon atom contributes
electrons and each of
the two oxygen atoms contributes two electrons for sharing to form
double covalent bond.
One
(e)
carbon atom forms
two
with
(f) Oxygen
and carbon
two
double covalent bonds
oxygen atoms.
atom
octet
achieves stable
atom
achieves
octet
electron arrangement
electron arrangement.
Steps to Explain the Formation of Covalent Bond for Covalent Compound
Predict formula
Determine the formula of the covalent
compound formed.
Draw the electron arrangement for the compound
formed based on the formula.
UNIT
5
(c) State the electron
arrangement achieved
by each atom in the
molecule (stable
duplet/octet electron
arrangement).
• Electrons are shared in pair.
• The number of pairs of electrons
shared is based on the number of
electrons needed by each atom to
achieve stable duplet/octet electron
arrangement.
(b) Based on the drawing of the electron
arrangement, state:
✽ the number of pairs of electrons
shared between atoms.
✽ the number of electrons of each
atom contribute for sharing in the
molecule and type of covalent bond
formed (single/double).
Comparing the Formation of Ionic and Covalent Bonds
Covalent Bond
State type of
element
involved.
Between metals (Groups 1, 2 and 13)
and non-metals (Groups 15, 16 and 17).
How bonds are
formed?
Electron is released by metal atoms and
received by non-metal atoms (electron
transfer).
What are the
type of particle
produced?
Metal atom forms
How to predict
the formulae?
Determine the coefficient of the charge of the
ions and criss cross.
© Nilam Publication Sdn. Bhd.
(a) State the electron
arrangement of each
atom and the number
of electrons needed
to achieve stable
duplet/octet electron
arrangement.
SP 5.3.2
Ionic Bond
positive
Non-metal atom forms
Explanation
• Write the electron arrangement of each atom.
• By referring to the number of valence electrons for each
atom, determine the number of electrons needed by each
atom to achieve stable duplet/octet electron arrangement.
• Cross the number of electrons needed to achieve stable
duplet/octet electron arrangement.
ion.
negative ion.
80
Between non-metals and
non-metals (Groups 14, 15, 16 and 17).
Pairs
of electrons are shared
by the same or different non-metals
atoms.
Neutral molecule .
Determine the number of electrons needed
to achieve stable duplet or octet electron
arrangement and criss cross.
MODULE • Chemistry FORM 4
5.4
HYDROGEN BOND
Define
electronegativity.
SK
5.4
Electronegativity is the strength of an atom in a molecule to attract electron
towards its nucleus. Examples of electronegative atoms are chlorine, oxygen and
nitrogen.
Remark:
Electronegativity generally increases from left to right across a period and decreases down a periodic table
group.
What is polar
molecule?
A polar molecule is a molecule containing polar bonds. Polar bonds form when there is a
difference between the electronegativity of the atoms participating in a bond.
Example:
Hydrogen chloride is a polar molecule. This is because the chlorine atom in the
electronegative
hydrogen chloride molecule is more
than hydrogen. The pair of
atom than with the hydrogen atom.
δ+
δ–
H
Cl
Give other
examples of polar
molecules.
Water (H2O), ammonia (NH3), sulphur dioxide (SO2) and hydrogen sulphide (H2S).
What is hydrogen
bond?
When hydrogen atom is covalently bonded to a highly electronegative atoms which is
oxygen, nitrogen or fluorine, hydrogen bond is an attraction between hydrogen atoms
with the electronegative atoms from other molecules.
SP 5.4.1
Remark:
The electronegative atom that is bonded to hydrogen atom has the following properties:
1 The atom should be electronegative in nature.
2 The atom should possess at least one lone pair of unshared electron.
3 The atom should be considerably small in size.
These conditions are fulfilled by only 3 atoms namely Nitrogen, Oxygen and Fluorine.
Water is the most
common example
for a compound
having hydrogen
bonds.
Explain how
hydrogen bond
forms in water.
SP 5.4.1
(a) When hydrogen is covalently bonded with the electronegative oxygen atom, the pair
electrons are more closely with the oxygen atom than with the hydrogen atom.
(b) This leads to the formation of partially positive charge (δ+) on hydrogen atom, H and
partially negative charge (δ–) on oxygen atom, O.
(c) The partially positive charged hydrogen atom is then attracted by the other partially
negative charged oxygen atom is known as hydrogen bond.
H
Hydrogen bond
H
δ+
O
δ–
H
δ+
O
δ–
H
δ–
Hydrogen bond
δ+
H δ+
H δ+
O
H
δ+
Covalent bond
O
H
81
H
O
δ+
δ– H
δ– O
H
H
5
chlorine
UNIT
electrons are more closely with
MODULE • Chemistry FORM 4
Explain how
hydrogen bond
forms in hydrogen
fluoride.
SP 5.4.1
(a) Fluorine is the more
electronegative
bond are more closely with the
(b) This leads to the formation of partially positive charge (δ+) on hydrogen atom, H
and partially negative charge (δ–) on fluorine atom, F.
(c) The partially positive charged hydrogen atom is then attracted by the other partially
negative charged fluorine atom is known as hydrogen bond .
Hydrogen bond
F δ–
Hδ+
Hδ+
F δ–
Covalent bond
Hδ+
Hδ+
F δ–
F δ–
Explain how
hydrogen bond
forms in ammonia.
atom, so the pair electrons shared in the
fluorine atom than with the hydrogen atom.
(a) Nitrogen is more
electronegative
than hydrogen, so pair electrons shared in the
nitrogen
bond are more closely with the
atom than with the hydrogen atom.
(b) This leads to the formation of partially positive charge (δ+) on hydrogen atom, H and
partially negative charge (δ–) on nitrogen atom, N.
(c) The partially positive charged hydrogen atom is then attracted by the other partially
negative charged nitrogen atom to form hydrogen bond .
Hydrogen bond
UNIT
H
Hδ+
H
Nδ–
5
H
How do hydrogen
bonds affect
boiling points?
SP 5.4.2
Covalent bond
Hδ+
Nδ–
H
H
Hδ+
Nδ–
H
The molecule held by the hydrogen bond has a higher boiling point than the molecule
held by the Van der Waals force.
Example 1:
Substance
Ethanol (C2H5OH)
Propane (C3H8)
Relative molecular mass
46
44
Boiling point / ºC
+78
–42
– Ethanol and propane have almost the same relative molecular mass and size.
– The boiling point of ethanol is higher than propane.
– Ethanol has a hydrogen atom attached to an oxygen atom as in a water molecule.
There are hydrogen bonds between ethanol molecules, more energy is needed to
break
the bonds before it boils.
– Propane is a non-polar molecule. Propane molecules are attracted by weak
Van der Waals force . There are no hydrogen bonds between propane molecules.
© Nilam Publication Sdn. Bhd.
82
MODULE • Chemistry FORM 4
Example 2:
Substance
Water (H2O)
Hydrogen sulfide (H2S)
Relative molecular mass
18
34
Boiling point / ºC
100
–60
higher
– The relative molecular mass of hydrogen sulfide is
boiling point of water is higher than hydrogen sulfide.
than water, but the
– There are hydrogen bonds between water (H2O) molecules, more energy is needed
to break the bonds before it boils.
– There is no hydrogen bond between hydrogen sulfide (H2S) molecules due to the
low electronegativity of sulphur atom.
Water acts as a polar solvent because it can be attracted to either the positive or negative
charge on a solute:
(a) The partially positive hydrogen atom on water molecule attracts other partially
negatively-charged regions of other molecules or negatively charged ionic
compound.
(b) The partially negative oxygen atom on water molecule attracts other partially
positively-charged regions of other molecules or positively charged ionic
compound.
Cl–
Na+
Cl–
Water
molecule
Na+
5
Water
molecule
Remark:
The explanation is to explain water as ionic compound solvent which will be learned in the section of
properties of ionic compound.
How do hydrogen
bonds affect
solubility in
water?
Molecules that can form hydrogen bond with water have a higher solubility in water.
Example:
polar
(a) Ethanol (C2H5OH) is a
molecule dissolves in water. Ethanol form
hydrogen bond with water molecules.
H
H H
| |
H–C–C–O–H
| |
H H
H
O
H
Hydrogen bond
O
H
(b) Other examples of polar molecules which form hydrogen bond with water
molecules are ammonia (NH3), hydrogen chloride (HCl), sugar (C6H12O6 ) and
methanol (CH3OH). The polarity of these molecules indicates that they will
dissolve in water.
Remark:
1 Solubility of ethanol in water will be studied in Form 5, carbon compound.
2 Non polar molecule like ethene (C2H4) is insoluble in water.
83
UNIT
Water is a polar
solvent. Explain.
MODULE • Chemistry FORM 4
Exercise
1
TP6
Cellulose and keratin are example of compounds that may form hydrogen bonding with other molecules. The
diagram shows the structural formula of cellulose and keratin.
Cellulose
H
H
|
O
H
O
H
|
O
O
O
O
O
Keratin
H 3C
CH3
H
O
O
O
HO
O
|
H
H
O
O
|
H
Cellulose is an organic compound mainly used to Keratin is a strong natural protein and it is the main
produce paper.
substance to form hair.
UNIT
5
Flipping the
paper is easier
when the
fingertips are
wet with water
compared to dry
fingertips.
Explain.
TP4
HOTS
polar
– The cellulose molecule that form the paper is
molecule.
Hydrogen
bonding
–
is formed when partially positive charged of hydrogen atom
in water is attracted to the partially negative charged of oxygen atom in cellulose.
– When fingertips are wet, there are hydrogen bond between water molecules and
cellulose in paper, thus it is easier to flip paper.
– When fingertips are dry, no hydrogen bond formed between water and cellulose
in paper, thus it is difficult to flip the paper.
Wet hair is
sticky compared
to dry hair.
Explain.
TP4
HOTS
– The keratin molecule that forms the outside layer of hair is the
– Hydrogen bonding is formed when partially positive
polar
molecule.
charged of hydrogen
atom in water is attracted to the partially negative charged of oxygen atom in
keratin.
– When hair is wet, there are hydrogen bond between the water molecule and
keratin in the outer layer of hair. As a result, the hair becomes sticky.
hydrogen bonding
– When hair is dry, there is no
formed between the water
molecule and keratin in the outer layer of hair. As a result, the hair is not sticky.
© Nilam Publication Sdn. Bhd.
84
MODULE • Chemistry FORM 4
2
TP5
The diagram shows arrangement of water molecules in liquid water and ice.
Hydrogen
bond
Ice (Solid)
Water (Liquid)
Hydrogen bonds are stable
Hydrogen bonds constantly
break and re-form
Explain the arrangement
of water molecules in
water (liquid).
In water (liquid), water molecules are closely held together by
hydrogen bond and move randomly.
Explain the arrangement
of water molecules in ice
(solid).
When water freezes, hydrogen bonds are stable, arranging the water
molecules far apart from each other. Hence, the volume of ice becomes
greater than that of the water.
What is the effect of
increase in the volume of
ice?
The effect of the increase in the volume of ice is that its density becomes
lower
than the density of water, thus ice becomes lighter than water.
Why does ice float on
water?
Ice floats because ice is less
dense
5
than water.
DATIVE BOND
SK
5.5
UNIT
5.5
What is dative
bond?
Dative bond or coordinate bond is a type of covalent bond between two atoms in which two
electrons are from one atom only.
Explain the
formation of
dative bond in
ammonium
ion, NH4+.
(a) The reaction between ammonia and hydrogen chloride will produce ammonium chloride.
SP 5.5.1
NH3 + HCl
NH4Cl
(b) Ammonium ion, NH4+, is formed by transferring a hydrogen ion (H+) from the hydrogen
chloride to the lone pair of electrons on the ammonia molecule.
Lone pair electrons
H+
H
H
N
Dative bond
+
H
H
Cl
H
Ammonia
H
N
H
+
Cl
H
Hydrogen
chloride
Ammonium ion,
NH4+
Chloride ion,
Cl–
NH3 + HCl → NH4+ + Cl–
(c) Ammonium ion is positively charged because only the hydrogen ion, H+ is transferred
from the chlorine atom to the nitrogen atom.
(d) The electron from hydrogen atom is left behind on the chlorine atom to form a negative
chloride ion.
85
MODULE • Chemistry FORM 4
Explain the
formation of
dative bond in
hydroxonium
ion, H3O+.
(a) Dissolving hydrogen chloride in water to make hydrochloric acid
H2O + HCl
H3O+ + Cl–
(b) Hydroxonium ion, H3O+ is formed by transferring of a hydrogen ion (H+) from the
hydrogen chloride to the lone pair of electrons on the water molecule.
SP 5.5.1
Lone pair electrons
Dative bond
H+
H
O
+
H
Cl
H
Water
Or
H
O
H
+
Cl
H
Hydrogen
chloride
Hydroxonium ion,
+
H 3O
Chloride ion,
Cl–
H2O + HCl → H3O+ + Cl–
(c) Hydroxonium ion, is positively charged because only the hydrogen ion, H+ is transferred
from the hydrogen chloride to the lone pairs on the oxygen atom.
Remark:
(i) Hydroxonium ion is H3O+ is formed by the combination of hydrogen ion, H+ from any acid and water
molecule.
(ii) If the hydrogen ion is written as H+ (aq), the "(aq)" represents the water molecule that the hydrogen ion is
attached to. When it reacts with an alkali for example, the hydrogen ion will be detached from the water
molecule again.
(Will be studied in the next topic, acid and base)
UNIT
5
5.6
METALLIC BOND
SK
5.6
What is valence
electron?
Valence electrons are the electrons in the outermost shell of an atom.
What is delocalised
electrons in metallic
atom?
In metal, atoms are packed closely together in regular arrangement. The metal is in
solid form. Metal atoms tend to lose their valence electrons and become positive ions.
The valence electrons from the metallic atom are free to move throughout the metal
structure. These mobile electrons are called delocalised electrons.
What is sea of
electrons?
Sea of electrons are the delocalised electrons that are free to move in the space
between metal atoms.
What is metallic
bond?
It is the strong electrostatic force between the sea of electrons and the positive metal
ions.
Explain how are
metallic bonds
formed.
– When the valence electrons released by the metallic atom, the atom become positive
metal ion.
– Metallic bonds are formed from the strong electrostatic attraction between
negatively charged sea of electrons and fixed, positively charged metal ions.
SP 5.6.1
Sea of electron
Valence
electron
Positive metal ion
Remark:
positive
– When an atom loses electrons, it become
ions.
– In metallic bonding, only the outer electrons are mobile. The positive metallic ions are
© Nilam Publication Sdn. Bhd.
86
immobile .
MODULE • Chemistry FORM 4
State the physical
properties of metal.
Explain.
Physical properties
SP 5.6.2
Explanation
High melting and boiling
points
A large amount of heat needed to overcome the
strong electrostatic force between sea of electrons with
the positive ions in metallic bonds.
Good conductor of
electricity
The
delocalised electron
are able to move freely to
delocalised
electrons can
conduct electricity. The
flow and carry the charge from the negative terminal to
the positive terminal when the electric current is applied.
Comparing The Formation of Ionic, Covalent and Metallic Bond
(Group 1, Between non-metals and
non-metals (Group 14, 15,
2 and 13) and non-metals
(Group 15, 16 and 17)
16 and 17)
metals
Between
How bonds are – Electrons is released by
formed?
metal atoms to form positive
ions
– Transfer of electrons to
achieve stable octet electron
arrangement
– Electrons are received by
non-metal atoms to form
negative ions
– Strong electrostatic force
between positive and
negative ions
Example of
electron
arrangement in
the particles
+
A
2–
E
–
Pairs
shared
of electrons are
by same or
non
-metal
different
atoms.
– Sharing of electrons to
achieve stable octet electron
arrangement.
– Two different structures of
covalent substances:
(i) Simple molecules
structure.
(ii) Weak Van der Waals
forces between simple
molecules
(iii) Many atoms bonded to
form a giant covalent
structures
Metallic Bond
Between metal atoms
– Metal atoms lose their
valence electrons to form
a sea of
delocalized electrons
– Strong electrostatic force
between sea of electrons
and metal ions
– Many metal ions bonded
together to form giant
lattice structure
+
A
Strong electrostatic
forces between ions
# Ionic bond is the strong
electrostatic force of
attraction between positively
charged ion and negatively
charged ion.
Strong covalent bond
between atoms in the
molecules
# Covalent bond is the shared
pairs of electrons between
atoms in a molecule.
87
Strong electrostatic force
between sea of electrons
and metal ions
5
State the type
of element
involved.
Covalent Bond
UNIT
Ionic Bond
MODULE • Chemistry FORM 4
5.7
SK
5.7
IONIC AND COVALENT COMPOUND
Experiment to Study the Difference Between Ionic Compound and Covalent Compound
To compare electrical conductivity of ionic compound To compare melting point of ionic
and covalent compound
compound and covalent compound
Manipulated variable:
Lead(II) bromide and naphthalene // Ionic and covalent
compounds
Manipulated variable:
Magnesium chloride and naphthalene//
Ionic and covalent compounds
Responding variable:
Electrical conductivity/deflection of ammeter pointer
Responding variable:
Melting point
Fixed variable:
Carbon electrodes
Fixed variable:
Amount of magnesium chloride and
naphthalene
Hypothesis:
Hypothesis:
Lead(II) bromide cannot conduct electricity in
solid state but can conduct electricity in molten
Magnesium chloride has a higher
SP 5.7.1
To compare solubility of ionic compound and
covalent compound in water and organic
solvent
Manipulated variable:
Magnesium chloride and naphthalene // Ionic and
covalent compounds
Responding variable:
Solubility of ionic and covalent compound in
water and organic solvent
Fixed variable:
Water and cyclohexane // water and organic solvent
Hypothesis:
state. Naphthalene cannot conduct electricity in
melting point than naphthalene
Magnesium chloride is soluble in water but
solid and molten states
Materials:
Magnesium chloride, naphthalene, water
insoluble in water but soluble in organic
Materials:
Lead(II) bromide, naphthalene
Apparatus:
Batteries, carbon electrodes, ammeter, Bunsen burner,
connecting wires, crucible, tripod stand
Apparatus:
Beaker, test tube, Bunsen burner, tripod
stand
Water
insoluble in organic solvent. Naphthalene is
solvent
Materials:
Magnesium chloride, naphthalene, distilled water,
cyclohexane
Apparatus:
Beaker, test tube, spatula
A
Distilled water
Carbon
electrode
UNIT
Lead(II)
bromide
Heat
Magnesium
chloride
5
Procedure:
1 Half spatula of magnesium chloride
and naphthalene powder are placed in
two different test tubes.
2 Both test tubes are heated in water
until the water boils.
3 The changes in physical state are
recorded.
Heat
Procedure:
1 A crucible is filled with lead(II) bromide powder
until it is half full.
2 Two carbon electrodes are dipped into lead(II)
bromide and carbon electrodes are connected to
batteries and ammeter using connecting wire.
3 The deflection on ammeter pointer is observed and
Observation:
recorded.
4 The lead(II) bromide powder is heated strongly until
Compound
it melts.
Magnesium
5 The deflection on ammeter pointer is observed and
chloride
recorded.
6 Steps 1 to 5 are repeated using naphthalene to replace
Naphthalene
lead(II) bromide.
Observation
No change.
Melts easily.
Liquid naphthalene
evaporate
Observation:
Compound
Naphthalene
Deflection of ammeter pointer
Solid
Molten
Lead(II) bromide
✗
✓
Naphthalene
✗
✗
Conclusion:
Magnesium
chloride
Procedure:
1 Half spatula of magnesium chloride and
naphthalene powder are placed in two different
test tubes.
2 About 5 cm3 distilled water is added into each
test tube.
3 The test tubes are shaken.
4 All observations are recorded.
5 Steps 1 to 4 are repeated by replacing water
with cyclohexane.
Observation:
Compound
Magnesium chloride has a higher
melting point than naphthalene.
Naphthalene
Solubility
Distilled water
Cyclohexane
Magnesium
chloride
Soluble
Insoluble
Naphthalene
Insoluble
Soluble
Conclusion:
Conclusion:
Lead(II) bromide cannot conduct electricity in solid
Magnesium chloride is soluble in water but
state but can conduct electricity in molten state.
insoluble in organic solvent. Naphthalene is
Naphthalene cannot conduct electricity in solid and
insoluble in water but soluble in organic
molten states.
© Nilam Publication Sdn. Bhd.
solvent.
88
89
+
_
+ _
+ _ +
_
+ _
+
+
_
amount of energy is needed to
overcome it.
overcome
Small
it. Giant molecules
amount of energy is
UNIT
5
such as silicon dioxide have very high melting
and boiling points.
needed to
molecules.
Low melting and boiling points because of the
weak “Van der Waals” force between
Weak Van der
Waals forces
between
molecules
High melting and boiling points
Compare and
explain
because positive ions and negative
melting and
boiling points. ions are attracted by strong
electrostatic force .
Large
Strong covalent bond between
atoms in the molecules
Carbon dioxide molecule, CO2
Weak Van der Waals forces (intermolecular
force) between molecule.
Strong electrostatic forces between
positive and negative ions
+
_
+
_
+
_
Sodium chloride, NaCl
Covalent compounds
State the type Strong electrostatic force between ions.
of forces
between
particles.
Example of
electron
arrangment/
structure
Ionic compounds
Comparing Physical Properties of Ionic and Covalent Compounds
structure before melting occurs.
high
– Very
melting point.
– Very strong carbon-carbon covalent bonds
broken
have to be
throughout the
Strong covalent bonds between atoms in
the giant structure
Silicon dioxide, SiO2
Carbon atoms
Covalent bond
between carbon atoms
Diamond
Giant covalent compound
MODULE • Chemistry FORM 4
© Nilam Publication Sdn. Bhd.
Most are soluble in water and
insoluble in organic solvent*. This is
because the polarisation of water
molecule. Water molecules have partially
positive end (the hydrogen end) and
partially negative end (the oxygen end).
Lead(II) bromide, PbBr2,
Sodium chloride, NaCl
Copper(II) sulphate, CuSO4
Compare
solubility in
water and
organic
solvent.
Example of
ionic and
covalent
compounds
the anode or cathode.
move . In
free
are not
to
molten or aqueous state, the ions are
free
to move to be attracted to
Compare and Cannot conduct electricity when in
explain
solid
form but is able to conduct
electrical
conductivity. electricity when in molten or
aqueous form. In solid form, the ions
5
90
Naphthalene, C8H10
Acetamide, CH3CONH2
Hexane, C6H14
* Organic solvents are covalent compounds
that exist as liquid at room temperature.
soluble
in water but
in
organic solvents* (example: ether, alcohol,
benzene, tetrachloromethane and propanone).
This is because covalent molecules and organic
solvents are both held together by weak Van der
Waals forces.
Insoluble
molten or aqueous state.
Cannot conduct electricity in all state.
Covalent compound is made up of neutral
molecules . No free moving ions in
Covalent compounds
UNIT
Ionic compounds
Diamond (carbon only)
Graphite (carbon only)
Silicone dioxide (silicone and oxygen)
– Insoluble in water and organic solvent.
– Attractions between solvent molecules and
carbon atoms is not strong enough to
overcome the strong covalent bonds in the
giant covalent structure.
there are delocalized electrons between
hexagonal layers.
there are no delocalized electrons because
all the valence electrons are used for
covalent bonds.
– Graphite can conduct electricity as
– Diamond and silicone dioxide
cannot conduct electricity at any state as
Giant covalent compound
MODULE • Chemistry FORM 4
Uses in daily
life
Magnesium
hydroxide, Mg(OH)2
is used in antacid to
reduce stomach acid
Common salt
Sodium chloride
(NaCl)
Salt
Baking powder
Sodium hydrogen
carbonate
(NaHCO3)
Lime
Calcium oxide
(CaO)
Potassium chloride is
used as fertiliser
FERTILISER
A
Urea,
(NH2)2CO
Ionic compounds
91
GA
R
VINEGAR
UNIT
5
Food stuffs
e.g. sugar
(glucose)
SU
Fire extinguishers
e.g. carbon dioxide
Butane, C4H10
in LPG gas for
cooking
Ethanol
C2H5OH used
in perfumes
Perfumes
Turpentine is a solvent
used for paint
Covalent compounds
Jewellery
Uses of diamond
Pencil
Uses of graphite
Cutting glass or
drilling rocks
Electrode in
batteries
Giant covalent compound
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
SPM PRACTICE
Subjective Questions
1
The table shows the melting points, boiling points and electrical conductivity of five substances, A to E.
Substance Melting point / ºC Boiling point / ºC
A
B
C
D
E
850
3 550
0
789
–98
2 100
4 830
100
1 447
–61
Electrical conductivity
in solid state
Conducts
Does not conduct
Does not conduct
Does not conduct
Does not conduct
Electrical conductivity
when dissolve in water
Does not conduct
Does not conduct
Does not conduct
Conducts
Does not conduct
(a) Which substance are solids at room temperature?
A, B, E
(b) Which substance is an ionic compound?
D
(c) Which substance could be diamond?
B
UNIT
(d) Which substance has delocalized electron?
A
5
2
The diagram on the left shows the carbon dioxide fire extinguisher
canister. Carbon dioxide is stored at high pressure in the liquid state in
the fire extinguisher.
(a) Carbon reacts with oxygen to produce carbon dioxide.
[Given that proton number for carbon is 6 and proton number for
oxygen is 8]
(i)
TP2
Fire extinguisher
canister contains liquid
carbon dioxide
State the type of bond present in this compound.
Covalent bond
(ii) Write the formula of the compound formed.
TP1
CO2
(iii) Explain how a compound is formed between element carbon and oxygen based on their electron
TP2
arrangement.
Carbon atom with electron arrangement 2.4 needs four electrons to achieve stable octet electron
arrangement. Oxygen atom with electron arrangement 2.6 needs two electrons to achieve stable
octet electron arrangement. One carbon atom share four pairs of electrons with two oxygen atoms
to form a molecule with the formula CO2. One carbon atom contributes four electrons and each of
the two oxygen atoms contributes two electrons for sharing to form double covalent bond. One
carbon atom forms two double covalent bond with two oxygen atoms. Carbon atom and oxygen
atom achieve stable octet electron arrangement that is 2.8.
© Nilam Publication Sdn. Bhd.
92
MODULE • Chemistry FORM 4
(iv) Draw the electron arrangement of carbon dioxide.
TP3
O
(v)
TP3
HOTS
C
O
Explain why carbon dioxide fire extinguishers are the only fire extinguisher recommended for fires
involving electrical equipment.
Carbon dioxide is safe to be used on and around electrical equipment. Carbon dioxide cannot
conduct electricity.
The diagram below shows the electron arrangement of compound A. Compound A is formed from the reaction
between element X and element Y.
+
–
X
TP2
Write the electron arrangement for atom of elements X and Y.
2.8.1
X:
Y:
2.8.7
(ii) Compare the size of atoms of elements X and Y. Explain your answer.
TP4
Atom Y is smaller than atom X. Atom X and atom Y have the same number of shells occupied with
electrons. The number of proton in the nucleus of atom Y is more than X. The strength of nuclei
attraction to the electrons in the shells of atom Y is stronger than atom X.
(b) How are X ion and Y ion formed from their respective atoms?
TP2
X ion: Atom X releases one electron
Y ion: Atom Y receives one electron
(c) (i)
TP2
Write the formula for compound A.
XY
(ii) Name type of bond in compound A.
TP2
Ionic bond
(iii) Write the chemical equation for the reaction between element X and element Y to form compound A.
TP3
2X + Y2 2XY
(d) Y can react with carbon to form a compound. Draw the electron arrangement for the compound formed.
TP3 [Given that proton number for carbon is 6]
93
5
(a) (i)
Y
UNIT
3
MODULE • Chemistry FORM 4
4
The table below shows the melting point and electrical conductivity of substances V, W, X and Y.
Substance Melting point (°C)
V
W
X
Y
(a) (i)
TP5
–7
80
808
1 080
Electrical conductivity
Solid
Cannot conduct electricity
Cannot conduct electricity
Cannot conduct electricity
Conduct electricity
Molten
Cannot conduct electricity
Cannot conduct electricity
Conduct electricity
Conduct electricity
Which of the substance is copper? Give reason for your answer.
Y. It can conduct electricity in solid and molten state.
(ii) Explain how copper conducts electricity in solid state.
TP5
Copper atoms release electrons to form free moving delocalized electrons.
When electric is applied, delocalized electrons flow and carry the charge from the negative terminal
to the positive terminal
(b) (i)
TP2
State the type of particles in substances V and W.
Molecule
(ii) Explain why substances V and W cannot conduct electricity in solid and molten state.
TP2
Substances V and W are made up of neutral molecules. No free moving ions in solid and molten
state.
(c) The boiling point of substance V is 59 °C. What is the physical state of substance V at room temperature?
TP2
Liquid
UNIT
(d) Draw the arrangement of particle V at room temperature.
5
TP3
(e) Explain why the melting and boiling points of substances V and W are low?
TP2
Van der Waals / intermolecular forces between molecules are weak. Small amount of heat energy is
required to overcome it.
(f) (i)
TP2
State the type of particle in substance X.
Ion
(ii) Explain why substance X cannot conduct electricity in solid but can conduct electricity in molten
TP2
state.
Ions are not freely moving // ions are in a fixed position in solid state. Ion can move freely in
molten state.
Chemical
Bond
© Nilam Publication Sdn. Bhd.
94
Objective
Questions
MODULE • Chemistry FORM 4
UNIT
6
ACIDS, BASES AND SALTS
Concept Map
Meaning
– Acid is a chemical substance which ionises in water to produce hydrogen ion.
– Bases is a chemical substance that reacts with acid to produce salt and water only.
– Alkali is a base that is soluble in water and ionises to hydroxide ion.
Explain the role of water in the
formation of hydrogen ions and
hydroxide ions
Chemical Properties
Acid
1 Acid + Metal ➝ Salt + Hydrogen
2 Acid + Metal carbonate ➝ Salt + Water +
Carbon dioxide
3 Acid + base ➝ Salt + Water (Neutralisation)
pH = –log [H+]
pH + pOH = 14
Example
pH value of acid
and alkali
Concentration of
hydrogen ion and
hydroxide ion
Base
1 Base + Ammonium salt ➝ Salt + Water +
Ammonia gas
2 Base + Acid ➝ Salt + Water (Neutralisation)
3 Base + Metal ion➝ Metal hydroxide
Based on ionisation of:
– Strong and weak acid /
alkali
– Monoprotic and
diprotic acids
Neutralisation
– Titration acid-alkali
– Application in daily life
The content standard (CS) for Acid and Bases is
designed in the following order to make it easier
for students to master the concepts during the
PdP process.
(6.1) Role of water → (6.3) Strength of acid and
alkali → (6.4) Chemical properties of acid and
alkali → (6.5) Concentration of aqueous solution
→ (6.6) Standard solution → (6.2) pH value →
(6.7) Neutralisation
Uses in daily life
Concentration of
acid and alkali in
mol dm–3 and g dm–3
Preparation standard
solution
95
pH of
solutions
used in daily
life
6
ACID AND
BASE
Can be classified
according to
Strength of acid
and alkali
– Strong acid
and strong
alkali
– Weak acid
and weak
alkali
Basicity of acid
– Monoprotic
acid
– Diprotic acid
UNIT
To compare and explain the
chemical properties of acid /
alkali in water, without
water or in other solvent
MODULE • Chemistry FORM 4
Concept Map
Meaning of salt
SALT
Solubility
Qualitative analysis
Soluble salt
Preparation
Reaction of
acids with:
(a) Alkali
(b) Metal
oxide
(c) Metal
carbonate
(d) Metal
Anion
Insoluble salt
Preparation Determining
formula
Double
decomposition
through
precipitation
Continuous
method
Cation
Confirmatory
test using
reagent
Confirmatory
test using
reagent
UNIT
(a) Silver nitrate and
nitric acid for
chloride ion
(b) Barium nitrate
and nitric acid for
sulphate ion
(c) Iron(II) sulphate
and sulphuric
acid for nitrate
ion
(d) Nitric acid and
lime water for
carbonate ion
(a) Sodium
hydroxide
solution
(b) Aqueous
ammonia
(c) Specific
reagents
6
Salt crystals
1 Characteristics of salt
crystals
2 To grow salt crystal
© Nilam Publication Sdn. Bhd.
