DK014

2021/2022
EDITION

3 in 1
*NOTES
*TUTORIAL QUESTION
*PRE-LAB

CDKH0EM14ISHTARYND1 BOOK

FARHANA.DR.ELMISHARLINA.KASBAH.ZANARINA.
KHAIRULASWARI.NOORASMAHAN.NOORNAJIHAH.
SHARIFAHFADTHYAH.SUHAINA.WANROSILAH.
CHEMISTRYUNIT.KOLEJMATRIKULASIKELANTAN.

NAME:____________________
TUTORAN:-----------------------

ACKNOWLEDGEMENT

SPECIAL THANKS AND APPRECIATION TO:
FARHANA BINTI UMANAN

KASBAH BIN ANDI DAHRAN@ASMAD
DR ELMI SHARLINA BINTI MD SUHAIMI

KHAIRUL ASWARI BIN AB RAHMAN
NOOR ASMAHAN BINTI ABDULLAH

WAN ROSILAH BIN WAN LLAH
SUHAINA BINTI MOHD YATIM
SHARIFAH FADTHYAH BINTI SYED BAHARUDDIN
NOOR NAJIHAH BINTI KAMARUDDIN
CHEMISTRY UNIT KELANTAN MATRICULATION COLLAGE
MATRICULATION DIVITION MINISTRY OF EDUCATION MALAYSIA

©chemistryunitkmkt2021

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TABLE OF CONTENT

CONTENT TITLE PAGE
CHAPTER 1 MATTER
ATOMS, MOLECULES AND IONS 1
• NOTE MOLE CONCEPT 12
• TUTORIAL CHEMICAL EQUATION AND STOICHIOMETRY
CHAPTER 2 ELECTRONIC CONFIGURATION 14
• NOTE PERIODIC TABLE 28
• TUTORIAL CHEMICAL BONDING
CHAPTER 3 CHEMICAL EQUATION 32
• NOTE REACTION KINETICS 40
• TUTORIAL
CHAPTER 4 42
• NOTE 56
• TUTORIAL
CHAPTER 5 60
• NOTE 77
• TUTORIAL
CHAPTER 6 81
• NOTE 94
• TUTORIAL
CHAPTER 7 99
• NOTE 119
• TUTORIAL
CHAPTER 8 122
• NOTE 132
• TUTORIAL
CHAPTER 9 135
• NOTE 153
• TUTORIAL
PRE-LAB MODULE DETERMINATION OF THE DENSITY OF WATER 157
• EXPERIMENT 1 STANDARD SOLUTION AND DETERMINING OF 159
• EXPERIMENT 2 THE CONCENTRATION OF ACID SOLUTION
QUANTITATIVE ANALYSIS OF BAKING SODA 162
• EXPERIMENT 3 MOLECULAR GEOMETRY 164
• EXPERIMENT 4

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CHAPTER 1: MATTER

GENERAL OVERVIEW

1.1 Definition and a) Define i. pure substance
classification matter (elements and
compounds)
b) Classify
matter into ii. mixtures (homogeneous
and heterogenous)

1.2 State of a) Explain i. arrangement of
matter the particles
ii. shape
1.0 general
MATTER properties iii. density

of iv.compressibility
solid,liquid

and gas

1.3 Elements a. Define element

1.4 Physical b.Identify the elements in the
and chemical periodic table

change a Decribe the physical and
chemical changes of mastter

1|Chapter 1 Chemistry1 DK014

1.1 DEFINITION AND CLASSIFICATION
➢ Matter is anything that occupies space and has mass; the physical material of

the universe e.g. air, water, animals, trees, and atoms.
➢ Matter may consist of atoms, molecules or ions.
➢ Atoms is the smallest particle of a chemical element that can exist.
➢ Molecules form when two or more atoms form chemical bonds with each other.

It doesn't matter if the atoms are the same or are different from each other.
➢ Ion is an electrically charged atom or group of atoms formed by the loss or

gain of one or more electrons, as a cation (positive ion) and anion (negative ion)

Bromine Water Gas in the conical flask Copper block

Example 1:

Iron, sodium chloride and water are made up of particles.

Substances Particles

Iron Atom

Sodium Chloride Ion

Water Molecule

Example 2 :

Classify all the following substances into atom, molecules and ion.

Magnesium sulphate Potassium Air

Oxygen gas Chlorine gas Charcoal

Substances Particles
Magnesium sulphate Ion
Potassium Atom
Air Molecule
Oxygen gas Molecule
Chlorine gas Molecule
Charcoal Atom

Matter can be divided into TWO types :
▪ Mixture (homogenous and heterogenous)
▪ Pure substances (elements and compunds)

✓ Mixture is a combination of two or more substances in which each
substances retains its own chemical identity.

✓ Mixture can be divided into two types :

Homogeneous Mixture
➢ The composition of a mixture, after sufficient stirring, is the same throughout

the solution. There are several examples of homogenous mixtures encountered in

everyday life.
➢ You can't pick out components of a homogeneous mixture or use simple

mechanical means to separate them. You can't see individual chemicals or

ingredients in this type of mixture. Only one phase of matter is present in a

homogeneous mixture.

2|Chapter 1 Chemistry1 DK014

Examples: soft drink, salt and water, air, rainwater, vinegar, dishwashing detergent
and steel.

Heterogeneous Mixture
➢ The individual components of a mixture remain physically separated
and can be seen as separate components.
➢ Usually, it's possible to physically separate components of a heterogeneous
mixture. For example, you can centrifuge (spin out) solid blood cells to
separate them from the plasma of blood. You can remove ice cubes from
soda. You can separate candies according to color.
➢ Examples :
Cereal in milk, vegetable soup, pizza, blood, ice in soda, salad dressing,

mixed nuts, bowl of coloured candies, oil and water, iron filings in sand and soil.

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Pure Substances
➢ Pure substances is each or which has a fixed composition and has consistent
properties throughout the sample.
➢ A pure substance participates in a chemical reaction to form predictable products.
In chemistry, a pure substance consists of only one type of atom, molecule, or
compound.
➢ Pure substances can be either elements or compounds.
➢ Compounds :
• A substances composed of two or more elements united
chemically in definite proportions.( fixed compositions)
• Can only be separated into their pure components (elements) by
chemical means.
• E.g. Water (H2O), Glucose (C6H12O6), Ammonia (NH3), Sodium
chloride (NaCl)
➢ Elements :
• An element is a substance that cannot be separated into simpler
substances by chemical means.
• In chemistry, an element is identified by its symbol, consist of one or
two letters, usually derived from the name of the element.
• Example: Gold (Au), Lead (Pb), Aluminium (Al)

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EXAMPLE 1 :
Classify each of the following as element, compund or mixture (homogenous or
heterogenous).

