This document provides an introduction and table of contents for a chemistry course book on Cambridge International AS and A Level Chemistry. It covers topics like the mass of atoms and molecules, relative atomic masses, isotopic masses, amount of substance, mole calculation, chemical formulae, solutions, gas volume calculations, and more. The document gives definitions and examples for these concepts. It also provides sample problems and homework questions related to chemical calculations involving moles, masses, and chemical equations.

1. Cambridge International AS and A Level
Chemistry
course book
Second Edition
By:
Mr. Farid Gul “Shinwari
Master in Chemistry
Contact me by
faridgul93@gmail.com
Class 11th

2. Table of Contents
Introduction
1. Mass of atoms and molecule
2. Relative atomic masses
3. Relative isotopic masses
4. Amount of substance
5. Mole calculation
6. Chemical formulae and chemical equations
7. Solutions and concentration
8. Calculation involving gas volumes

3. Introduction
For thousands of years, people have heated rocks and
distilled plant juices to extract materials. Over the past
two centuries, chemists have learnt more and more
about how to get materials from rocks, from the air and
the sea and plants. They have also found out the right
conditions to allow these materials to react together to
make new products such as medicines, plastics, dye,
engineering materials etc,
When we make a new substance it is important to mix
the reactants in the correct amounts in order to avoid
their wastage and quality.
In order to do this we need to know about relative
masses of atoms and molecules and how these are used
in chemical calculations.

4. Mass of atoms and molecules
Relative atomic masses (Ar):
The mass of an atom of an element compared with the mass of
one atom of carbon-12 taken as 12 units is called relative
atomic mass
OR
the relative atomic mass of an element is the number of times
one atom of that element is heavier than 1/12th of one atom of
carbon-12
Explanation:
Atoms of different elements have different masses. When we
perform chemical calculations we need to know the correct
mass of atoms.
For this purposes we compare this mass with the mass of the
same number of standard atoms that is carbon-12.

5. Mass of atoms and molecules
Relative isotopic mass
the mass of an atom of an isotope compared with one-twelfth
of the mass of an atom of carbon-12
For example, the relative isotopic mass of carbon-13 is 13
Example;
Carbon-12 and Carbon-13
Chlorine-35 and Chlorine-37

6. Mass of atoms and molecules
What is the relative atomic mass of chlorine?
If we know both the natural abundance of every isotope of an element
and their isotopic masses, then we can calculate the relative
atomic mass of the element correctly
In general
R.A.M. = (R.I.M.1 x abundance1) + (R.I.M.2 x abundance2)
100
75.5% of chlorine atoms are chlorine-35
24.5% of chlorine atoms are chlorine-37
Relative atomic mass = (75.5 x 35 + 24.5 x 37)
100
= 35.5

7. Mass of atoms and molecules
Molecular mass
The sum of atomic masses of all the atoms present in a
molecule is called molecular mass.
Example; Molecular mass of methane (CH4) = 12 + (4)(1) = 16
Hence molecular mass of methane = 16 a.m.u.
Relative molecular mass
Relative molecular mass of a compound is the number of times
one molecule of it is heavier than 1/12th of C-12.
Compounds Relative molecular mass
Hydrochloric acid 36.5
Oxygen molecule 32

8. Mass of atoms and molecules
Formula unit
The smallest repeating unit of an ionic compound is called
formula unit.
Example,
Calcium Chloride CaCl2
Cupper sulphate, CuSO4

9. Mass of atoms and molecules
Formula mass
The sum of atomic masses of all the atoms present in a
formula unit of an ionic compound is called formula mass.
Example;
Formula mass of sodium chloride(NaCl) = 23+35.5 = 58.5
Relative Formula mass
Relative formula mass is the number of times formula mass
heavier than 1/12th of one atom of carbon-12
Ionic compound Relative formula mass
Calcium carbonate 100

10. Mass of atoms and molecules
Question H.W.
1. Use the periodic table to calculate the relative formula
masses of the following:
a. Calcium Chloride, CaCl2
b. Cupper sulphate, CuSO4
c. Magnesium Nitrate, Mg(NO3) 2.6H2O
d. Sodium chloride

11. Hydrated and anhydrous compounds
Hydrated compounds
Compounds which contains a definite number of moles of
water in their structure is called hydrated compound.
For example;
CuSO4.5H2O
CoCl2.6H2O
Anhydrous compounds
Compounds which contains no moles of water in their
structure is called anhydrous compound.
For example;
CuSO4
CoCl2.