Uses in daily life
1 In food preparation
2 In agriculture
3 In medical
substances
96
Action of heat
on salt
Some salts decompose when
they are heated:
Salt → metal oxide + gas
(Colour of residue refers to
certain cation)
(Gas identification refers to
certain anion / cation)
MODULE • Chemistry FORM 4
6.1
THE ROLE OF WATER IN SHOWING ACIDIC AND ALKALINE PROPERTIES
Define acid
SP 6.1.1
Hydrogen chloride
gas is a covalent
compound exist in
the form of molecule.
When the gas is
bubbled into water,
hydrochloric acid is
produced. Explain.
SK
6.1
Acid is a chemical substance which ionises in water to produce hydrogen ion, H+.
– As hydrogen chloride dissolves in water, hydrogen chloride molecule ionises to
hydrogen ion and chloride ion in aqueous solution. This aqueous solution is called
hydrochloric acid.
I
When dissolve in water
– Ionisation equation:
HCl (aq)
Hydrochloric acid +
+
H+ (aq)
+
Hydrogen ion
I
−
Cl– (aq)
Chloride ion
– An aqueous hydrogen ion, H+(aq) is actually the hydrogen ion combined with water
molecule to form hydroxonium ion H3O+. However this ion can be written as H+.
HCl(g) H
+ H2O(l)
Hydrogen
chloride
H3O+
Hydroxonium ion
+
H3O+ (aq)
+
Hydroxonium
ion
H+(aq)
+
Hydrogen ion
Cl– (aq)
Chloride
ion
H2O
The ionisation of hydrochloric
acid is represented as:
H2O
HCl(aq)
H+ (aq) + Cl– (aq)
Remark:
Hydroxonium ion, H3O+ is formed by the transfer of a hydrogen ion (H+) from the hydrogen chloride to
the lone pair of electrons on the water molecule. The type of bond formed is of dative bond (studied in
UNIT 5: chemical bond)
Basicity of an acid is the number of ionisable hydrogen atom per molecule of an acid
molecule in an aqueous solution.
– Monoprotic: One acid molecule ionises to one hydrogen ion.
– Diprotic: One acid molecule ionises to
two
hydrogen ions.
– Triprotic: One acid molecule ionises to
three
hydrogen ions.
Hydrochloric acid is monoprotic acid because one molecule of hydrochloric acid
ionises to one hydrogen ion.
Examples of acid and their basicity
Ionisation of acid
Number of hydrogen ions
produced per molecule
of acid
Basicity of acid
HNO3 (aq)
Nitric acid
–
H+(aq)
+ NO3 (aq)
Hydrogen ion Nitrate ion
One
Monoprotic
H2SO4 (aq)
Sulphuric acid
2–
2H+(aq)
+ SO4 (aq)
Hydrogen ion Sulphate ion
Two
Diprotic
H3PO4 (aq)
Phosphoric acid
3–
+ PO4 (aq)
3H+(aq)
Hydrogen ion Phosphate ion
Three
Triprotic
*CH3COOH (aq)
Ethanoic acid CH3COO–(aq) +
H+(aq)
Ethanoate ion Hydrogen ion
One
Monoprotic
Not all hydrogen atoms in ethanoic acid are ionisable
97
6
SP 6.1.2
Acid tastes sour, corrosive, turns moist blue litmus paper to red and conduct electricity
in aqueous solution state.
UNIT
What are the
physical properties
of acid?
What is basicity of
acid?
MODULE • Chemistry FORM 4
Define base
Bases is a chemical substance that reacts with acid to produce salt and water only.
SP 6.1.1
Example of base
Most bases are metal oxide or metal hydroxide which are ionic compound. Example of
bases are magnesium oxide, zinc oxide, sodium hydroxide and potassium hydroxide.
(a) Copper(II) oxide (a base) reacts with sulphuric acid to produce copper(II) sulphate
(a salt) and water.
CuSO4 +
H 2O
CuO + H2SO4
(b) Zinc hydroxide (a base) reacts with hydrochloric acid to produce zinc chloride (a
salt) and water.
ZnCl2
2H2O
Zn(OH)2 + 2HCl
+
Define alkali
Alkali is a base that is soluble in water and ionises to hydroxide ion, OH–.
SP 6.1.1
Example
(a) Sodium hydroxide dissolves in water and ionises to hydroxide ion.
NaOH(aq)
Na+ (aq) + OH– (aq)
(b) Ammonia solution is obtained by dissolving ammonia molecule in water, ionisation
occur to produce a hydroxide ion, OH–.
NH3(g) + H2O(l)
NH4+ (aq) + OH– (aq)
(c) Other examples of alkalis are barium hydroxide and calcium hydroxide.
What are the
physical properties
of alkali?
Alkali tastes bitter, slippery and turns moist red litmus paper to blue and conduct
electricity in aqueous solution state.
Exercise
UNIT
Complete the following table:
6
Name
TP2
Soluble base (alkali)
Formula
Ionisation equation
Insoluble base
Name
Formula
Sodium
oxide
Na2O
Na2O(s) + H2O
2NaOH(aq)
NaOH(aq)
Na+ (aq) + OH– (aq)
Copper(II)
oxide
CuO
Potassium
oxide
K2O
K2O(s) + H2O
2KOH(aq)
KOH(aq)
K+ (aq) + OH– (aq)
Copper(II)
hydroxide
Cu(OH)2
Ammonia
NH3
Zinc
hydroxide
Zn(OH)2
Sodium
hydroxide
NaOH
NaOH(aq)
Na+ (aq) + OH– (aq)
Aluminium
oxide
Al2O3
Potassium
hydroxide
KOH
KOH(aq)
K+ (aq) + OH– (aq)
Lead(II)
hydroxide
Pb(OH)2
Barium
hydroxide
Ba(OH)2
Ba2+ (aq) + 2OH– (aq)
Magnesium
hydroxide
Mg(OH)2
NH3(g) + H2O
Ba(OH)2(aq)
NH4+ (aq) + OH– (aq)
Bases that can dissolve in water (soluble bases) are known as alkali
© Nilam Publication Sdn. Bhd.
98
MODULE • Chemistry FORM 4
Uses of Acid, Bases and Alkali in Daily Life
Uses of acid
Uses
Acid
Sulphuric acid
To make fertiliser, detergents, paint,
synthetic polymer and electrolyte in
lead-acid accumulator
Production of sulphuric acid in industry
Hydrochloric
acid
As a cleansing agent in a toilet cleaner, to
clean metal before electroplating
Hydrochloric acid react with oxide layer on the
surface of metal
Nitric acid
To manufacture fertilisers, explosive
substances, dyes and plastic
Ethanoic acid
To make vinegar and as a preservative for
pickles
Preservative will be studied in the Form 5
syllabus : Chemical for Consumers (Food
additive)
Benzoic acid
To preserve food
Preservative will be studied in the Form 5
syllabus: Chemical for Consumers (Food
additive)
–
Uses of base and alkali
Alkali / Base
Uses
Remark
Ammonia
To make fertilizer, nitric acid, grease
remover and keep latex in liquid state
Coagulation of latex will be studied in the
Form 5 syllabus: Polymer
Calcium hydroxide
(lime)
To neutralize acidic soil.
To make cement and lime water.
–
Magnesium
hydroxide
To make toothpaste and antacid (to
neutralise excess acid in the stomach)
–
Sodium hydroxide
To make soap and detergents
Aluminium
hydroxide
To make antacid (to neutralise excess
acid in the stomach)
Soap and detergent will be studied in the
Form 5 syllabus: Chemical for Consumers
(soap and detergent)
99
–
6
2
Remark
UNIT
1
MODULE • Chemistry FORM 4
Role of Water and the Properties of Acid
How do you explain
the role of water in the
formation of hydrogen
ion which causes acidic
properties?
– Acid molecules ionise in water to form hydrogen ions. The presence of
hydrogen ions is needed for the acid to show its acidic properties.
Acid does not show acidic
properties without water
or dissolved in organic
solvent. Explain.
molecules
– Acid will remain in the form of
in two conditions:
(a) Without the presence of water for example dry hydrogen chloride gas and
*glacial ethanoic acid.
(b) Acid is dissolved in *organic solvent for example solution of hydrogen
chloride in methylbenzene and ethanoic acid in propanone.
* Glacial ethanoic is pure ethanoic acid
* Organic solvent is covalent compound that exist as liquid at room
temperature such as propanone, methylbenzene and trichloromethane
Example 1: Hydrochloric chloride gas dissolved in propanone and in water.
Hydrogen chloride gas dissolved in propanone
SP 6.1.3
Hydrogen chloride gas dissolved in water
Propanone
Water
UNIT
Hydrogen chloride molecules in propanone do not
ionise .
Hydrogen chloride molecule in water
hydrogen ion and chloride ion:
ionises
to
6
HCl(aq) ➝ H+(aq) + Cl–(aq)
Example 2: Reaction between zinc powder with hydrogen chloride in propanone and hydrogen chloride in water.
Zinc powder with hydrogen chloride
in propanone
– No bubbles.
– Hydrogen chloride in propanone does not react with zinc.
Hydrogen chloride gas
dissolved in propanone
– Hydrogen chloride molecules in propanone do not ionise.
Hydrogen chloride exist as molecule only, there are
no hydrogen ions present.
– Hydrogen chloride in propanone does not show acidic
properties.
Zinc powder
© Nilam Publication Sdn. Bhd.
100
MODULE • Chemistry FORM 4
Zinc powder with hydrogen chloride
in water
– Bubbles are released.
– Hydrogen chloride in water react with zinc.
ionises
– Hydrogen chloride molecule in water
ions
Hydrogen
present.
Hydrogen chloride
gas dissolved in water
.
– Hydrogen chloride in water (hydrochloric acid) shows acidic
properties.
Zinc powder
Example 3: Glacial ethanoic acid and ethanoic acid dissolved in water.
Glacial ethanoic acid
Ethanoic acid dissolved in water
Water
molecule
(i) Glacial ethanoic acid does not
metal, base or metal carbonate.
(ii) Glacial ethanoic acid does not
litmus paper to red.
.
only, no
react with
turn blue
free moving ions, glacial ethanoic
– There are no
acid cannot conduct electricity (non electrolyte).
ionise
– Ethanoic acid molecules
partially in
water to produce hydrogen ion:
CH3COOH (aq)
CH3COO– (aq) + H+ (aq)
– Hydrogen ions are present.
react
(i) Ethanoic acid
or metal carbonate.
blue
(ii) Ethanoic acid turn
red
with metal, base
6
– Glacial ethanoic acid exist as
hydrogen ions present:
ionise
litmus paper to
UNIT
Glacial ethanoic acid molecules do not
.
free moving ions, ethanoic acid can
– There are
conduct electricity (electrolyte).
Role of Water and Properties of Alkali
How do you explain the role of water in the
formation of hydroxide ion which cause
alkaline properties?
Alkali dissolves
hydroxide ions.
Alkaline does not show alkaline properties
without water or dissolved in organic
solvent. Explain.
Without water or in organic solvents, no hydroxide ions are
not shown
produced, so the alkaline properties are
.
101
and
ionises
in water to produce
MODULE • Chemistry FORM 4
Example 1: Sodium hydroxide without water and sodium hydroxide dissolved in water.
Solid sodium hydroxide without water
SP 6.1.3
Sodium hydroxide dissolved in water
(sodium hydroxide solution)
Water
– Sodium ions and hydroxide ions are attracted by
strong electrostatic force in solid sodium hydroxide.
– Sodium hydroxide do not ionise to hydroxide ion.
(i) Solid sodium hydroxide does not react with
ammonium salt.
(ii) Solid sodium hydroxide
red
litmus paper to
does not
turn
blue
.
free moving ions, solid sodium
– There are no
cannot
hydroxide
conduct electricity (non
electrolyte).
– Sodium hydroxide ionise in water to produce
hydroxide ion.
NaOH (aq) ➝ Na+ (aq) + OH– (aq)
– Hydroxide ions are present:
(i) Sodium hydroxide solution
acid and ammonium salt.
reacts
(ii) Sodium hydroxide solution turns
red
litmus paper to
.
– There are
solution
with
blue
free moving ions, sodium hydroxide
can
conduct electricity (electrolyte).
Example 2: Ammonia gas dissolved in propanone and ammonia dissolved in water.
Ammonia gas dissolved in propanone
Ammonia gas dissolved in water
UNIT
6
Propanone
Water
– Ammonia
ionise
molecules
in
propanone
do
not
.
– Ammonia in propanone exist as molecule only,
no hydroxide ions present.
(i) Ammonia in propanone does not react with
ammonium salt.
(ii) Ammonia in propanone
litmus paper to blue.
– There are no
propanone
electrolyte).
free moving
cannot
© Nilam Publication Sdn. Bhd.
does not
turn red
ions, Ammonia in
conduct electricity (non
102
ionise partially
– Ammonia molecules
in water
to produce hydroxide ions.
NH3 (g) + H2O (l)
NH4+(aq) + OH– (aq)
– Hydroxide ions are present.
(i) Ammonia aqueous
ammonium salt.
reacts
(ii) Ammonia aqueous turns
blue
paper to
.
with acid and
red
litmus
– There are free moving ions, ammonia aqueous
can conduct electricity (electrolyte).
103
II
Oxalic acid
+ water
Blue litmus
paper
accepted .
Blue litmus paper turns red
II
Conclusion: Hypothesis is
No change
Effect on blue litmus
paper
I
Test tube
Observation:
UNIT
6
Oxalic acid in water shows acidic
property
Oxalic acid without water does
not show acidic property
Inference
Procedure:
1 Two test tubes are labelled and placed on a test tube rack.
2 About 1 spatula of oxalic acid powder is poured into each test tube.
3 About 2 cm3 distilled water is added to test tube II and the mixture is shaken.
4 A piece of dry blue litmus paper is dipped into each test tube.
5 Any changes are observed and recorded.
I
Blue litmus
paper
Oxalic acid
powder
Aim : To show that water is needed for an acid to show its acidic properties.
Problem statement : Does an acid need water to show its acidic properties?
Manipulated variable : Presence of water
Responding variable : Color change of blue litmus paper
Constant variable : Oxalic acid
Hypothesis : Oxalic acid without water do not shows acidic properties/
do not change blue changes blue litmus paper to red. Oxalic
acid in water shows acidic properties / changes blue litmus
paper to red
Materials : Oxalic acid powder, distilled water and blue litmus paper
Apparatus : Test tube, test tube rack
SP 6.1.3
Barium hydroxide
accepted .
Red litmus paper turns
blue
No change
Effect on red litmus
paper
Conclusion: Hypothesis is
Aqueous
solution
Without
water
Barium
hydroxide
Observation:
Aqueous solution of barium
hydroxide shows alkaline property
Dry barium hydroxide does not
show alkaline property
Inference
Procedure:
1 One spatula of dry barium hydroxide powder is placed in a dry watch glass.
2 A piece of dry red litmus paper is dipped into the dry barium hydroxide
powder.
3 Colour change of red litmus paper is observed and recorded.
4 A few drops of distilled water are added to dissolve dry barium hydroxide
powder in the watch glass.
5 Colour change of red litmus paper is observed and recorded.
Red litmus paper
Aim : To show that water is needed for an alkali to show its alkaline properties.
Problem statement : Does an alkali need water to show its alkaline properties?
Manipulated variable : Presence of water
Responding variable : Colour change of red litmus paper
Constant variable: Barium hydroxide
Hypothesis : Barium hydroxide without water does not change colour of red
litmus paper/does not show alkaline properties. Barium
hydroxide added with water changes red litmus paper to blue /
shows alkaline properties.
Materials : Barium hydroxide, distilled water, red litmus paper
Apparatus : Watch glass, dropper
Experiment to Show That the Presence of Water is Essential for the Acid and Alkali to Show the Acidity and Alkaline Properties
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
6.2
SK
6.3
THE STRENGTH OF ACID AND ALKALI
What is meant by
strong acid and
weak acid?
Explain with
examples.
SP 6.3.1
SP 6.3.2
(a) A strong acid is an acid that ionises completely in water to produce high
concentration of hydrogen ion, H+.
Example:
Hydrochloric acid is a strong acid ionises completely in water to produce high concentration
of hydrogen ions and chloride ions,
HCl (aq)
H+ (aq) + Cl– (aq)
Hydrochloric acid ionises
completely to produce high
concentration of hydrogen ions, H+
Water
Other examples of strong acid:
Strong acid
Ionisation equation
Nitric acid, HNO3
HNO3(aq)
Sulphuric acid, H2SO4
H2SO4 (aq)
H+ (aq) + NO3– (aq)
2H+ (aq) + SO42– (aq)
Particles present
H+ and Cl–
H+ and SO42–
(b) A weak acid is an acid that partially ionises in water to produce low concentration
of hydrogen ion, H+.
UNIT
Example:
Ethanoic acid is a weak acid ionises partially in water to produce low concentration of
hydrogen ions and ethanoate ions,
CH3COOH (aq)
H+ (aq) + CH3COO– (aq)
6
An aqueous solution ethanoic acid ionises
partially to produce low concentration of
hydrogen ions, H+. The majority remains as
molecules, CH3COOH.
Water
Other example of weak acid:
Weak acid
Carbonic acid,
H2CO3
© Nilam Publication Sdn. Bhd.
Ionisation equation
H2CO3 (aq)
104
2H+ (aq) + CO32– (aq)
Particles present
H2CO3, H+ and
CO32–
MODULE • Chemistry FORM 4
SP 6.3.2
Example:
Sodium hydroxide solution is a strong alkali ionises completely in water to produce high
concentration of hydroxide ions and sodium ions:
NaOH (aq)
Na+ (aq) + OH– (aq)
All sodium hydroxide ionises completely to
produce high concentration of hydroxide ion
Water
Other examples of strong alkali:
Strong alkali
Ionisation equation
Potassium
hydroxide, KOH
KOH(aq)
Barium hydroxide,
Ba(OH)2
Ba(OH)2 (aq)
K+ (aq) + OH– (aq)
Ba2+ (aq) + 2OH– (aq)
Particles present
K+ and OH–
Ba2+ and OH–
(b) A weak alkali is an alkali that partially ionises in water to produce low concentration
of hydroxide ion, OH–.
Example:
Ammonia aqueous is a weak alkali ionises partially in water to produce low concentration
of hydroxide ions:
NH3 (g) + H2O (l)
NH4+ (aq) + OH– (aq)
An aqueous ammonia ionizes partially to
produce low concentration of hydroxide ions,
OH– and ammonium ions, NH4+. The majority
remains as ammonia molecules, NH3
Water
105
6
SP 6.3.1
(a) A strong alkali is an alkali that ionises completely in water to produce high
concentration of hydroxide ion, OH–.
UNIT
What is meant by
strong alkali and
weak alkali?
Explain with
examples.
MODULE • Chemistry FORM 4
6.3
1
CHEMICAL PROPERTIES OF ACID AND ALKALI
Describe chemical properties of acid:
Chemical properties
1
Acid + Metal
Hydrogen
Salt +
* Acids react with metals that
are more electropositive than
hydrogen in Electrochemical
Series, acids do not react with
copper and silver (type of
reaction is displacement, the
metals that are placed above
hydrogen in Electrochemical
Series can displace hydrogen
from acid)
* Application of the reaction:
– Preparation of soluble salt (Topic
Salt)
– Preparation of hydrogen gas in
determination of the empirical
formula of copper(II) oxide
(Topic Chemical Formula and
Equation)
2
Acid + Metal carbonate
Salt + Water + Carbon
dioxide
UNIT
6
* Application of the reaction:
– Preparation of soluble salt
(Topic Salt)
– Confirmatory test for anion
carbonate ion in qualitative
analysis of salt (Topic Salt)
© Nilam Publication Sdn. Bhd.
SK
6.4
SP 6.4.1
Example of experiment
Observation
Magnesium + Hydrochloric
acid
– The grey
solid
dissolves.
– Colourless
gas bubbles
are released.
When a
burning
wooden
splinter is
placed at the
mouth of
the test tube,
‘pop sound’
is produced.
Chemical equation:
Mg + 2HCl
– The white
solid
dissolves.
Hydrochloric acid Lime water
– Colourless
gas bubbles
are released.
When the
gas is
Calcium carbonate
passed
(a) About 5 cm3 of dilute
through
hydrochloric acid is
lime water,
poured into a test tube.
the lime
(b) One spatula of calcium
water turns
carbonate powder is
chalky.
added into the acid.
(c) The gas released is
passed through lime
water as shown in the
diagram.
(d) The observations are
recorded.
Chemical equation:
CaCO3 + 2HCl
Burning
wooden splinter
Hydrochloric acid
Magnesium powder
(a) About 5 cm3 of dilute
hydrochloric acid is
poured into a test tube.
(b) One spatula of
magnesium powder is
added to the acid.
(c) A burning wooden
splinter is placed at the
mouth of the test tube.
(d) The observations are
recorded.
Calcium carbonate +
Hydrochloric acid
106
Remark
MgCl2 + H2
Inference:
– Magnesium
reacts
with
hydrochloric acid.
– Hydrogen gas is
released.
CaCl2 + H2O + CO2
Inference:
– Calcium carbonate
reacts
with
hydrochloric acid.
– Carbon dioxide
gas is released.
MODULE • Chemistry FORM 4
Acid + Base / Alkali
Water
Salt +
Copper(II) oxide +
Sulphuric acid
– The black
solid
dissolves.
– The
colourless
solution
turns blue.
Sulphuric acid
* Acid neutralises base/alkali
* Application of the reaction:
– Preparation of soluble salt
(Topic Salt)
Copper(II) oxide
(a) Sulphuric acid is
poured into a beaker
until half full.
(b) The acid is warmed
gently.
(c) One spatula of
copper(II) oxide
powder is added to the
acid.
(d) The mixture is stirred
with a glass rod.
(e) The observations are
recorded.
CuSO4 + H2O
Inference:
– Copper(II) oxide
reacts
with
sulphuric acid.
– The blue solution is
copper(II) sulphate .
Write the chemical formulae for the following compounds:
Compounds
Hydrochloric acid
Chemical formulae
Compounds
HCl
Chemical formulae
Magnesium oxide
MgO
Nitric acid
HNO3
Calcium oxide
CaO
Sulphuric acid
H2 SO4
Copper(II) oxide
CuO
Ethanoic acid
CH3COOH
Lead(II) oxide
PbO
Sodium hydroxide
NaOH
Sodium nitrate
NaNO3
Potassium hydroxide
KOH
Potassium sulphate
K2 SO4
Calcium hydroxide
Ca(OH)2
Barium hydroxide
Ba(OH)2
Sodium carbonate
Na2CO3
Sodium chloride
NaCl
Magnesium hydroxide
Mg(OH)2
Magnesium
Mg
Ammonium sulphate
(NH4)2SO4
Zinc
Zn
Sodium
Na
Hydroxide ion
OH–
Sodium sulphate
Na2 SO4
Carbon dioxide
CO2
Hydrogen gas
H2
CuCO3
Sodium oxide
Na2O
Copper(II) carbonate
Water
Calcium carbonate
H2O
Magnesium nitrate
107
CaCO3
Mg(NO3)2
6
2
Chemical equation:
CuO + H2SO4
UNIT
3
MODULE • Chemistry FORM 4
3
Ionic equation:
What is ionic
equation?
Ionic equation is an equation that shows particles that change during chemical reaction.
How to write ionic
equation?
Example:
(i) Reaction between sulphuric acid and sodium hydroxide solution:
Write balanced equation:
H2SO4 + 2NaOH
Na2SO4 + 2H2O
Write the formula of all the particles in the reactants and products:
2H+ + SO42– + 2Na+ + 2OH–
2Na+ + SO42– + 2H2O
Remove all the particles in the reactants and products which remain
unchanged:
2H+ + SO42– + 2Na+ + 2OH–
2Na+ + SO42– + 2H2O
Ionic equation:
2H+ + 2OH–
2H2O
H+ + OH–
H 2O
(ii) Reaction between zinc and hydrochloric acid :
Write balanced equation:
2HCl + Zn
ZnCl2 + H2
Write the formula of all the particles in the reactants and products:
2H+ + 2Cl– + Zn
Zn2+ + 2Cl– + H2
Remove all the particles in the reactants and products which remain
unchanged:
2H+ + 2Cl– + Zn
Zn2+ + 2Cl– + H2
Ionic equation:
2H+ + Zn
Zn2+ + H2
4
Write the chemical equations and ionic equations for the following reactions:
UNIT
Reactant
6
Hydrochloric acid and
#
magnesium oxide
MgO + 2HCl
MgCl2 + H2O
Hydrochloric acid and sodium
hydroxide
HCl + NaOH
NaCl + H2O
Hydrochloric acid and magnesium
Hydrochloric acid and #calcium
carbonate
Sulphuric acid and zinc
Sulphuric acid and #zinc oxide
Sulphuric acid and #zinc carbonate
Nitric acid and #copper(II) oxide
Nitric acid and sodium hydroxide
#
Chemical equations
2HCl + Mg
MgCl2 + H2
2HCl + CaCO3
CaCl2 + CO2
+ H2O
H2SO4 + Zn
H2SO4 + ZnO
ZnSO4 + H2
ZnSO4 + H2O
H2SO4 +ZnCO3
ZnSO4 + CO2
+ H2O
2HNO3 + CuO
Cu(NO3)2 +
H2O
HNO3 + NaOH
NaNO3 + H2O
Ionic equations
2H+ + MgO
H+ + OH–
2H+ + Mg
Mg2+ + H2O
H2O
Mg2+ + H2
2H+ + CaCO3
Ca2+ + CO2 +
H2O
2H+ + Zn
2H+ + ZnO
Zn2+ + H2
Zn2+ + H2O
2H+ + ZnCO3
Zn2+ + CO2 +
H2O
2H+ + CuO
H+ + OH–
Cu2+ + H2O
H2O
Ions in magnesium oxide, calcium carbonate, zinc oxide, zinc carbonate and copper(II) oxide cannot be
separated because the compounds are insoluble in water and the ions do not ionise.
© Nilam Publication Sdn. Bhd.
108
MODULE • Chemistry FORM 4
Describe chemical properties of alkali:
SP 6.4.2
Write the balanced chemical equation
for the reaction
Chemical properties
2
3
Alkali + Acid Salt + Water
* Alkali neutralises acid.
* Application of the reaction:
– Preparation of soluble salt (Topic Salt)
(a) Potassium hydroxide and sulphuric acid:
Alkali + Ammonium salt Salt +
Water + Ammonia gas
*Ammonia gas is released when alkali is heated
with ammonium salt. Ammonia gas has pungent
smell and turns moist red litmus paper to blue.
* Application of the reaction:
– Confirmatory test for cations ammonium in
qualitative analysis of salt (Topic Salt)
(a) Ammonium chloride and potassium hydroxide:
Alkali + Metal ion Insoluble metal
hydroxide
* Most of the metal hydroxides are insoluble.
* Hydroxides of transition element metals are
coloured.
* Application of the reaction:
– Confirmatory test for cations in qualitative
analysis of salt (Topic Salt)
(a) 2OH–(aq) + Mg2+(aq)
Mg(OH)2(s)
(white precipitate)
(b) 2OH–(aq) + Cu2+(aq)
Cu(OH)2(s)
(blue precipitate)
H2SO4 + 2KOH
K2SO4 + 2H2O
(b) Barium hydroxide and hydrochloric acid:
2HCl + Ba(OH)2
KOH + NH4Cl
BaCl2 + H2O
KCl + H2O + NH3
(b) Ammonium sulphate and sodium hydroxide:
2NaOH + (NH4)2SO4
Na2SO4 + 2H2O + 2NH3
Exercise
1
The table below shows two experiments for the reaction between magnesium with hydrogen chloride in solvent
X and solvent Y.
Set-up of apparatus
Experiment I : Magnesium ribbon
with hydrogen chloride in solvent X.
Experiment II : Magnesium ribbon with
hydrogen chloride in solvent Y.
Observation
Bubbles of colourless gas are released
and magnesium ribbon dissolves
No bubble of gas
(a) Name possible substances that can be solvent X and solvent Y.
TP1
Solvent X : Water
Solvent Y : Propanone/ methylbenzene / trichloromethane
(b) Name the gas released in Experiment I.
TP5
Hydrogen gas
(c) Compare observation in Experiment I and Experiment II. Explain your answer.
TP4
reacts
Hydroyen chloride in solvent X in Experiment I
with magnesium. Hydrogen chloride in
solvent Y in Experiment II
Hydroyen chloride in solvent
does not react
ionises
with magnesium.
to H : HCl
+
H+ + Cl–. H+ ions react with magnesium atom
to produce hydrogen molecule: Mg + 2H+ Mg2+ + H2. Hydrogen chloride in solvent Y remains in the
form of molecule . No hydrogen ion present.
109
6
1
UNIT
5
© Nilam Publication Sdn. Bhd.
110
The presence of hydrogen ions
enable acid to show its acidic
properties:
(a) It changes blue litmus
paper to red
(b) Its pH value is less than 7
(c) It reacts with metal,
base/ alkali and metal
carbonate
2
Alkali dissolves and ionises in
water to form hydroxide ions.
Example:
NaOH (aq)
Na+ (aq) +
–
OH (aq)
The presence of hydroxide
ions enable alkali to show its
alkaline properties:
(a) It changes red litmus
paper to blue
(b) Its pH value is more
than 7
(c) It reacts with acid and
ammonium salt
1
2
Alkali
They conduct electricity because there are free moving ions
Acid dissolves and ionises in
water to form hydrogen ions.
Example:
HCl (aq)
H+ (aq) + Cl– (aq)
Acid /Alkali dissolve in water
1
6
Acid
UNIT
Without the presence of
hydrogen ion, acid does not
show its acidic properties:
(a) It does not change blue
litmus paper to red
(b) Its pH value is 7
(c) It does not react with
metal, base/alkali and
metal carbonate
2
2
1
Without the presence of
hydroxide ion, alkali does not
show its alkali properties:
(a) It does not change red
litmus paper to blue
(b) Its pH value is 7
(c) It does not react with
acid and ammonium salt
Alkali does not ionise without
water or in organic solvent.
There are no hydroxide ions
present
Alkali
• organic solvent is covalent compound that exists as liquid at room
temperature such as propanone, methylbenzene and trichloromethane
They do not conduct electricity because there are no free moving ions
Acid remains in the form of
molecule without water or in
organic solvent. There are no
hydrogen ions present.
1
Acid
Acid /Alkali without water or in organic solvent
Role of Water in Showing the Properties of Acid and Alkali
MODULE • Chemistry FORM 4
→
H+ (aq)
+ Cl– (aq)
111
Acid
: HNO3 (aq) →
→
H+ (aq) + NO3– (aq)
H+ (aq) + Cl– (aq)
Example:
Ethanoic acid : CH3COOH (aq)
UNIT
6
CH3COO– (aq) + H+ (aq)
2–
+
(iii) Sulphuric acid
: H2SO4 (aq) → 2H (aq) + SO4 (aq)
partially ionises
(b) A weak acid is an acid that
in water
+
low
concentration
of
hydrogen
ion,
H
to produce
(ii) Nitric acid
(i) Hydrochloric acid : HCl (aq)
to produce
Example:
in water
high concentration of hydrogen ion, H+
Example:
Ammonia solution : NH3 (g) + H2O (l)
NH4+ (aq) + OH– (aq)
in water to
K+ (aq) + OH– (aq)
Na+ (aq) + OH– (aq)
partially ionises
(b) A weak alkali is an alkali that
low
concentration
of
hydroxide
ion, OH–
produce
(ii) Potassium hydroxide: KOH (aq) →
(i) Sodium hydroxide: NaOH (aq) →
produce
Example:
in water to
3 Base/alkali + Metal ion → Metal
hydroxide
2 Base/alkali + Acid → Salt + Water
1 Base/alkali + Ammonium salt →
Salt + Water + Ammonia gas
high concentration of hydroxide ion, OH-
Base /
Alkali
CHEMICAL PROPERTIES :
Example:
Sodium hydroxide dissolves in water and
ionises to hydroxide ion.