(a) Platinum
(b) Table salt
(c) Soy sauce
(d) Sugar
(e) Silver
(f) Milk
(g) Aluminium

ANSWER :

Substance Classification

a) Platinum Element
b) Table salt Compound
c) Soy sauce Homogenous mixture
d) Sugar Compound
e) Silver Element
f) Milk Compound
g) Aluminium Element

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EXAMPLE 2 :
Classify the following as homogenous or heterogenous mixture.

a. Maple syrup
b. Sea water
c. Melted rocky road
d. Ice cream
e. Wine
f. Gasoline
g. Batter for chocolate chips cookies

ANSWER : Classification
Substance
Homogenous mixture
a. Maple syrup Homogenous mixture
b. Sea water Heterogenous mixture
c. Melted rocky road Heterogenous mixture
d. Ice cream Homogenous mixture
e. Wine Homogemous mixture
f. Gasoline Heterogenous mixture
g. Batter for chocolate chips

cookies

EXERCISE 1:
Classify the following substances as a homogeneous mixture, heterogeneous mixture,
compound or element.
i) Vinegar ii) Liquefied nitrogen iii) salad dressing iv) Salt

Substance Classification

EXERCISE 2:
Classify each of the following substances as an element, a compound, a homogeneous
mixture, or a heterogeneous mixture:

salt, pure water, soil, salt water, pure air, carbon dioxide, gold and bronze.

Substance Classification

1.2 STATES OF MATTER
Three states of matter :
▪ Solid
▪ Liquid
▪ Gas

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General Properties of Solid, Liquid and Gas

States of matter Solid Liquid Gas
Characteristic

Arrangement of ▪ Close packed ▪ Less closely ▪ Very far apart
particles and orderly packed and not
manner and not in
Shape and volume orderly manner
▪ Have very ▪ More empty contact
Density little space ▪ A lot of empty
Compressibility between space between
particles particles space

▪ has a definite ▪ has a distinct ▪ No fixed shape
shape and a volume or volume
definite volume independent of
its container but Low
High has no specific High
Almost none shape

High
Very low

EXAMPLE :
What are the three states of matter? Give an example of each.

Three states of matter are solid, liquid and gas. Example of solid are building, book, car
Example of liquid are water and oil. Example of gas are oxygen gas, hydrogen gas,
chlorine gas.

EXERCISE
Complete the table below :

States of matter Liquid Gas
Characteristic

Arrangement of ▪ Less closely
particles
packed and not
Shape and volume orderly manner
▪ More empty
Density space between
Compressibility
Example particles

▪ has a definite Low
shape and a High
definite volume

Almost none

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1.3 ELEMENTS

➢ A substance that cannot be separated into simpler substances by any
chemical means.

➢ Arranged in periodic table in order of increasing proton number.
➢ There are 118 elements have been identified.
➢ 90 elements occur naturally on Earth.

Example: Gold, aluminium, lead, oxygen, carbon.
➢ 28 elements have been created by scientists e.g. Technetium, americium,

seaborgium.
➢ most of the elements can be classified in the periodic table .

Who invented Periodic Table?

The scientists involved are: Antoine Lavoisier, Johann W.Dobereiner, John Newlands,
Lothar Meyer, Dimitri Medeleev and H.J. G. Moseley.

Lothar Meyer Dimitri Mendeleev Henry Mosely

Complied a periodic table Published the first version Rearranging the elements
of 56 elements based on of the periodic table in in the periodic table by their
regular repeating pattern 1869. Arranged elements atomic numbers.
of physical properties. according to increasing
atomic masses.

All the elements are arranged in order of increasing proton number in Periodic
Table

The Modern Periodic Table consist of 118 elements, 7 rows (called period), 18 columns
(called group) and 4 blocks (s,p,d,f).

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• A vertical column of elements is called a group. There are group 1 to group 18
• A horizontal row of elements is known as a period. There are 7 periods in the periodic

table.( Period 1 to Period 7).

Group

Period

DIFFERENT BETWEEN ATOM AND MOLECULE

An atom can be defined as the smallest constituent particle of an element which
showcases independent existence.
Example: Ne, O

A molecule can be defined as the combinations of two or more atoms which are held
together by chemical bonds. A molecule is the smallest portion of a substance which
showcases all the properties of the substance. On breaking down a molecule further, we
see properties of the constituent elements.
Example: O2, HCl

COMPARISON ATOM AND MOLECULE

Factor Atom molecule
Definition
Most fundamental and smallest part Two or more atoms chemically
Example
that can exist of an element. bonded together.
Oxygen – O Oxygen – O2
Phosphorus – P Phosphorus – P4
Sulphur – S Sulphur – S8
Hydrogen – H Water – H2O

Structure Smallest particle with properties of Combination of two or more atoms.
an element.
Stability An atom may not always be stable Molecules are formed to attain
in nature due to the presence of stability.
Constituent electrons in the outer shells.
Elements Protons, Electrons & Neutrons Two or more atoms of the same or
Reactivity different elements
Except for the noble elements, Compared to a molecule, the level
atoms of all elements showcase a of reactivity is less as some of the
certain level of reactivity. valence points are filled by
electrons of combined elements.

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EXAMPLE :
Give the chemical symbol or name for the following elements below :

a. Helium
b. Platinum
c. Cobalt
d. Tin
e. Sb
f. Pb
g. Br
Answer:
a. Helium - He
b. Platinum -Pt
c. Cobalt -Co
d. Tin -Sn
e. Sb -Antimony
f. Pb - Lead
g. Br -Bromine

EXERCISE :
1. Give the chemical symbol or name for the following elements below :
a. Rubidium -
b. Tungsten –
c. Caesium-
d. Hydrogen-
e. Xe-
f. Kr-
g. Carbon-
h. Hg-

1.4 PHYSICAL AND CHEMICAL CHANGES

Physical Change

• Changes of physical state of a substance that does not change its chemical

composition.

• Same substance before and after the change

• No new substances are formed when the state of a substance changes.

• A change of state is interconvertible.

• There is a change in the composition of the substances in question; in a physical

change there is a difference in the appearance, smell, or simple display of a sample

of matter without a change in composition.

• Some common changes (but not limited to) are texture,colour,temperature, shape

and change of state.

Example 1 :

Formation of dry ice

CO2(s) CO2(g)

Example 2 :
➢ Aluminium foil is cut in half.
➢ Clay is molded into a new shape.
➢ Butter melts in warm toast.
➢ Water evaporates from the surface of

the ocean.
➢ A juice box in the freezer freezes.
➢ Ice melting.

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Chemical Change

• Substance is transformed into a chemically different substance.
• A formation of a new substance through a chemical reaction.
• Change in matters that alter the chemical composition of a substances.
• The follow are all indicators of chemical reactions :

a. Change in temperature .Example is the process of neutralisation.
b. Change in colour. Example is metal rusting
c. Noticiable odour . Example is a rotten egg.
d. Formation of precipitate. Example is reaction between soluble carbonate reacts

with Barium metal. Precipitate of Barium carbonate is formed.
e. Formation of bubble gas . Example of proses during electrolysis, formation of

bubbles gas oxygen and hydrogen gas is formed.

EXERCISE 1
1. Label each of the following as either a physical process or a chemical process.
a. Crushing a metal can - ________________
b. Production of urine in the kidneys -___________________
c. Melting piece of chocolate- _______________________
d. Burning fossils fuels -_______________________
e. Discharging a battery -_________________________
f. Burning candle -_________________________

EXERCISE 2

Determine the following process involve physical or chemical change. Give your reason.