12. Hydrated and anhydrous compounds
Water of crystallization
Specific number of water molecules associated with a
crystal structure of a compound is called water of
crystallization.
When writing chemical formula for hydrated compounds we
show water of crystallization separated from main formula
by dot.
HOME WORK BOOK PAGE 41
Question:1 a.b.c.d

13. 3.3 Accurate relative atomic mass
Mass spectrometry
An analytical tool which is used to
measure the atomic mass of each isotope
present in an element, or molecular mass
Of molecule.
It also find out %age of each isotope
in an element.
Working
Relative isotopes abundance
The proportion of one particular isotope
in a mixture of isotopes, usually
expressed as a percentage.
The heights of the peaks in a mass spectrum show the
percentage of each isotope present.

14. 3.3 Accurate relative atomic mass
Mass spectrometry

15. Amount of substance
The Mole and Avogadro Number (NA)
Mole
The amount of substance that contains 6.022×1023 particles.
Mole can be represented by symbol ‘mol’ ‘ n’
Examples
1 mole of sodium (23g) = 6.022×1023 atoms
1 mole of carbon dioxide(44g) = 6.022×1023 molecules
Calculating mole
The number of mole of a substance can be find out by using the
following equation
Number of mole (mol) = mass of substance(g)/molar mass
Molar mass:
The mass of one mole of a substance is represented by symbol M

16. Amount of substance
WORKED EXAMPLE,1
Calculate the amount of substance in moles in10.7g of
Sulphur atoms.
Solution, As we know that mol = mass/molar mass
moles of Sulphur atoms in 10.7g = 10.7g/32 = 0.333mol
So moles of Sulphur atoms in 10.7g = 0.3333mol
HOME WORK
Question 3 page 46
a. Use these Ar values (Fe=58, N=14 ,O=16, S=32) to calculate
the amount of substance in moles in each of the following:
i. 10.7g of Sulphur atoms
ii. 64.2g of Sulphur molecules (S8)
iii. 60.45g of anhydrous iron nitrate Fe(NO3)3.

17. Amount of substance
Calculating mass from mole equation
To find out the mass of a substance present in a given number
of moles, we need to rearrange the equation
Number of mole (mol) = mass of substance(g)/molar mass
mass of substance(g)= Number of mole (mol) × molar mass
Example; 2
What mass of sodium hydroxide(NaOH), is present in 0.25 mol
of sodium hydroxide?
Solution

18. Amount of substance
Example 1, P=6
Calculating mole How many moles of sodium chloride are
present in 117g of sodium chloride (NaCl)?
(Ar value: Na = 23, Cl = 35.5)
Solution

19. Amount of substance
Avogadro’s number (6.022 x 1023)
The number of atoms, molecules, formula unit in one mole of a
substance is called Avogadro's number.
It is represented by symbol NA
NA = 6.022 x 1023
Examples;
1 mole of carbon = 6.022 x 1023 C atoms
1 mole of water = 6.022 x 1023 molecules
1 mole of sodium chloride = 6.022 x 1023 Formula unites
Number of particles= Number of moles × Avogadro number
N = n × NA or N = m/M × NA

20. Amount of substance
Question 3
Example; 2
calculate the total number of atoms in 7.10g of chlorine
atoms(Ar value: Cl=35.5).
Solution,
Number of moles = mass of substance(g)/molar mass
Number of mole (mol) = 7.10/35.5 = 0.2mol
Number of particles= Number of moles × Avogadro number
Number of particles= 0.2mol× 6.022 x 1023 = 12.044 x 1022
Total No. of Cl. Atoms in 7.10g = 12.044 x 1022

21. Mole calculation
Reacting masses
The chemical equation shows us ratio of moles of the
reactants and products (stoichiometry).
For example, In the reaction
Fe2O3 + 3CO 2Fe + 3CO2
In the above reaction 1 mole of Fe2O3 reacts with
3moles of CO to form 2moles of Fe & 3moles of CO2.
The ratio of this equation is 1:3:2:3.