NaOH (aq) → Na+ (aq) + OH– (aq)
ionises to hydroxide ion.
• Most bases are metal oxide or metal hydroxide
Example : CuO, MgO, Pb(OH)2
• Alkali is a base that is soluble in water and
with acid to produce salt and water only.
MEANING :
• Base is a chemical substance that reacts
STRONG ALKALI & WEAK ALKALI :
ionises completely
(a) A strong alkali is an alkali that
Acid + Base / Alkali
➝ Salt + Water
CHARACTERISTICS :
red
1 Change
litmus
blue
paper to
2 pH is more than 7
3 Taste slippery
Complete the following To Compare
the Acid and Base/Alkali
CHARACTERISTICS :
blue
1 Change
litmus
red
paper to
2 pH is less than 7
sour
3 Taste
STRONG ACID & WEAK ACID :
ionises completely
(a) A strong acid is an acid that
2 Acid + Metal carbonate → Salt +
Water + Carbon dioxide
Acid
+ Base/alkali → Salt + Water
3
1 Acid + Metal → Salt + Hydrogen
CHEMICAL PROPERTIES :
Hydrochloric acid → Hydrogen ion + Chloride ion
Example :
HCl (aq)
water to produce hydrogen ion.
MEANING :
Acid is a chemical substance which ionises in
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
6.4
CONCENTRATION OF AQUEOUS SOLUTION
SK
6.5
What is a solution?
A solution is a homogeneous mixture formed when a solute is dissolved in a solvent.
For example copper(II) sulphate solution is prepared by dissolving copper(II) sulphate
powder (solute) in water (solvent).
What is
concentration?
Concentration of a solution is the quantity of solute in a given volume of solution
which is usually 1 dm3 of solution.
(a) Mass of solute in gram per 1 dm3 solution, g dm–3.
SP 6.5.1
Concentration of solution (g dm–3) =
Mass of solute in gram (g)
Volume of solution (dm3)
(b) Number of moles of solute in 1 dm3 solution, mol dm–3.
Concentration of solution (mol dm–3) =
Mass mole of solute (mol)
Volume of solution (dm3)
UNIT
6
What is molarity?
The concentration in mol dm–3 is called molarity or molar concentration. The unit
mol dm–3 can be represented by ‘M’.
What is the
relationship between
number of moles with
molarity and volume
of solution?
Number of mole of solute (mol)
Volume of solution (dm3)
Number of mole of solute (mol) = Molarity × Volume (dm3)
n = MV
Mv
n=
1 000
n = Number of moles of solute
M =Concentration in mol dm–3 (molarity)
V = Volume of solution in dm3
v = Volume of solution in cm3
Can concentration in
mol dm–3 be
converted to g dm–3
and vice versa?
The concentration of a solution can be converted from mol dm–3 to g dm–3 and vice versa.
How to solve
numerical problems
involving molarity of
acid and alkali?
Example:
What is the concentration of the following solution in g dm–3 and mol dm–3?
(a) 10 g sodium hydroxide, NaOH in 500 cm3 solution.
(b) 0.5 mol potassium hydroxide, KOH in 2.5 dm3 solution.
SP 6.5.2
Molarity =
× molar mass of the solute
mol dm–3
÷ molar mass of the solute
Solution:
500 cm3
= 0.5 dm3
1 000
10 g
Concentration of solution =
= 20 g dm–3
0.5 dm3
Relative formula mass of NaOH = 23 + 16 + 1 = 40,
molar mass of NaOH = 40 g mol–1
20 g dm–3
Concentration of solution =
= 0.5 mol dm–3
40 g mol–1
0.5 mol
(b) Concentration of solution =
= 0.2 mol dm–3
2.5 dm3
Relative formula mass of KOH = 39 + 16 + 1 = 56,
Molar mass of KOH = 56 g mol–1
Concentration of solution = 0.2 mol dm–3 × 56 g mol–1 = 11.2 g mol–1
(a) Volume of solution =
© Nilam Publication Sdn. Bhd.
112
d dm–3
MODULE • Chemistry FORM 4
1
TP2
The molarity of sodium hydroxide solution is
2 mol dm–3. What is the concentration of the
solution in g dm–3?
[Relative atomic mass: Na = 23, O = 16, H = 1]
Answer:
3
SP 6.5.2
80 g dm–3
Calculate the molarity of the solution obtained
when 14 g of potassium hydroxide is dissolved
in distilled water to make up 500 cm3 solution.
[Relative atomic mass: K = 39, H = 1, O = 16]
Answer:
Calculate the molarity of a solution which is
prepared by dissolving 0.5 mol of hydrogen
chloride, HCl in distilled water to make up
250 cm3 solution.
2 mol dm–3
4
0.5 mol dm–3
How much of sodium hydroxide in gram should
be dissolved in water to prepare 500 cm3 of
0.5 mol dm–3 sodium hydroxide solution?
[Relative atomic mass: Na = 23, O = 16, H = 1]
Answer:
10 g
UNIT
Answer:
2
6
Exercise
6.5
STANDARD SOLUTION
SK
6.6
What is a standard
solution? SP 6.6.1
Standard solution is a solution in which its concentration is accurately known.
How to prepare a
standard solution?
The steps taken in preparing a standard solution are:
1 Calculate the mass of solute needed to produce the required volume and molarity.
2 The solute is weighed.
3 The solute is completely dissolved in distilled water and then transferred to a
volumetric flask that is partially filled with distilled water.
4 Distilled water is added until the calibration mark of the volumetric flask and the
flask is inverted to make sure thorough mixing.
SP 6.6.2
113
MODULE • Chemistry FORM 4
Example:
Preparing 100 cm3 of 2.0 mol dm–3 standard sodium hydroxide solution.
1
2
3
4
5
SP 6.6.2
Calculate mass sodium hydroxide
2 × 100
Number of mol NaOH / =
= 0.2 mol
1 000
Mass of NaOH = 0.2 mol × 40 g mol–1 = 8.0 g
8.0 g of solid sodium hydroxide is weighed in a dry weighing bottle.
8.0 g of solid sodium hydroxide in the weighing bottle is transferred into
a beaker containing 25 cm3 of distilled water. The mixture is stirred to
dissolve the solid.
The solution from the beaker is then carefully poured into a 100 cm3
volumetric flask through a filter funnel.
The weighing bottle and the beaker are rinsed with a small amount of
distilled water and poured into the volumetric flask.
Solid sodium
hydroxide
ON
OFF
g
g
ON
OFF
Glass
rod
Filter
funnel
100 cm3
volumetric
flask
6
Distilled water is poured into the volumetric flask until the calibration
mark.
Distilled water
Calibration
mark
7
The volumetric flask is then closed with a stopper and inverted a few
times to get homogenous solution.
Calibration
mark
UNIT
2 mol dm–3
6
Preparation of a Solution by Dilution
SP 6.6.3
What is dilution
method?
Dilution is a process of diluting a concentrated solution by adding a solvent such as
water to obtain a diluted solution.
How to prepare a
standard solution by
dilution?
Adding water to the standard solution lowered the concentration of the solution.
Since no solute is added, the amount of solute in the solution before and after dilution
remains unchanged:
Number of mol of solute before dilution
= Number of mole of solute after dilution
M1V1 M2V2
=
1 000 1 000
Therefore,
M1V1 = M2V2
M1 =
V1 =
M2 =
V2 =
© Nilam Publication Sdn. Bhd.
Initial concentration of the solute
Initial volume of the solution in cm3
Final concentration of the solute
Final volume of the solution in cm3
114
MODULE • Chemistry FORM 4
Example:
Preparing 100 cm3 of 0.2 mol dm–3 of sodium hydroxide solution from 2 mol dm–3 of sodium hydroxide solution
using dilution method. SP 6.6.3
1
3
Calculate the volume of 2 mol dm–3
sodium hydroxide solution required.
M1V1 = M2V2
2 mol dm–3 × V1 = 0.2 mol dm–3 × 100 cm3
V1 = 10 cm3
Transfer 10 cm3 of 2 mol dm–3
sodium hydroxide solution into
100 cm3 volumetric flask.
2
Use pipette to draw 10 cm3 of
sodium hydroxide solution from
stock solution (2 mol dm–3 sodium
hydroxide solution).
4
Add distilled water until
the calibration mark.
Calibration
mark
Volumetric
flask
5
Distilled water
The volumetric flask is then closed with a stopper and inverted a few times to
get homogenous solution.
Calibration
mark
2 mol dm–3
TP2
If 300 cm3 water is added to 200 cm3 hydrochloric
acid, 1 mol dm–3. What is the resulting molarity of
the solution?
2
Calculate the volume of nitric acid, 1 mol dm–3
needed to be diluted by distilled water to obtain
500 cm3 of nitric acid, 0.1 mol dm–3.
UNIT
1
SP 6.6.3
6
Exercise
Answer:
6.6
PH VALUE
What is pH scale?
SP 6.2.1
What is measured by
each pH values?
0.4 mol dm–3
Answer:
50 cm3
SK
6.2
The pH is a scale of numbers ranging between 0 to 14 to measure the degree of
acidity and alkalinity of an aqueous solution.
Each pH value is a measure of the concentration of hydrogen ions, H+ or hydroxide
ions, OH–.
115
MODULE • Chemistry FORM 4
Example 1:
Calculate pH value for:
(a) An acid with the
concentration of
hydrogen
= 3.5 × 10–3 mol dm–3
(b) 0.01 mol dm–3 of
hydrochloric acid.
SP 6.2.2
(a) pH = –log [H+]
= –log (3.5 × 10–3)
= 2.46
(b) Hydrochloric acid is a strong acid, ionises completely in water to produce
hydrogen ion.
HCl
→
0.01 mol dm–3
H+
+
0.01 mol dm–3
Cl–
pH = –log [H+]
= –log (0.01) = –log (10–2)
=2
UNIT
6
Example 2:
pH value of nitric acid is
3.30. Calculate the
concentration of
hydrogen ion in the acid.
pH = –log [H+]
3.3 = –log [H+]
[H+] = antilog(–3.3)
[H+] = 5.0 × 10–4
What is measured by
each pOH value?
Each pH value is a measure of concentration of hydroxide ions, OH– in a solution.
pOH = –log [OH–]
where [OH–] = concentration of the hudroxide ion, OH– in mol dm–3.
Example 3:
Calculate the pOH
value of an alkali if the
concentration of OH–
ion is 0.0001 mol dm–3.
pOH = –log [OH–]
pOH = –log (0.0001) = –log (10–4)
pOH = 4
What is the
relationship of pH and
pOH?
The sum of pH and pOH is always 14.
pH + pOH = 14
Example 4:
What are the pOH and
the pH value of
potassium hydroxide,
0.0125 mol dm–3?
Potassium hydroxide is a strong alkali, ionises completely to produce hydroxide ion.
KOH
→ K+ +
OH–
–3
0.0125 mol dm
0.0125 mol dm–3
pOH = −log [OH−] = −log 0.0125
= −(−1.903) = 1.903
The pH can be obtained from the pOH value:
pH + pOH = 14.00
pH = 14.00 − pOH = 14.00 − 1.903 = 12.10
What is the
relationship of the
concentration of
hydrogen and
hydroxide ions with pH
value?
pH
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14
–7
+
0
–1
–2
–3
–4
–5
–6
–8
–9
[H ] 10 10 10 10 10 10 10 10 10 10 10–10 10–11 10–12 10–13 10–14
[OH–] 10–14 10–13 10–12 10–11 10–10 10–9 10–8 10–7 10–6 10–5 10–4 10–3 10–2 10–1 100
Remark: Here it helps to rewrite the concentration as 1.0 × 10–4
pH < 7:
• Acidic solution.
• The higher the concentration of
hydrogen ion, H+, the lower the
pH value.
pH = 7
Neutral
Remark:
[H+] = concentration of hydrogen ion in mol dm–3
[OH–] = concentration of hydroxide ion in mol dm–3
pH = –log(10–x) = –(–x)log10 = x(1) = x and
pOH = –log(10–y) = –(–y)log10 = y(1) = y
© Nilam Publication Sdn. Bhd.
116
pH > 7:
• Alkaline solution.
• The higher the concentration
of hydroxide ion, OH–, the
higher the pH value.
MODULE • Chemistry FORM 4
pH of solutions used in
daily life. Complete the
table.
Solution
pH value pOH
[H+]
[OH–]
–3
(mol dm ) (mol dm–3)
Acidic / alkaline /
neutral
1
13
10–1
10–13
Acidic
Lime juice
2
12
10–2
10–12
Acidic
Carbonated
drink
3
11
10–3
10–11
Acidic
Vinegar
3
11
10–3
10–11
Acidic
Orange juice
4
10
10–4
10–10
Acidic
Coffee
5
9
10–5
10–9
Acidic
Tea
5
9
10–5
10–9
Acidic
Milk
6
8
10–6
10–7
Acidic
Distilled water
7
7
10–7
10–7
Neutral
Toothpaste
8
6
10–8
10–6
Alkaline
Blood
8
6
10–8
10–6
Alkaline
Detergent
10
4
10–10
10–4
Alkaline
Household
cleaner
11
3
10–11
10–3
Alkaline
6
Gastric juice
State the colours of the
acid-base indicators in
acidic, neutral and
alkaline solution.
The pH of an aqueous solution can be measured by:
(a) pH meter
(b) Acid-base indicator
Indicator
UNIT
How to measure the
pH value of a solution?
Colour
Acid
Neutral
Alkali
Litmus solution
Red
Purple
Blue
Methyl orange
Red
Orange
Yellow
Colourless
Colourless
Pink
Red
Green
Purple
Phenolphthalein
Universal
indicator
117
MODULE • Chemistry FORM 4
Exercise
1
TP2
Stomach acid is a solution of hydrochloric.
The concentration of hydrogen ion in the acid
is 1.2 × 10–3 mol dm–3. What is the pH value of
stomach acid?
pH = −log [H+]
= −log [1.2 × 10–3]
= −(−2.92) = 2.92
Answer:
3
2
What is the hydroxide ion concentration in a
solution that has a pOH value of 5.70?
5.70 = –log [OH–]
–5.70 = log [OH–]
[OH–] = 10–5.70 = 2.00 × 10–6 mol dm–3
2.92
Answer:
Blood has a pH of 7.3 (slightly alkaline). Calculate
the concentration of hydrogen ions and hydroxide
ions in the blood.
pH = −log [H+] = 7.3
log [H+] = −7.3
[H+] = antilog −7.3
[H+] = 5 × 10–8 mol dm–3
pOH + pH = 14
pOH = 14 – 7.3 = 6.7
pOH = –log [OH–] = 6.7
log [OH–] = –6.7
[OH–] = 2.00 × 10–7 mol dm–3
4
2 × 10–6 mol dm–3
Water exposed to air contains carbonic acid,
H2CO3 due to the reaction between carbon dioxide
and water:
CO2(aq) + H2O(l)
H2CO3(aq)
The concentration of hydrogen ion air-saturated
water caused by the dissolved CO2 is
2.0 × 106 mol dm–3. Calculate the pH of the
solution.
pH = −log [H+]
= −log (2.0 × 10–6)
= 5.70
UNIT
6
[H+] = 5 × 10–8 mol dm–3
–
–7
–3
Answer: [OH ] = 2.00 × 10 mol dm
Answer:
5.70
The Relationship between pH Value and Concentration of Acid and Alkali
How do you relate the
pH with the molarity
of an acid and an
alkali?
The pH value of an acid or an alkali depends on the concentration of hydrogen ions
or hydroxide ions:
The higher the concentration of hydrogen ions in acidic solution, the lower the
pH value.
The higher the concentration of hydroxide ions in alkaline solution, the higher
the pH value.
What are the factors
that can affect the
concentration of
hydrogen and
hydroxide ions of acid
and alkali?
The pH value of an acid or an alkali depends on:
(a) The strength of acid or alkali
– the degree of ionisation or dissociation of the acid and alkali in water.
(b) Molarity of acid or alkali
– the concentration of acid or alkali in mol dm–3.
(c) Basicity of an acid
– the number of ionisable hydrogen atom per molecule of an acid molecule in
an aqueous solution.
© Nilam Publication Sdn. Bhd.
118
MODULE • Chemistry FORM 4
Experiment to investigate the relationship between pH values with the molarity of solution:
SP 6.2.3
Problem statement
: How does the concentration of an acid and an alkali affect the pH value?
Manipulated variable : Concentration of sodium hydroxide solution and hydrochloric acid.
Responding variable : pH value
Constant variable
: Sodium hydroxide solution and hydrochloric acid
Hypothesis : The higher the concentration of sodium hydroxide solution, the higher the pH value. The higher the
concentration of hydrochloric acid, the lower the pH value.
Materials : Hydrochloric acid of 0.1 mol dm–3, 0.01 mol dm–3, 0.001 mol dm–3 and 0.0001 mol dm–3
Sodium hydroxide solution of 1 mol dm–3, 0.1 mol dm–3, 0.01 mol dm–3, 0.001 mol dm–3 and 0.0001
mol dm–3
Apparatus : 100 cm3 beakers, pH meter
1.21
pH meter
Hydrochloric acid
Procedure:
1 About 30 cm3 of 1.0 mol dm–3 hydrochloric acid is poured into a dry beaker.
2 A clean and dry pH meter is dipped into the hydrochloric acid as shown in the diagram.
3 The value of pH meter is recorded.
4 Steps 1 to 3 are repeated by replacing 1.0 mol dm–3 hydrochloric acid with 0.1 mol dm–3, 0.01 mol dm–3,
0.001 mol dm–3 and 0.0001 mol dm–3 hydrochloric acid.
5 The experiment is repeated by replacing hydrochloric acid with sodium hydroxide solutions of different
concentration.
1.0
0.1
0.01
0.001
0.0001
0
1
2
3
4
Molarity of NaOH (mol dm–3)
1.0
0.1
0.01
0.001
0.0001
pH value
14
13
12
11
10
pH value
2
Sodium hydroxide solution
Conclusion:
Hypothesis is accepted. The higher the concentration of sodium hydroxide solution, the higher the pH value. The
higher the concentration of hydrochloric acid, the lower the pH value.
119
UNIT
Molarity of HCl (mol dm–3)
6
Result:
1 Hydrochloric acid
UNIT
6
© Nilam Publication Sdn. Bhd.
2.00
0.01 mol dm–3 HCl
1.00
0.1 mol dm–3 HCl
I
120
H+ + Cl–
0.1 mol dm–3
H+
+ Cl–
0.1 mol dm–3
– The pH value of 0.1 mol dm–3 of
lower
hydrochloric acid is
than
–3
0.01 mol dm of hydrochloric acid.
0.1 mol dm–3 hydrochloric acid is higher
than 0.01 mol dm–3 hydrochloric acid.
– Concentration of hydrogen ion in
HCl
0.01 mol dm–3
– 0.01 mol dm–3 of hydrochloric acid ionises
–3
to form 0.1 mol dm
hydrogen ion:
HCl 0.1 mol dm–3
Compare and – Hydrochloric acid is a strong acid
explain the
ionises
completely in water to
concentration
hydrogen ion.
of ion
–3
hydrogen, H+ – 0.1 mol dm of hydrochloric acid ionises to
–3
and pH value
form 0.1 mol dm
hydrogen ion:
pH meter
reading
Experiment
2–
ionises
H+
+ Cl–
0.05 mol dm–3
0.05 mol dm–3
– The pH value of 0.05 mol dm–3 of sulphuric
lower
acid is
than 0.05 mol dm–3 of
hydrochloric acid.
(higher than)
hydrochloric acid.
– Concentration of hydrogen ion in 0.05 mol
double of
dm–3 sulphuric acid is
HCl
0.01 mol dm–3
– 0.05 mol dm–3 of hydrochloric acid ionises
completely in water to form
0.05 mol dm–3 hydrogen ion:
– Hydrochloric acid is a strong monoprotic
acid.
H2SO4 2H + SO4
0.1 mol dm–3
0.05 mol dm–3
+
completely in water to form
0.1 mol dm–3 hydrogen ion:
– 0.05 mol dm–3 of sulphuric acid
diprotic
0.05 mol dm–3 HCl
0.05 mol dm–3 H2SO4
– Sulphuric acid is a strong
acid.
1.30
1.00
II
0.1 mol dm–3 CH3COOH
3.45
weak
acid ionises partially in
H+ + Cl–
0.1 mol dm–3
less than
0.1 mol dm–3
H+
+ CH3COO–(aq)
– The pH value of 0.1 mol dm–3 of hydrochloric acid
lower
than of 0.1 mol dm–3 of ethanoic acid.
– Concentration of hydrogen ion in 0.1 mol dm–3 of
higher
hydrochloric acid is
than of 0.1 mol dm–3
of ethanoic acid.
0.1 mol dm–3
CH3COOH(aq)
– 0.1 mol dm–3 of ethanoic acid ionises to produce less
–3
than 0.1 mol dm
hydrogen ion:
ion.
water to produce lower concentration of hydrogen
– Ethanoic acid is a
HCl 0.1 mol dm–3
– 0.1 mol dm–3 of hydrochloric acid ionises completely in
–3
water to form 0.1 mol dm
hydrogen ion:
– Hydrochloric acid is a strong acid ionises
completely in water to produce high hydrogen ion.
0.1 mol dm–3 HCl
1.00
III
Example:
The diagram below shows the reading of pH meter for different types and concentration of acids. The aim of the experiment is to investigate the relationship between
concentration of hydrogen ions with the pH value. Compare the concentration of hydrogen ions and the pH value of the following acids. Explain your answer.
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
Exercise
1
The table below shows the pH value of a few substances.
Substance
pH value
Ethanoic acid 0.1 mol dm
3
–3
Hydrochloric acid 0.1 mol dm
1
Glacial ethanoic acid
7
–3
(a) (i)
TP1
What is meant by weak acid and strong acid?
Weak acid : An acid that partially ionises in water to produce low concentration of hydrogen
ion, H+.
Strong acid : An acid that ionises completely in water to produce high concentration of hydrogen
ion, H+.
(ii) Between ethanoic acid and hydrochloric acid, which acid has the higher concentration of H+ ion?
TP4
Explain your answer.
higher
– Hydrochloric acid has
concentration of H+ ions than ethanoic acid.
– Hydrochloric acid is a strong acid which ionises
higher
concentration of H+ ions:
H2O
which ionises
H2O
H+(aq) + Cl–(aq)
partially in water to produce
lower
CH3COO– (aq) + H+ (aq)
(iii) Why do ethanoic acid and hydrochloric acid have different pH value?
TP2
– The higher the concentration of H+ ions, the lower the pH value.
higher , the pH value is
– The concentration of H+ ions in hydrochloric acid is
– The concentration of H+ ions in ethanoic acid is
lower
, the pH value is
lower
higher
.
.
(b) Glacial ethanoic acid has a pH value of 7 but a solution of ethanoic acid has a pH value less than 7. Explain
TP4 the observation.
molecules
ionise
– Glacial ethanoic acid
do not
. Glacial ethanoic acid consists of only
neutral . No hydrogen ions
CH3COOH molecules . The CH3COOH molecules are
present. The pH value of glacial ethanoic acid is 7.
– Ethanoic acid ionises partially in water to produce ethanoate ions and hydrogen ions causes
the solution to have acidic property. The pH value of the solution is less than 7.
2
The table shows the pH value of a few solutions.
Solution
pH
(a) (i)
TP2
P
1
Q
3
R
5
S
7
Which solution has the highest concentration of hydrogen ion?
Solution P
(ii) Which solution has the highest concentration of hydroxide ion?
Solution U
121
T
11
U
14
6
CH3COOH(aq)
in water to produce
UNIT
HCl(aq)
– Ethanoic acid is a weak acid
concentration of H+ ions:
completely
MODULE • Chemistry FORM 4
(b) Which of the following solution could be
TP2
(i)
Q
0.01 mol dm–3 of hydrochloric acid?
(ii) 0.01 mol dm–3 of ethanoic acid?
R
(iii) 0.1 mol dm–3 ammonia aqueous?
T
(iv) 1 mol dm–3 of hydrochloric acid?
P
(v)
U
1 mol dm–3 sodium hydroxide solution?
S
(vi) 1 mol dm–3 potassium sulphate solution?
(c) (i)
TP3
State two solutions which react to form a neutral solution.
P/Q/R and T/U
Hydrochloric acid/ethanoic acid with ammonia aqueous/sodium hydroxide solution.
(ii)
3
TP4
Which solutions will produce carbon dioxide gas when calcium carbonate powder is added?
P/Q//Hydrochloric acid/ethanoic acid
(a) Compare the number of mol of H+ ions which are present in 50 cm3 of 1 mol dm–3 of sulphuric acid and 50 cm3
of 1 mol dm–3 of hydrochloric acid. Explain your answer.
Acid
Calculate
number of
hydrogen ion,
H+
50 cm3 of 1 mol dm–3 of sulphuric acid 50 cm3 of 1 mol dm–3 of hydrochloric acid
Number of mol of sulphuric acid
50 × 1
=
1 000
= 0.05 mol
Number of mol of hydrochloric acid
50 × 1
=
1 000
= 0.05 mol
2H+ + SO42–
H2SO4
From the equation,
H+ + Cl–
HCl
From the equation,
UNIT
1 mol H2SO4 :
6
0.05
mol H2SO4 :
2
mol H+
0.1
mol H+
1 mol HCl :
0.05
mol HCl :
Compare the
The number of H+ ion in 50 cm3 of 1 mol dm–3 of sulphuric acid is
number of
number of H+ in 50 cm3 of 1 mol dm–3 of hydrochloric acid.
hydrogen ions
Explanation
1
mol H+
0.05
mol H+
twice
the
diprotic
acid whereas hydrochloric acid is monoprotic acid.
1 mol of sulphuric ionises to 2 mol of H+ ions whereas 1 mol of hydrochloric acid
ionises to 1 mol of H+ ions. The number of H+ ions in both acid with the same
Sulphuric acid is
volume and concentration is
acid.
doubled
in sulphuric acid compared to hydrochloric
(b) Suggest the volume of 1 mol dm–3 of hydrochloric acid that has the same number of H+ with 50 cm3 of
TP2 1 mol dm–3 of sulphuric acid.
100 cm3
© Nilam Publication Sdn. Bhd.
122
MODULE • Chemistry FORM 4
6.7
NEUTRALISATION
What is neutralisation?
SK
6.7
Neutralisation is the reaction between an acid and a base to form only salt and water:
SP 6.7.1
Acid + Base
Salt + Water
Example:
HCl (aq) + NaOH (aq)
2HNO3 (aq) + MgO (s)
What happens during
neutralisation
reaction?
NaCl (aq) + H2O (l)
Mg(NO3)2 (aq) + H2O (l)
During neutralisation, the acidity of an acid is neutralised by an alkali. At the same
time, the alkalinity of an alkali is neutralised by an acid. The hydrogen ions in the
acid react with hydroxide ions in the alkali to produce water:
H+ (aq) + OH– (aq)
Application of neutralisation in daily life:
SP 6.7.3
Application
Agriculture
H2O (l)
Example
soda lime
limestone
1 Acidic soil is treated with powdered
(calcium oxide, CaO),
(calcium carbonate, CaCO3) or ashes of burnt wood.
acidic
2 Basic soil is treated with compost. The
gas from the decomposition of
neutralises
compost
the alkalis in basic soil.
1 Acidic gases emitted by industries are neutralised with
before the gases are released into the air.
soda lime
(calcium oxide, CaO)
ammonia solution /
2 Organic acid produced by bacteria in latex is neutralises by
ammonium hydroxide and prevents coagulation.
3 Effluents from the electroplating industry contain acids such as sulphuric acid. It is treated by
adding lime to neutralise it before it is discharged into rivers and streams.
Health
1 Excess acid in the stomach is neutralised with its anti-acids that contain bases such as
aluminium hydroxide , calcium carbonate and
magnesium hydroxide .
2 Toothpastes contain
bases
(such as magnesium hydroxide) to neutralise the acid
produced by bacteria in the mouth.
Baking powder (sodium hydrogen carbonate) is used to cure acidic bee stings.
3
Vinegar (ethanoic acid) is used to cure alkaline wasp sting.
4
123
UNIT
Industries
6
3 The acidity of water in farming is controlled by adding soda lime (calcium oxide, CaO).
4 Neutralisation reaction between acid and alkali can produce fertiliser. For example,
ammonium nitrate, ammonium sulphate and urea.
MODULE • Chemistry FORM 4
Exercise
The diagram below shows two different fertilisers used by the farmers
onto the crops so that the crops can grow faster and bigger and thus,
their crop yields can be increased.
By using your knowledge in chemistry, determine the best fertiliser to
be used by the farmers. TP5
Answer:
FERTILISER
A
Urea,
(NH2)2CO
FERTILISER
B
Ammonium nitrate,
NH4NO3
Urea, (NH2)2CO
Percentage of N =
28
× 100% = 46.67%
60
Ammonium nitrate, NH4NO3
28
Percentage of N =
× 100% = 35%
80
Urea, (NH2)2CO is the best fertiliser because it contains higher percentage of nitrogen by mass.
An Acid-base Titration
SP 6.7.2
UNIT
It is a technique used to determine the volume of an acid required to neutralise a fixed
volume of an alkali with the help of acid-base indicator.
What is end point?
– When the acid has completely neutralised the given volume of an alkali, the
titration has reached the end point.
– The end point is the point in the titration at which the indicator changes colour.
– The commonly used indicators are phenolphthalein and methyl orange.
– The volume of acid obtained from the titration can be used to calculate the
concentration of alkali.
6
What is acid-base
titration?
Example: SP 6.7.2
Aim
: To determine the concentration of hydrochloric acid using a standard solution of sodium hydroxide
solution using acid-base titration.
Apparatus : Pipette and pipette filler, 250 cm3 conical flask, burette, retort stand and white tile.
Material : X mol dm–3 of hydrochloric acid, 1 mol dm–3 of sodium hydroxide solution, phenolphthalein
X mol dm–3 of hydrochloric acid
1 mol dm–3 of sodium hydroxide
solution + phenolphthalein
© Nilam Publication Sdn. Bhd.
124
MODULE • Chemistry FORM 4
Procedure:
1 Fill a burette with an X mol dm–3 of hydrochloric acid and record initial burette reading.
2 Use a pipette to measure 100 cm3 of 1 mol dm–3 of sodium hydroxide solution and transfer into a conical flask.
3 Add 2 to 3 drops of phenolphthalein to the sodium hydroxide solution.
4 Place the conical flask on top of a white tile.
5 Add hydrochloric acid drop by drop into sodium hydroxide solution and swirl the conical flask.
6 Continue adding acid until the purple colour of phenolphthalein change and record final burette reading.
7 Repeat the titration twice.
Result:
Titration
I
II
III
Final burette reading (cm3)
V2
V4
V6
Initial burette reading (cm3)
V1
V3
V5
V 2 – V1 = x
V 4 – V3 = y
V 6 – V5 = z
Volume of hydrochloric acid (cm3)
Calculation:
SP 6.7.3
The average volume of hydrochloric acid =
x+y+z
3
= q cm3
Write a balanced equation.
Write the information from the titration above the equation.
M = 1 mol dm–3 M= ?
V = 100 cm3
V = q cm3
NaOH (aq) + HCl ➝ NaCl + H2O
Calculate mol of sodium hydroxide solution by using the formula:
Mv
n=
1 000
Number of moles
hydroxide
100 × 1
=
= 0.1
1 000
Use the mole ratio of the substances involved to find the number of
moles of hydrochloric acid.