Process Physical /chemical Reason

change

You blow dry your wet hair

In baking biscuits and other quick

breads, the baking powder reacts

to release carbon dioxide bubbles.

A straight piece of wire is coiled to

form a spring.

In a fireworks show, the fireworks

explode giving off heat and light.

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TUTORIAL 1

1. Match the following terms. A mixture with no visible separation
Matter between its components.
Element A substances that cannot be separated
into simpler substances by chemical
Compound means.
Matter that has distinct properties and
Mixture composition that does not vary from
Homogeneous Mixture sample to sample.
Anything that occupies space and possess
Pure Substances mass.
Heterogeneous Mixture A combination of 2 or more substances in
which the substances retain their identity
(chemical properties).
A mixture in which the components can be
visibly distinguished.
Substance that contains two or more
elements chemically combined in a fix
proportion.

2. Classify the following substances as a homogeneous mixture, heterogeneous

mixture, compound or element.

(a) Helium gas (b) Solid Sodium chloride

(c) Gold (d) Oxygen gas

(e) Water (f) Air

(g) Concrete (h) Lead (II) sulphate

(i) Chocolate chip cookies (j)Pure alcohol

(k) Sea water (l) Limestone

3. a) What are the three states of matter? Give an example of each.

b) Complete the following table to describe the characteristics of the three

states of matter.

States of matter Characteristic Solid Liquid Gas

Arrangement of particles

Shape and volume

Density

Compressibility

4. a) What is the difference between a physical change and a chemical change?
b) Classify each of the following as a chemical or physical change.
i.Paper ignites when placed in a flame.
ii.Charcoal burns leaving ashes.
iii.Sugar dissolves in water.
iv.Dry ice forms a vapor "cloud" when exposed to the air.
v.Formation of an ice cube from liquid water.
vi.Heating a metal until it is red hot.
vii.Drying clothes.

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5. Identify the following as physical (P) or chemical (C) changes.
a) NaCl (Table Salt) dissolve in water
b) An apple is cut
c) Baking soda reacts with vinegar
d) Alcohol evapoarates.
e) Milk sours.
f) Pancakes cook
g) A tire is inflated
h) Paper towel absorbs water
i) Ag (silver) tarnishes
j) Heat changes water to steam
k) Fe (iron) rusts
l) Ice melts
m)Sugar dissolves in water
n)Grass grows
o)Food is digested
p)Wood rots

Multiple Choices Questions

1. There are three main states of matter: liquid, solid, and gas. All have different
physical properties. Which statement describes the physical state of a solid but not of
a liquid or a gas?
A. It has its own shape.
B. It does not have its own shape.
C. It takes the shape of its container.
D. It changes shape with temperature.

2. At room temperature helium does not have a definite shape or volume. Which state is
it in?
A. gas
B. solid
C. water
D. liquid

3 Which of the following is an example of physical change?
A. Mixing baking soda and vinegar together, and this causes bubbles and foam.
B. A glass cup falls from the counter and shatters on the ground.
C. Lighting a piece of paper on fire and the paper burns up and leaves ashes.
D. Baking a birthday cake for your mother.

4. Which of the following is an example of chemical change?
A. Filling up a balloon with hot air.
B. Taking a glass of water and freezing it by placing it in the freezer.
C. A plant collecting sunlight and turning it into food.
D. Your cat ripping up your homework.

5. Which change can be easily be reversed?
A. Chemical Change
B. Physical Change
C. Both a physical and chemical change
D. Neither a physical or chemical change

Revised by,
Noor Asmahan Abdulllah
Zanarina Binti Sapiai

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1CHAPTER 2: ATOMS, MOLECULES AND IONS

GENERAL OVERVIEW

2.1 Proton a) Sub-atomic Proton
number, nucleon particles.
Electron
number and b) Definition
isotope. Neutron
c) Isotopic
Proton number
Notation (Z)

Nucleon
number (A)

Isotopes

2.0 ATOMS, 2.2 Atomic and Relative atomic mass
MOLECULE Molecular Masses (Ar)
AND IONS
Relative molecular
mass (Mr)

2.3 Nomenclature of a) Definition Cations
inorganic compound Anions
Cations
b) IUPAC nomenclature Anions

Salts

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2.1 Proton number, nucleon number and isotope.

- - is the smallest particle in an element that can participate in a chemical
- reaction.

- are composed of electrons, protons and neutrons

- Example : Na, Mg, Al …

SUBATOMIC PARTICLES Atom is
➢ An atom made up of: electrically
• Proton neutral

• Neutron

• Electron

➢ An atomic nucleus consists of protons and neutrons

Neutron Electron
Proton

➢ The magnitude of charge posses by proton is equal to that of an electron, but
the sign of the charge is opposite

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Particle Mass Charge Charge
(g) (Coulombs) (units)

Electron (e-) 9.1 x 10-28 -1.6 x 10-19 -1

Proton (p+) 1.67 x 10-24 +1.6 x 10-19 +1

Neutron (n) 1.67 x 10-24 0 0

PROTON NUMBER (Z)
➢ Number of protons in the nucleus of atom of an element.
➢ Also called atomic number

Example :
All carbon atoms (Z = 6) have 6 protons
All oxygen atoms (Z = 8) have 8 protons
All uranium atoms (Z = 92) have 92 protons

NUCLEON NUMBER (A)
➢ Total number of protons and neutrons in the nucleus of an atom of an element.
➢ Nucleon number also called mass number.

Nucleon Number = Number of protons + number of neutrons

Number of Neutrons = Nucleon Number (A) – Proton Number (Z)

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ISOTOPE

➢ Two or more atoms of the same element having same proton number but
different nucleon number.
@

➢ Two or more atoms of the same element having same number of protons but
different number of neutrons.

Atomic Symbol @ Isotopes Notation

The symbol for an atom: A X Element Symbol
Nucleon Number Z
Proton Number 3

Example: 1 H 2 1H
1
1H (Tritium)
(Protium)
(Deuterium)
235
U238
92 U
92

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EXAMPLE 1:
Give the number of protons, neutrons, electrons and charge in each of the following
species:

Answer

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EXAMPLE 2:
Write the appropriate notation for each of the following species :

4 A
2

3 B +
1

12C

D14 3−

7

EXERCISE 1:
Consider an atom of 32P.

a) How many protons, neutrons and electron does this atom contain?
b) What is the symbol of the atom obtained by removing five electron to 32P?
c) What is the symbol of the atom obtained by adding one neutron to 32P?

2.2 Atomic and Molecular Masses

RELATIVE MASS
➢ The mass of an atom relative to another atom can be determined experimentally
using mass spectrometer.
➢ Currently, carbon-12 isotope is used as standard to measure relative atomic mass.
➢ Relative atomic mass is dimensionless.

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Relative atomic mass, Ar
➢ A mass of one atom of an element compared to one twelfth mass of one atom of

carbon-12 atom.
Relative atomic mass = mass of one atom of an element (amu)

12

1 x mass of one C atom (amu)
12
➢ Relatives atomic mass is a ratio, hence, it has no unit.

Relative Molecular mass, Mr
➢ A mass of one molecule of a compound compared to one twelfth mass of one atom

of carbon-12 atom.
➢ Sum of the Relative Atomic mass, Ar of each atom within the molecule.