22. Mole calculation
Example;1
Magnesium burns in oxygen to form magnesium oxide.
2Mg + O2 2MgO
We can calculate the mass of oxygen needed to react with 1 mole of magnesium. We
can calculate the mass of magnesium oxide formed.
Step1. Write the balanced equation.
Step2. Multiply each formula mass in g by the related stoichiometry number in the
equation.
2Mg + O2 2MgO
2×24g 1×32g 2×(24g+16g)
48g 32g 80g
From this calculation we can deduce that:
32g of oxygen are needed to react exactly with 48g of magnesium to form
80g of magnesium oxide.

23. Mole calculation
Example;2
Iron oxide reacts with carbon monoxide to form iron and
carbon dioxide.
Fe2O3 + 3CO 2Fe + 3CO2
Calculate the maximum mass of iron produced when
798g of iron oxide is reduced by excess carbon
monoxide.(Ar values, Fe = 58.8, O = 16)
Step1 Fe2O3 + 3CO 2Fe + 3CO2
Step2 1mole Fe2O3 2moles Fe
(2×56)+(3×16) 2×56
159g Fe2O3 111g
Step3 798g 111/159×798 = 557g Fe

24. Mole calculation
Question,5
a. Sodium react with excess of oxygen to form sodium peroxide
Na2O2.
2Na + O2 Na2O2
Calculate the maximum mass of sodium peroxide formed
when 4.60g of sodium is burnt in excess oxygen.

25. Mole calculation
Percentage composition by mass
Percentage( %age ) composition means the number
of parts by mass of an element in 100 parts by mass
of a compound.
% of an element = Atomic mass__ × 100
Molecular mass
The % age composition of water can be calculated as:
% of Hydrogen = 2 × 100 = 200 = 11.11 %
18 18
% of Oxygen = 16 ×100 = 1600 = 88.88%
18 18

26. Mole calculation
Percentage composition by mass
Example,1
Calculate the composition by mass of iron in iron oxide.
Fe2O3 (Ar values Fe = 56, O = 16).
Example,2
Calculate the composition by mass of carbon in ethanol.
C2H5OH (Ar values C = 12, H = 1, O = 16).

27. Percentage calculation
Percentage yield
In many chemical reactions, especially organic reactions, not
all the reactants are changed to the products you want.
This is because there are other reactions carried out at the
same time; reactants or products are lost to the atmosphere;
or reactions does not go to completion.
The percentage yield tells you how much of a particular
product you get from the reactants compared with the
maximum theoretical amount that you can get.
Percentage yield = actual yield/predicted yield × 100
The actual yield is the moles or mass of product obtained by
experiment.
The predicted yield is the moles or mass of product obtained
by calculation if no side products are formed and all of a
specific reactant is converted to a specific product.

28. Chemical formula
The collection of symbols of each element present in a
molecule of a substance is called chemical formula.
Oxygen O2
Hydrogen H2
Ammonia NH3
Glucose C6H12O6
Water H2O

29. Chemical formula

30. Chemical formula
Empirical Formula
The chemical formula which shows simplest ratio b/w the atoms
present in a molecule of a compound is called Empirical Formula.
Examples
 All Ionic compounds are represented by their E.F.
 The Empirical and molecular formula for simple inorganic
molecules are often the same.
 Organic molecules often have different Empirical and
molecular formula
Compounds Emp. formula Molecular formula
Water H2O H2O
Hydrogen peroxide HO H2O2
Sulphure dioxide SO2 SO2
Butane C2H5 C4H10
cyclohexane CH2 C6H12

31. Chemical formula
Empirical Formula
Question: p = 11 H/W
Write the empirical formula for the following compounds
Compounds Emp. formula Molecular formula
Hydrazine N2H4
octane C8H18
Benzene C6H6
Ammonia NH3
Cyclo pentane C5H10

32. Chemical formula
Empirical Formula
Deduce empirical formula by using %age composition:
The E.F. can be find out by the steps given below.
Step 1. Note the % by mass of elements
Step 2. divide by Ar values of elements
Step 3. Divide by the lowest figure.

33. Chemical formula
Empirical Formula
Example 1: P=52
A compound of carbon and hydrogen contains 85.7% C
and 14.3% H by mass. (Ar values C=12, H=1)
Deduce the empirical formula of this hydrocarbon.
Solution C H
Step 1. % by mass of element 85.7 14.3
Step 2.÷by Ar values of element 85.7/12=7.142
14.3/1=14.3
Step 3.÷by the lowest figure 7.142/7.142 =1
14.3/7.142= 2
Empirical formula is CH2.