From the equation,
1 mol NaOH : 1 mol HCl
0.1 mol NaOH : 0.1 mol HCl
Convert the mole of hydrochloric acid into the unit required by
using the formula:
Mv
n=
1 000
Mv
or n = MV
1 000
= Number of moles of solute
= Concentration in mol dm–3 (molarity)
= Volume of solution in dm–3
= Volume of solution in cm3
n=
n
M
V
v
125
UNIT
6
sodium
M×q
1 000
0.1 × 1 000
M=
mol dm–3
q
0.1 mol =
Remark:
of
MODULE • Chemistry FORM 4
6.8
SALT, CRYSTALS AND USES OF SALT IN DAILY LIFE
What is salt?
SK
6.8
A salt is a compound formed when the hydrogen ion in an acid is replaced with
metal ion or ammonium ion. Example: Sodium chloride, copper(II) sulphate,
potassium nitrate and ammonium sulphate.
SP 6.8.1
Write the formulae of the salts in the Table of Salts below by replacing hydrogen ion in sulphuric acid, hydrochloric
acid, nitric acid and carbonic acid with metal ions or ammonium ion:
Table of Salts
UNIT
6
Metal
ion
Sulphate salt
(from H2SO4)
Chloride salt
(from HCl)
Nitrate salt
(from HNO3)
Carbonate salt
(from H2CO3)
Na+
Na2SO4
NaCl
NaNO3
Na2CO3
K+
K2SO4
KCl
KNO3
K2CO3
NH4+
(NH4 )2SO4
NH4Cl
NH4NO3
(NH4 )2CO3
Mg2+
MgSO4
MgCl2
Mg(NO3 )2
MgCO3
Ca2+
CaSO4
CaCl2
Ca(NO3 )2
CaCO3
Al3+
Al2(SO4 )3
AlCl3
Al(NO3 )3
Al2(CO3 )3
Zn2+
ZnSO4
ZnCl2
Zn(NO3 )2
ZnCO3
Fe2+
FeSO4
FeCl2
Fe(NO3 )2
FeCO3
Pb2+
PbSO4
PbCl2
Pb(NO3 )2
PbCO3
Cu2+
CuSO4
CuCl2
Cu(NO3 )2
CuCO3
Ag+
Ag2SO4
AgCl
AgNO3
Ag2CO3
Ba2+
BaSO4
BaCl2
Ba(NO3 )2
BaCO3
Physical Characteristics of Salt Crystals
How are salt crystals
formed?
Salt crystals are formed when a saturated salt solution is cooled down.
Example of crystals
Sodium chloride crystals
What are the
characteristics of
crystals?
Copper(II) sulphate crystals
Calcium fluoride crystals
Crystals of the same type of salt have the following characteristics:
(a) Fixed geometrical shape. Their sizes could be different. Fast crystallisation
will produce small crystals, slow crystallisation will produce bigger crystals
(b) Flat faces, straight edges and sharp angels
(c) Fixed angle between adjacent faces
Remark:
Crystals of the different salts have different geometric shape like cubes, rhombus, cuboid and prism.
© Nilam Publication Sdn. Bhd.
126
MODULE • Chemistry FORM 4
How are the soluble
salts purified by
recrystallisation?
– The insoluble impurities in salt solution are removed by filtration whereas the
soluble are removed by recrystallisation.
– Steps in recrystallisation process includes:
(i) The soluble salt is dissolved in a suitable solvent, usually water.
(ii) The aqueous solution is then heated to evaporate until the solution saturated.
(iii) When the hot saturated solution is allowed to cool, the salt reappears as pure
crystals, leaving behind the impurities in the solvent.
(iv) The crystals are obtained as a residue through ltration.
How to grow
copper(II) sulphate
crystals?
– Growing copper(II) sulphate crystals is done by hanging a small copper(II)
sulphate crystal in a saturated copper (II) sulphate solution.
– Let the water evaporate slowly to get a larger copper(II) sulphate crystal.
How to grow
copper(II)
sulphate crystal
Uses of Salt in Daily Life
Glass rod
Nylon thread
Small copper(II)
sulphate crystals
Saturated copper (II)
sulphate solution
SP 6.8.3
Examples:
Agriculture
Medical
substances
Other uses
Uses
Sodium chloride, NaCl
As food seasoning and food preservative in salted
fish
Monosodium glutamate (MSG)
To enhance the taste of food
Sodium hydrogen carbonate (NaHCO3)
As baking powder in cake and bread
Sodium benzoate (C6H5COONa)
Preservative in food such as tomato sauce, oyster
sauce and jam
Sodium nitrate (NaNO3)
Preservative in processed meat such as burger and
sausages
– Nitrate salt such as potassium nitrate
(KNO3), sodium nitrate (NaNO3)
– Ammonium salt such as ammonium
sulphate [(NH4)2SO4], ammonium
nitrate (NH4NO3)
Fertiliser
Copper(II) sulphate (CuSO4) and
iron(II) sulphate (FeSO4)
Pesticides
Calcium carbonate (CaCO3) and
calcium hydrogen carbonate (CaHCO3)
Antacid to reduce acidity in the stomach of gastric
patient
Calcium sulphate (CaSO4)
To make plaster of Paris that is used to support
fractured bones
Barium sulphate, BaSO4
Enables the intestine of suspected stomach cancer
patients to be seen clearly in X-ray film
Potassium manganate (KMnO4)
As antiseptic to kill germ
Silver bromide (AgBr)
To produce black and white photographic film
Tin(II) fluoride, SnF2
Added to toothpaste to prevent tooth decay
127
6
Food
preparation
Salt
UNIT
Application
MODULE • Chemistry FORM 4
6.9
PREPARATION OF SALTS
SK
6.9
What are the classification
of salt?
(i) Soluble salt
State general rules of
solubility for salt in water.
(i) All salts K+, Na+ and NH4+ are soluble.
(ii) All nitrate salts are soluble.
(iii) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3.
(iv) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4.
(v) All chloride salts are soluble except PbCl2 and AgCl.
* Based on the solubility of the salts in water, shade the insoluble salts in the
Table of Salts in page 126.
Plan an experiment to study
the solubility of various salt
in water.
Aim: To study solubility of various salts in water.
Problem statement: Are all salts of sulphate, chloride, nitrate and carbonate
soluble in water?
Manipulated variable: Sulphate salts, chloride salts, nitrate salts, carbonate
salts
Responding variable: Solubility of salts in water
Constant variable: Volume of distilled water, quantity of salt
Hypothesis:
(i) All nitrate salts are soluble.
(ii) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3.
(iii) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4.
(iv) All chloride salts are soluble except PbCl2 and AgCl.
Material: Powder of nitrate, sulphate, chloride and carbonate salts for
sodium, potassium, ammonium, barium, silver, lead, calcium and magnesium,
distilled water
Apparatus: Test tubes, Bunsen burner, test tube holder, spatula, measuring
cylinder
(ii) Insoluble salt
UNIT
6
Procedure:
1 Measure and pour 5 cm3 of distilled water into a test tube.
2 Add half spatula of sodium nitrate into the test tube. Shake the test tube.
3 Record the observation.
4 Repeat steps 1- 3 for all salts stated above.
Observation:
(Draw table to record the observations)
What are the methods to
prepare soluble salt?
I Using titration method
II By adding excess of solid reactant to acid
SP 6.9.2
What is the method to
prepare insoluble salt?
III By the precipitation method
SP 6.9.3
© Nilam Publication Sdn. Bhd.
128
MODULE • Chemistry FORM 4
1
Preparing Soluble Salt
I
Using titration method
When do we use this
method?
SP 6.9.2
When both reactants are aqueous solutions:
Acid (aq) + Alkali (aq) ➝ Salt (aq) + Water
+
+
Salts of Na , K and NH4+ are prepared by using titration method. Identify these salts
from the Table of Salts on page 126.
*Alkali is a base that soluble in water and ionises to hydroxide ion.
Remark:
If one of the aqueous reactants become excess, it is difficult to remove it by any physical methods (unlike
insoluble solid that can be removed by filtration, see this method of adding excess of solid later). Hence
titration allows the adding acid to alkali until end point with the aid of indicator. The right amount of
aqueous solutions react together without any one of them become excess.
II By adding excess of solid reactant to acid
When do we use this
method?
One of the reactants is insoluble metal, carbonate or insoluble base.
(i) Acid + metal (s) ➝ salt + hydrogen
(ii) Acid + metal oxide (s) ➝ salt + water
(iii) Acid + metal carbonate (s) ➝ salt + water + carbon dioxide
Soluble salts which are not salts of Na+, K+ and NH4+ are prepared by adding excess
of solid reactant to acid. Identify these salts from the Table of Salts on page 126.
Exercise
UNIT
6
Remark:
A solid reactant can be added until excess to the fix volume of acid. It is removed by filtration.
TP2
Choose pair of reactants that require the use of titration method or by adding excess solid.
Titration method
(3 / 7)
By adding excess of solid reactant
(3 / 7)
CuO (s) + H2SO4 (aq)
7
3
NaOH (aq) + HCl (aq)
3
7
KOH (aq) + HNO3 (aq)
3
7
Zn (s) + HCl (aq)
7
3
Reactants
129
MODULE • Chemistry FORM 4
2
Preparing Insoluble Salt
SP 6.9.3
III By precipitation method
State the name of the
reaction to produce an
insoluble salt.
Double decomposition reactions.
What type of reactants
are needed for this
method?
The precipitate of insoluble salt is formed when two different solutions that contain
the cation and anion of the insoluble salt are mixed. The precipitate is then obtained
by filtration.
Give the chemical
equation for the
precipitation of
lead(II) sulphate.
Pb(NO3)2 (aq) + Na2SO4 (aq) ➝ PbSO4(s) + 2NaNO3 (aq)
Remark:
How to deduce the correct reactants from a given insoluble salt: lead(II) sulphate.
Pb2+
SO42–
Nitrate Pb2+ because all nitrate salts
are soluble.
Sulphate of NH4+ / Na+ / K+
because all are soluble
It is advisable to suggest the reactants in which one is nitrate salt and the other one salt of ammonium
/ potassium / sodium as they are always soluble.
Write the ionic
equation for the
reaction.
Suggest the suitable
reactants and write the
ionic equation for the
following insoluble
salts.
Pb2+ + SO42– ➝ PbSO4
UNIT
6
Insoluble salt
Solution of
cation
Solution of anion
Lead(II) iodide,
PbI2
Lead(II) nitrate
Potassium iodide
Barium sulphate,
BaSO4
Barium nitrate
Ammonium
sulphate
Silver chloride,
AgCl
Silver nitrate
Sodium chloride
Calcium
Calcium nitrate
carbonate, CaCO3
Potassium
carbonate
Ionic equation
Pb2+ + 2I–
PbI2
Ba2+ + SO42–
BaSO4
AgCl
Ag+ + Cl–
Ca2+ + CO32–
CaCO3
Remark:
Lead(II) iodide and lead(II) chloride are exceptional insoluble salts.
1 Lead (II) iodide is insoluble yellow solid but dissolves in hot water and forms a yellow solid again
when cooled.
2 Lead (II) chloride is insoluble white solid but dissolves in hot water and forms a white solid again
when cooled.
How Salt
is Made
© Nilam Publication Sdn. Bhd.
130
131
–
–
–
–
Method III
UNIT
6
Insoluble salt
The salt is prepared by precipitation
method.
(Double decomposition reaction)
– Mix two solutions containing cations
and anions of insoluble salts.
– Stir with glass rod.
– Filter using filter funnel.
– Rinse the residue with distilled
water.
– Dry the residue by pressing it
between filter papers.
SP 6.9.3
Evaporate the filtrate until it becomes a saturated solution.
Dip in a glass rod, if crystals are formed, the solution is saturated.
Cooled at room temperature.
Filter and dry the salt crystals by pressing them between filter papers.
– Add metal/metal oxide/metal carbonate powder until excess
into a fixed volume of the heated acid.
– Filter the mixture to remove the excess metal/metal oxide/
metal carbonate.
– A titration is conducted to determine the
volume of acid needed to neutralise a fixed
volume of an alkali with the aid of an
indicator.
– The same volume of acid is then added
to the same volume of alkali without any
indicator to obtain pure and neutral salt
solution.
Method II
Other than K+, Na+, NH4+
SP 6.9.2
The salt is prepared by reacting acid with insoluble metal/metal
oxide/metal carbonate:
– Acid + Metal
Salt + Hydrogen (Displacement reaction)
– Acid + Metal oxide
Salt + Water (Neutralisation reaction)
– Acid + Metal carbonate
Salt + Water + Carbon dioxide
Method I
Salts K+, Na+, NH4+
Soluble salt
PREPARATION OF SALT
Garam disediakan berdasarkan keterlarutannya sebagaimana yang ditunjukkan pada carta aliran di bawah:
The salt is prepared by titration method of
acid and alkali using an indicator.
– Acid + Alkali
Salt + Water
(Neutralisation Reaction)
1
Describe the Preparation of Soluble and Insoluble Salt
MODULE • Chemistry FORM 4
© Nilam Publication Sdn. Bhd.
132
➊
100
3
1
➋
Heat
Acid
powder
into the acid
mol dm of any acid
–3
The filtrate is
salt solution
Salt crystals
➌
➍
Heat
➁
Saturated
salt solutions
• Evaporate the salt solution until
saturated solution is formed.
• The salt solution is poured into
evaporating dish .
➂
➃
Residue is
salt crystals
• Filter the miture to separate
the salt crystals .
➏
➎
Salt crystals
• Cool it at room temperature until salts crystals
are formed.
.
The residue is
metal /metal oxide/
metal carbonate .
• Filter the mixture to
separate excess metal/
metal oxide/metal carbonate
with the salt solution .
• Dry the salt crystals by
pressing them between filter
papers.
• Measure and pour
cm of
and pour into a beaker.
• Add metal/metal oxide/metal carbonate
gently .
and heat
Heat
Excess of metal/
metal oxide/
metal carbonate
• Stir the mixture with a
glass rod .
• Add metal / metal oxide / metal
carbonate powder to the acid
excess .
until
Method II:
Soluble salt except K+, Na+ and NH4+
6
Steps to Prepare Soluble Salt
UNIT
2
➀
Alkali
Acid
• Measure and pour
50 cm3 of
1 mol dm–3 of
any alkali into a
conical flask. Add
a few drops of
phenolphthalein.
• 1 mol dm–3 of any
acid is titrated to the
alkali until neutral by
using an indicator.
The volume of acid
used is recorded.
• Repeat the titration
indicator to get
pure and neutral
salt solution.
Method I:
Soluble salt of K+, Na+
and NH4+
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
Steps to Prepare Insoluble Salt
Method III: Preparation of Insoluble XnYm Salt by Double Decomposition Reaction
solution contains Xm+ cation into a
beaker.
Precipitate of
XnYm salt
is formed.
The residue is
XnYm salt.
100
2 Measure and pour
cm3
1
of
mol dm–3 of aqueous
solution contains Yn– anion into
another beaker.
3 Mix both solutions and stir the
glass rod
with
.
4 Filter the
mixture
the precipitate with
XnYm
residue is
Salt
X nY m
5 Press the precipitate
to dry it.
133
mixture
rinse
and
distilled water . The
salt.
between
6
100
1 Measure and pour
cm3
1
of
mol dm–3 of aqueous
SP 6.9.3
UNIT
3
filter papers
MODULE • Chemistry FORM 4
Exercise
1
TP2
Complete the following table by writing “S” for soluble salts and “IS” for insoluble salts. Write all the possible
chemical equations to prepare soluble salts and two chemical equations for insoluble salts.
Salt
“S” / “IS”
Chemical equations
UNIT
S
Zn + 2HCl
ZnCl2 + H2
ZnCO3 + 2HCl
ZnCl2 + CO2 + H2O
ZnO + 2HCl
ZnCl2 + H2O
Sodium nitrate
S
NaOH + HNO3
NaNO3 + H2O
Silver chloride
IS
AgNO3 + KCl
AgNO3 + NaCl
AgCl + KNO3
AgCl + NaNO3
Copper(II) sulphate
S
CuO + H2SO4
CuSO4 + H2O
CuCO3 + H2SO4
CuSO4 + CO2 + H2O
Lead(II) sulphate
IS
Pb(NO3)2 + K2SO4
Pb(NO3)2 + Na2SO4
Aluminium nitrate
S
2Al + 6HNO3
2Al(NO3)3 + 3H2
Al2O3 + 6HNO3
2Al(NO3)3 + 3H2O
Al2(CO3)3 + 6HNO3
2Al(NO3)3 + 3CO2 + 3H2O
Lead(II) chloride
IS
Pb(NO3)2 + 2KCl
Pb(NO3)2 + 2NaCl
Magnesium nitrate
S
Mg + 2HNO3
Mg(NO3)2 + H2
MgO + 2HNO3
Mg(NO3)2 + H2O
MgCO3 + 2HNO3
Mg(NO3)2 + CO2 + H2O
Potassium chloride
S
KOH + HCl
Lead(II) nitrate
S
PbO + 2HNO3
Pb(NO3)2 + H2O
PbCO3 + 2HNO3
Pb(NO3)2 + CO2 + H2O
Barium sulphate
IS
BaCl2 + K2SO4
BaCl2 + Na2SO4
6
Zinc chloride
© Nilam Publication Sdn. Bhd.
134
PbSO4 + 2KNO3
PbSO4 + 2NaNO3
PbCl2 + 2KNO3
PbCl2 + 2NaNO3
KCl + H2O
BaSO4 + 2KCl
BaSO4 + 2NaCl
MODULE • Chemistry FORM 4
The diagram below shows the set-up of apparatus to prepare soluble salt Y.
Nitric acid
25 cm3 of 1 mol dm–3 potassium
hydroxide solution + phenolphthalein
Phenolphthalein is used as an indicator in a titration between nitric acid and potassium hydroxide solution.
25 cm3 of nitric acid completely neutralises 25 cm3 of 1 mol dm–3 potassium hydroxide solution. The experiment
is repeated by reacting 25 cm3 of 1 mol dm–3 potassium hydroxide solution with 25 cm3 nitric acid without
phenolphthalein. Salt Y is formed from the reaction.
(a) Name salt Y.
TP2
Potassium nitrate
(b) Write a balanced equation for the reaction that occurs.
TP3
HNO3 + KOH
KNO3 + H2O
(c) Calculate the concentration of nitric acid.
TP3
Number of moles of KOH = 1 ×
25
= 0.025 mol
1 000
From the equation,
1 mol KOH : 1 mol HNO3
0.025 mol KOH : 0.025 mol HNO3
Concentration of HNO3, M
0.025 = M ×
25
1 000
6
M = 1 mol dm–3
UNIT
2
(d) Why the experiment is repeated without phenolphthalein?
TP1
To get pure and neutral salt solution Y.
(e) Describe briefly how a crystal of salt is obtained from the salt solution.
TP6
The salt solution is poured into an evaporating dish. The solution is heated to evaporate the solution until
one third of its original volume // a saturated solution formed. The saturated solution is allowed to cool
until salt crystals Y are formed. The crystals are filtered and dried by pressing them between filter papers.
(f) Name two other salts that can be prepared with the same method.
TP2
Potassium / sodium / ammonium salt. Example: potassium nitrate, sodium sulphate.
(g) State the type of reaction in the preparation of the salts.
TP1
Neutralisation
135
MODULE • Chemistry FORM 4
3
The following are the steps in preparation of dry copper(II) sulphate crystals.
Step I: Copper(II) oxide powder is added a little at a time with constant stirring to the heated 50 cm3 of
1 mol dm–3 sulphuric acid until some of it no longer dissolve.
Step II: The mixture is filtered.
Step III: The filtrate is poured into an evaporating dish and heated to evaporate the solution until one third
of its original.
Step IV: The salt solution is allowed to cool at room temperature for the crystallisation to take place.
Step V: The crystals formed are filtered and dried by pressing them between filter papers.
(a) (i)
TP4
State two observations during Step I.
– Black solid dissolve
– Colourless solution turns blue
(ii) Write a balance chemical equation for the reaction that occur in Step I.
CuO + H2SO4
CuSO4 + H2O
(iii) State the type of reaction in the preparation of the salts.
Neutralisation
(b) Why is copper(II) oxide powder added until some of it no longer dissolve in Step I?
To make sure that all sulphuric acid has reacted.
TP2
UNIT
(c) What is the purpose of heating in Step III?
To evaporate the water and make the copper(II) sulphate solution to become saturated.
TP2
6
(d) What is the colour of copper(II) sulphate?
Blue
TP3
(e) What is the purpose of filtration in
(i) Step II?
– To remove the excess copper(II) oxide.
TP3
– To obtain copper(II) sulphate solution as a filtrate.
(ii) Step V?
To obtain copper(II) sulphate crystals as a residue.
(f) Draw a labelled diagram to show the set-up of apparatus used in Step II and Step III.
TP5
Excess of
copper(II)
oxide
Filter paper
Copper(II) sulphate
solution
© Nilam Publication Sdn. Bhd.
136
Copper(II) sulphate
solution
Heat
MODULE • Chemistry FORM 4
(g) Can copper powder replace copper(II) oxide in the experiment? Explain your answer.
TP5
Cannot. Copper is less electropositive than hydrogen in the Electrochemical Series, copper cannot
displace hydrogen from the acid.
(h) Name other substance that can replace copper(II) oxide to prepare the same salt. Write a balance chemical
TP3 equation for the reaction that occur.
Substance : Copper(II) carbonate
Balance equation : CuCO3 + H2SO4
The diagram below shows the flow chart for the preparation of zinc carbonate and zinc sulphate through reactions
I and II.
Zinc nitrate
Reaction I
Zinc carbonate
Reaction II
Zinc sulphate
(a) Based on the flow chart above, classify the above salt to soluble salt and insoluble salt.
TP2
Soluble salt : Zinc nitrate, zinc sulphate
Insoluble salt : Zinc carbonate
(b) (i)
TP2
State the reactant for the preparation of zinc carbonate from zinc nitrate in reaction I.
Sodium carbonate solution / potassium carbonate solution / ammonium carbonate solution
(ii) State the type of reaction that occurs in reaction I.
Double decomposition
(iii) Describe the preparation of zinc carbonate from zinc nitrate in the laboratory through reaction I.
TP6
50 cm3 of 1 mol dm–3 zinc nitrate solution is measured and poured into a beaker. 50 cm3 of 1 mol dm–3
sodium carbonate solution is measured and poured into another beaker. The mixture is stirred with
residue is rinsed with distilled water. The white precipitate is dried by pressing it between filter papers.
(iv) Write the chemical equation for the reaction in (b)(iii).
TP3
Zn(NO3)2 + Na2CO3
ZnCO3 + 2NaNO3
(c) (i)
TP3
State the reactant for the preparation of zinc sulphate from zinc carbonate in reaction II.
Sulphuric acid
(ii) Describe laboratory experiment to prepare zinc sulphate from zinc carbonate through reaction II.
TP6
50 cm3 of 1 mol dm–3 of sulphuric acid is measured and poured into a beaker and is heated. The white
precipitate from reaction I/zinc carbonate powder is added to the acid until in excess. The mixture is
stirred with a glass rod. The excess white precipitate is filtered out. The filtrate is poured into an
evaporating dish. The salt solution is gently heated until saturated. The hot saturated salt solution is
allowed to cool for crystals to form. The crystals formed are filtered and dried by pressing it between
sheets of filter papers.
(iii) Write the chemical equation for the reaction in (c)(ii).
TP3
ZnCO3 + H2SO4
ZnSO4 + H2O + CO2
137
6
a glass rod and a white solid, zinc carbonate (ZnCO3) is formed. The mixture is filtered and the
UNIT
4
CuSO4 + H2O + CO2
MODULE • Chemistry FORM 4
Constructing Ionic Equation for the Formation of Insoluble Salt
SP 6.9.4
How to construct ionic equation for the
formation of insoluble salt?
The ionic equation for the formation of insoluble salt can
be constructed if the number of moles of anion and cation
to form 1 mol of insoluble salt are known.
State the type of the experiment to determine
the ionic equation for the formation of
insoluble salt.
The continuous variation method.
Example of experiment:
Problem statement : How to construct ionic equation for the formation of lead(II) chromate precipitate?
Aim : To construct ionic equation for the formation of lead(II) chromate precipitate
Manipulated variable : Volume of lead(II) nitrate solution
Responding variable : Height of precipitate
Constant variable : Volume and concentration of potassium chromate(II) solution, concentration of lead(II)
nitrate solution, size of test tube
Hypothesis : When the volume of lead(II) nitrate solution increases, the height of precipitate will increase
until all potassium chromate(VI) has reacted.
Apparatus : Test tubes of same size, measuring cylinder//burette, test tube rack, stopper, ruler
–3
Materials : 0.5 mol dm of lead(II) nitrate and potassium chromate(VI) solution
5 cm3 of 0.5 mol dm–3 of potassium chromate(VI) solution is poured into every test tube as shown in the
following diagram:
UNIT
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
5 cm3
6
5 cm3
3
1 cm3 of 0.5 mol dm1–3cm
lead(II)
solution
test
tube,
27cm
second
test tube and so
3
3
3
3
2 nitrate
cm3
3 cm3 is 4added
cm3 to 5the
cmfirst
6 cm
cm3 to the
8 cm
on until 8 cm3 to the eighth test tube as shown in the following diagram:
1 cm3
2 cm3
3 cm3
4 cm3
5 cm3
Precipitate / Mendakan
Precipitate
© Nilam Publication Sdn. Bhd.
138
6 cm3
7 cm3
8 cm3
MODULE • Chemistry FORM 4
Procedure:
(a) Eight test tubes of same size are labelled from 1 to 8 and placed on a test tube rack.
(b) 5.00 cm3 of 0.5 mol dm-3 potassium chromate(VI) solution is measured and poured into each of the test tubes.
(c) 1 cm3 of 0.5 mol dm-3 lead(II) nitrate solution is added to the first test tube, 2 cm3 to the second test tube and
so on until 8 cm3 to the eighth test tube.
(d) The test tubes are stoppered and shaken well.
(e) The test tubes are left aside for one hour on a test tube rack.
(f) The height of precipitate in each test tube is measured.
(g) The colour of the solution above the precipitate in each test tube is recorded.
(h) A graph of height of precipitate against volume of lead(II) nitrate solution added is plotted.
Result:
1
2
3
4
5
6
7
8
Volume of
0.5 mol dm–3
of potassium
chromate(VI)
solution / cm3
5.00
5.00
5.00
5.00
5.00
5.00
5.00
5.00
Volume of
0.5 mol dm–3
of lead(II)
nitrate
solution / cm3
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
Height of
precipitate /
cm
0.5
0.8
1.2
1.6
2.0
2.0
2.0
2.0
Colour of
solution
above
precipitate
Yellow
Yellow
Yellow
Yellow
Excess of
potassium
Excess of
potassium
Excess of
potassium
Chemicals in Excess of
the solution potassium
above
chromate(VI)
precipitate
and
potassium
Ions present
in the solution
above the
precipitate
Colourless Colourless Colourless Colourless
Potassium Excess of
lead(II)
nitrate
chromate(VI) chromate(VI) chromate(VI)
nitrate
formed.
and
and
and
and
potassium
potassium
potassium
potassium
Excess of
lead(II)
Excess of
lead(II)
nitrate
nitrate
and
potassium
and
potassium
nitrate
nitrate
nitrate
nitrate
nitrate
nitrate
nitrate
formed.
formed.
formed.
formed.
formed.
formed.
formed.
NO3-, K+
CrO42-
NO3-, K+
CrO42-
NO3-, K+
CrO42-
NO3-, K+
CrO42-
139
NO3-, K+
Pb2+,
NO3-, K+
Pb2+,
NO3-, K+
Pb2+,
NO3-, K+
UNIT
6
Test tube
MODULE • Chemistry FORM 4
1
TP4
2
TP4
Complete the following table:
Observation
Inference
The height of precipitate in
test tubes 1 to 4 increases.
The solution above
precipitate is yellow in test
tubes 1 to 4.
• The increase in volume of lead(II) nitrate added increases the mass of
precipitate formed in test tubes 1 to 4.
• There are excess of potassium chromate(VI)
solution in test tubes
The height of precipitate
remains constant in test
tubes 5 to 8.
The solution above
precipitate is colourless in
test tubes 5 to 8.
• In test tube 5, potassium chromate(VI) solution has completely reacted
with lead(II) nitrate solution. All chromate(VI) ions and all lead(II) ions
have reacted. The solution above precipitate is potassium nitrate
.
1 to 4.
• In test tubes 6 to 8, there are excess of
lead(II) nitrate solution is added.
lead(II)
ions as more
The graph below is obtained when the height of precipitate is plotted against the volume of lead(II) nitrate
solution.
Height of precipitate (cm)
2
0
1
2
3
4
5
6
7
8
Volume of lead(II) nitrate /cm3
9
(a) What is the colour of lead(II) nitrate solution and potassium chromate(VI) solution?
Lead(II) nitrate solution
: colourless
UNIT
Potassium chromate(VI) solution : yellow
6
(b) (i)
State the name of the precipitate formed.
Lead(II) chromate
(ii) What is the colour of the precipitate?
Yellow
(c) (i)
Based on the above graph, what is the volume of lead(II) nitrate solution needed to completely react
with potassium chromate(VI) 5 cm3 of solution?
5 cm3
(ii) Calculate the number of moles of lead(II) ions in 5.0 cm3 of 0.5 mol dm–3 lead (II) nitrate solution.
Ionisation equation of lead(II) nitrate solution:
Pb2+ + 2NO3–
Pb(NO3)2
Number of moles of Pb(NO3)2 = 5.0 × 0.5 = 0.0025 mol
1 000
From the ionisation equation,
1
mol Pb(NO3)2 :
0.0025
© Nilam Publication Sdn. Bhd.
mol Pb(NO3)2 :
1
mol Pb2+
0.0025
mol Pb2+
140
MODULE • Chemistry FORM 4
(iii) Calculate the number of mol of chromate(VI) ions in 5.0 cm3 of 0.5 mol dm–3 potassium chromate(VI)
solution.
Ionisation equation of potassium chromate(VI) solution:
2K+ + CrO4 2-
K2CrO4
From the ionisation equation,
1
mol K2CrO4:
1
mol CrO42-
0.0025
mol K2CrO4:
0.0025
mol CrO42-
(iv) Calculate the number of moles of chromate(VI) ions reacts completely with one mol of lead(II) ions.
0.0025
1
mol Pb2+ : reacts completely with
0.0025
mol CrO42-
1
mol CrO42-
mol Pb2+ :
(d) Write the ionic equation for the formation of the precipitate.
Pb2+ + CrO42- ➝ PbCrO4
Solving Numerical Problems Involving the Salt Preparation
Relationship between mol of substance with mass, volume of gas, volume of solution and concentration of
solution:
Mass in gram
MV
n = 1 000
Number of mol (n)
× 24 dm3 mol–1
÷ 24 dm mol
3
–1
Volume of
gas in dm3
Example:
50 cm3 of 2 mol dm–3 sulphuric acid is added to excess of copper(II) oxide powder. Calculate the mass of copper(II)
sulphate formed in the reaction.
[Relative atomic mass: H = 1, O = 16, Cu = 64, S = 32]
Write a balanced equation
Write the information from the question above the
equation
M = 2 mol dm–3
V = 50 cm3
CuO(aq) + H2SO4
Convert the given quantity into moles by using the
method shown in the chart above.
Number of moles of sulphuric acid
2 × 50
=
1 000
= 0.1 mol
From the equation,
1 mol H2SO4 : 1 mol CuSO4
0.1 mol H2SO4 : 0.1 mol CuSO4
Use the mole ratio of the substances involved to find
the number of moles of other substance.
Remark:
The coefficient of each formula shows the number of moles of
reactants react and products formed.
Convert the mole into the quantity required in the
question by using the method shown in the chart
above.