Relative atomic mass = mass of one molecule of a compound (amu)
1 x mass of one 12C atom (amu)
12

➢ Relatives molecular mass is a ratio, hence, it has no unit.
Example :
What is the relatives molecular mass for sodium chloride, NaCl?
Mr (NaCl) = Ar (Na) + Ar (Cl)

= 23.0 + 35.5
= 78.5

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2.3 Nomenclature of inorganic compound

CATION
➢ Ion with a positive charge.
➢ If a neutral atom loses one or more electrons it becomes a cation

ANION

➢ ion with a negative charge
➢ If a neutral atom gains one or more electrons it becomes an anion.

NOTES !!

➢ The proton number, Z, is the nuclear charge and also the number of electrons
in a neutral atom of the element.

No. of proton = no. of electron 0 charge (neutral)

No. of proton > no. of electron + charge (cation)
No. of proton < no. of electron (atom lost electrons)

- charge (anion)
(atom gained electrons)

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EXAMPLE 1:

Give the chemical symbol for :

a) the ion of Titanium with 22 protons, 26 neutrons and 19 electrons
b) the ion of Sulphur that has 16 neutrons and 18 electrons.

Answer:

a) Ti3+
b) S2-

EXERCISE :

Give the names and charge of the cation and anion in each of the following compounds :
a) CaO
b) Na2SO4
c) Fe(NO3)2
d) KClO4
e) Cr(OH)3

IUPAC NOMENCLATURE

Cation Anion Salt

Cation

1. Metals ions with one stable oxidation state.

➢ Group 1, 2 and 13
➢ Naming of the metal followed by the word ‘ion’
➢ Example: Na+ is sodium ion

Al3+ is aluminium ion

Zn2+ is zinc ion

2. Metals ions with more than one stable oxidation state.

➢ Metals from block-d and block-p
➢ Naming of the metal followed by Roman numerals corresponding to the oxidation

state, enclosed in brackets
➢ Example:

Cu+ is copper(I) ion Cu2+ is copper(II) ion
Fe2+ is iron(II) ion Fe3+ is iron(III) ion

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3. Cations formed form molecules composed of nonmetal atoms have names
that end in ‘–ium’

➢ Example : NH4+ is ammomium ion

H3O+ is hydronium ion

COMMON CATIONS

CHARGE FORMULA NAME FORMULA NAME
1+ H+ hydrogen ion
Li+ lithium ion NH4+ Ammonium ion
K+ potassium ion
Cs+ cesium ion
Ag+ silver ion

2+ Mg2+ Magnesium ion Co2+ Cobalt(II) ion

Ca2+ Calcium ion Cu2+ Copper(II) ion

Sr2+ Strontium ion Mn2+ Manganese(II) ion

Ba2+ Barium ion Hg22+ Mercury(I) ion

Cd2+ Cadmium ion Hg2+ Mercury(II) ion

Ni2+ Nickel (II) ion

Pb2+ Lead (II) ion

Sn2+ Tin (II) ion

3+ Al3+ Aluminium ion Cr3+ Chromiun (III) ion

B3+ Boron ion Ar3+ Arsenic(III)

N3+ Nitrogen ion Fe3+ Iron(III)

P3+ Phosphorus ion Co3+ Cobalt(III)

Ni3+ Nickel(III)

Sb3+ Antimony(III)

4+ C4+ Carbon Pb4+ Lead(IV)
Mn4+ Manganese(IV)
Si4+ Silicon

Sn4+ Tin(IV)

5+ N5+ Nitrogen Sb5+ Antimony(V)

P5+ Phosphorus Bi5+ Bismuth(V)

As5+ Arsenic(V)

Table 1

NOTE:

Certain elements such as Copper, Iron, Tin, and others exhibit more than one oxidation

number. In these cases, the name of the element is followed by the oxidation number in

Roman numerals in parenthesis.

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Anion

1. Names of monoatomic anions are formed by replacing the ending of the name of
the element in ‘ –ide’.

➢ Example : H- is hydride ion
O2- is oxide ion
N3- is nitride ion
Cl- is chloride ion

➢ A few polyatomic anions also have names ending in ‘-ide’.

Example : OH- is hydroxide ion
CN- is cyanide ion
O22- is peroxide ion

1. Names of anions of oxo acids (oxygen containing acids) end in ‘–ate’ and ‘–ite’.

➢ The -ate is used for the most common or representative oxyanion of an element.
➢ The -ite is used for an oxyanion that has the same charge but one O atom fewer.

➢ Example : NO3- is nitrate ion SO42- is sulfate ion
NO2- is nitrite ion SO32- is sulfite ion

2. Anions derived by adding H+ to an oxyanion are named by adding as a prefix
the word hydrogen or dihydrogen.

➢ Example : HCO3- is hydrogen carbonate ion
H2PO4- is dihydrogen phosphate ion

COMMON ANIONS

CHARGE FORMULA NAME FORMULA NAME
1- H- Hyride ion CH3COO- Acetate ion
F- Fluoride ion Nitrate ion
2- Br- Bromide ion NO3- Permanganate ion
I- Iodide ion MnO4- Thiocyanate ion
3- CN- Cyanide ion SCN- Hydroxide ion

O2- Oxide ion OH-
S2- Sulfide ion
CO32- Carbonate ion
N3- nitride ion CrO42- Chromate ion
P3- Phosphide ion Cr2O72- Dichromate ion
SO42- Sulfate ion
Table 2
PO43- Phosphate ion
PO33- Phosphite ion

24 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

Common Elements Having More Than One Oxidation Number

Elements with oxidation Elements with oxidation

numbers of +1 and +2: numbers of +2 and +3:

Copper Cu Chromium Cr

Mercury Hg Iron Fe

Cobalt Co

Nickel Ni

Elements with oxidation Elements with oxidation

numbers of +2 and +4: numbers of +3 and +5:

Manganese Mn Arsenic As

Tin Sn Antimony Sb

Lead Pb Bismuth Bi

Salt

➢ Salts are ionic compounds which, when dissolved in water, break up completely
into ions.

➢ They arise by the reaction of acids with bases, and they always contain either a
metal cation or a cation derived from ammonium (NH4+).

Nomenclature of Salt

1. Salts are named as follows: ‘name of cation’ <space> ‘name of anions’
Example :

➢ The total oxidation
number of the first, or
positive, part of the
compound must be equal
but opposite in charge to
the total oxidation number
of the second, or negative,
part of the compound.

➢ The algebraic sum of the
oxidation numbers of the
elements and polyatomic
ions in a chemical
compound is zero.

25 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

Example 1: Write the formula for potassium bromide.

Step 1:
➢ Looking at the list of oxidation numbers in Table 1 and Table 2, it is found that
potassium, K, has an oxidation number of +1 and bromide (the combined form of
bromine), Br, has an oxidation number of -1.
➢ Writing the symbol of the positive element or polyatomic ion first, the formula is:
KBr

Step 2:
➢ The algebraic sum of the oxidation numbers is +1 + (-1) = 0
➢ Thus, the positive and negative oxidation numbers match and the formula of
potassium bromide is correct as written above.
➢ The formula of potassium bromide is interpreted to mean that a molecule of the
compound contains one atom of potassium and one atom of bromine.