34. Chemical formula
Empirical Formula
Question H/W
Example 2: P=52
The composition by mass of a hydrocarbon is 10%
hydrogen and 90% carbon. Deduce the empirical
formula of this hydrocarbon.

35. Chemical formula
Molecular Formula
The chemical formula which shows the actual number
of atoms present in a molecule of a compound is
called Molecular Formula.
Compounds Molecular formula
Water H2O
Hydrogen peroxide H2O2
Sulphure dioxide SO2
Butane C4H10
cyclohexane C6H12

36. Chemical formula
Molecular Formula
Deduce the molecular formula:
In order to deduce molecular formula we need to know;
 The relative formula mass of the compound.
The empirical formula.

37. Chemical formula

38. Chemical formula

39. Oxidation state
The apparent charge on an atom in a molecule or
compound is called oxidation state or number.
For example;
Oxidation state of Na+1 , Ca+2 , Al+3 are +1, +2, +3
And Br-1 , O-2 are -1 and -2.
Importance of oxidation number
1. It is useful in naming compounds
2. In writing chemical formula
3. In balancing chemical equation.
The colors of solution changes with the change of
oxidation state such as;
Cr+2 = Blue, Cr+3 = Green, Cr+6 Orange

40. Rules for assigning oxidation number
1. the oxidation number of all the elements in free state is zero.
For example; H2, Cl2 , O2 , Na and Fe
2. The oxidation number of group IA,IIA and IIIA is +1,+2,+3
3. The oxidation number of halogen atom is -1.
4. The oxidation number of hydrogen is +1
5.The oxidation number of oxygen is -2 in its compounds.
But in peroxides is -1 e.g Na2O2 , CaO2
and in superoxide is -1/2 e.g KO2 , RbO2
6. The sum of the oxidation number of all the atoms in a
compound is zero.

41. Electrochemistry
Example:1
Calculate the oxidation number of Cr in K2Cr2 O7
Oxidation number of K = +1
Oxidation number of O = -2
Oxidation number of Cr =?
Putting these values in formula K2Cr2 O7
2(+1) +2(x) + 7( -2) = 0
2 + 2x (-14) = 0
2x + 2 -14 = 0
2x -12 = 0 =>
2x = 12 by dividing 2 on both sides we get
x = 12 = 6
2
X is Cr whose oxidation number is 6

42. Calculate the oxidation number
Example 2
Find the oxidation state of metals in the following species.
A. FeO B. KMnO4 C. Na2Cr2O7 D. HgOH
Solution
A. x + -2 = 0 → x = +2
B. +1 + x + 4(-2) = 0 → x = +7
C. 2(+1) + 2x + 7(-2) = 0 → x = +6
D. x + (-2) + 1 = 0 → x = + 1

43. Calculate the oxidation number
Example 3
Find the oxidation state of indicated atoms in the following
species.
A. PO4
-3 B. CO3
-2 C. K2CrO4 D. NH4
+1
Solution
A. x + 4(-2) = -3 → x = +5
B. x + 3(-2) = -2 → x = +4
C. 2(+1) + x + 4(-2) = 0 → x = +6
D. x + 4(+1) = +1 → x = -3

44. Chemical formula
How to write chemical formula for ionic compound
1) Write the correct symbols of ions side by side.
2)Write the +ve ions on left hand side &-ive ions on right hand side.
3) Balanced the no. of +ve charges by the no. of –ve charges.
4) Write the valency on the right top side of each ions.
5) Write the polyatomic radicals in brackets.
6)Cross the valencies to the lower right of each radicals.
7) if the valencies are equal then don't cross it.