?g
CuSO4(aq) + 2H2O(l)
Mass of CuSO4
= 0.1 mol × [64 + 32 + (16 × 4)] g mol–1
= 16 g
141
6
Solution concentration
in mol dm–3 (M) and
volume in cm3 (V)
× (RAM/RMM/RFM) g mol–1
UNIT
÷ (RAM/RMM/RFM) g mol–1
MODULE • Chemistry FORM 4
Exercise
1
TP3
27.66 g of lead(II) iodide is precipitated when 2.0 mol dm–3 of aqueous lead(II) nitrate solution is added to an
excess of aqueous potassium iodide solution. Calculate the volume of aqueous lead(II) nitrate solution used.
[Relative atomic mass: I = 127, Pb = 207]
M = 2 mol dm–3
V = ? cm3
Pb(NO3)2(aq) + 2KI(aq)
27.66 g
PbI2(s) + 2KNO3(aq)
Number of moles of PbI2 =
From the equation,
27.66
(207 + 2 × 127)
= 0.06 mol
1 mol PbI2 : 1 mol Pb(NO3)2
0.06 mol PbI2 : 0.06 mol Pb(NO3)2
n mol
0.06 mol
Volume of Pb(NO3)2 = M mol dm–3 = 2 mol dm–3 = 0.03 dm3 = 30 cm3
2
TP3
Zinc oxide powder is added to 100 cm3 of 2 mol dm–3 nitric acid to form zinc nitrate. Calculate
(i) the mass of zinc oxide that has reacted.
(ii) the mass of zinc nitrate produced. [Relative atomic mass: H = 1, O = 16, N = 14, Zn = 65]
M = 2 mol dm–3
V = 100 cm3
(i) 2HNO3(aq) + ZnO(s)
Zn(NO3)2(aq) + H2O(l)
100 × 2
Number of moles of HNO3 =
= 0.2 mol
1 000
From the equation,
2 mol HNO3 : 1 mol ZnO
0.2 mol HNO3 : 0.1 mol ZnO
Mass of ZnO = 0.1 × [65 + 16] = 8.1 g
UNIT
6
(ii) From the equation,
2 mol HNO3 : 1 mol Zn(NO3)2
0.2 mol HNO3 : 0.1 mol Zn(NO3)2
Mass of Zn(NO3)2 = 0.1 mol × [65 + [14 + (16 × 3)] × 2] g mol–1
= 0.1 × 189
= 18.9 g
3
TP3
200 cm3 of 1 mol dm–3 barium chloride solution reacts with 100 cm3 of 1 mol dm–3 silver nitrate solution.
Calculate the mass of precipitate produced. [Relative atomic mass Ag = 108, Cl = 35.5]
M = 1.0 mol dm–3
M = 1.0 mol dm–3
V = 200 cm3
V = 100 cm3 BaCl2
+ 2AgNO3
?g
2AgCl + Ba(NO3)2
1 × 200
= 0.2 mol (excess)
1 000
1 × 100
Number of moles of silver nitrate =
= 0.1 mol
1 000
Number of moles of barium chloride =
From the equation,
1 mol BaCl2 : 2 mol AgNO3 : 2 mol AgCl
0.2 mol BaCl2 (excess) : 0.1 mol AgNO3 : 0.1 mol AgCl
Mass of AgCl = 0.1 mol × [108 + 35.5] g mol–1 = 14.35 g
© Nilam Publication Sdn. Bhd.
142
MODULE • Chemistry FORM 4
6.10 ACTION OF HEAT ON SALT
What are the effect of heat
on salts?
SP 6.10.1
1
Common gas identification:
Gas
SK
6.10
Some salts decompose when they are heated:
gas
metal oxide
Salt (Colour of residue refer + (Gas identification refers to
certain anion/cation)
to certain cation)
SP 6.10.2
Observation / Test
Inference
Nitrogen
dioxide,
NO2
– Brown gas.
– Place a moist blue litmus paper at the mouth of the
boiling tube, blue litmus paper turns red.
– Nitrogen dioxide gas is produced
by heating nitrate salt.
– Nitrate ion, NO3– present.
Oxygen,
O2
– Colourless gas.
– Put a glowing wooden splinter into boiling tube, the
glowing wooden splinter is relighted.
– Oxygen gas is produced by
heating nitrate or chlorate(V) salt.
– Nitrate ion, NO3– present or
ClO3– ion present.
Carbon
dioxide,
CO2
– Colourless gas.
– Pass the gas through lime water, lime water turns
chalky.
– Draw the set-up of apparatus to conduct the test:
– Carbon dioxide gas is produced
by heating carbonate salt.
– Carbonate ion, CO32– present.
Calcium
carbonate
Heat
– Ammonia gas is produced by
heating ammonium salt with
alkali.
– Ammonium ion NH4+ present.
Hydrogen
gas, H2
– Colourless gas
– Placed burning wooden splinter at the mouth of the
test tube
– The gas burns with “pop” sound
– Hydrogen gas is produced
Chlorine
gas, Cl2
– Greenish yellow gas
– Place moist blue litmus paper at the mouth of test tube
– Blue litmus paper turns red, it is then bleached
– Chlorine gas is produced
Hydrogen
chloride
gas, HCl
– Colourless gas
– Dip a glass rod into
concentrated ammonia
solution and place it near
the mouth of the test tube
– White fumes are formed
– Hydrogen chloride gas is
produced by heating of sodium
chloride with concentrated
sulphuric acid
Glass rod
dip in
concentrated
ammonia
Concentrated H2SO4
NaCl
– Colourless gas with pungent smell
Sulphur
– Bubble gas through acidified potassium dichromate(
dioxide gas,
III)
SO2
– The orange solution turns green
143
– Sulphur dioxide gas is produced
UNIT
– Colourless gas with pungent smell.
– Place a moist red litmus paper at the mouth of the
boiling tube, red litmus paper turns blue.
Ammonia,
NH3
6
Lime water
MODULE • Chemistry FORM 4
2
Action of heat on nitrate and carbonate salts.
Cation
SP 6.10.2
Nitrate (NO3–)
Carbonate (CO32–)
Decompose to oxygen gas and metal nitrite
when heated
Does not decompose when heated
K+
2KNO3
White solid
2KNO2
+ O2
White solid
–
Na+
2NaNO3
White solid
2NaNO2 + O2
White solid
–
Decompose to oxygen gas, nitrogen dioxide
gas and metal oxide when heated
Decompose to carbon dioxide gas and metal oxide
when heated
Ca2+
2Ca(NO3)2
White
solid
2CaO + 4NO2 + O2
White
Brown
solid
fume
CaCO3
White
solid
CaO
+
White
solid
Mg2+
2Mg(NO3)2
White
solid
2MgO
White
solid
+
MgCO3
White
solid
MgO
White
solid
Al3+
4Al(NO3)3
White
solid
2Al2O3
White
solid
+
Zn2+
UNIT
Pb2+
6
Cu2+
4NO2 + O2
Brown
fume
12NO2 + 3O2 2Al2(CO3)3
Brown
White
fume
solid
2Zn(NO3)2
White
solid
2ZnO +
4NO2 + O2
Yellow when Brown
hot, white
gas
when cold
2Pb(NO3)2
White
solid
2PbO +
4NO2 + O2 PbCO3
Brown when Brown
White
hot, yellow
gas
solid
when cold
2Cu(NO3)2
Blue
solid
2CuO
Black
solid
+
4NO2 + O2
Brown
fume
6CO2
Turn lime
water chalky
CO2
Turn lime
water chalky
PbO
+
Brown when
hot, yellow
when cold
CO2
Turn lime
water chalky
CuCO3
Green
solid
CuO
Black
solid
Sulphate salts are more stable, they are not easily decompose when heated.
4
Chloride salts do not decompose except NH4Cl: NH4Cl(s)
144
+
ZnO
+
Yellow when
hot, white
when cold
3
© Nilam Publication Sdn. Bhd.
+ CO2
Turn lime
water chalky
2Al2O3
White
solid
ZnCO3
White
solid
CO2
Turn lime
water chalky
NH3(g) + HCl(g)
+
CO2
Turn lime
water chalky
MODULE • Chemistry FORM 4
SP 6.10.2
Observation
Inference / conclusion
A white salt is heated.
– Brown gas is released, the gas turns moist
blue litmus paper red.
– Residue is yellow when hot and white when
cold.
–
A green salt is heated.
– Colourless gas released, the gas turns lime
water chalky.
– Residue is black.
–
A white salt is heated.
– Colourless gas released, the gas turns lime
water chalky.
– Residue is brown when hot and yellow when
cold.
–
A white salt is heated.
– Colourless gas released, the gas turns lime
water chalky.
– Residue is yellow when hot and white when
cold.
–
A blue salt is heated.
– Brown gas is released, the gas turns moist
blue litmus paper red.
– Residue is black.
–
Nitrogen dioxide
present.
– The residue is
ion present.
– The white salt is
Nitrate
gas released.
zinc
Zinc
oxide.
zinc nitrate
ion
.
Carbon dioxide
gas released. Carbonate ion
present.
– The residue is copper(II) oxide. Copper(II)
ion present.
– The green salt is copper(II) carbonate .
Carbon dioxide
gas released. Carbonate
present.
lead(II)
Lead(II)
– The residue is
oxide.
ion present.
– The white salt is lead(II) carbonate .
Carbon dioxide
present.
gas released.
– The residue is
ion present.
zinc
– The white salt is
zinc carbonate
Carbonate
ion
Zinc
oxide.
.
Nitrogen dioxide gas released.
present.
– The residue is copper(II) oxide.
ion present.
copper(II) nitrate
– The blue salt is
Nitrate
ion
Copper(II)
.
A white salt is heated.
– Brown gas is released, the gas turns moist
blue litmus paper red.
– Residue is brown when hot and yellow when
cold.
– Nitrogen dioxide gas released. Nitrate ion present.
– The residue is lead(II) oxide. Lead(II) ion present.
lead(II) nitrate
– The white salt is
.
A white salt is heated.
– Colourless gas released, the gas turns lime
water chalky.
– Residue is white
–
Carbon dioxide
gas released. Carbonate
present.
– The possible residues are CaO/MgO/Al2O3.
– From the above table, action of heat on salt can be used to identify
lead(II) carbonate
zinc nitrate
zinc carbonate
,
,
,
copper(II) carbonate .
ion
ion
lead(II) nitrate
copper(II) nitrate
– Confirmatory test for other cations and anions is carried out by Confirmatory Tests for Anions and Cations.
145
6
Complete the following table:
UNIT
5
,
and
MODULE • Chemistry FORM 4
6.11
QUALITATIVE ANALYSIS
SK
6.11
What is qualitative
analysis of salt?
Qualitative analysis of a salt is a chemical technique to identify the ions present in a
salt.
What is the
preliminary
examination on salt?
The preliminary examination is on the physical properties such as colour and
solubility, indicate the possibility the presence of certain cations, anions or metal
oxide.
SP 6.11.2
How are qualitative
analysis being done to
identify salts?
UNIT
SP 6.11.2
Solid
Aqueous
Salts / Cation / Metal oxide
White
Colourless
K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Pb2+, NH4+
Green
Insoluble
CuCO3
Light green
Light green
Fe2+, example: FeSO4, FeCl2, Fe(NO3)2
Blue
Blue
Cu2+,
example: CuSO4, Cu(NO3 )2 and CuCl2
Brown
Brown
Fe3+
Black
Insoluble
CuO
Yellow when hot,
white when cold
Insoluble
ZnO
Brown when hot,
yellow when cold
Insoluble
PbO
The qualitative analysis consists of the following steps:
(a) Observe the physical properties on salt.
(b) The action of heat on salts.
(c) Prepare aqueous solution of salts and conduct confirmatory test for cation and
anion that present.
6
Confirmatory Tests for Cations
SP 6.11.1
List the cations.
Ca2+, Zn2+, Al3+, Pb2+, Cu2+, Fe2+, Fe3+, NH4+
What are the two
steps for testing
cations?
Step 1: Add a few drops of sodium hydroxide solution or ammonia solution to the
aqueous solution of salt and observe.
Step 2: Add excess of sodium hydroxide solution or ammonia solution and observe.
What are the
essential
observations?
Three main observations:
1 Is there a precipitate formed in a few drops of testing reagent?
2 If yes, what is the colour of the precipitate?
3 Does the precipitate dissolve in excess testing reagent?
© Nilam Publication Sdn. Bhd.
146
MODULE • Chemistry FORM 4
SP 6.11.1
Salt solution
Add an excess of
sodium hydroxide
solution/ ammonia
solution
White/coloured precipitate
(Insoluble metal hydroxide)
Precipitate dissolves
(Precipitate is soluble in
excess of sodium
hydroxide solution/
ammonia solution)
Precipitate does not
dissolve (Precipitate is
insoluble in excess of
sodium hydroxide
solution/ammonia
solution)
A
1
2
3
4
5
6
7
8
Using sodium hydroxide, NaOH:
2 cm3 of calcium nitrate solution is poured into a test tube.
A few drops of sodium hydroxide solution are added into the test tube using dropper.
The test tube is shaken.
Observation on whether the precipitate is produced and its colour is recorded.
If a precipitate is formed, sodium hydroxide solution is added continuously until no further changes occurred.
The test tube is shaken.
Observation on whether the precipitate dissolved in excess sodium hydroxide solution is recorded.
Steps 1 – 7 are repeated by replacing calcium nitrate solution with aluminium nitrate, copper(II) sulphate,
iron(II) sulphate, iron(III) sulphate, lead(II) nitrate, magnesium nitrate, zinc nitrate and ammonium chloride
solutions.
B
1
2
3
4
5
6
7
8
Using ammonia solution, NH3:
2 cm3 of calcium nitrate solution is poured into a test tube.
A few drops of ammonia solution are added into the test tube using dropper.
The test tube is shaken.
Observation on whether the precipitate is produced and its colour is recorded.
If a precipitate is formed, ammonia solution is added continuously until no further changes occurred.
The test tube is shaken.
Observation on whether the precipitate dissolved in excess ammonia solution is recorded.
Steps 1 – 7 are repeated by replacing calcium nitrate solution with aluminium nitrate, copper(II) sulphate,
iron(II) sulphate, iron(III) sulphate, lead(II) nitrate, magnesium nitrate, zinc nitrate and ammonium chloride
solutions.
147
6
Sodium hydroxide
solution/ ammonia
solution
UNIT
Confirmatory Tests for Cations
MODULE • Chemistry FORM 4
Result:
Cations
Ca2+
Mg2+
Zn2+
Al3+
Pb2+
Sodium hydroxide solution
small
excess
amount
White
Insoluble
precipitate
in excess
White
Insoluble
precipitate
in excess
White
Soluble
precipitate
in excess
White
Soluble
precipitate
in excess
White
Soluble
precipitate
in excess
Ammonia solution
small
excess
amount
No change
No change
White
precipitate
White
precipitate
White
precipitate
White
precipitate
Insoluble
in excess
Soluble
in excess
Insoluble
in excess
Insoluble
in excess
Fe2+
Green
precipitate
Insoluble
in excess
Green
precipitate
Insoluble
in excess
Fe3+
Brown
precipitate
Insoluble
in excess
Brown
precipitate
Insoluble
in excess
Blue
precipitate
Soluble
in excess
No change
No change
Cu2+
NH4+
Blue
Insoluble
precipitate
in excess
No change. A gas that changes
red litmus paper to blue is
released when heated.
Confirmatory test with other
reagent
–
–
–
Refer to page 150
Refer to page 150
Add a few drops of potassium
hexacyanoferrate(III) solution, dark
blue precipitate is formed
• Add a few drops of potassium
hexacyanoferrate(II) solution, dark
blue precipitate is formed
• Add a few drops of potassium
thiocyanate solution, blood red
colouration is formed
–
Add a few drops of Nessler’s reagent,
brown precipitate is formed
UNIT
(a) Reaction with small amount until excess of sodium hydroxide solution: (refer to the above table)
6
Solution
Add a little sodium
Pungent smell, moist red litmus paper turn to blue
hydroxide solution
contains:
Heat
Ca2+, Mg2+,
NH4+
NH4+
3+
2+
Al , Zn ,
Pb2+, Fe2+, Fe3+,
No precipitate
Cu2+ (blue),
Cu2+, NH4+
Fe2+ (green),
Fe3+ (brown)
Precipitate
Coloured precipitate
formed
Soluble
Zn2+, Al3+, Pb2+
Pb2+, Al3+,
Zn2+, Ca2+,
White
Add excess sodium
Mg2+
precipitate
hydroxide solution
Ca2+, Mg2+
Insoluble
© Nilam Publication Sdn. Bhd.
148
MODULE • Chemistry FORM 4
(b) Reaction with small amount until excess of ammonia solution:
Ca2+
No precipitate
Solution
contains:
Ca2+, Mg2+,
Add a
Al3+, Zn2+, Pb2+, little
Fe2+, Fe3+, Cu2+ ammonia
solution
Cu (blue),
Fe2+ (green),
Fe3+ (brown)
2+
Add excess
aqueous
ammonia
Soluble
Insoluble
Precipitate
formed
Cu2+
Fe2+, Fe3+
Coloured
precipitate
Pb , Al ,
Zn2+, Mg2+
2+
White precipitate
3+
Add excess
aqueous
ammonia
Soluble
Zn2+
Mg2+,
Insoluble Al3+, Pb2+
(c) Conclusion of the confirmatory test for colourless white cations:
(i) Zn2+: White precipitate, soluble in excess of sodium hydroxide and ammonia solution
(ii) Mg2+: White precipitate, insoluble in excess of sodium hydroxide and ammonia solution
NH4+: No precipitate with sodium hydroxide solution and pungent smell released when heated, gas
released changes red litmus paper to blue
(v)
(d) Conclusion of the confirmatory test for coloured cations.
(i) Cu2+: Blue precipitate insoluble in excess of sodium hydroxide solution and soluble in excess
ammonia solution
(ii) Fe2+: Green precipitate, insoluble in excess of sodium hydroxide and ammonia solution
(iii) Fe3+: Brown precipitate, insoluble in excess of sodium hydroxide and ammonia solution
149
UNIT
(iv) Ca2+: White precipitate insoluble in excess of sodium hydroxide and no precipitate with ammonia
solution
6
(iii) Al3+: White precipitate, soluble in excess of sodium hydroxide and insoluble in excess ammonia
Pb2+ solution
MODULE • Chemistry FORM 4
(e) All cations can be identified with confirmatory test using sodium hydroxide solution and ammonia
solution except Al3+ and Pb2+.
(f) To differentiate between Al3+ and Pb2+:
– Al3+ and Pb2+ are differentiated by double decomposition reaction. An aqueous solution containing
SO42–/ Cl–/ I– anion is used to detect the presence of Al3+ and Pb2+.
– Precipitate is formed when solution containing SO42–/ Cl–/ I– added to Pb2+.
– No precipitate when solution containing SO42–/ Cl– / I– added to Al3+.
(g) Write the ionic equations for the formation of precipitates:
Al3+ and Pb2+
Add sodium
sulphate solution
No change
Add potassium
iodide solution
Add sodium
chloride solution
White precipitate
No change
Pb2+
Al3+
Pb2+ + SO42–
Pb2+ + 2I–
PbSO4
White precipitate
UNIT
Al3+
Pb2+
6
Pb2+ + 2Cl–
PbCl2
SP 6.11.1
List the anions.
CO32–, Cl–, SO42–, NO3–
What are the essential
observation?
The observation could be one of the following:
• Colour of precipitate
• Gas released
© Nilam Publication Sdn. Bhd.
Pb2+
Al3+
No change
Confirmatory Tests for Anions
Yellow
precipitate
150
PbI2
151
Ionic equation:
CO32– + 2H+ → 2H2O + CO2
Conclusion:
Carbonate ion present.
Inference:
• The gas is carbon dioxide .
Observation:
• Effervescence occurs.
• Lime water turns chalky.
Sodium carbonate
Lime
water
4 cm3 of dilute hydrochloric
acid is added into the test
tube.
The gas given off is passed
through lime water.
Acid
2
1
UNIT
Ionic equation:
Ag+ + Cl– → AgCl
Conclusion:
Chloride ion present.
6
silver
Ba2+ + SO42– → BaSO4
Ionic equation:
Conclusion:
Sulphate ion present.
barium
Inference:
• The precipitate is
sulphate .
White
precipitate
Dilute hydrochloric acid / nitric
acid is added into the test tube until
no further changes
2 cm3 of barium chloride / barium
nitrate solution is added into the
test tube.
Inference:
• The precipitate is
chloride .
2
1
Observation:
• A white precipitate is formed.
White
precipitate
Dilute nitric acid is added into
the test tube until no further
changes.
2 cm3 of silver nitrate solution
is added into the test tube.
Observation:
• A white precipitate is formed.
2
1
SP 6.11.1
2 cm3 of solution that contains anion Xn– is poured into a test tube.
Confirmatory Tests for Anions
Brown ring
2 cm3 of dilute sulphuric acid is
added into the test tube followed
by 2 cm3 of iron(II) sulphate
solution. The test tube is shaken.
The test tube is slanted and held
with a test tube holder.
A few drops of concentrated
sulphuric acid are dropped
along the wall of the test tube
and is held upright.
Inference:
Nitrate ion present.
Observation:
• Brown ring is formed.
3
2
1
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
SPM PRACTICE
Subjective Questions
1
TP3
50 cm3 of 1 mol dm–3 sodium hydroxide solution is neutralised by 25 cm3 of sulphuric acid. Calculate the
concentration of sulphuric acid in mol dm–3 and g dm–3. [RAM: H = 1, S = 32, O = 16]
M = 1 mol dm–3 M = ?
v = 50 cm3
v = 25 cm3
Na2SO4 + 2H2O
50
Number of moles of NaOH = 1 ×
= 0.05 mol
1 000
From the equation,
2 mol NaOH : 1 mol H2SO4
0.05 mol NaOH : 0.025 mol H2SO4
2NaOH
H2SO4
+
n mol
v dm3
= 0.025 mol = 1 mol dm–3
25
dm3
1 000
Concentration of H2SO4 =
(
)
Concentration of H2SO4
= 1 mol dm–3 × (2 × 1 + 32 + 16 × 4) g mol–1
= 98 g dm–3
2
UNIT
TP3
Calculate the volume of 2 mol dm–3 sodium hydroxide needed to neutralise 100 cm3 of 1 mol dm–3 hydrochloric acid.
6
M = 2 mol dm–3
v = ? cm3
M = 1 mol dm–3
v = 100 cm3 NaOH
HCl
+
Number of moles of HCl = 1 ×
From the equation,
NaCl
100
= 0.1 mol
1 000
1 mol HCl : 1 mol NaOH
0.1 mol HCl : 0.1 mol NaOH
n mol
­
M mol dm–3
0.1 mol
=
2 mol dm–3
= 0.05 dm3
= 50 cm3
Volume of NaOH =
© Nilam Publication Sdn. Bhd.
152
+
H2O
MODULE • Chemistry FORM 4
3
TP4
Experiment I
1 mol dm–3 of nitric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
Experiment II
1 mol dm–3 of sulphuric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
Compare the volume of acids needed to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution in
Experiment I and Experiment II. Explain your answer.
Answer:
Calculation
NaOH + HNO3
Experiment I
NaNO3 + H2O
100
1 000
= 0.1 mol
Mol of NaOH = 1 ×
From the equation,
1 mol NaOH :
From the equation,
2 mol NaOH :
1
1 mol dm–3 × v
=
1 000
v=
4
Na2SO4 + 2H2O
100
1 000
= 0.1 mol
Mol of NaOH = 1 ×
mol HNO3
0.1 mol NaOH : 0.1 mol HNO3
Mv
Mol of HNO3 =
1 000
M= Concentration of HNO3
v = Volume of HNO3 in cm3
Comparison
and
explanation
2NaOH + H2SO4
1
mol H2SO4
0.1 mol NaOH : 0.05 mol H2SO4
Mv
Mol of H2SO4 =
1 000
M = Concentration of H2SO4
v = Volume of H2SO4 in cm3
1 mol dm–3 × v
= 0.05 mol
1 000
3
v = 50 cm
0.1
mol
100 cm3
doubled
– The volume of acid needed in Experiment I is
of Experiment II.
diprotic
monoprotic
– Sulphuric acid is
acid while nitric acid is
.
– One mol of sulphuric acid ionises to two mol of H+ ions, one mol of nitric acid
ionises to one mol of H+ ions.
– The number of H+ ions in the same volume and concentration of both acids is
doubled
in sulphuric acid compared to hydrochloric acid.
The diagram below shows the apparatus set-up for the titration of potassium hydroxide solution with sulphuric
acid.
0.5 mol dm–3 sulphuric acid
50 cm3 of 1 mol dm–3 potassium
hydroxide solution + methyl orange
153
6
Balanced
equation
Experiment I
UNIT
Experiment
MODULE • Chemistry FORM 4
0.5 mol dm–3 sulphuric acid is titrated to 50 cm3 of 1 mol dm–3 potassium hydroxide solution and methyl orange
is used as indicator.
(a) (i) Name the reaction between sulphuric acid and potassium hydroxide.
TP1
Neutralisation
(ii) Name the salt formed in the reaction.
TP2
Potassium sulphate
(b) Suggest an apparatus that can be used to measure 25.0 cm3 of potassium hydroxide solution accurately.
TP1
Pipette
(c) What is the colour of methyl orange
TP1
(i) in potassium hydroxide solution?
Yellow
(ii) in sulphuric acid?
Red
(iii) at the end point of the titration?
Orange
(d) (i)
TP3
Write a balanced equation for the reaction that occurs.
2KOH + H2SO4
K2SO4 + 2H2O
(ii) Calculate the volume of the 0.1 mol dm–3 sulphuric acid needed to completely react with 50 cm3 of
TP3
0.1 mol dm–3 potassium hydroxide.
Number of moles of KOH = 0.1 ×
50
= 0.005 mol
1 000
UNIT
From the equation,
6
2 mol KOH : 1 mol H2SO4
0.005 mol KOH : 0.0025 mol H2SO4
n mol
Volume of H2SO4 =
M mol dm–3
0.0025 mol
=
0.1 mol dm–3
= 0.025 dm3
= 25 cm3
(e) (i)
TP4
The experiment is repeated with 0.1 mol dm–3 hydrochloric acid to replace sulphuric acid. Predict the
volume of hydrochloric acid needed to neutralise 50.0 cm3 potassium hydroxide solution.
50 cm3 // double the volume of sulphuric acid
(ii) Explain your answer in (e)(i).
– Hydrochloric acid is a monoprotic
acid whereas sulphuric acid is a
diprotic
– The same volume and concentration of both acids, hydrochloric acid contains
number of mole of H+ ions as in sulphuric acid.
© Nilam Publication Sdn. Bhd.
154
acid.
half
the
MODULE • Chemistry FORM 4
5
(a) Substance A is white in colour. When A is strongly heated, brown gas, B and gas C are released. Gas C
TP4 lighted a glowing wooden splinter. Residue D which is yellow in colour when hot and white when cold is
formed.
(i) Name substances A, B, C and D.
A: Zinc nitrate
B: Nitrogen dioxide
C: Oxygen
D: Zinc oxide
(ii) Write the chemical equation when substance A is heated.
2Zn(NO3)2
2ZnO + 4NO2 + O2
(b) A colourless solution E gives the following results when a few series of tests are conducted.
TP5
S1 –Add sodium hydroxide solution, a white precipitate is formed. The precipitate is soluble in excess
sodium hydroxide solution.
S2 –Add ammonia solution, a white precipitate is formed. The precipitate is insoluble in excess ammonia
solution.
S3 – Add potassium iodide solution, a yellow precipitate F, is formed.
(i)
What are the possible cations present in substance E as a result of S1 test?
Pb2+, Al3+, Zn2+
(ii) What are the possible cations present in solution E as a result from S1 and S2 tests?
Pb2+, Al3+
(iii) What is the ion present in E after S3 test has been done? Write an ionic equation for the formation of
substance F.
2+
Ion present: Pb
The table below shows the colour of five solutions labelled A, B, C, D and E added with small amount until
excess of sodium hydroxide solution and ammonia solution.
Solution
A
B
C
D
E
Colour
Blue
Colourless
Light green
Colourless
Colourless
With sodium hydroxide solution
Blue precipitate insoluble in excess
White precipitate soluble in excess
Dirty green precipitate
White precipitate soluble in excess
White precipitate insoluble in excess
With ammonia solution
Blue precipitate soluble in excess
White precipitate soluble in excess
Dirty green precipitate
White precipitate insoluble in excess
White precipitate insoluble in excess
(a) State the cations present in:
A: Cu2+
B: Zn2+
C: Fe2+
2+
E: Mg
(b) State another test to identify C.
Add potassium hexacyanoferrate(III) solution, dark blue precipitate formed
(c) What are the possible cations present in solution D?
Al3+, Pb2+
155
6
6
TP4
PbI2
UNIT
2+
–
Ionic equation: Pb + 2I
MODULE • Chemistry FORM 4
(d) Describe briefly a test that can differentiate the cations present in solution D.
Add a few drops of potassium iodide / sodium chloride / sodium sulphate solution into 1 cm3 of solution
D. Yellow/ white precipitate formed, lead(II) ion / Pb2+ present. No precipitate, aluminium ion / Al3+
present.
7
TP4
The diagram below shows the flow chart of changes that took place beginning from solid M. Solid M is a zinc
salt. When solid M is heated strongly, it decomposes into solid Q which is yellow when hot and white when
cold.
Reaction I
Reaction II
Solid M
Add dilute nitric acid
Heat
Solid Q + carbon dioxide gas
Solution S
+
Carbon dioxide gas
Reaction III + Magnesium
Zinc metal + Magnesium nitrate solution
(a) (i)
State one chemical test for carbon dioxide gas.
Passed the gas through lime water, lime water turns chalky
(ii) Draw a diagram of the apparatus set-up to carry out reaction I.
UNIT
6
Solid M
Heat
Lime water
(b) Name solids M and Q.
M: Zinc carbonate
Q: Zinc oxide
(c) State the observations made when excess ammonia solution is added to solution S.
White precipitate, soluble in excess of ammonia solution.
(d) (i)
TP3
Write the chemical equation for reaction II.
ZnCO3 + 2HNO3
© Nilam Publication Sdn. Bhd.
Zn(NO3)2 + H2O + CO2
156
+
Water
MODULE • Chemistry FORM 4
(ii) For reaction II, calculate the volume of carbon dioxide gas released at room condition if 12.5 g solid
M decomposes completely. [Relative atomic mass: C = 12, O = 16, Zn = 65, 1 mole of gas occupies
24 dm3 at room condition]
12.5
= 0.1 mol
125
From the equation,
1 mol M : 1 mol CO2
0.1 mol M : 0.1 mol CO2
Volume of CO2 = 0.1 mol × 24 dm3 mol–1 = 2.4 dm3
Number of moles of solid M =
(e) Name reaction III.
Displacement reaction
(f) Describe a chemical test to determine the presence of anion in the magnesium nitrate solution.
2
– About
cm3 of magnesium nitrate solution is poured into a test tube.
– 2 cm3 of dilute
sulphuric acid
is added to the solution followed by 2 cm3 of
shaken
.
iron(II) sulphate
solution. The mixture is
slanted
– The test tube is
and held with a test tube holder.
– A few drops of concentrated sulphuric acid are dropped along the wall of the test tube and is held
upright.
nitrate
– A brown ring is formed between two layers. Anion present is
ion.
You are given zinc chloride crystals. Describe how you would conduct a chemical test in the laboratory to
identify the ions present in zinc chloride crystals.
– The
Dissolve
half spatula of zinc chloride crystals in 10 cm3 of
distilled water .
solution
is poured in three test tubes.
sodium hydroxide
solution
– Add a few drops of
solution to zinc chloride
until
excess
white
precipitate
sodium
hydroxide
.A
soluble in excess of
solution formed.
ammonia
solution
– Add a few drops of
solution to another zinc chloride
until
excess
white
precipitate
ammonia
.A
soluble in excess of
solution formed. Ions
zinc ions
present are
.