KBr

Example 2 : Write the formula of iron(II) bromide.

Step 1 :
➢ In this example, it is found that the oxidation number of iron, Fe, is +2 (as indicated
by the Roman numeral) and the oxidation number of bromide, Br, is -1.
➢ If the formula of iron(II) bromide is written as:
FeBr

Step 2 :
➢ the algebraic sum of the oxidation numbers of the elements in this formula is +2 + (-
1) = +1.
➢ There is an excess of the positive oxidation number and the addition of a second
bromide ion will be needed to make the sum zero.
➢ Thus, for the formula:
FeBr

Step 3 :
➢ the algebraic sum of the oxidation numbers is +2 + [2 x (-1)] = 0. This formula, as
written, is in an inconvenient form since the formula of bromide appears twice. In
order to simplify the formula, a subscript is used to indicate the number of bromine
atoms required. In this example, two bromine atoms are needed, so the proper
formula for iron(II) bromide is written as:
FeBr2

Formula (Salt) Name
RbBr rubidium bromide
KI3 potassium triiodide
sodium carbonate
Na2CO3 sodium Hydrogen carbonate
NaHCO3 sodium perchlorate
NaClO4 aluminium sulphate
Al2(SO4)3 copper (I) chloride
copper (II) chloride
CuCl iron (II) chloride
CuCl2 iron (III) chloride
FeCl2
FeCl3

26 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

MnSO4 manganese (II) sulphate
NaH2PO4 sodium dihydrogen phosphate
boron phosphide
BP carbon tetrachloride
CCl4 silver iodide
AgI lithium fluoride
LiF sodium nitrate
NaNO3 calcium fluoride
mercury(I) sulfide
CaF2 iron(II) sulfate
cobalt(III) oxide
Hg2S
FeSO4 Example of Salt:
Co2O3

2. Hydrate salts are named as follows: name of salt <space> multiplying prefix
(hydrate)
Example : CuSO4.5H2O is Copper (II) sulphate pentahydrate
AlK(SO4)2.12H2O is Aluminium potassium sulphate dodecahydrate

Exercise :
1.

Name the following compounds.

a) PbI2 b) FeSO4
c) Ag2CO3 d) NaCN

e) Cu(NO3)2 f) K2C2O4
g) HgCl h) CCl4

2. Write formulas for the following compounds

a) ammonium sulfide
b) magnesium phosphate
c) sodium iodate
d) chromium(III) chloride
e) potassium permanganate
f) zinc bromide
g) boron triiodide

27 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

TUTORIAL 2

SECTION A

1.Smaller particles in an atom are called
A. atomic particles
B. subatomic particles
C. smaller particles
D. neutral particles

2. Electrons orbiting around nucleus bear
A. positive charge
B. negative charge
C. no charge
D. neutral charge

3. Neutrons carry
A. positive charge
B. negative charge
C. neutral charge
D. no charge

4. An electrically charged particle which is formed when an atom gains or loses electron is
called
A. ion
B. charge
C. formula
D. neutron

5.The current atomic mass scale is based on the adoption of
A. exactly 16 as the average relative atomic mass of naturally occurring oxygen.
B. exactly 12 as the average relative atomic mass of naturally occurring carbon.
C. exactly 16 as the relative atomic mass of the 16O isotope of oxygen.
D. exactly 12 as the relative atomic mass of the 12C isotope of carbon.

28 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

6. Positive ions are formed when an atom
A. Gain electron
B. Loses protons
C. Gain proton
D. Loses electron

7. Identify the ions present in (NH4)2Cr2O7.
A. N3-, H+, Cr3+ and O2-
B. N3-, H-, Cr3+ and O2-
C. NH4+ and Cr2O72-
D. NH3 and H2Cr2O7

8. Calcium reacts with fluorine to form an ionic solid. The correct formula for this solid is
A. CF
B. CaF2
C. Ca2F
D. C2F

9. What is the charge on a titanium ion in the compound Ti3(PO4)4
A. +3
B. +2
C. -3
D. +4

10. Consider the following three compounds:

I) Fe(CH3CO2)3 II)Ba(NO3)2 III) Li2O.

Select the answer below that contains the correct names for compounds I – III.

A. I) Iron(III) acetate II) Barium nitrate III) Lithium oxide

B. I) Iron triacetate II) Barium(II) nitrate III) Dilithium oxide

C. I) Iron(II) acetate II) Barium(II) nitrite III) Lithium hydroxide

D. I) Iron acetate II) Barium dinitrate III) Lithium oxide

29 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

SECTION B

1. Define:
i. proton number
ii. nucleon number

2. Give the number of protons, neutrons, electrons and charge in each of the following

species: 23
i.
ii. A

iii. 11

B87 2+

38

C-127

53

3. An element X with a nucleon number of 65 has 35 neutrons in the nucleus.
The ion formed by this element has 28 electrons. Write the isotopic notation of
(i) Atom X
ii) Ion X

4. Magnesium has three isotopes with mass numbers 24, 25 and 26.
(a) Write the isotopic notation for each.
(b) How many neutrons are in an atom of each isotope
(c) Describe the similarities and differences between these three kinds of atoms of
Magnesium.

5. Define:
i) relatives atomic mass
ii) relative molecular mass.

6. Which of these compounds would you expect to be ionic; N2O, Na2O, CaCl2, SF4?.
Explain.

7. Fill in the blanks in the following table.

Chemical formula IUPAC Name
copper(II) ion
Salt Cu(NO3)2 chromium(III) chloride
chloride ion
Cation

Anion NO3-

Salt

Cation

Anion

Salt Ba(ClO4)2
Cation Ba2+

Anion

30 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

Salt ammonium sulphate
Cation
Anion SO42-

8. TABLE 1 shows the atomic structure of four particles, represented by the letter A to D.

The particles are atoms and ions

TABLE 1

Particle Electrons Protons Neutrons

A6 6 6

B 12 12 12

C 10 12 12

D6 6 8

Use the letter A to D to answer the following questions

(i) Which two particles are atom and ion of the same element? Explain

(ii) Which two particles are isotopes of the same element? Explain.

Revised by:
Noor Najihah Binti Kamaruddin
Zanarina Binti Sapiai

31 |C h a p t e r 2 C h e m i s t r y 1 D K 0 1 4

CHAPTER 3.0: MOLE CONCEPT

GENERAL OVERVIEW

a) mass of C-12
Define
mole Avogadro's
constant (NA)

3.1 Avogadro b) Relate room
Number and Molar Avogadro's condition
number with
Mass molar mass and Standard
molar volume of Temperature
Pressure (STP)
gas at

a) Define empirical formula
molecular formula
3.2
3.0 Empirical b) Determine mass
empirical and composition
MOLE and
CONCEPT Molecular molecular combustion data
Formula formula from

3.3 molarity
Concentration
a) Define density

% by mass

b) Perform calculations on
molarity and %by mass
*include dilution

32 | Chapter 3 Chemistry 1 DK014

3.1 AVOGADRO NUMBER AND MOLAR MASS

▪ What is a MOLE?
A mole is a counting unit used by chemists to express the number of atoms or
molecules in a sample.