45. Chemical formula
Write the chemical formula for the molecules of the
following compounds.
Magnesium Chloride
Step 1. Magnesium chloride.
Step 2. Mg Cl
Step 3. Mg+2 Cl-1
Step 4. Mg+2 Cl-1
Step5. Formula = MgCl2

46. Chemical formula
Step 1. Aluminum sulphate
Step 2. Al SO4
Step 3. Al+3 SO4
-2
Step 4. Al+3 SO4
-2
Step 5. = Al2(SO4)3

47. Chemical equation
Representation of a chemical reaction in terms of symbols
and formula is called a chemical equation.
in general,
Example
When zinc metal reacts with Hydrochloric acid, Zinc
Chloride (ZnCl2) is formed and H2 is given out.
Zns + 2HClaq ZnCl2(aq) + H2
Symbol State of substances
S Solid
l Liquid
g Gas
aq Aqueous solution

48. Chemical equation
1. Write down the correct symbols/ formulae of the
reactants and products.
2. When the number of atoms of an element is
different on both sides, a suitable number is put
before the formula.
3. Start with relatively small numbers.
4. Work with one element at one time.
5. The diatomic molecules are balanced at the last
like H2, N2, and O2 etc.

49. 1. CaCl2(aq) + Na2CO3(aq) CaCO3(aq) + NaCl(aq)
Reactants Products
1 Ca 1 Ca
2 Cl 1 Cl
2 Na 1 Na
1 C 1 C
3 O 3 O
So on the right hand side putting 2 before NaCl to balance
the equation.
CaCl2(aq) + Na2CO3 (aq) CaCO3 + 2NaCl(aq)
Balancing chemical equation

50. 2. N2(aq) + H2 --------------------> NH3(g)
Reactants ---------------------> Products
2N 1N
2H 3H
Putting 2 with NH3 on right hand side to
balance N.
Putting 3 as coefficient before H2 on left
side balances H atom as well.
N2(aq) + 3H2 --------------------> 2NH3(g)
Now,
2N 2N
6H 6H
Balancing chemical equation

51. Solution
Homogeneous mixture of solute and solvent is called
solution.
Example,
Solute + Solvent Solution
Sugar + water Sugar solution.
Solution and concentration

52. Concentration
The amount of solute present in a given quantity of solution.
Concentrations are usually expressed in term of molarity,
Molarity
The number of moles of solute in one liter of solution.
Molarity(Concentration) = mol / liter(dcm3)
Solution and concentration

53. Calculating the concentration of a solution
when performing calculation involving concentration in moldm-3
you need to:
 Change mass in grams to mole.
 Change cm3 to dm-3 by dividing the number of cm3 by 1000.
Solution and concentration

54. Calculating the concentration of a solution
Solution and concentration

55. Calculating the concentration of a solution
Solution and concentration

56. Calculating mass of a substance in a solution
Solution and concentration
We often need to calculate the mass of a substance present in
a solution of known concentration and volume.
To do this we need to;
 Multiplying moles of solute by molar mass.
Mass of solute =number of moles(mol) × molar mass(g/mol).
m = n ×M

57. Calculating mass of a substance in a solution
Solution and concentration

58. Calculating number of moles of solute in a solution
Solution and concentration
The number of moles of a solute dissolved in a solution can
be calculated by using concentration (molarity) equation.
M(concentration) = Number of moles / Volume(L or dm3)
In order to calculate number of moles lets rearranged the eq.
Number of moles= M(concentration) × Volume(L or dm3)

59. Solution and concentration
Titration
Titration is a process where a solution of known
concentration is used to determine the concentration of
an unknown solution.
Ma× Va = Mb × Vb
Standard solution
A solution whose volume and Concentration are
known is called titrant or Standard solution.
End-point or Equivalence point
The neutralization point (color change) is known as an
end-point.

60. Calcualtions involving gas volumes
Using the molar gas volume
Avogadro's hypothesis:
It states that equal volumes of all gases at the same
temperature and pressure contain equal numbers of particles.
Examples;
volume of 1 mole of methane = 24dm3
volume of 1 mole of Carbon dioxide = 24dm3
volume of 1 mole of hydrogen = 24dm3

61. Calculations involving gas volumes
Using the molar gas volume
Calculating volume of gas at S.T.P.
Volume of gas = moles of gas × 24 L/mol
Moles of gas = Volume of gas/ 24 L/mol
Mass = moles × molar mass
Example;1
Calculate the volume of 0.40mol of nitrogen at r.t.p.
Solution,
volume in dm3 (L) = 24 × moles of gas
volume = 24 × 0.40
= 9.6 dm3 (L)

62. Calculations involving gas volumes
Using the molar gas volume
Calculating mass of gas at S.T.P.

63. Calculations involving gas volumes
Using the molar gas volume
Calculating volume of gas at S.T.P.