3
nitric
acid
– About 2 cm of dilute
is added to 2 cm3 of zinc chloride solution followed by 2 cm3 of
silver nitrate solution. White precipitate formed. Ions present are chloride ions.
Objective
Questions
157
6
–
UNIT
8
TP6
MODULE • Chemistry FORM 4
UNIT
RATE OF REACTION
7
Concept Map
Average
rate of
reaction
Daily
activities
Industry
processes
Rate of
reaction at any
given time
Measuring rate of
reaction
Measurement of the change in
quantity of reactant or product
per unit time
Examples in
Application
Meaning
RATE OF
REACTION
Factors affect rate of reaction
Experiments on the effect of
Can be explained using
7
Temperature
Concentration
Catalyst
The larger
the size of
solid
reactant, the
less total
surface area
expose to
collision
The higher
the
temperature,
the higher
the kinetic
energy of
particles
The higher
the
concentration
of solution,
the higher the
number of
particles per
unit volume
The
presence of
catalyst
lower the
activation
energy of
the reaction
Related to
Collision
Theory
Effective
collision
Related to
Frequency
of effective
collision
Cause
Rate of reaction
increases
© Nilam Publication Sdn. Bhd.
158
Happen if
UNIT
Size
Particles
possess
activation
energy
Particles
collide at the
correct
orientation
MODULE • Chemistry FORM 4
7.1
DETERMINE RATE OF REACTION
SK
7.1
State the meaning of rate
of reaction. SP 7.1.2
The rate of reaction is a measurement of the change in quantity of reactant or
product per unit time.
State the relationship
between rate of reaction
and time.
high
1 The rate of reaction is
within a short period of time.
low
2 The rate of reaction is
within a long period of time.
if the reaction occurs
fast
if the reaction occurs
slowly
3 The rate of reaction is inversely proportional to time:
1
Rate of reaction ∝
Time taken
Give examples of slow
reactions.
SP 7.1.1
How to determine rate of
reaction?
– Rusting of iron in the air.
– Photosynthesis.
– Fermentation of fruit juice to form alcohol.
Rate of reaction can be determined by calculating the rate of chemical change or
measured quantity in a chemical change per unit time.
SP 7.1.4
Rate of reaction =
How to identify the
change in quantity of
reactant/product for
measuring rate of
reaction?
Give example.
SP 7.1.3
What are the possible
unit for the rate of
reaction?
Change in quantity of reactant/product
Time taken for the change to occur
The change in amount of reactant or product in any reaction which is chosen for
the purpose of measuring rate of reaction must be observable and measurable.
Example:
(a) Decrease in the mass of reactant.
(b) Increase in the mass of product.
(c) Increase in volume of gas released.
(d) Formation of precipitate as a product.
Units for the rate of reaction depends on the unit for the reactant or product of the
reaction. The possible units are:
(a) g s–1 or g min–1 for increase in mass of product or decrease in mass of reactant
(b) cm3 s–1 or cm3 min–1 for increase of volume gas released
(c) s–1 or min–1 for fixed amount of precipitate formed in different experiments
159
7
SP 7.1.1
– Reaction of marble chip with hydrochloric acid.
– Reaction of magnesium with sulphuric acid.
– Reaction of potassium with water.
– Burning of fuel.
UNIT
Give examples of fast
reactions.
MODULE • Chemistry FORM 4
How to measure the
observable changes when
a reaction produce gas?
SP 7.1.3
Observable
changes
Chemical reaction
Reaction between
magnesium and
hydrochloric acid:
Decrease in
mass of
magnesium
Method of measuring the
observable changes
Hydrochloric
acid
Magnesium
100 g
Reading from the balance is recorded
in every 30 seconds.
Mg(s) + 2HCl(aq) →
MgCl2(aq) + H2(g)
Increase in
volume of
hydrogen
Method I
Hydrochloric
acid
Water
Magnesium
Hydrogen gas is collected by water
displacement in a burette. The volume
of hydrogen gas collected is recorded
every 30 seconds.
Method II
Hydrochloric acid
Magnesium
UNIT
The rate of reaction is measured by the
volume of gas collected in the gas
syringe per unit time.
*This apparatus set-up can also be used
to measure the increase in volume of
other gases that are insoluble for
example oxygen, hydrogen and
carbon dioxide.
7
How to measure the
observable change when
a reaction produces
precipitate?
SP 7.1.3
Observable
changes
Chemical reaction
Reaction between
sodium thiousulphate
and hydrochloric acid:
Na2S2O3(aq) +
2HCl(aq) →
2NaCl(aq) + H2O(l) +
SO2(g) + S(s)
© Nilam Publication Sdn. Bhd.
Formation of
sulphur as a
precipitate
*Volume of sulphur
dioxide gas, SO2
cannot be
measured by water
displacement
because sulphur
dioxide is soluble
in water.
160
Method of measuring the
observable changes
Sodium
thiosulphate
solution +
hydrochloric
acid
Amount of solid sulphur
formed is measured by the
time taken for the mark ‘X’
placed under the conical
flask can no longer be seen.
Water
161
Time / s
t1
The rate of reaction at t1 second
= The gradient of tangent to the
curve at t1 s
D y cm3
=
Dxs
Dx
Dy
Volume of carbon dioxide gas / cm3
Sketch of graph:
Calcium
carbonate
Hydrochloric
acid
Rate of reaction is measured by
volume of gas released in every
30 seconds by water
displacement in a burette.
Example
Time / s
Time / s
UNIT
7
t1
Rate of reaction for sodium thiosulphate
solution at temperature T1 °C
1
=
= y s–1
t1 s
T1
thiosulphate solution / °C
t1
Rate of reaction for sodium thiosulphate
solution with concentration M1 mol dm–3
1
=
= x s–1
t1 s
(b) Temperature of sodium
M1
thiosulphate solution / mol dm–3
Sketch of graph:
(a) Concentration of sodium
Sodium thiosulphate
solution +
hydrochloric acid
Rate of reaction is measured by time taken
for the formation of precipitate.
Example
3
Water
Time / s
1 23 45 67 8
Average rate of reaction in the first 4
minutes
(V – 0) cm3
=
= x cm3 s–1
(4 – 0) s
V
Volume of carbon dioxide gas / cm
Sketch of graph:
Calcium
carbonate
Hydrochloric
acid
Water
Time / s
1 23 45 67 8
Average rate of reaction in the fourth
minute
(V – V1) cm3
= 2
= y cm3 s–1
(4 – 3) s
V1
V2
Volume of carbon dioxide gas / cm3
Sketch of graph:
Calcium
carbonate
Hydrochloric
acid
Average rate of reaction within
certain period of time:
Rate of reaction is measured by
volume of gas released in every 30
seconds by water displacement in a
burette.
Example
Average rate of reaction
Average rate of reaction from 0
second:
Rate of reaction is measured by
volume of gas released in every 30
seconds by water displacement in a
burette.
Example
Measurement of Rate of Reaction
SP 7.1.4
Rate of reaction at any given time/temperature/concentration
How to determine the rate of reaction from the graph?
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
Exercise
1
SP 7.1.5
An experiment is carried out to determine the rate of reaction of 20 cm3 of 0.5 mol dm–3 hydrochloric acid with
excess calcium carbonate. The results are shown below.
Time / s
Volume of CO2 /
cm3
(a) (i)
TP3
0
15
30
45
60
75
90
105
120
135
150
165
0.00 10.00 16.00 22.00 27.00 31.50 36.00 39.50 42.00 44.00 44.00 44.00
Write a chemical equation for the above reaction.
CaCO3 + 2HCl → CaCl2 + H2O + CO2
(ii) State the observable and measurable changes in the experiment.
TP3
Increase in volume of carbon dioxide/decrease in mass of calcium carbonate
(iii) State the meaning of the rate of reaction for the above reaction.
TP3
Change in volume of carbon dioxide gas in one second/change in mass of calcium carbonate in one
second.
(iv) Draw an apparatus set-up to measure rate of reaction in the given reaction.
TP3
Hydrochloric acid
UNIT
7
Calcium carbonate
© Nilam Publication Sdn. Bhd.
162
Water
MODULE • Chemistry FORM 4
(b) Draw the graph of the volume of carbon dioxide gas collected against time.
Volume of CO2 / cm3
50
40
30
7
20
10
UNIT
TP4
0
30
60
90
120
163
150
180
Time / s
MODULE • Chemistry FORM 4
(c) From the graph, determine:
(i) the average rate of reaction in the first minute.
TP3
Total volume of carbon dioxide gas collected in the first minute
=
Time taken for the change to occur
27
=
60
= 0.45 cm3 s–1
(ii) the average rate of reaction in the second minute
TP3
=
Total volume of carbon dioxide gas collected between first minute and the second minute
Time taken for the change to occur
= 42 – 27
60
= 0.25 cm3 s–1
(iii) the time when the reaction has completed
TP3
135 s
(iv) the average rate of reaction for overall reaction
TP3
Total volume of carbon dioxide collected
=
Time taken for the change to occur
44
=
135
= 0.326 cm3 s–1
UNIT
(v) the rate of reaction at 30 seconds
TP3
= the gradient of the graph at 30 seconds
7
= 0.405 ± 0.1 cm3 s–1
(vi) the rate of reaction at 105 seconds
TP3
= the gradient of the graph at 105 seconds
= 0.217 ± 0.1 cm3 s–1
(d) Compare the rate of reaction at 30 seconds and 105 seconds. Explain your answer.
TP4
Rate of reaction at 30 seconds is higher than at 105 seconds because the concentration of hydrochloric
acid decreases as time increases.
© Nilam Publication Sdn. Bhd.
164
MODULE • Chemistry FORM 4
Sketch a curve for volume of
hydrogen gas collected against
time for the reaction between
excess of zinc powder with 50 cm3
of 1 mol dm–3 hydrochloric acid.
The tangents on the curve at t1, t2
and t3 are shown.
Volume of hydrogen / cm3
Tangent on the
curve at t1, t2 and
t3 respectively
SP 7.1.4
0 t1
Write the balanced equation for
the reaction.
Calculate the volume of hydrogen
gas collected in the experiment at
room conditions.
V
t2
Time / min
t3
Zn + 2HCl → ZnCl2 + H2
From the equation,
2 mol of HCl : 1 mol H2
0.05 mol HCl : 0.025 mol H2
Volume of H2
= 0.025 mol × 24 dm3 mol–1
= 0.6 dm3
= 600 cm3
Compare the gradient of the curve
at t1 and t2. Explain your answer.
The gradient of tangent on the curve at t2 is lower than t1. The
rate of reaction at t2 is lower than at t1. The rate of reaction
decreases as the time increases because
mass
of zinc and
concentration
What is the gradient at t3? Explain
your answer.
of hydrochloric acid
decreases .
zero , the rate of
The gradient of tangent on the curve at t3 is
zero . The reaction is completed at t3. All
reaction at t3 is
reacted
hydrochloric acid has
because zinc powder used is in
excess . At t3, maximum volume of hydrogen gas is collected.
The maximum volume of hydrogen gas collected is
Sketch a curve for mass of zinc
against time.
Mass of zinc / g
Time / s
165
600 cm3
.
7
TP3
TP4
Excess of zinc powder is added to 50 cm3 of 1 mol dm–3 hydrochloric acid. The volume of hydrogen gas
collected and time taken are recorded. Complete the following table. SP 7.1.3
UNIT
2
MODULE • Chemistry FORM 4
Sketch a curve for concentration
of hydrochloric acid against time.
Concentration of hydrochloric acid / mol dm–3
Time / s
3
The diagram below shows a conical flask containing calcium carbonate powder and hydrochloric acid. The
experiment is set up to study the rate of reaction by measuring the mass loss of the reacting conical flask.
Cotton wool
Hydrochloric acid
Calcium carbonate
100 g
(a) (i)
TP3
Electronic balance
Write a balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid.
CaCO3 + HCl → CaCl2 + H2O + CO2
(ii) What could have caused the mass loss of the conical flask in the reaction?
The carbon dioxide gas released from the reaction between hydrochloric acid and calcium carbonate
(b) What is the function of cotton wool?
TP3
To prevent any liquid from escaping from the flask.
(c) Describe how this set up of apparatus can be used to measure the rate of reaction.
TP3
– The gas release causes the mass of conical flask decreases as the reaction progresses
UNIT
– The reading of electronic balance are taken at regular intervals of time
– The rate reaction is calculated by the total amount of mass loss of the conical flask per unit time taken
7
(d) The sketch of graph below shows the results of the experiment when the reading is plotted.
TP4
Mass of conical flask / g
A
B
C
80
Time / s
State and describe the rate of reaction in the following points shown in the graph.
(i) Point A
The gradient of the graph is the steepest. The rate of reaction is the highest at the start.
© Nilam Publication Sdn. Bhd.
166
MODULE • Chemistry FORM 4
(ii) Point B
The gradient of the graph becomes less steep. The rate of reaction decreases.
(iii) Point C
The graph is a horizontal straight line. The reaction has stopped.
(e) State one factor that affect the rate of reaction from point A to B.
Concentration of hydrochloric acid decreases as time increases.
(f) Sketch a graph to show the decrease in mass against time on the graph below.
Decrease in mass / g
Time / s
80
(g) The chemical equation below shows a neutralization reaction.
TP5 NaOH(aq) + HCl(aq) → NaCl(aq) + H O(l)
2
Can the rate of reaction for the reaction measured by the mass loss of the reacting conical flask? Explain
your answer.
Cannot. The reaction does not produce any gas. No gas escaping from the flask to cause the decrease in
7.2
1
FACTORS THAT AFFECT THE RATE OF REACTION
UNIT
7
mass in the content of conical flask.
SK
7.2
The rate of reaction is affected by:
(a) Size of solid reactant
(b) Concentration of solution (for the reactant used in the form of solution)
(c) Temperature of solution at which the reaction occurs
(d) Presence of catalyst (for a particular reaction)
(e) Pressure of gas reactant
Rate of
Reaction
167
168
Hypothesis:
Manganese(IV) oxide increases the rate of
decomposition of hydrogen peroxide.
Manipulated variable:
The presence of manganese(IV) oxide
Responding variable:
The rate of reaction
Fixed variables:
Volume and concentration of hydrogen
peroxide solution
Apparatus:
Test tubes, measuring cylinder, retort stands
and clamp, filter funnel, spatula, electronic
balance, beaker.
Hypothesis:
The smaller the size of calcium carbonate,
the higher the rate of reaction.
Manipulated variable:
The size of calcium carbonate
Responding variable:
The rate of reaction
Fixed variables:
Mass of calcium carbonate chips, volume
and concentration of hydrochloric acid, the
temperature of the reaction mixture.
Apparatus:
Conical flask, basin, delivery tube and
rubber stopper, retort stand and clamp,
measuring cylinder, burette, stop watch and
weighing balance.
peroxide?
dilute hydrochloric acid?
rate of decomposition of hydrogen
chips affect the rate of its reaction with
B
Problem statement:
How does manganese(IV) oxide affect the
A
Manganese(IV)
oxide
Problem statement:
How does the size of calcium carbonate
Water
Hydrogen
peroxide
solution
Glowing
wooden
splinter
Catalyst
Hypothesis:
When the temperature of sodium
thiosulphate solution increases, the rate of
its reaction with sulphuric acid increases.
Manipulated variable:
Temperature of sodium thiosulphate solution
Responding variable:
The rate of reaction
Hypothesis:
When the concentration of sodium
thiosulphate solution increases, the rate of
its reaction with sulphuric acid increases.
Manipulated variable:
Concentration of sodium thiosulphate
solution
concentration of dilute sulphuric acid, size
of conical flask.
Apparatus:
100 cm3 conical flask, 50 cm3 and 5 cm3
measuring cylinder, stop watch,
thermometer.
volume and concentration of dilute
sulphuric acid, temperature of mixture, size
of conical flask.
Apparatus:
100 cm3 conical flask, 50 cm3 and 5 cm3
measuring cylinder, stop watch, white paper
with mark ‘X’.
thiosulphate solution, volume and
Fixed variables:
Volume of sodium thiosulphate solution,
Fixed variables:
Volume and concentration of sodium
reaction with sulphuric acid?
reaction with sulphuric acid?
Responding variable:
The rate of reaction
thiousulphate solution affect the rate of its
‘X’ mark
thiousulphate solution affect the rate of its
Heat
Paper
Sodium thiosulphate Conical
solution +
flask
sulphuric acid
Problem statement:
How does the temperature of sodium
Sodium
thiosulphate
solution
Temperature
Problem statement:
How does the concentration of sodium
Sodium thiosulphate
solution +
Sulphuric acid
Concentration of solution
X
Calcium carbonate
Hydrochloric
acid
Size of solid reactant
SP 7.2.1
43
42
41
40
39
38
37
36
35
7
© Nilam Publication Sdn. Bhd.
34
UNIT
Planning of Experiments to Study the Factor that Affect the Rate of Reaction
MODULE • Chemistry FORM 4
169
3
Volume of gas / cm3
Burette reading / cm 3
Time / s
B Small calcium carbonate chips
Volume of gas / cm3
Burette reading / cm
Time / s
A Large calcium carbonate chips
Tabulation of data:
Procedure:
1 A basin is filled with water until half full.
2 A burette full with water is inverted into
the basin.
3 It is then clamped vertically using retort
stand.
4 5 g of large calcium carbonate chips is
weighed and put inside the conical flask
as shown in the diagram.
5 50 cm 3 of hydrochloric acid 0.2 mol dm
–3
is measured and poured into the conical
flask.
6 The conical flask is closed immediately
with a stopper fitted with delivery tube as
shown in the diagram.
7 A stop watch is started immediately.
8 The volume of gas released is recorded
for every 30 seconds.
9 Steps 1 to 8 are repeated using 5 g small
calcium carbonate chips to replace 5 g of
large calcium carbonate chips.
10 The graphs of volume of carbon dioxide
against time for the two experiments are
plotted on the same axes.
Materials:
Large and small calcium carbonate chips,
0.2 mol dm–3 of hydrochloric acid.
Observation
UNIT
7
Mass of manganese(IV) oxide before
reaction
= ......g
Mass of manganese(IV) oxide after reaction
= ......g
B
A
Test tube
Tabulation of data:
Procedure:
1 Two test tubes are labelled A and B
respectively.
2 5 cm3 of 20-volume of hydrogen
peroxide solution is measured and poured
separately into test tubes A and B.
3 1 g of manganese(IV) oxide powder is
weighed and added to test tube B.
4 A glowing wooden splinter is quickly
placed at the mouth of each test tube.
Observe the changes.
5 At the end of the reaction, the mixture in
test tube B is filtered to separate the
manganese(IV) oxide powder and the
residue is rinsed with distilled water.
6 Manganese(IV) oxide powder is pressed
between to filter papers to dry it. Dry
manganese(IV) is weighed again.
7 All observations are recorded.
Materials:
20-volume of hydrogen peroxide solution,
manganese (IV) oxide powder, filter paper,
glowing wooden splinter
1
–1
time / s
Time taken for ‘X’ to
disappear/s
Concentration of
Na2S2O3 solution /
mol dm –3
Volume of water /
cm3
Volume of Na2S2O3 /
cm3
Tabulation of data:
0 10 20 30 40
50 40 30 20 10
Procedure:
1 50 cm3 of sodium thiosulphate solution is
measured and poured into a conical flask.
2 The conical flask is placed on top of a piece
of paper with a mark ‘X’ at the centre.
3 5 cm3 of sulphuric acid 1 mol dm–3 is
measured and poured quickly and
carefully into the conical flask. Swirl the
conical flask at the same time start the
stop watch.
4 The mark ‘X’ as shown in the above
diagram is observed.
5 The stop watch is stopped immediately
when the mark ‘X’ is no longer visible.
6 The time taken for the mark ‘X’ is no
longer visible is recorded.
7 Steps 1 to 6 are repeated using different
volumes of sodium thiosulphate solution
with different volumes of distilled water
as shown in the table.
8 Graphs of concentration of sodium
thiosulphate against time and
concentration
of sodium thiosulphate against 1 are
time
plotted.
Materials:
0.2 mol dm–3 sodium thiosulphate solution,
1 mol dm–3 sulphuric acid, distilled water.
1
–1
time / s
Time taken for
‘X’ to
disappear/ s
Temperature of
Na2S2O3,
55 50 45 40 35 30
solution / °C
Tabulation of data:
Procedure:
1 50 cm3 of sodium thio­sul­phate solution is
measured and poured into a conical flask.
2 The temperature of this solution is
measured using thermometer and recorded.
3 The conical flask is placed on top of a
piece of paper with a mark ‘X’ at the
centre.
4 5 cm3 of sulphuric acid is measured and
poured quickly and carefully into the
conical flask. Swirl the conical flask at the
same time start the stop watch.
5 The mark ‘X’ as shown in the above
diagram is observed.
6 The stop watch is stopped immediately
when the mark ‘X’ is no longer visible.
7 The time taken for the mark ‘X’ is no
longer visible is recorded.
8 Steps 1 to 7 are repeated using same
volume and con­centration of sodium
thiosulphate solution but heat­ed gently to
a higher temperature of 35°C, 40°C,
45°C, 50°C and 55°C. All other
conditions remain unchanged.
9 A graph of temperature against time is
plotted and temperature against 1 are
time
plotted.
Materials:
0.2 mol dm–3 sodium thiosulphate solution,
1 mol dm–3 sulphuric acid, distilled water.
MODULE • Chemistry FORM 4
7
Volume of carbon dioxide / cm3
Sketch of the graph
(b) The gradient of tangent at t1 for experiment I is
greater than experiment II.
Interpretation and conclusion
SP 7.2.1
© Nilam Publication Sdn. Bhd.
170
t1
Diagram 2
Time / s
Experiment II
Experiment I
Volume of CO2 / cm3
Tangent at t1 for
experiments I and II
hydrochloric acid in experiments I and II is equal .
5 Conclusion:
The rate of reaction of the small calcium carbonate
chips is higher than the larger calcium carbonate
chips.
3 Since calcium carbonate used is in excess, all
hydrochloric acid has reacted .
The number of mole of hydrochloric acid in both
experiments
50 × 0.2
=
= 0.01 mol
1 000
4 The volume of hydrogen gas collected for both
experiments is equal because number of mol of
(c) The rate of reaction at t1 for experiment I is
higher than experiment II.
1 From the sketch of graph in Diagram 1:
(a) The average rate of reaction for the first t1 seconds
in experiment I
Experiment I
V1 cm3
V1
Experiment II:
=
= X cm3 s–1
Experiment II
t
3
–3
1s
V2
50 cm of 0.2 mol dm hydrochloric acid + excess of
(b) The average rate of reaction for the first t1 seconds
large calcium carbonate chips.
in experiment II
Balanced equation:
V2 cm3
Time / s
=
= Y cm3 s–1
t
1
t1 s
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Diagram 1
The value of V1 is larger than V2
Observable changes to measure rate of reaction:
⇒ The average rate of reaction for the first t1
• Total volume of carbon dioxide gas
Volume of carbon dioxide collected in every 30 seconds
seconds in experiment I is higher than
released for the first t1 seconds in
experiment II.
experiment
I
=
V
1
by water displacement in the burette
2
From
the
sketch of graph in Diagram 2:
• Total volume of carbon dioxide gas
* The size of CaCO3 is changed in both experiments. The
released for the first t1 seconds in
(a) The tangent at t1 in experiment I is steeper
volume and concentration of HCl are kept constant.
experiment II = V2
than experiment II.
Experiment I:
50 cm3 of 0.2 mol dm–3 hydrochloric acid + excess of
small calcium carbonate chips.
(a) Factor: Size of reactant
Intepretation of data and conclusion for the experiments to investigate factors that affect the rate of reaction.
UNIT
2
MODULE • Chemistry FORM 4
171
SO2(g) + S(s)
UNIT
7
Experiment:
50 cm3 of 0.2 mol dm–3 sodium thiosulphate solution at
30°C + 5 cm3 of 1.0 mol dm–3 sulphuric acid. Experiment is
repeated using 50 cm3 of 0.2 mol dm–3 sodium thiosulphate
solution at 35°C, 40°C, 45°C and 50°C respectively +
5 cm3 of 1.0 mol dm–3 sulphuric acid.
Balanced equation:
Na2S2O3(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) +
(c) Factor: Temperature of reaction mixture
Observable changes to measure rate of reaction:
Time taken for the mark ‘X’ placed under the conical flask
to disappear from view. Fixed quantity of solid sulphur
is formed in every experiment.
* The concentration of Na2S2O3(aq) is changed in all
experiments. The volume and temperature of sulphuric
acid are kept constant.
SO2(g) + S(s)
Experiment:
45 cm3 of 0.2 mol dm–3 sodium thiosulphate solution +
5 cm3 of 1.0 mol dm–3 sulphuric acid. Experiment is
repeated four more times using 0.2 mol dm–3 sodium
thiosulphate solution diluted with different volume of
distilled water.
Balanced equation:
Na2S2O3(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) +
(b) Factor: Concentration of solution
Diagram 5
Temperature of sodium
thiosulphate solution / °C
Time / s
Sketch of the graph
Diagram 4
1
/ s–1
time
Concentration of sodium
thiosulphate solution / mol dm–3
Diagram 3
Time / s
Concentration of sodium
thiosulphate solution / mol dm–3
Sketch of the graph
1
3
2
1
to appear and the faster for the mark ‘X’ to disappear.
From the sketch of graph in Diagram 5:
(a) As the temperature of sodium thiosulphate solution
decreases, a longer time is needed for the mark ‘X’
to disappear from view.
(b) Temperature of sodium thiosulphate solution is
inversely
proportional to time taken for the
mark ‘X’ to disapear.
(c) The higher the temperature, the shorter is the
yellow
time taken for the
sulphur precipitate
Interpretation and conclusion
rate of reaction .
Conclusion:
The higher the concentration of sodium thiosulphate
solution, the higher is the rate of reaction .
From the sketch of graph in Diagram 4:
As the concentration of sodium thiosulphate increases,
1
1
the value of time increases. time represents the
to appear and the faster for the “X” sign to disappear.
the mark ‘X’ to disappear from view.
(b) Concentration of sodium thiosulphate solution is
inversely proportional to time taken for the
mark ‘X’ to disapear.
(c) The higher the concentration, the shorter is the
time taken for the yellow sulphur precipitate
From the sketch of graph in Diagram 3:
(a) As the concentration of sodium thiosulphate
solution decreases, a longer time is needed for
Interpretation and conclusion
MODULE • Chemistry FORM 4
UNIT
7
© Nilam Publication Sdn. Bhd.
172
Effect of quantity of manganese(IV)
oxide on the rate of decomposition
of hydrogen peroxide
Observable changes to measure rate of reaction:
Ignition of glowing wooden splinter.
Test tube B:
MnO2
2H2O2
H2O + O2
2H2O2 → H2O + O2
Balanced equation:
Test tube A:
Test tube B:
10 cm3 of 20-volume of hydrogen peroxide decomposed
with the presence of 5 g of manganese(VI) oxide.
Test tube A:
10 cm3 of 20-volume of hydrogen peroxide decomposed
without manganese(VI) oxide.
(d) Factor: Presence of catalyst
1
/ s–1
time
The glowing wooden
splinter glows dimly
The glowing wooden
splinter ignites brightly
A
B
2
1
Conclusion:
The higher the temperature of sodium thiosulphate
higher
solution, the
is the rate of reaction .
3
The glowing wooden splinter ignites with bright flame,
fast
hydrogen peroxide decomposes very
in
test tube B with the presence of manganese(IV) oxide.
MnO2
2H2O2
H 2O + O 2
increase
4 Manganese(IV) oxide can
the rate of
decomposition of hydrogen peroxide in test tube B to
produce more oxygen gas which ignites the glowing
wooden splinter.
5 Mass of manganese(IV) oxide powder remains
unchanged during experiment.
6 Manganese(IV) oxide acts as a catalyst .
Conclusion:
Manganese(IV) oxide increases the rate of
decomposition of hydrogen peroxide.
2H2O2 → H2O + O2
Oxygen gas is released during decomposition of
hydrogen peroxide. The quantity of oxygen gas
released is observed in the ignition of the glowing
wooden splinter.
The glowing wooden splinter glows dimly in test tube
slow
A, hydrogen peroxide decomposes very
without the presence of manganese(IV) oxide.
Interpretation and conclusion
From the sketch of graph in Diagram 6:
As the temperature of sodium thiosulphate solution
1
increases . 1
increases, the value of time
time
represents the rate of reaction.
2
Mass of manganese(IV) oxide before
reaction = 5 g
3
Mass of manganese(IV) oxide after
reaction = 5 g
Observation
Test tube
Sketch of the graph
Diagram 6
Temperature of sodium
Observable changes to measure rate of reaction:
Time taken for the mark ‘X’ placed under the conical flask thiosulphate solution / °C
disappear
to
from view. Fixed quantity of solid
sulphur is formed in every experiment.
MODULE • Chemistry FORM 4
3
V
Time
(minute/
second)
(b) The gradient
of the curve
(a) The maximum
quantity of product
Quantity of product (g/mol/cm3)
UNIT
7
An increase in the quantity of catalyst chemical reaction increases the rate of reaction, the
greater .
gradient of the curve becomes
The presence of catalyst in certain chemical reaction increases the rate of reaction, the
greater .
gradient of the curve becomes
The presence of
catalyst
the rate of reaction, the
The higher the temperature of a solution, the
greater
is the gradient of the curve.
The temperature of
reaction mixture
higher
The higher the concentration of a solution, the higher the rate of reaction, the greater is the
gradient of the curve.
The concentration
of solution
the rate of reaction, the
The smaller the size of a solid reactant, the
greater
is the gradient of the curve.
higher
Effect on the gradient of the curve
SP 7.2.1
The size of solid
reactant
Factor
(b) The gradient of the curve: It depends on the factors that affect the rate of reaction.
The curve for graph of amount of product formed against time in a chemical reaction consists of two parts:
(a) The maximum quantity of product: It depends on the number of mol of reactants react in the chemical reaction.
MODULE • Chemistry FORM 4
173
UNIT
7
© Nilam Publication Sdn. Bhd.
174
burette.
displacement in the
30 seconds by water
1.0 mol dm–3
1.0 mol dm–3
Concentration of H2SO4
size
of zinc and
30°C
40°C
Temperature
is double of
the
gradient
of the curve for experiment I is
greater
than experiment II.
higher
⇒ Initial rate of reaction in experiment I is
than experiment II because the
temperature of reaction mixture in experiment I
higher
than experiment II,
⇒ The rate of reaction in experiments I and II is not affected by
concentration of sulphuric acid.
Powder
II
collected in every
Type of zinc
Powder
Experiment
Compare rate of reaction:
experiment I
50 × 1
= 0.05 mol
1 000
⇒ The maximum volume of hydrogen gas collected in
experiment II .
(b) Number of mol of H2SO4 in experiment II =
Zinc is in excess in experiments I and II, the volume of hydrogen gas collected is not
affected by the quantity of zinc.
Volume of H2 collected depends on number of mol of H2SO4
100 × 1
(a) Number of mol of H2SO4 in experiment I =
= 0.1 mol
1 000
I
3
SP 7.2.1
Number of mol of reactant / Quantity of product / Factor
Observable changes to
measure rate of reaction:
Volume of hydrogen gas
ZnSO4(aq) + H2(g)
Balanced equation:
Zn(s) + H2SO4(aq) →
Experiment II:
Excess of zinc powder + 50 cm3
of 1.0 mol dm–3 sulphuric acid at
30°C
Experiment I:
1
Excess of zinc powder + 100 cm3
of 1.0 mol dm–3 sulphuric acid at 2
40°C
(a) Experiment
Exercise: Sketch the graph of volume of gas produced against time for following experiments.