● Definition mole in terms of mass of carbon-12
The amount of substance that contain as many elementary particles
(atoms/molecules/ions) as there are atoms in exactly 12.000g of 12C

1 mol of C-12 atoms has a mass of exactly 12.000 grams

● Definition mole in terms of Avogadro's constant (NA)
1 mol of particle contain 6.02 x 1023 particles.

1 mol 12C atoms = 6.02 x 1023 12C atoms
1 mol H2O molecules = 6.02 x 1023 H2O molecules
1 mol NO3- ions = 6.02 x 1023 NO3- ions

*particles can be atoms, molecules or ions

● What is MOLAR MASS?
⮚ The mass (in grams) of one mole of any substance
⮚ Unit: grams per mole (g/mol @ gmol-1)
⮚ The molar mass of any substance is numerically equal to its formula weight
in a.m.u.
⮚ Example:
Molar mass of helium = 4.0 gmol-1
Molar mass of sodium = 23.0 gmol-1
⮚ Relationship between mol and molar mass as follows;

( )
, = ( / )

● What is MOLAR VOLUME?
⮚ is a volume occupied by 1 mol of gas
⮚ 2 conditions - room temperature
- Standard temperature and pressure (STP)

Standard temperature Standard pressure
0oC 1 atm
273.15 K 760 torr
101325 N m-2
101325 Pa

33 | Chapter 3 Chemistry 1 DK014

⮚ At STP, 1 mol of any gas occupies 22.4 L
⮚ At room temperature, 1 mol of any gas occupies 24.0 L

Figure 1: Relationship between mol, mass, number of particles and volume of gas

EXAMPLE 1:
Calculate the number of moles of

a) 60.2 x 1020 atom of argon
b) 3.01 x 1023 molecule of oxygen

Answer:
a) 1 mol Argon contains 6.02 x 1023 atoms

x mol Argon contains 60.2 x 1020 atoms

x mol = 60.2 x 1020 atoms x 1 mol = 0.01 mol
6.02 x 1023

b) 1 mol oxygen contains 6.02 x 1023 molecules
x mol oxygen contains 3.01 x 1023 molecule

x mol = 3.01 x 1023 atoms x 1 mol = 0.50 mol
6.02 x 1023

EXAMPLE 2:
How many moles of He in 6.46 g of He? [ molar mass He= 4.00 gmol-1 ]

Answer:

( )
= ( / )

6.46
= 4.00 / = 1.615

EXAMPLE 3:
Calculate the volume occupied by 19.61 g of N2 at STP. [ molar mass N= 14.00 gmol-1 ]

Answer:

19.61
= 2(14.00 ) = 0.700

34 | Chapter 3 Chemistry 1 DK014

At STP:
1 mol of N2 occupies 22.4 L
0.700 mol of N2 occupies 22.4 L x 0.700 mol = 15.69 L
1 mol

EXERCISE:
1. Calculate the number of H atoms in 0.350 mol of C6H12O6

2. Calculate mass (g) of 0.433 mol of calcium nitrate, Ca(NO3)2?

3. What is the volume (in liters) occupied by 49.8g of HCl gas at room temperature?

Answer:
1. 2.53 X 1024 H atoms

2. 71.1 g

3. 32.8 L

3.2 EMPIRICAL AND MOLECULAR FORMULA
▪ DEFINITION OF EMPIRICAL FORMULA

⮚ formula that shows the simplest ratio of all elements in a molecule

▪ DEFINITION OF MOLECULAR FORMULA
⮚ formula that shows the actual number of atoms of each element in a molecule

Example;

EMPIRICAL FORMULA MOLECULAR FORMULA
H2O H2O
CH2O
NH2 C6H12O6
N2H4

The relationship between empirical formula & molecular formula is;
Molecular formula = n (Empirical formula)

EXAMPLE 4:

A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar
mass is 56 g mol-1. Determine the empirical formula and molecular formula of the compound.

Answer:
Assume mass of hydrocarbon = 100 g

Element C H
Mass (g)
Mol (mol) 85.7 14.3

Simplest ratio 85.7 = 7.142 14.3 = 14.3

12.00 1.00


7.142 14.3
7.142 = 1 7.142 = 2.00

Thus, empirical formula is CH2
Empirical formula mass of CH2 = 1(12.0) + 2(1.0) = 14.0

Molecular formula = n (Empirical formula)

35 | Chapter 3 Chemistry 1 DK014

56
= 14 = 4
Thus, molecular formula is (CH2)4 = C4H8

EXAMPLE 5:
0.974 g sample of compound A, is composed of C, H and O was burnt in excess of oxygen
producing 2.464 g of CO2 and 0.432 g of H2O. Determine the empirical formula of the
compound

Answer:
mol CO2 = 2.464 g / 44 gmol-1 = 0.056 mol

1 mol CO2 contains 1 mol C
0.056 mol CO2 contains 0.056 mol C
Mass C = 0.056 mol x 12 g/mol = 0.672 g

Mol H2O= 0.432 g / 18 gmol-1 = 0.024 mol
1 mol H2O contains 2 mol H
0.024 mol H2O contains 0.048 mol H
Mass H = 0.048 mol x 1 g/mol = 0.048 g

Mass C + H + O = 0.974 g
Mass O = 0.974 g - 0.672 g - 0.048 g = 0.254 g

Element C H O
Mass (g)
Mol (mol) 0.672 0.048 0.254

Simplest ratio 0.672 0.048 = 0.048 0.254 = 0.016

12.00 1.00 16.0

= 0.056

0.056 0.048 0.016
0.016 = 3.5 0.016 = 3.0 0.016 = 1.0

3.5 x 2 = 7 3.0 x 2 = 6 1x2=2

Thus , empirical formula is C7H6O2

TIPS:
Never round off values close to whole number in order to get a simple ratio, but multiply
the value by a factor until you get a whole number

EXERCISE:
1. During physical activity, lactic acid (molar mass = 90.08 g/mol) forms in muscle tissue

and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0
mass % C, 6.7 mass % H and 53.3 mass % O. Determine empirical formula and
molecular formula of lactic acid.

2. Caproic acid, which is responsible for the foul odor of dirty socks. Combustion of 0.225g
sample of this compound produces 0.512 g CO2 and 0.209 g H2O. What is the empirical
formula of caproic acid. Caproc acid has a molar mass of 116g/mol, what is the molecular
formula?

36 | Chapter 3 Chemistry 1 DK014

Answer:
1. Empirical formula = CH2O, molecular formula = C3H6O
2. Empirical formula = C3H6O, molecular formula = C6H12O2

3.3 CONCENTRATION
▪ MOLARITY
⮚ Molarity (M) defined as the number of moles of solute dissolved in one liter
of solution
( )
, = ( )
⮚ Unit for molarity: mol L−1 @ mol dm−3 @ M

▪ DENSITY
⮚ Density is defined as the amount of mass in a unit volume of substance

( )
, = ( )
⮚ Unit for density: gmL−1 @ g/cm3

● % BY MASS
⮚ The ratio of the mass of a solute to the mass of the solution, multiplied by 100
percent

% = + 100
% = 100



❖ ADDITIONAL INFORMATION
● What is solute?