Time / s
Experiment II
Experiment I
Volume of hydrogen / cm3
Sketch of the graph
MODULE • Chemistry FORM 4
100 × 0.5
= 0.05 mol
1 000
Sketch of the graph
175
II
III
water displacement in the
burette.
Chips
Chips
Chips
Type of CaCO3
higher
0.5 mol dm–3
0.5 mol dm–3
1.0 mol dm–3
Concentration of HCl
of the curve for experiment I is
greater
equal
of HCl in experiment I is
than experiment II.
than experiment II, the
UNIT
7
⇒ Initial rate of reaction in experiment II is
to experiment III because the
concentration of HCl in experiments II and III is the
same , the gradient
same .
of the curve in experiments II and III is the
gradient
concentration
⇒ The rate of reaction in experiments I, II and III is not affected by size of calcium
carbonate.
higher
⇒ Initial rate of reaction in experiment I is
than experiment II because the
I
collected in every 30 seconds by
Experiment
Compare rate of reaction:
⇒ The maximum volume of carbon dioxide gas in experiment I is double of
experiment II .
Time / s
Experiment II
Experiment I
3
⇒ The maximum volume of carbon dioxide gas in experiment III is double of Volume of carbon dioxide / cm
experiment I .
Experiment III
(c) Number of mol of HCl in experiment III =
25 × 0.5
= 0.0125 mol
1 000
Volume of CO2 collected depends on the number of mol of HCl
25 × 1.0
(a) Number of mol of HCl in experiment I =
= 0.025 mol
1 000
2
(b) Number of mol of HCl in experiment II =
Calcium carbonate is in excess in experiment I, II and III, the volume of carbon dioxide
gas collected is not affected by the quantity of calcium carbonate.
Number of mol of reactant / Quantity of product / Factor
1
Observable changes to measure
3
rate of reaction:
Volume of carbon dioxide
CaCl2(aq) + H2O(l) + CO2(g)
Balanced equation:
CaCO3(s) + 2HCl(aq) →
Experiment III:
Excess calcium carbonate chips
and 100 cm3 of 0.5 mol dm–3
hydrochloric acid
Experiment II:
Excess calcium carbonate chips
and 25 cm3 of 0.5 mol dm–3
hydrochloric acid
Experiment I:
Excess calcium carbonate chips
and 25 cm3 of 1.0 mol dm–3
hydrochloric acid
(b) Experiment
MODULE • Chemistry FORM 4
© Nilam Publication Sdn. Bhd.
Observable changes to
measure rate of reaction:
Volume of hydrogen gas
collected in every
30 seconds by water
displacement in the
burette.
Balanced equation:
Mg(s) + 2HCl(aq) →
MgCl2(aq)
Experiment II:
Excess of magnesium ribbon +
200 cm3 of 1.0 mol dm–3
hydrochloric acid
4
3
2
1
200 × 1
= 0.2 mol
1 000
176
Ribbon
II
higher
1.0 mol dm–3
1.0 mol dm–3
Concentration of HCl
than experiment II because the total
higher
surface area of magnesium powder in experiment I is
than magnesium
ribbon in experiment II, the gradient of the curve for experiment I is greater than
experiment II.
Powder
I
Initial rate of reaction in experiment I is
Type of Mg
Experiment
Compare rate of reaction:
⇒ The maximum volume of hydrogen gas collected in experiment II is double of
experiment I .
(b) Number of mol of HCl in experiment II =
Magnesium is in excess in experiments I and II, the volume of hydrogen gas collected is
not affected by the quantity of magnesium.
Volume of H2 collected depends on the number of mol of HCl.
100 × 1
(a) Number of mol of HCl in experiment I =
= 0.1 mol
1 000
7
Experiment I:
Excess of magnesium powder +
100 cm3 of 1.0 mol dm–3
hydrochloric acid
Number of mol of reactant / Quantity of product / Factor
UNIT
(c) Experiment
Time / s
Experiment I
Experiment II
Volume of hydrogen / cm3
Sketch of the graph
MODULE • Chemistry FORM 4
177
Observable changes to
measure rate of reaction:
Volume of hydrogen gas
collected in every 30 seconds by
water displacement in the
burette.
Zn + H2SO4 → ZnSO4 + H2
Experiment III:
Zn + H2SO4 → ZnSO4 + H2
Balanced equation:
Experiments I and II:
Experiment III:
Excess of zinc granules and
100 cm3 of 0.5 mol dm–3
sulphuric acid
Experiment II:
Excess of zinc granules and
100 cm3 of 0.5 mol dm–3
sulphuric acid + 5 cm3 of
copper(II) sulphate solution
Experiment I:
Excess of zinc granules and
100 cm3 of 1.0 mol dm-3
sulphuric acid + 5 cm3 of
copper(II) sulphate solution
(d) Experiment
3
2
1
Granules
III
to
than experiment II.
than experiment II, the gradient
than experiment II because the
–
Copper(II) sulphate
Copper(II) sulphate
Presence of catalyst
equal
is double of
UNIT
7
higher
⇒ Initial rate of reaction in experiment II is
than experiment III because the
catalyst copper(II) sulphate is present in experiment II. The gradient of the curve in
experiment II is greater than experiment III.
concentration of H2SO4 in experiment I
of the curve for experiment I is greater
higher
higher
0.5 mol dm–3
0.5 mol dm
–3
1.0 mol dm–3
Concentration of H2SO4
⇒ Initial rate of reaction in experiment I is
Granules
Granules
I
II
Type of zinc
Experiment
Compare rate of reaction:
⇒ The maximum volume of hydrogen gas collected in experiment II is
experiment III.
⇒ The maximum volume of hydrogen gas collected in
experiment II .
100 × 0.5
= 0.05 mol
1 000
(c) Number of mol of H2SO4 in experiment III =
experiment I
100 × 0.5
= 0.05 mol
1 000
(b) Number of mol of H2SO4 in experiment II =
Zinc is in excess in experiments I, II and III, the volume of hydrogen gas collected is not
affected by the quantity of zinc.
Volume of H2 depends on the number of mol of H2SO4.
100 × 1
(a) Number of mol of H2SO4 in experiment I =
= 0.1 mol
1 000
Number of mol of reactant / Quantity of product / Factor
Time / s
Experiment III
Experiment II
Experiment I
Volume of hydrogen / cm3
Sketch of the graph
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
7.3
APPLICATION OF FACTORS THAT AFFECT THE RATE OF REACTION IN
DAILY LIFE
Give examples of the
application of factor of
size that affect the rate
of reaction in daily
activities.
SP 7.3.1
(a) Burning of charcoal – When food is cooked with smaller pieces of charcoal,
faster
the food cooked
. The smaller pieces of charcoal have a larger
total
exposed surface area. Hence, smaller pieces charcoal burns faster
to produce
more
heat.
total
(b) Cooking of smaller pieces of food – The
surface area on a smaller
absorb
larger . The food can
cut pieces of food is
more heat. Hence,
the time taken for the food to be cooked is
Give examples of the
application of factor of
temperature that
affect the rate of
reaction in daily
activities.
SP 7.3.1
shorter
.
(a) Storage of food in a refrigerator – When the food is kept in refrigerator, the
lower
food lasts longer. The
temperature in the refrigerator slows down
bacteria . The
bacteria
the activity of the
produces less toxin ,
lower .
hence the rate of decomposition of food is
(b) Cooking food in a pressure cooker – The high pressure in pressure cooker
increases the boiling point of water to a temperature above 100ºC. The
kinetic energy of the particles in the food is increase/higher . Hence time
taken for the food to be cooked is shorter . Thus the food cooked
at a higher temperature in a pressure cooker.
Give examples of the
application of factor of
concentration that
affect the rate of
reaction in daily
activities.
SP 7.3.1
SK
7.3
faster
(a) Statue, bridge or building made from iron deteriorate more rapidly than in less
polluted air.
– In a polluted atmosphere, the concentration of sulphur dioxide is high.
Sulphur dioxide dissolve in rain water to form high concentration of acid
rain.
– High concentration of acid corrodes building, monuments and statues
faster
made from marble (calcium carbonate)
because calcium
UNIT
carbonate react with acid to produce salt, water and carbon dioxide.
– Iron rust faster with the high concentration of acid rain.
7
(b) Iron structures facing seafront rust faster. Sea breeze contains high
concentration of dissolved salts. The high concentration dissolved salts
becomes better electrolyte. An electrolyte will increase the electrical conductivity
of water. Hence, the rate corrosion of iron increases.
Give examples of
factor of catalyst in
industrial processes
and daily life.
SP 7.3.1
(a) Haber process is an industrial process to manufacture ammonia gas . The
optimum conditions to run the process at a higher rate are:
(i) The temperature is 400ºC – 500ºC.
(ii) The pressure is 200 – 300 atm.
iron
(iii) The catalyst is
, Fe.
Chemical equation:
N2(g) + 3H2(g)
© Nilam Publication Sdn. Bhd.
178
400ºC – 500ºC
Fe, 200 – 300 atm
2NH3(g)
MODULE • Chemistry FORM 4
(b) Contact process is an industrial process to manufacture sulphuric acid. The
optimum conditions to run the process at a higher rate are:
(i) The temperature is 450ºC.
(ii) The pressure is 1 atm.
(iii) The catalyst is vanadium(V) oxide , V2O5.
Chemical equation:
2SO2 + O2
V2O5, 450ºC
1 atm
2SO3
(c) Ostwald process is an industrial process to manufacture
platinum .
catalyst used for this process is
nitric acid
. The
(d) A catalytic converter is a device that uses platinum as a catalyst to convert three
harmful compounds in car exhaust into harmless compounds. In a catalytic
converter, the catalyst converts:
– carbon monoxide into carbon dioxide.
– hydrocarbons into carbon dioxide and water.
– nitrogen oxides back into nitrogen and oxygen.
THE COLLISION THEORY
During a chemical reaction,
1 Chemical bonds in the reactants are broken.
2 New bonds in the products are formed.
According to the collision theory, a chemical reaction can only occur when a
reacting particles collide with one another with certain amount of kinetic
energy.
It is the minimum amount of energy needed for the reacting particles to react.
SP 7.4.2
Are all chemical reactions
have the same activation
energy, Ea?
Define effective collision.
SP 7.4.1
What happens when the
energy of collision between
particles less than activation
energy, Ea?
SP 7.4.2
No, different chemical reactions have different activation energy.
The collisions that lead to a chemical reaction and result in the formation of
products.
The reactants particles just bounce off each other and no reaction occur.
Bounce
Collision
Reactant A
⇒
⇒
⇒
⇒
Below Ea
Reactant B
Collision energy of particles < activation energy.
The chemical bonds in the reactants are not broken.
No reaction.
The collision is an ineffective collision.
179
No new products
formed
7
The energy changes are
always caused by two
processes that occur during
chemical reaction when
reactants change to products.
What are these two
processes?
How can a reaction take place
according to the collision
theory?
What is activation energy, Ea?
SK
7.4
UNIT
7.4
MODULE • Chemistry FORM 4
What happens when the
collision between particles
does not occur at the correct
orientation?
SP 7.4.2
The reactants particles just bounce off each other and no reaction occur.
Collision
Wrong orientation
Reactant B
Reactant A
Bounce
No new products
formed
⇒ The chemical bonds in the reactants are not broken. No reaction.
⇒ The collision is an ineffective collision.
What happens when the
energy of collision between
the reacting particles have
equal or more than the
activation energy, Ea and
collide at correct orientation?
SP 7.4.2
What are the conditions for a
reaction to occur?
SP 7.4.2
How to show the energy
changes of reacting particles
and the activation energy, Ea
in a reaction?
Achieves Ea
Correct orientation
Collision
Reactant B
Reactant A
⇒
⇒
⇒
⇒
New products are
formed
Collision energy of particles ≥ activation energy.
The chemical bonds in the reactants are broken.
Reaction occurs.
The collision is an effective collision.
For a reaction to occur, the collision must satisfy two conditions:
(a) The colliding particle must have enough energy i.e equal or more than a
minimum amount of energy known as *activation energy, Ea.
*The activation energy differs in different chemical reaction. The lower
the activation energy, the higher the rate of reaction.
(b) The colliding particles must also have the right orientation of collisions.
The energy changes of the reacting particles and the activation energy of a
reaction is shown in an energy profile diagram.
(a) * Exothermic reaction
Energy
SP 7.4.3
Ea
Reactants
UNIT
Products
Reaction path
7
(b) * Endothermic reaction
Energy
Ea
Products
Reactants
Reaction path
Ea – The minimum energy the reactant particles must possess before the
collision between them can result in a chemical reaction.
*Exothermic and endothermic reactions will studied in form 5,
Thermochemistry.
© Nilam Publication Sdn. Bhd.
180
MODULE • Chemistry FORM 4
How to relate the frequency
of effective collision with the
rate of reaction?
SP 7.4.2
How to change the frequency
of effective collision?
The effective collisions will result in chemical reaction. When frequency of
effective collision increases, the rate of reaction will also increase.
Remark:
Frequency of collision is the number of collisions in one second. When the frequency of collision between
particles of reactants increases, the frequency of effective collisions between particles will also increase.
Frequency of effective collision can be changed by changing the following:
1 The frequency of collision between particles (number of collisions per
unit time).
Or
2 The activation energy of the chemical reaction.
State factors that can change
the frequency of effective
collision.
SP 7.4.2
1
2
3
4
Size of solid reactant.
Concentration of reacting solution.
Temperature of reaction mixture.
The presence of catalyst.
Remark:
Factors
Affect
Frequency
of collision
or activation
energy
Affect
Frequency
of effective
collision
Affect
Rate of
reaction
Relationship Between Frequency of Effective Collision with Factors Affecting Rate of Reaction
How does the change in size
of solid reactant affect rate
of reaction?
The smaller the size of solid reactant, the larger total surface area exposed
to collision that allows more collision between reacting particles.
Explain how this factor
affects the rate of reaction.
– A chemical reaction takes place on the surface of a solid reactant as it is
where collision between reacting particles take place.
– When the size of solid reactant is larger, the area exposed to collisions is
smaller because collisions are limited on the surface of solid. Thus, the total
surface area at which reaction occur is smaller.
– When the solid reactant is broken into smaller size, the area exposed to
collision is larger when surface area in the inner part of the solid becomes
available for collisions. Hence, the total surface area at which reaction occur
becomes larger.
Particle of
substance A
Particle of
substance B
When the size
substance B decreases
When the size of solid
substance B is large, only outer
layer of reacting particles B can
react with reacting particles A
When size of solid substance B is
broken into smaller size, more
reacting particles B can react
with reacting particles A
– This will increase the frequency of collision between particles, thus,
frequency of effective collisions between particles increases which leads
to an increase of rate of reaction.
181
7
Size of Solid Reactant
UNIT
1
MODULE • Chemistry FORM 4
2
Concentration
How does the
concentration of solution
affect rate of reaction?
– Concentration is the number of solutes dissolved in a given volume of solution.
– A higher concentration means that the number of particles of solute per unit
volume is also higher, and vice versa.
Explain how this factor
affects the rate of
reaction.
– When the concentration of solution is higher, the number of particles per unit
volume increases.
– This will increase the frequency of collision between particles, thus, frequency
of effective collisions between particles increases.
Particle of
substance A
Particle of
substance B
When the concentration
increases
At a lower concentration, the number
of particles of A and B in a unit
volume is lower, thus the frequency
of collision between particles is low.
At a higher concentration, the number
of particles of A and B in a unit volume
is higher, thus the frequency of collision
between particles is high.
– This leads to an increase of rate of reaction.
Explain why the
monoprotic acid and
diprotic acid have
different rate of reaction
when the concentration
of the acids is the same.
– For diprotic acid, 1 mol dm–3 of the acid ionises to 2 mol dm–3 of hidrogen ions,
H+
Example:
H2SO4
1 mol dm–3
2H+ + 2 mol dm–3
SO42–
– For monoprotic acid, 1 mol dm–3 of the acid ionises to 1 mol dm–3 of hidrogen
ions, H+
Example:
HCl
1 mol dm–3
H+ + 1 mol dm–3
Cl–
UNIT
– Hence, when the same concentration of sulphuric acid and hydrochloric acid are
used, the concentration of hydrogen ions in sulphuric acid is double.
– The rate of reaction using sulphuric acid is higher than hydrochloric acid.
7
3
Temperature
How does the
temperature of solution
affect rate of reaction?
The higher the temperature of a substance, the higher the kinetic energy of the
reacting particles and vice versa.
Explain how this factor
affects the rate of
reaction.
– When the temperature increases, the kinetic energy of reacting particles increases
and the particles move faster.
– This will increase the frequency of collision between particles, thus, frequency
of effective collisions between particles increases.
– This leads to an increase of rate of reaction.
© Nilam Publication Sdn. Bhd.
182
MODULE • Chemistry FORM 4
Catalyst
Define catalyst.
It is a substance that change the rate of reaction without itself being chemically
changed in the reaction.
How does catalyst
affect the rate of
reaction?
Catalyst provides alternative pathway of a reaction at lower activation energy for the
reacting particles to react.
Explain how this
factor affects the rate
of reaction.
1 In a reaction, if the reacting particles collide with energy lower than the activation
energy, there will be no reaction.
2 When catalyst is added to the same reaction, its provides alternative pathway with
a lower activation energy for the reacting particles to react.
3 With this lower activation energy, more reacting particles will have enough energy
to react when they collide.
4 This will increase the frequency of effective collisions between particles which
leads to an increase of rate of reaction.
Energy
Uncatalysed reaction pathway
Catalysed
reaction
pathway
Ea
Reactants
Ea′
Products
Reaction path
Remark:
Catalyst provide lower activation energy for the reacting particles to collide effectively.
State the
characteristics of
catalyst.
1 A catalyst does not change the amount of products of a reaction.
2 Catalyst is specific to a particular reaction, different chemical reactions need
different catalyst.
3 Catalyst remains unchanged chemically during a reaction.
4 Catalyst may undergo physical change during a reaction.
5 Only a small amount of catalyst needed to achieve big increase in rate of reaction.
6 More amount of catalyst used can further increase the rate of reaction.
7 Catalyst in powder form can further increase the rate of reaction.
183
7
– Ea : The minimum energy the reactant particle must possess before collision between
them can result in a chemical reaction.
– Ea′ : The lower activation energy in the presence of a catalyst. Catalyst is a substance
that increases the rate of a chemical reaction without itself undergoing any
chemical change.
UNIT
4
MODULE • Chemistry FORM 4
Conclusion SP 7.2.1 SP 7.4.1
Steps to explain factors that affect the rate of reaction based on collision theory:
Size of Reactant
The smaller the size of
reactant,
the total surface area
exposed to collision is
larger.
Concentration of
Reactant
Temperature of
Reaction Mixture
The higher the
concentration of
reactants,
the number of particles
The higher the
temperature,
the kinetic energy of
in a unit
higher.
(Ea′).
The reacting particles
move faster.
More colliding particles
activation
achieve
reacting particles is
volume is higher.
The frequency of
The frequency of
collision between particles
effective
UNIT
The rate of reaction
Catalyst provides an
alternative path of
reaction which needs
lower activation energy
energy
increases.
collision between* particles increases.
increases
7
*State the particles that collide based on ionic equation.
© Nilam Publication Sdn. Bhd.
Catalyst
184
.
.
Compare factor that affects the rate of reaction in both experiments.
Example:
(a) Factor size of solid reactant (for reaction between different size of calcium
carbonate and hydrochloric acid)
The size of calcium carbonate chips in experiment I is smaller than experiment II
(b) Factor concentration (for the reaction between different concentration of
hydrochloric acid with zinc)
The concentration of hydrochloric acid in experiment I is higher than experiment
II
(c) Factor temperature (for the reaction between different temperature of
hydrochloric acid and zinc)
The temperature of reaction mixture in experiment I is higher than experiment II
(d) Factor catalyst (for the reaction between zinc and sulphuric acid)
Copper(II) sulphate present as a catalyst for the reaction between zinc and
sulphuric acid in experiment I but not in experiment II
185
7
Compare rate of reaction in both experiments
Rate of reaction in experiment I is higher than experiment II.
5
UNIT
Compare frequency of effective collision between* particles in both experiments.
(a) Factor size solid reactant
The frequency of effective collision between calcium carbonate and hydrogen
ion in experiment I is higher than experiment II
(b) Factor concentration
The frequency of effective collision between hydrogen ions and zinc atoms
in experiment I is higher than experiment II
(c) Factor temperature
The frequency of effective collision between hydrogen ions and zinc atoms in
experiment I is higher than experiment II
(d) Factor catalyst
The frequency of effective collision between hydrogen ions and zinc atoms in
experiment I is higher than experiment II
4
• Analyse the condition for both experiments given in the form of description/balanced
equation/diagram to identify factor that affects the rate of reaction.
1
Compare frequency of collision between number of *particles in
both experiments.
(a) Factor size solid reactant
The frequency of collision between calcium carbonate and
hydrogen ion in experiment I is higher than experiment II
(b) Factor concentration
The frequency of collision between hydrogen ions and zinc
atoms in experiment I is higher than experiment II
(c) Factor temperature
The frequency of collision between hydrogen ions and zinc atoms
in experiment I is higher than experiment II
(d) **Factor catalyst
*State the particles that collide in the reaction based on the ionic equation
**Catalyst does not increase the frequency of collision between particles
3
Compare how the factor affects the rate of reaction in both
experiments.
(a) Factor size solid reactant
The total surface area of calcium carbonate chips in experiment I is
larger than experiment II
(b) Factor concentration
The number of hydrogen ions per unit volume in experiment I is
higher than experiment II
(c) Factor temperature
The kinetic energy of hydrogen ion in experiment I is higher than
experiment II
(d) Factor catalyst
• Copper(II) sulphate lower the activation energy for the reaction
between zinc and sulphuric acid in experiment I
• More colliding particles achieve the activation energy
2
Steps to Compare And Explain Factor that Affects Rate of Reaction Between Any Two Experiments
MODULE • Chemistry FORM 4
MODULE • Chemistry FORM 4
Exercise
1
TP3
Complete the following table.
Chemical equation
Ionic equation
Mg + 2HCl → MgCl2 + H2
Mg + 2H+ → Mg2+ + H2
Magnesium atom
and hydrogen ion
(ii) Mg + 2HNO3 → Mg(NO3)2 + H2
Mg + 2H+ → Mg2+ + H2
Magnesium atom
and hydrogen ion
(iii) Zn + H2SO4 → ZnSO4 + H2
Zn + 2H+ → Zn2+ + H2
Zinc atom and
hydrogen ion
(iv) Zn + 2CH3COOH → (CH3COO)2Zn + H2
Zn + 2H+ → Zn2+ + H2
Zinc atom and
hydrogen ion
(v) CaCO3 + 2HCl → CaCl2 + H2O + CO2
CaCO3 + 2H+ → H2O + CO2 +
Ca2+
Calcium carbonate
and hydrogen ion
(i)
(vi) 2H2O2 → 2H2O + O2
–
(vii) Na2S2O3 + H2SO4 → Na2SO4 + H2O +
SO2 + S
2
TP4
*Particles that collide
in the reaction
S2O32– + 2H+ → H2O + SO2 + S
Hydrogen peroxide
molecules
Thiosulphate ion
and hydrogen ion
Compare rate of reaction in Experiment I and Experiment II. Explain based on collision theory.
Experiment I
Experiment II
Reactant
20 cm3 of 0.5 mol dm–3 hydrochloric 20 cm3 of 0.5 mol dm–3
acid + excess of calcium carbonate
hydrochloric acid + excess of
powder at 30°C
calcium carbonate chips at 30°C
UNIT
7
Balanced equation
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Ionic equation
CaCO3 + 2H+ → H2O + CO2 + Ca2+
Calculate volume of carbon
dioxide gas released in both
experiments at room
conditions. 1 mol of gas
occupies 24 dm3 at room
conditions
0.5
Number of moles of HCl = 20 × 1 000 = 0.01 mol
From the equation
2 mol HCl : 1 mol CO2
0.01 mol HCl : 0.005 mol CO2
Volume of CO2 = 0.005 mol × 24 dm3 mol–1
= 0.12 dm3
= 120 cm3
Factor that affects rate of
reaction
Size of calcium carbonate in experiment I is smaller than experiment II.
Compare how the factor
affects rate of reaction
The total surface area of calcium carbonate exposed to collision in
experiment I is higher than experiment II.
Compare the frequency of
collision between
* particles
The frequency of collisions between calcium carbonate and hydrogen ions
in experiment I is higher than experiment II.
© Nilam Publication Sdn. Bhd.
186
MODULE • Chemistry FORM 4
Experiment I
Experiment II
Compare the frequency of
effective collisions between
*particles
The frequency of effective collisions between calcium carbonate and
hydrogen ions in experiment I is higher than experiment II.
Compare rate of reaction
Rate of reaction in experiment I is higher than experiment II.
Sketch of the graph
Volume of carbon dioxide gas / cm3
Experiment I
Experiment II
Time / s
* particles – S
tate the type of particles (atom/ion/molecule) that collide based on the ionic equation for the
reaction.
Complete the following table to compare and explain the rate of reaction in Experiment I and Experiment II
based on collision theory.
Experiment I
Experiment II
Reactant
20 cm of 0.5 mol dm hydrochloric
acid + excess of magnesium powder at
30°C
20 cm of 0.5 mol dm–3 sulphuric acid
+ excess of magnesium powder at
30°C
Balanced equation
Mg + 2HCl → MgCl2 + H2
Mg + H2SO4 → MgSO4 + H2
Ionic equation
Mg + 2H+ → Mg2+ + H2
Mg + 2H+ → Mg2+ + H2
Calculate volume of gas
released in each
experiment at room
condition
Number of moles of HCl
20 × 0.5
=
= 0.01 mol
1 000
Number of moles of H2SO4
20 × 0.5
=
= 0.01 mol
1 000
From the equation
2 mol HCl : 1 mol H2
0.01 mol HCl : 0.005 mol H2
Volume of H2
= 0.005 mol × 24 dm3 mol–1
= 0.12 dm3
= 120 cm3
From the equation
1 mol H2SO4 : 1 mol H2
0.01 mol H2SO4 : 0.001 mol H2
Volume of H2
= 0.01 mol × 24 dm3 mol–1
= 0.24 dm3
= 240 cm3
–3
Ionisation equation of
HCl → H+ + Cl–
acid and concentration of 1 mol of hydrochloric acid ionise to
H+ ion
1 mol H+
1 mol dm–3 hydrochloric acid ionise to
1 mol dm–3 H+
3
H2SO4 → 2H+ + SO42–
1 mol of sulphuric acid ionise to
2 mol H+
1 mol dm–3 sulphuric acid ionise to
2 mol dm–3 H+
Compare the
Concentration of hydrogen ion, H+ in experiment II is double of experiment I.
concentration of reactant
Compare how the factor
affects rate of reaction
The number of hydrogen ion per unit volume in experiment II is double of
experiment I.
187
7
3
UNIT
3
TP4
MODULE • Chemistry FORM 4
Experiment I
Experiment II
Compare the frequency
of collision between
*particles
Frequency of collisions between hydrogen ions and magnesium atoms in
experiment II is higher than experiment I.
Compare the frequency
of effective collisions
between *particles
Frequency of effective collisions between hydrogen ions and magnesium atoms
in experiment II is higher than experiment I.
Compare rate of reaction Rate of reaction in experiment II is higher than experiment I.
Sketch of the graph
Volume of hydrogen gas / cm3
Experiment II
240
Experiment I
120
Time / s
Compare the gradient
and amount of product
for the curve in both
experiments. Explain.
(i) The gradient of the curve for experiment II is
because the rate of reaction in experiment II is
greater
higher
than experiment I
than experiment I.
double
(ii) The volume of hydrogen gas collected in experiment II is
of
experiment I. Sulphuric acid is diprotic acid while hydrochloric acid is
monoprotic acid. One mol of sulphuric acid ionises to
one
H+ ions, one mol of hydrochloric acid ionises to
two
mol of
mol of H+ ions.
(iii) The number of H+ ions in the same volume and concentration of both acids
is double in sulphuric acid.
UNIT
* particles – State the type of particles (atom/ion/molecule) that collide based on the ionic equation for the
reaction.
7
4
TP4
Complete the following table to compare and explain how does copper(II) sulphate solution affect the rate of
reaction.
Experiment I
Excess
of zinc
granules
Experiment II
Water
50 cm3 of 1.0 mol dm–3
hydrochloric acid
Balanced equation
© Nilam Publication Sdn. Bhd.
Zn(s) + 2HCl(aq) → ZnCl2(aq) +
H2(g)
188
Excess
of zinc
granules
Water
50 cm3 of 1.0 mol dm–3 hydrochloric
acid + Copper(II) sulphate solution
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
CuSO4
MODULE • Chemistry FORM 4
Experiment I
Ionic equation
Experiment II
Zn + 2H+ → Zn2+ + H2
Number of moles of HCl =
Calculate volume of hydrogen
gas released in experiments I
and II at room conditions.
1 mol of gas occupies 24 dm3
at room conditions
50 × 1
= 0.05 mol
1 000
From the equation
2 mol HCl : 1 mol H2
0.05 mol HCl : 0.025 mol H2
Volume of H2 = 0.025 mol H2 × 24 dm3 mol–1
= 0.6 dm3
= 600 cm3
Compare the factor that
affects rate of reaction
Reaction between zinc and hydrochloric acid in experiment I is without
catalyst while copper(II) sulphate added in experiment II acts as a catalyst.
Compare how the factor
affects rate of reaction
Copper(II) sulphate lowers the activation energy for the reaction between
zinc and hydrochloric acid in experiment II.
Compare the frequency of Frequency of effective collisions between zinc atom and hydrogen ion in
effective collisions between experiment II is higher than experiment I.
*particles
Compare rate of reaction
Rate of reaction in experiment II is higher than experiment I.
Sketch of the graph for
experiment I and experiment
II
Volume of hydrogen gas / cm3
Experiment II
Compare the gradient and
amount of product for the
curve in both experiments.
Explain.
1
The gradient of the curve for experiment II is greater than experiment
I because the rate of reaction in reaction experiment II is higher than
experiment I.
2
The volume of hydrogen gas released in experiment I is equal to
experiment II. The volume and concentration of hydrochloric acid in
experiments I and II are similar. Copper(II) sulphate in experiment II
that is used as a catalyst does not affect the total volume of hydrogen
gas produced.
189
UNIT
Time / s
7
Experiment I
MODULE • Chemistry FORM 4
SPM PRACTICE
Subjective Questions
1
The table below shows the data for three experiments that have been carried out to determine the effect of
catalyst on the decomposition of hydrogen peroxide to water and oxygen.
Volume of H2O2 / cm3
Concentration of H2O2 / mol dm–3
Mass of MnO2 / g
I
20
1.0
–
II
20
1.0
1.0
III
20
1.0
2.0
(a) Write the chemical equation for the decomposition of hydrogen peroxide.
TP3
2H2O2 → 2H2O + O2
(b) Draw the set-up of apparatus for the experiment.
TP3
Manganese(IV) oxide
Water
H2O2
(c) Between experiments I, II and III, which has the highest rate of reaction? Explain your answer.
TP4
Experiment III. The amount of catalyst used in experiment III is more than experiment II.
UNIT
(d) Explain how does manganese(IV) oxide affect the rate of decomposition of hydrogen peroxide by using
TP4 collision theory.
• Manganese(IV) oxide provides an alternative path with lower activation energy for the decomposition
7
of hydrogen peroxide
• The frequency of effective collisions between hydrogen peroxide molecules increases.
• The rate of decomposition of hydrogen peroxide increases.
(e) Calculate the volume of oxygen gas released in experiment II at room temperature and pressure.
TP3
Number of moles of H2O2 =
20 × 1
= 0.02 mol
1 000
From the equation,
2 mol H2O2 : 1 mol O2
0.02 mol H2O2 : 0.01 mol O2
Volume of O2 = 0.01 mol × 24 dm3 mol–1
= 0.24 dm3
= 240 cm3
© Nilam Publication Sdn. Bhd.
190
MODULE • Chemistry FORM 4
(f) Sketch the graph of volume of oxygen gas against time for experiments I, II and III.
TP4
Volume of oxygen gas / cm3
III
II
I
Time / s
(g) State one factor other than concentration and catalyst that can affect the rate of decomposition of hydrogen
TP2 peroxide.
Temperature of hydrogen peroxide.
A student carried out three sets of experiment to investigate the factors affecting the rate of reaction. The table
below shows the information and the result of the experiment.
Time taken to collect
Set
Reactants
maximum volume of gas / s
I
3 g magnesium ribbon and 50 cm3 of 1 mol dm–3 hydrochloric acid
100
3
–3
II
3 g magnesium powder and 50 cm of 1 mol dm hydrochloric acid
60
3
–3
3 g magnesium powder and 50 cm of 1 mol dm hydrochloric acid
III
30
and copper(II) sulphate solution
(a) Write a chemical equation to show the reaction between magnesium and hydrochloric acid.
TP3
Mg + 2HCl → MgCl2 + H2
7
(b) Calculate the number of mole of
TP3 (i) Magnesium
[Relative atomic mass of Mg = 24]
Number of moles Mg = 3 = 0.125 mol
24
UNIT
2
(ii) Hydrochloric acid
Number of moles HCl = 1 × 50 = 0.05 mol
1 000
(c) Calculate the maximum volume of hydrogen gas produced at room condition.
TP3
[1 mole of gas occupies the volume of 24 dm3 at room condition]
From the equation:
2 mol HCl : 1 mol H2
0.05 mol HCl : 0.025 mol H2
Volume of hydrogen gas = 0.025 × 24 dm3 = 0.6 dm3 = 600 cm3
(d) What is the average rate of reaction for
TP3
(i) Set I?
Rate of reaction = 0.6 = 0.006 dm3 s–1 // 600 = 6 cm3 s–1
100
100
191
MODULE • Chemistry FORM 4
(ii) Set II?
Rate of reaction = 0.6 = 0.01 dm3 s–1 // 600 = 10 cm3 s–1
60
60
(iii) Set III?
Rate of reaction = 0.6 = 0.02 dm3 s–1 // 600 = 20 cm3 s–1
30
(e) (i)
TP4
30
The diagram below shows the curve obtained for set I when the volume of hydrogen gas liberated
against time is plotted. On the same axes, sketch the curve that you would expect to obtain if the
experiment is repeated using 3 g magnesium ribbon and 50 cm3 of 2 mol dm–3 hydrochloric acid.
Volume of hydrogen gas / dm3
1.2
0.6
Time / s
(ii) Explain how you obtain the curve in (e)(i).
1 Initial rate of reaction of the experiment is higher because the concentration of hydrochloric acid
is higher, therefore the curve is steeper.
2 Hydrochloric is the limiting factor // Magnesium is in excess.
3 Maximum volume of hydrogen gas collected is double because the number of mole of
hydrochloric acid is double.
UNIT
(f) (i)
Compare the rate of reaction in Set II and Set III.
TP4
Rate of reaction in Set III is higher than Set II.
7
(ii) By using the collision theory, explain the difference in the rate of reaction between Set II and Set III.
– Reaction between magnesium and hydrochloric acid in Set II is without catalyst while copper(II)
sulphate presence in Set III as a catalyst.
– Copper(II) sulphate lower the activation energy for the reaction between magnesium and
hydrochloric acid in Set III.
– Frequency of effective collisions between magnesium atom and hydrogen ion in Set III is higher
than Set II.
© Nilam Publication Sdn. Bhd.
192
MODULE • Chemistry FORM 4
(iii) Sketch energy profile diagram which shows an activation energy for the reactions in Set II and Set III.
Energy
– Ea : The activation energy in Set II.
– Ea′: The activation energy in Set III.
Ea
Ea′
Mg + 2HCl
MgCl2 + H2
Reaction path
The diagram below shows a sketch of curve I and curve II for volume of carbon dioxide gas collected against
time. Curve I is for the reaction in Experiment I between 2.0 g marble chips added to 50 cm3 of 1 mol dm–3
hydrochloric acid at room temperature.
Volume of carbon dioxide gas / cm3
II
I
Time / s
(a) Write a chemical equation to show the reaction between marble chips and hydrochloric acid.
TP3
CaCO3 + 2HCl → CaCl2 + H2O + CO2
(b) State two ways to obtain curve II for the reaction of 50 cm3 of 1 mol dm–3 hydrochloric acid at room
TP3
temperature.
(ii) The experiment is carried out a 40 °C (higher than room temperature)
(c) Explain each of your answer in (b).
TP4
7
(i) 2.0 g marble powder is used to replace 2.0 g marble chips
UNIT
3
(i) 2.0 g marble powder in Experiment II has a larger total surface area than 2.0 g marble chips in
Experiment I, hence initial rate of reaction of Experiment II is higher than Experiment I. Maximum
volume of carbon dioxide gas collected in Experiment I and II are the same because the quantity of
marble and hydrochloric acid are the same in both experiments.
(ii) Experiment II is conducted at higher temperature compared to Experiment I, hence intial rate of
reaction Experiment II is higher than Experiment I. Maximum volume of carbon dioxide gas collected
in Experiment I and II are the same because the quantity of marble and hydrochloric acid are the same
in both experiments.
193
MODULE • Chemistry FORM 4
(d) Which of the two reactants in Experiment I is in excess? Explain your answer.
TP5
[Relative atomic mass: C = 12, O = 16, Ca = 40]
2g
Number of moles CaCO3 =
= 0.02 mol
100 g mol–1
Number of moles HCl = 50 × 1 = 0.05 mol
1 000
From the equation:
1 mol of CaCO3 react with 2 mol of HCl
0.02 mol of CaCO3 react with 0.04 mol of HCl
→ Hydrochloric acid is in excess
4
The diagram below shows the set up of apparatus to study the effect of concentration on the rate of reaction.
Hydrochloric acid
Magnesium ribbon
In each experiment, x g magnesium ribbon is allowed to react with excess y mol dm–3 hydrochloric acid. The table
shows the mixing of different volumes of hydrochloric acid y mol dm–3 and water in experiment A, B and C.
Experiment
A
B
C
Volume of hydrochloric acid y mol dm–3 / cm3
50
400
200
Volume of water / cm3
350
0
200
UNIT
The result of these experiments are plotted in the same graph shown below.
Volume of hydrogen gas / cm
7
Curve I
Curve II
Curve III
100
50
0
20
50
80
Time / s
(a) What is meant by concentration?
TP1
It is the amount of solute in a given volume of solution
(b) Which of the experiments are represented by:
TP5
B
(i) Curve I:
(ii) Curve II:
© Nilam Publication Sdn. Bhd.
194
C
(iii) Curve III:
A
MODULE • Chemistry FORM 4
(c) Why is the volume of hydrogen gas the same in all experiments?
TP2
Magnesium is the limiting factor. The volume of hydrogen gas released is limited by the number of mol
of magnesium present.
(d) By using collision theory, compare rate of reaction in experiment A and experiment B. Explain your
TP4 answer.
• Rate of reaction of experiment B is higher than experiment A
• Concentration of hydrochloric acid in experiment B is higher than experiment A
• Number of hydrogen ions per unit volume in experiment B is higher than A
• Frequency of collision between hydrogen ion and magnesium atom in experiment B is higher than A
• Frequency of effective collision between hydrogen ion and magnesium atom in experiment B is higher
than A
(e) (i)
TP3
Write a balanced chemical equation for the reaction between magnesium and hydrchloric acid.
Mg + 2HCl → MgCl2 + H2
(ii) Calculate the mass of magnesium that reacts in all three experiments.
[Relative atomic mass : Mg = 24, 1 mol of gas occupy 24 dm3 at room conditions]
Volume of hydrogen = 100 cm3 = 0.1 dm3
Mol of hydrogen gas =
0.1 dm3
= 0.00417 mol
24 dm3 mol–1
UNIT
7
From the equation:
1 mol of H2 : 1 mol of Mg
0.00417 mol H2 : 0.00417 mol of Mg
Mass of Mg = 0.00417 mol × 24 g mol–1 = 0.1 g
Objective
Questions
195
MODULE • Chemistry FORM 4
UNIT
MANUFACTURED
SUBSTANCES IN INDUSTRY
8
Concept Map
MANUFACTURED
SUBSTANCE IN INDUSTRY
Alloy
Meaning
Alloy is a mixture
of two or more
elements with a
certain fixed/
specific
composition. The
major component
in the mixture is a
metal.
Purpose of
alloying
– To increase
hardness and
strength
– To resist
corrosion
– To improve
appearance
UNIT
Composition,
properties and
uses of alloy
8
Alloy:
• Bronze
• Brass
• Steel
• Stainless steel
• Duralumin
• Pewter
Glass
Ceramic
Major component
From sand:
Silicon and
Oxygen
An inorganic
non-metallic solid.
It is made up of
either metal or
semi metal
compounds.
General
Properties
General
Properties
– Hard but brittle
– Chemically
inert
– Transparent
– Good electric
and heat
insulator
– Noncompressible
– Hard and strong
– Chemically
inert
– High melting
and boiling
points
– Good electric
and heat
insulator
– Breakable
Compositions,
properties and
uses of different
types of glass
Type of glass
• Fused glass
• Borosilicate
glass
• Soda-lime glass
• Lead glass
Meaning
Type of ceramic
Traditional
ceramic
Advanced
ceramic
Component
From clay:
Aluminosilicate
(Al2O3.2SiO2.
2H2O)
Component
Non-organic
substance such
as oxide,
carbide and
nitride
Example, properties
and uses
© Nilam Publication Sdn. Bhd.
196
Composite
Material
Meaning
Combination of
two or more nonhomogeneous
materials.
Component
Matrix materials
and reinforce
materials
Properties
Composite
materials have
properties that are
superior to those
of the original
components
Example,
properties
and uses
• Superconductor
• Reinforced
concrete
• Fiberglass
• Photochromic
glass
MODULE • Chemistry FORM 4
8.1
ALLOY AND ITS IMPORTANCE
What is the meaning of alloy?
SP 8.1.1
mixture of two or more elements with a certain
Alloy is a
fixed/specific composition. The major component in the mixture is a
metal
Relate the arrangement of
atoms in pure metals to their
ductile and malleable
properties.
SP 8.1.2
SK
8.1
.
Force
Pure metals
Pure metal is made up of one type of
Atoms in pure metals are all the same
The same
size
atoms
.
size
.
atoms are orderly arranged in
layers
.
force
When
is applied to the pure metal, layers of atoms
slide
easily over one another.
Draw the arrangement of
atoms in
(a) Bronze (90% copper and
10% tin)
(b) Steel (99% iron and 1% of
carbon)
[Relative atomic mass:
Cu = 64, Sn = 119, Fe = 56,
C = 12]
(a) Bronze
Explain why an alloy is
stronger than its pure metal in
terms of the arrangement of
atoms in metals and alloys.
Atoms of other element added to the pure metal to make an alloy consists
of atoms different in size.
Carbon
Tin
Iron
Copper
disrupt
These atoms
the orderly arrangement of atoms in pure metal.
force
When
is applied to an alloy, the presence of foreigns atoms
layers of atoms from
sliding
.
State three reasons why pure
(a) To increase the strength and hardness of pure metals.
metals are alloyed before used. (b) To increase the resistance to corrosion of pure metals.
SP 8.1.3
(c) To improve the appearance of a pure metal.
197
8
reduce
UNIT
SP 8.1.2
(b) Steel
MODULE • Chemistry FORM 4
Flow chart shows the composition, properties and uses of some alloys.
ALLOY
Major component
COPPER
IRON
Type of alloy
Type of alloy
BRONZE
(90% Cu, 10% Sn)
– Hard and strong,
does not corrode,
(shiny surface)
– Uses:
Building statue or
monuments,
medals, swords
and artistic
materials.
BRASS
(70% Cu, 30% Zn)
– Hard and
strong.
– Uses:
Musical
instrument and
kitchenware.
STEEL
(98% Fe, 2% C)
– Hard and
strong.
– Uses:
Construction
of building
and bridge and
railway tracks.
ALUMINIUM
CUPRONICKEL
(75% Cu, 25% Ni)
– Shiny, hard and does
not corrode.
– Uses:
Making coins
TIN
Type of alloy
DURALUMIN
(93% Al, 3% Cu, 3% Mg,
1% Mn)
– Light and strong.
– Uses:
Building body of aeroplane
and bullet train.
Experiment to compare the hardness of alloy and pure metal.
(a) Aim: To compare the hardness of brass and copper
SP 8.1.2
UNIT
8
(b) Hypothesis: Brass is harder than copper
(c) Manipulated variable: Copper and brass blocks
(d) Responding variable: Hardness of the copper and brass blocks
(e) Fixed variable: 1 kg weight
(f) Apparatus: Retort stand and clamp, 1 kg weight, string, metre ruler
Materials: Steel ball, copper block, brass block
© Nilam Publication Sdn. Bhd.
198
STAINLESS STEEL
(73% Fe, 18% Cr,
8% Ni, 1% C)
– Shiny, strong and
does not rust.
– Uses:
Making cutlery and
surgical instrument.
Type of alloy
PEWTER
(95% Sn, 3.5% Sb,
1.5% Cu)
– Luster, shiny and
strong.
– Uses:
Making souvenirs.
MODULE • Chemistry FORM 4
(g) Procedure:
(i) A steel ball bearing is tapped onto a copper block.
(ii) A 1 kg weight is hung at a height of 50 cm above the
copper block as shown in the diagram.
(iii) Drop the 1 kg weight onto the steel ball.
(iv) Measure the diameter of the dent formed on the copper
block with a ruler.
(v) Repeat the experiment three times on other parts of the
same copper block.
(vi) Steps (i) to (v) are repeated using a brass block to replace
the copper block.
Set-up of the apparatus:
Retort stand
String
1 kg weight
Steel ball
Cellophane
tape
Copper block
(h) Results:
Experiment
1
2
3
Average diameter / cm
Diameter of dent on copper block / cm
a
b
c
a+b+c
=x
3
Diameter of dent on brass block / cm
d
e
f
d+e+f
=y
3
(i) Discussion:
The average diameter of dent on copper, x is larger than the average diameter of dent on brass, y.
(j) Conclusion:
Brass is harder than copper // alloy is harder than pure metal.
Experiment to compare the resistant to corrosion between alloy and its pure metal.
SP 8.1.2
(a) Aim: To compare the resistant to corrosion between iron and stainless steel
(b) Problem statement: Is steel more resistant to corrosion than iron?
(c) Hypothesis: Stainless steel is more resistant to corrosion than iron
UNIT
8
(d) Manipulated variable: Iron nail and stainless steel nail
(e) Responding variable: Rusting of the nail
(f) Fixed variable: Volume and concentration of sodium chloride solution
(g) Apparatus: Test tube, test tube rack, measuring cylinder 10 cm3
Aim: Stainless steel nail, iron nail, sodium chloride solution 0.5 mol dm–3, sand paper
Sodium chloride
solution
Stainless steel nail
Test tube A
Iron nail
Test tube B
199
MODULE • Chemistry FORM 4
(h) Procedure:
1 A stainless steel nail and an iron nail are cleaned with a sand paper to remove any rust on the surface.
2 The test tubes, labelled A and B are filled with 5 cm3 sodium chloride solution 0.5 mol dm3.
3 A stainless steel nail is put in test tube A and an iron nail is put in test tube B.
4 The test tubes are left aside for four days.
5 Observations are recorded after four days.
(i) Observation:
Types of nail
Observation
Stainless steel nail
No change
Iron nail
Brown solid is formed around the nail
(j) Conclusion:
Iron rusts while stainless steel does not rusts/Stainless steel is more resistant to corrosion than iron
8.2
SK
8.2
COMPOSITION OF GLASS AND ITS USES
Name the element which forms the
major component of glass.
Silicon
List the property of glass.
dioxide, SiO2 which exist naturally in sand.
Transparent, hard but brittle, non-porous, heat insulator, electric
insulator, resistant to chemical, easy to clean, and can withstand
compression.
SP 8.2.1
How to make different type of glass?
Different types of glass with different properties are formed by
heating the silica together with other chemicals.
What are the type of glass, composition, properties and their uses?
Types of glass
Soda lime
UNIT
Borosilicate
8
Fused glass
Composition
SP 8.2.1
Special Properties
Uses
Silicon dioxide,
Sodium carbonate,
Calcium carbonate
– Good chemical durability
High
–
termal expansion but cannot
Silicon dioxide,
Boron oxide,
Sodium oxide,
Aluminum oxide
– Good chemical durability
Low
–
thermal expansion
Silicon dioxide
– Optically transparent
– Good chemical durability
withstand
heat
heat
– Resistant to
high temperature
–
.
when heated to
Low
thermal expansion
temperature
– Can be heated to high
resistance to thermal shock
Lead glass
Silicon dioxide,
Sodium oxide,
Lead(II) oxide
© Nilam Publication Sdn. Bhd.
– High refractive index and
– Glittering appearance
200
and
density
Making flat glass,
electrical bulbs, mirrors
and glass containers
Making cookware and
laboratory glassware
such as boiling tubes
and beakers.
Laboratory glassware,
lenses, telescope
mirrors and optical
fibres.
Tableware, crystal
glassware and
decorative glassware.
MODULE • Chemistry FORM 4
8.3
COMPOSITION OF CERAMICS AND ITS USES
What is ceramic?
SP 8.3.1
SK
8.3
A ceramic is an inorganic non-metallic solid. It is made up of either metal or
non-metal compounds that have been shaped and then hardened by heating to
high temperatures.
Give examples of
compounds that formed
ceramics. SP 8.3.1
– Metal compounds such as aluminium oxide, Al2O3 and magnesium oxide,
MgO
– Semi metal compounds such as boron nitride, BN, and silicon carbide, SiC
How atoms of elements
bonded in a ceramic?
Explain how the bonds
affect its properties.
– Atoms of the elements in a ceramic are bonded by strong ionic and covalent
bonds
– Since both ionic and covalent bonds are stronger than metallic bond, ceramic
materials are stronger and harder than metals.
– The strong ionic and covalent bonds in ceramic also determines high melting
point and chemical stability of ceramic materials.
– Unlike metals, there are no free moving electrons in ceramics. Therefore,
ceramics cannot conduct electricity and heat.
– Due to the combined ionic and covalent bonding in ceramic, the particles
cannot slide easily over each other. The ceramic breaks when too much force
is applied.
List the general
properties of ceramic.
(a)
(b)
(c)
(d)
(e)
(f)
SP 8.3.1
Hard and strong
Chemically inert and non-corrosive
Good insulator of heat
High heat resistant and remain stable under high temperature
Good insulator of electric
Breakable
What are the
Ceramics are classified as:
classification of ceramics? (i) Traditional ceramics
(ii) Advance ceramic
201
UNIT
What is advanced
– Advanced ceramics are being developed in a variety forms. They include
ceramics? Give examples.
oxide ceramic and non-oxide ceramics such as carbide and nitride.
SP 8.3.2
(i) Examples of oxide ceramics are alumina (aluminium oxide, Al2O3) and
zirconia (zirconium oxide, ZrO2).
(ii) Examples of non-oxide ceramics are silicone carbide (SiC) and silicone
nitrade (SiN).
– Advanced ceramics have the properties of high heat resistance and abrasion,
very inert chemically and super-conductivity.
– Example of uses and properties of advanced ceramics:
(a) Alumina and zirconia are chemically inert materials, high fracture
toughness, high density, high hardness and wear resistance. Zirconia is
used as dental implants and alumina is used as bone substitutes in
orthopedic operations such as hip and knee replacement
8
What is traditional
– Traditional ceramics are made by heating clay such as kaolin to a very high
ceramics? Give examples.
temperature. The main composition of kaolin is aluminosilicate
(Al2O3.2SiO2.2H2O)
– When water and clay are mixed, the clay is soft and can be molded to any
desired shape and size. The molded clay is then heated to a high temperature.
– Example of the uses of traditional ceramics are to make bricks, plates, pottery
and cement.
MODULE • Chemistry FORM 4
(b) Silicone carbide:
(i) Silicone carbide is very hard and strong, it is used to make a disc cutter.
(ii) Silicone carbide also has high resistance to heat and resistant to
thermal shock. It is used to make disc brake.
(c) Tungsten carbide (WC) is very hard and resistant to abrasion, it is used to
make tungsten carbide rings. Unlike rings made of traditional metals
such as gold, silver and platinum, tungsten carbide is extremely hard. The
only thing that can scratch a tungsten carbide ring is a diamond.
Remark:
A disc cutter is a power tool used for cutting hard materials, ceramic tile, metal, concrete, and
stone for example.
Complete the table for
the properties and uses
of ceramics.
SP 8.3.2
Ceramic
Kaolin
Kaolin
Oxide
ceramic
Alumina
8.4
Classification
of ceramic
Traditional
ceramics
Traditional
ceramics
Advance
ceramics
Properties
Hard and strong
Chemically inert
and non-corrosive
Have high melting
point and good
insulator of heat,
remain stable
under high
temperature
Good insulator
electric
Advance
ceramics
COMPOSITE MATERIALS AND ITS IMPORTANCE
What are composite
materials?
SP 8.4.1
Uses
Building materials
such as cement, tiles,
bricks, roof and toilet
bowl.
Kaolin
Insulation such as
lining of furnace, wall
of nuclear reactor and
engine parts.
Electric insulator in
electrical items such
as electric plugs, oven
and electric cables.
SK
8.4
UNIT
8
– Composite materials are formed by combining two or more non-homogeneous
materials.
– Most composites material are made up of just two materials which are matrix
or binder materials and reinforcement materials.
– The matrix or binder surrounds and binds together the reinforcement material (a
cluster of fibers or fragments of a much stronger materials) to form composite
material:
+
Reinforce material
=
Matrix material
Composite material
– Example of natural composite material is wood. Wood is made of long fibers
of cellulose. These fibers are held together by another plant polymer called
lignin:
cellulose fiber
⇒ The reinforce material is
⇒ The matrix material is
© Nilam Publication Sdn. Bhd.
202
lignin
MODULE • Chemistry FORM 4
What are the difference
between composite
materials and their
component? SP 8.4.3
Composite materials have properties that are
original components.
superior
Examples of composite materials, their components, special properties and the uses
Composite materials
Reinforced concrete
(i) Concrete is reinforced
with steel reinforcement
and wire netting.
(ii) Concrete is a matrix
material and steel bars,
wires or netting are
reinforce materials.
Components
Properties of
component
Steel bars or – High
steel wire
tensile
netting
strength
– Easily
corrode
Concrete
– Strong but
brittle
– Weak in
tension
– Resistant to
corrosion
Fiber glass
(i) Produced when glass
fibers are added with
plastics
(ii) Glass fibers are
reinforce materials
and plastic is a
matrix material
Glass fiber
Brittle, strong
and hard
Plastic
Soft, flexible
and low
density
Fiber optic
(i) Core – glass fibers are
reinforce material
(ii) Cladding – glass/plastic
is a matrix material
(iii) Protective jacket –
plastic is a matrix
material
Plastic
– Lower
refractive
index
– Flexible
– Nonconductor
Glass fiber
– High
refractive
index
– Hard
– Nonconductor
Concrete
Protective jacket
SP 8.4.2
Properties of
composite material
Example of uses
– Very strong and
able to withstand
tensile forces
– Does not crack
easily
– Can be molded
into any shape
– Resistant to
corrosion
Construction of
building, roads,
bridges and oil
platforms.
– High tensile
strength
– Hard, light and
strong
– Low density
– Chemically inert
– Easily molded in
thin layers and
still strong
– Heat and electric
insulator
Making water storage
tanks, small boats,
helmets, motor vehicle
bodies, car bumper,
racket strings and
fishing rod.
– Thinner, hence
easily bent and
lighter
– Chemically more
stable than wires
– Data is
transmitted
digitally
– High transmission
capacity
– Less susceptible
to interference
– In
telecommunication,
where telephone
substations are
linked by fiber optic
– Domestic cable
television network
– Video cameras
– Linking computers
within local area
network (LAN)
– As laser beam to
perform surgery
– In endoscopy, an
instrument to
examine internal
part of the body
Cladding
Core
203
8
Steel netting
UNIT
Steel rod
than those of the
MODULE • Chemistry FORM 4
Properties of
component
Properties of
composite material
Composite materials
Components
Photochromic glass
(i) Produced by embedding
crystals of silver
chloride in a glass/
transparent polymer.
(ii) Glass/transparent
polymers as the
matrix and silver
chloride is the
reinforce material.
Glass/
Transparent
polymers
– Transparent –
– Does not
–
sensitive
towards
light
Silver
chloride or
silver
bromide
– Sensitive to
intensity of
light
– Absorb UV
light
Superconductors
A mixture of barium oxide,
copper(II) oxide, yttrium
oxide can be made in ceramic
called perovskite, YBCO
Copper(II)
oxide
High
electrical
resistance
Barium
carbonate
Transparent
Darken when
exposed to bright
light and becomes
clear when
exposed to dim
light
Conduct electricity
with no resistance
when it is cooled at
low temperature
Yttrium
oxide
Example of uses
– Optical lens in
glasses
– Car windshield
– Light intensity
meters
– Used in medical
magnetic-imaging
devices (MRI)
– Generators and
transformers
– Computer parts
– Bullet train
SPM PRACTICE
Subjective Questions
1
The table shows the properties of two different glasses.
Glass X
The components are silica, sodium carbonate,
boron oxide and aluminium oxide.
Glass Y
The components are silica, sodium carbonate
and calcium carbonate.
(a) Name type of glass X and Y.
TP2
UNIT
Glass X: Borosilicate glass
Glass Y: Soda lime glass
8
(b) The diagram shows a glass that is used on oven.
TP4
HOTS
P
Glass X can be used to make part P. What will happen if glass Y is used to make part P? Give one reason.
Glass Y will crack because glass Y cannot withstand high temperature.
© Nilam Publication Sdn. Bhd.
204
MODULE • Chemistry FORM 4
2
Fiber optics can be used to replace copper wire for data transmission.
Copper wire
Fiber optic
(a) Name three structures that can form fiber optic.
TP1
Core, cladding and protective jacket
(b) How does the fiber optic send data and information?
TP2
Data is transmitted digitally at high transmission capacity.
(c) Compare and contrast fiber optic and copper wire in domestic cable television network.
TP4
Both fiber optic and copper wire can transmit data. Fiber optic can transmit larger capacity of data and
not affected with electromagnetic interference.
The table shows the examples and component of four types of manufactured substances in industry.
Type of manufactured
substances
Example
P
Reinforced concrete to build building
Cement, sand, small pebbles and steel
Q
Medal made from bronze
Copper and metal S
Decorative glassware made from R
type of glass
Silicon dioxide, sodium oxide, lead(II)
oxide
Ceramic
Silicon carbide
–
(a) State the name of P, Q, R and S.
TP2
P: Composite materials
R: Lead crystal
Q: Alloy
S: Tin
(b) (i)
TP2
State the use of reinforced concrete.
To make framework of buildings and bridges.
(ii) What is the advantage of using reinforced concrete compared to concrete?
Reinforced concrete can withstand higher pressure/support heavier loads/stronger/higher tensile
strength than concrete.
205
8
Glass
Component
UNIT
3
MODULE • Chemistry FORM 4
(c) (i)
TP3
Draw the arrangement of particles in
Pure copper
Bronze
Copper
Copper
Tin
(ii) Bronze is harder than pure copper. Explain.
TP4
Atoms of pure copper metal are same size. They arranged orderly in layers. Layers of atoms easily
slide over each other when external force is applied on them. The size of tin atoms which are bigger
than copper in bronze disrupt the orderly arrangement of copper atoms. The presence of tin atoms
reduce the layers of metal atoms from sliding when force is apply.
(d) (i)
TP2
Name the category of ceramic for silicon carbide.
Modern ceramic
(ii) State the use of silicon carbide and the properties with respective to its use.
To make cutter disc. It can withstand high heat resistance and can withstand thermal shock.
UNIT
8
Objective
Questions
© Nilam Publication Sdn. Bhd.
206
7
6
5
4
3
2
1
12
223
Francium
Fr
87
133
Cesium
Cs
55
85.5
Rubidium
Rb
37
39
Potassium
K
19
23
Sodium
Na
39
57
56
89
226
Radium
227
257
Rutherfordium
Rf
104
178.5
Hafnium
72
Hf
91
Zirconium
40
Zr
48
Titanium
22
Ti
4
Key:
Actinides
Thorium
232
Th
90
Cerium
140
Ce
58
260
Dubnium
105
Db
181
Tantalum
73
Ta
93
Niobium
41
Nb
51
Vanadium
V
23
5
55
52
43
Tc
44
Ru
56
Iron
26
Fe
8
60
59
92
91
61
Pm
265
Hassium
108
Hs
190
Osmium
76
Os
101
Metal
Proactinium
231
Pa
Uranium
238
U
Neptunium
237
93
Np
Ds
110
195
Platinum
78
Pt
106
Palladium
46
Pd
59
Nickel
28
Ni
10
111
Rg
Gold
197
79
Au
Silver
108
47
Ag
Copper
64
29
Cu
11
62
Sm
266
Plutonium
244
94
Pu
64
Gd
272
Semi-metal
Americium
243
95
Am
Curium
247
96
Cm
Europium Gadolinium
152
157
63
Eu
271
Meitnerium Darmstadtium Roentgenium
109
Mt
192
Iridium
77
Ir
103
Rhodium
45
Rh
59
Cobalt
27
Co
9
Praseodymium Neodymium Promethium Samarium
141
144
147
150
Nd
262
Pr
262
Bohrium
107
Bh
186
Rhenium
75
Re
98
Scaborgium
Sg
106
184
Tungsten
74
W
96
Molybdenum Technetium Ruthenium
42
Mo
Chromium Manganese
Mn
25
7
Name of the element
Proton number
Transition elements
Cr
24
6
Relative atomic mass
Symbol of the element
Lanthanides
Actinium
Ac
139
88
Ra
137
Lanthanum
La
Barium
Ba
89
Yttrium
88
Strontium
Y
45
38
Sr
40
Scandium
21
Sc
3
Calcium
20
Ca
24
Magnesium
Mg
9
11
7
Beryllium
Be
4
2
Lithium
Li
3
1
Hydrogen
H
1
1
32
49
98
Cf
Dysprosium
162.5
66
Dy
Thallium
204
81
Tl
Indium
115
99
Es
Holmium
165
67
Ho
Lead
207
82
Pb
Tin
119
50
Sn
Germanium
73
Non-metal
Berkelium Californium Einsteinium
247
249
254
97
Bk
Terbium
159
65
Tb
Ununbium
285
112
Uub
Mercury
201
80
Hg
Cadmium
112
In
48
Cd
Gallium
70
Ge
31
Ga
Silicone
28
14
Si
Carbon
12
6
C
14
Aluminium
27
13
Al
Zinc
65
30
Zn
12
5
B
13
Boron
11
The Periodic Table of Elements
10
85
101
Md
Thulium
169
69
Tm
Polonium
210
102
No
Ytterbium
173
70
Yb
Astatine
210
At
84
Po
Iodine
127
I
53
Bromine
80
35
Br
Chorine
35.5
17
Cl
Fluorine
19
Lr
103
Lutetium
175
71
Lu
Radon
222
86
Rn
Xenon
131
54
Xe
Krypton
84
36
Kr
Argon
40
18
Ar
Neon
20
Ne
9
F
Helium
4
17
Tellurium
128
52
Te
Selenium
79
34
Se
Sulphur
32
S
16
Oxygen
16
8
O
16
Fermium Mendelevium Nobelium Lawrencium
253
256
254
257
100
Fm
Erbium
167
68
Er
Bismuth
209
83
Bi
Antimony
122
51
Sb
Arsenic
75
33
As
Phosphorus
31
P
15
Nitrogen
14
7
N
15
He
2
18