Solute is substance present in the smaller amount

● What is solvent?
Substance present in the larger amount

● What is solution?
Homogenous mixture of two or more substances

37 | Chapter 3 Chemistry 1 DK014

● DILUTION

⮚ Procedure for preparing less concentrated solution from a more concentrated
solution

⮚ moles of solute before dilution = moles of solute after dilution

MiVi = MfVf

Mi = initial molarity Mf = final molarity

Vi = initial volume Vi = final volume

dilution

EXAMPLE 6:
A matriculation student prepared a solution by dissolving 5.528 x 10-3 mol of sodium
carbonate, Na2CO3 in 250.0 cm3 of water. Calculate its molarity.

Answer:

( )
, = ( )

, = 5.528 10 − 3

0.25 ( )

= 0.0221 mol L-1

EXAMPLE 7:
A sample of 0.892 g of potassium chloride (KCl) was dissolved in 54.6 g of water. What is
the % by mass of KCl in this solution?

Answer:


% = 100

0.892
% = 0.892 + 54.6 100

= 1.61 %

38 | Chapter 3 Chemistry 1 DK014

EXAMPLE 8:
A 15.00−mL sample of 0.450 M K2CrO4 is diluted to 100.00 mL. What is the concentration

of the new solution?

Answer:

MiVi = MfVf

( ) = 0.450 (15.00 ) = 0.0675
100.00

EXAMPLE 9:
A sample of commercial concentrated hydrochloric acid is 11.8 M HCl and has a
density of 1.190 g/mL. Calculate the % by mass.

Answer:

( )
11.8 = ( )

( )
1.190 / = ( )

Assume volume of solution = 1.0 L,

11.8 = ( )

1.0

mole of HCl = 11.8 mol
mass HCl = 11.8 mol x (1 + 35.5) g/mol = 430.7 g
mass of solution = 1.190 / x 1000 ml = 1190 g

430.7
% = 1190 100

= 36.2 %

EXERCISE

1. A student prepared a solution of NaCl by dissolving 1.461 g of NaCl in a 250 mL
volumetric flask. What is the molarity of this solution. (molar mass Na = 23gmol-1, Cl =
35.5gmol-1)

2. How would you prepare 50.0mL of 0.2M HNO3 form a stock solution of 4.0 M HNO3 ?

3. A sample of 6.44g of naphathalene is dissolved in 80.1g of benzene. Calculate the mass
% of naphthalene in this solution.

4.An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.11g/mL. Calculate
its molarity.

Answer:

1.0.100M NaCl
2. 0.0025 L = 2.5mL
Transfer 2.5 mL of HNO3 solution into a 50.0mL volumetric flask and add sufficient water
until reach the calibrated mark.
3. 7.44%
4. 9.79 M

39 | Chapter 3 Chemistry 1 DK014

TUTORIAL 3

SECTION A
1. 0.224 L of H2 gas at STP is equivalent to

A. 1 mol
B. 1 g
C. 6.02 x 1023 molecules
D. 0.01 mol

2. 11.2 L of a gas at STP weighs 14 g. The gas could not be
A. N2
B. CO
C. N2O
D. B2H6

3. 5 L of 0.4M H2SO4 contain
A. 2.0 mol of H2O
B. 0.4 mol of H2SO4
C. 5.0 mol of H2SO4
D. 2.0 mol of H2SO4

4. A compound of Vanadium and Oxygen is found to be 56.04 percent by weight
Vanadium. What is the empirical formula of the compound?
A. V2O
B. VO2
C. V2O5
D. V2O3

5. Which of the following could be an empirical formula?
A. BaO2
B. H2O2
C. C2H4
D. C6H12O6

SECTION B
1. Define mole in terms of

(a) The mass of C-12

(b) Avogadro’s constant, NA

2. Calculate the number of moles for:

(a) 1.806×1022 molecules of nitrogen.

(b) 10.0 L O2 gas at STP.

40 | Chapter 3 Chemistry 1 DK014

3. Determine the number of:
(a) atoms in 5.0 g of silver, Ag
(b) molecules in 25 g of methane, CH4
(c) carbon atoms in 0.50 mol of ethane, C2H4

4. Quinine C20H24N2O2, is a compound extracted from cinchona tree which is traditionally
used to treat malaria. If given a 1.08 g of quinine sample, calculate:
(a) the molecular mass of quinine.
(b) the number of moles of quinine.
(c) the number of molecules of quinine.
(d) the number of hydrogen atoms.
(e) the mass of carbon atoms in gram.

5. Elemental analysis of a sample an ionic compound gave the following results; 2.82 g of
Na, 4.35 g of Cl and 7.83 g of O. What is the empirical formula of the compound?

6. A compound consists of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its
empirical formula? If the molecular weight of this compound is 180 g/mol , determine its
molecular formula.

7. The combustion of 11.5 g ethanol (CXHYOZ) produced 22.0 g CO2 and 13.5 g H2O.
Determine the empirical formula of ethanol from this experiment.

8. Define the molarity. What is the molarity of the solution, if 15 g NaCl occupy a volume of
75 mL

9. The effectiveness of a nitrogen fertilizer is determined mainly by its mass % N. Rank the
following fertilizers in terms of their effectiveness:
Potassium nitrate, ammonium nitrate, ammonium sulfate

10. 8.00 mass % aqueous solution of ammonia has a density of 0.9651 g/ml. Calculate the
molarity of NH3

Revised by:
Sharifah Fadthyah Binti Syed Baharuddin
Zanarina Binti Sapiai

41 | Chapter 3 Chemistry 1 DK014

CHAPTER 4: CHEMICAL EQUATION AND STOICHIOMETRY

GENERAL OVERVIEW

4.0 CHEMICAL 4.1 Chemical Determine chemical mass and
EQUATION AND equation oxidation equation by mole
STOICHIOMETRY number of inspection
4.2 element in a volume gas
Stoichiometry chemical method at RTP and
formula
Write and redox STP
balance equation by
ion-electron volume and
Calculate concentration
amount of method
reactant and of solution
calculation
product involve

Define: limiting
reactant
Perform
calculation theoretical
involving: yield

actual yield

percentage
yield

limiting
rectant

percentage
yield

42 | Chapter 4 Chemistry 1 DK014

4.1 CHEMICAL EQUATION

Oxidation Numbers
▪ A positive or negative whole number assigned to an element in a molecule or ion on the basis

of a set of formal rules; to some degree it reflects the positive or negative character of that atom.
▪ Each atom in a neutral subtance or ion is assigned an oxidation number
▪ Rules for assigning oxidation number:

1. For an atom in its elemental form, the oxidation number is always zero. Thus,
oxidation number of Na atom is 0 ,each H atom in H2 molecule is 0, each Cl atom in Cl2
molecule is 0, and each P atom in P4 molecule is 0.

2. For any monoatomic ion, the oxidation number equals the ionic charges. Thus,
oxidation number of K in K+ ion is +1 and oxidation number of S in S2- ion is -2.

3. Nonmetals usually have negative oxidation number, although they sometimes be
positive:
i) Oxidation number of oxygen is usually -2 (except in peroxide, oxidation number
of oxygen is -1)
ii) Oxidation number of hydrogen is +1 when bonded to nonmetals and -1 when
bonded to metals.
iii) Oxidation number of fluorine is -1. Oxidation number of other halogens is -1
except in oxyanions (halogen combine with oxygen), they have positive oxidation
state.

4. In a neutral compound, the total oxidation number of each atoms that made up the
molecule is zero.

5. The oxidation number of polyatomic ions, the total oxidation number of each atom that
made up the ion is equal to the net charge of the ion.

EXAMPLE 1:
What is the oxidation number of chromium, Cr in;
a. Cr metal
b. Cr3+
c. CrO3
d. Cr2O72-

Answer:
a. Oxidation number of Cr atom in its elemental form is 0

b. Oxidation number of Cr in Cr3+ = +3

c. Oxidation number of CrO3 = 0

Cr + 3(-2) = 0
Cr = +6

d. Oxidation number of Cr2O7 = -2
2 (Cr) + 7(-2) = -2
Cr = +6

43 | Chapter 4 Chemistry 1 DK014

Balancing chemical equation (inspection method)
▪ Chemical equation is a representation of a chemical reaction using the chemical formulas of

the reactants and products.
▪ The chemical equation need to be balanced as to determine the amount of product that can

be made or the amount of reactant that is required.
▪ For example, when gas hydrogen burns, it react with oxygen in air to form water. The

chemical equation for this reaction is:

2 H2 + O2  2 H2O
o We read the + sign as “react with” and the arrow as “produces”
o The chemical formula at left-hand side of the arrow is reactant and right-hand side is

product.
o The number in the front of the formulas, called coefficient
o Because atoms are neither created or destroyed in any reaction, a balance reaction

must have equal number of atoms of each element on each side of the arrow (law of

conservation of mass).
▪ How to construct a balanced chemical equation:

1. Write down the unbalanced equation. Write the correct formulae for the reactants on the

left-hand side of the arrow and products on the right-hand side.

2. Balance element/atom by determining the coefficients that provide equal numbers of each

type of atom on the both sides of the equation
o Balance the metallic element, followed by non-metallic atoms
o Balance the hydrogen and oxygen atoms

3. Check to ensure that the total number of atoms of each element is the same on both sides

of equation.

EXAMPLE 2:
Balance this equation.
Fe2O3 + HCl → FeCl3 + H2O

Answer:

Step 1 Fe2O3 + HCl → FeCl3 + H2O
Write down the unbalanced equation
Fe2O3 + HCl → 2 FeCl3 + H2O
Step 2 Fe2O3 + 6 HCl → 2 FeCl3 + 3 H2O
Balance element/atom by determining the
coefficients that provide equal numbers of each Element Left Right
type of atom on the both sides of the equation Fe 2 2
o Balance the metallic element, followed by Cl 6 6
O 33
non-metallic atoms H 66
o Balance the hydrogen and oxygen atoms

Step 3
Check to ensure that the total number of atoms
of each element is the same on both sides of
equation

44 | Chapter 4 Chemistry 1 DK014

Balancing redox equation (ion-electron method)
▪ Redox reaction is a reaction that involves both reduction and oxidation reactions. It

involves transfering of electrons from one reactant to another and the oxidation states of
these atoms change.
▪ There are additional requirement in balancing redox reaction: the gains and losses electrons
must be balanced. If a substance loses a certain number of electron during the reaction,
another subtance must gain that same number of electron. (Electrons are neither created nor
destroyed in any chemical reaction.)

Half-reaction
▪ Half-reaction is refering to the equations that show either oxidation or reduction alone.
▪ Oxidation is a process in which a substance loses one or more electrons; results an increase

in an oxidation number. Substance that being oxidised is called reducing agent.
▪ Reduction is is a process in which a substance gains one or more electrons; results a

decrease in an oxidation number. Substance that being reduced is called oxidising agent.

▪ In the overall redox reaction, the number of electrons lost in the oxidation half-reaction must
equal to the number or electrons gained in the reduction half-reaction.

▪ How to balance redox reaction in acidic aqueous solution:
1. Separate the equation into two half-reactions.
2. Balance each half-reaction.
o Balance the elements other than O and H
o Add H2O to balance O atom
o Add H+ to balance H atoms
o Add e- to balance the charges
3. Multiply each half-reaction by an integer, so that number of electron lost in one half-
reaction equals the number gained in the other.
4. Combine the two half-reactions and cancel out the species that appear on both sides of
the equation.
5. Check to ensure the number of atoms and the charges are balanced.

▪ How to balance redox reaction in basic aqueous solution:
1. Balance the half-reactions as if they occurred in acidic solution
2. Count the number of H+ in each half-reaction, and then add the same number of OH- to
each side of the half-reaction. (H+ + OH- → H2O). the resulting water molecules can be
canceled as needed.

45 | Chapter 4 Chemistry 1 DK014

EXAMPLE 3:

Complete and balance this equation by ion-electron method:
Cr2O72- (aq) + Cl- (aq) → Cr3+ (aq) + Cl2 (g) (acidic solution)

Answer: +6 reduction +3

Cr2O72- (aq) + Cl- (aq) → Cr3+ (aq) + Cl2 (g)

-1 oxidation 0

Step 1 Oxidation: Cl- (aq) → Cl2 (g)
Separate the equation into two Reduction: Cr2O72- (aq) → Cr3+ (aq)
half-reactions
Step 2 2Cl- (aq) → Cl2 (g)
Balance each half-reaction. 2Cl- (aq) → Cl2 (g) + 2e-
o Balance the elements other
Cr2O72- (aq) → 2Cr3+ (aq)
than O and H Cr2O72- (aq) → 2Cr3+ (aq) + 7H2O (l)
o Add H2O to balance O 14H+ (aq) + Cr2O72- (aq) → 2Cr3+ (aq) + 7H2O (l)
6e- + 14H+ (aq) + Cr2O72- (aq) → 2Cr3+ (aq) + 7H2O (l)
atom
o Add H+ to balance H

atoms
o Add e- to balance the

charges

Step 3 (2Cl- (aq) → Cl2 (g) + 2e- )  3
Multiply each half-reaction by 6Cl- (aq) → 3Cl2 (g) + 6e-
an integer, so that number of
electron lost in one half- 6Cl- (aq) → 3Cl2 (g) + 6e-
reaction equals the number 6e- + 14H+ (aq) + Cr2O72- (aq) → 2Cr3+ (aq) + 7H2O (l)
gained in the other. 6Cl- (aq) + 14H+ (aq) + Cr2O72- (aq) → 3Cl2 (g) + 2Cr3+ (aq) + 7H2O (l)
Step 4
Combine the two half-reactions Element/charge Left Right
and cancel out the species that Cl 6 6
appear on both sides of the H 14 14
equation. Cr 2 2
Step 5 O 7 7
Check to ensure the number of +6 +6
atoms and the charges are Charge
balanced.

EXAMPLE 4:

Complete and balance this equation by ion-electron method:
CN- (aq) + MnO4- (aq) → CNO- (aq) + MnO2 (s) (basic solution)

Answer: oxidation +4

+2

46 | Chapter 4 Chemistry 1 DK014

+7 reduction +4