GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
Grade 11
SESSION 11
(LEARNER NOTES)
MOLE CONCEPT, STOICHIOMETRIC CALCULATIONS
Learner Note: The mole concept is carried forward to calculations in the acid and base
section, as well as in the chemical equilibrium. It is important to know this section well.
Question 1: 5 minutes (Taken from DoE Nov 2007)
(Revise calculations of molecular mass. Remember chemical calculations need
formulae, substitution and an answer with the correct unit.)
The compound NaHCO3 is commonly known as baking soda. A recipe requires 1,6 g of
baking soda, mixed with other ingredients, to bake a cake.
1.1
1.2
Calculate the number of moles of NaHCO3 used to bake the cake.
How many atoms of oxygen are there in1,6 g baking soda?
(3)
(4)
(7)
Question 2: 10 minutes (Taken from MED Nov 2009)
(The sum of all the percentages equals…yes, 100. Therefore, in a 100 g of the
substance the ratio will be the same. Learn the “method” the steps are always
repeated. The elements are mostly given in the same order as they appear in the
formula)
One of the active ingredients in vinegar is Ethanoic acid. Ethanoic acid has a molecular mass
of 60 g.mol-1 and the following percentage composition
39,9 % carbon
6,7 % hydrogen
53,4 % oxygen
2.1
2.2
2.3
Define the concept empirical formula
Determine the empirical formula of Ethanoic acid
What is the molecular formula of Ethanoic acid?
(2)
(5)
(3)
(10)
Question 3: 10 minutes (Adapted from MED Nov 2009)
The contact process is given by the equation below.
SO2 (g) + O2 (g) → SO3 (g)
3.1
Balance the chemical equation
(2)
In an investigation 256 g SO2 reacts with 80 g O2 in a reaction vessel.
3.2
Calculate the number of moles of each reactant present at the start of the reaction(5)
3.3
Identify the limiting reagent in the reaction and justify your answer.
(2)
(Limiting reagents are frequently asked – the one that limits the reaction is the one that
will be used up first. You must first work out the number of moles represented by the
given masses of the reactants, then determine the limiting reagent by using the moll
ratio)
3.4
Calculate the mass of SO3 produced in the reaction
(4)
(13)
Page 1 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
Grade 11
SESSION 11
(LEARNER NOTES)
Question 4: 5 minutes
(It is a common mistake to interpret the “dot” as a multiply function. This is the water
that is trapped in the crystal during crystallisation and must be ADDED to the mass of
the ionic compound.)
4.1
Calculate the percentage water of crystallisation in CuSO4 . 5 H2O
4.2
Calculate the concentration of a 250 ml solution of sodium hydroxide if 10 g of the
solute is dissolved.
(4)
(8)
SECTION B:
(4)
SOLUTIONS AND HINTS
Question 1
Learner Note: Emphasize the correct use of formulae and layout. Use the correct
unit
1.1
M (NaHCO3)
= 23 + 1 + 12 + 3(16)
= 84 g.mol-1 
m
M
1,6
n =

84
n =
= 0,02 mol  (rounded to 2 decimal places)
1.2
Each atom has three oxygen atoms, there is 0,02 mol of atoms
1 mol = 6,023 x 1023 particles
therefore 0,02 mol  x 3 atoms x 6,023 x 1023 particles = 3,44 x 1022 oxygen
atoms
Page 2 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
Grade 11
SESSION 11
(LEARNER NOTES)
Question 2
Learner Note: If the steps are followed, the questions become relatively simple.
Remember we are working out what the ratio of the elements are in the molecule
therefore when you determine the number of mol the element mass is usedregardless of whether the element is diatomic!
2.1
Empirical formula – the simplest ratio of atoms in a molecule
Elements
In 100g
Convert mass to
mol
m
n =
M
C
39,9 g
H
6,7 g
39,9
12
6, 7
1
= 3,325 mol
= 6,7 mol
3.325
3.325
6,7
3.325
3.3375
3.325
=1
= 2,01
=1
1
2
Divide by
smallest answer
Ratio of
elements in the
empirical
formula
O
53,4 g

53,4
16
= 3,3375 mol
1


Empirical formula CH2O 
2.3
M (CH2O)
= 12 + 2(1) + 16
= 30 g.mol-1 
Molecular mass is double empirical formula mass therefore molecular formula is
C2H4O2 
Question 3
(It is important to balance chemical equations before doing the calculations. The only
time where the balancing coefficients are used will be in the ratio step.)
3.1
2 SO2 (g) + O2 (g) → 2 SO3 (g)
3.2
n =
m
M
n =
256
64
n = 4 mol SO2
n =
m
M
n =
80
32
n = 2,5 mol O2
Page 3 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
3.3
3.4
Grade 11
SESSION 11
(LEARNER NOTES)
A ratio of 2 mol SO2 is needed for 1 mol O2 according to reaction. Therefore 2,5 mol
O2 needs 5 mol SO2 to react completely, the SO2 is therefore the limiting reagent
4 mol SO2 reacts and 4 mol SO3 is produced
n =
m
M
4 =
m

80
m = 320 g SO3 made 
Question 4
4.1
M (CuSO4 . 5 H2O)
= 63,5 + 32 + 4(16) + 5 (1 + 1 + 16)
= 159,5 + 90
= 249,5 g.mol-1 
90
x 100 
249 ,5
= 36 % water 
% water =
4.2
c =
c =
m
MV
10
40x0,25
c = 1 mol.dm-3
Page 4 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
Grade 11
SESSION 11
(LEARNER NOTES)
SECTION C: ADDITIONAL CONTENT NOTES
The Mole
The mole is one of the base units of the SI and is the base unit of the amount of matter of
substance.
A mole of any substance always has the same number of atoms.
Definition: A mole of any substance is that amount of substance which contains as many
elementary particles as there are atoms in 12g of carbon-12.
The amount of matter or substance is not the same as the mass of the substance. While one
mole of a substance contains the same number of particles, their masses are not the same.
The formula used to calculate the number of moles in a substance:
m
In symbols
n=
M
Avogadro’s Number
One mole of any substance is approximately equal to 6,02x10 23 elementary particles.
This very large number is known as Avogadro’s number or Avogadro’s constant and has the
symbol NA or L.
Therefore: Number of particles = Avogadro’s number x number of moles
In symbols Np = NA x n
Molar Volumes of Substances
Molar volume is the volume of one mole of a substance and can be measured in dm 3/mol.
One mole of any gas occupies a volume of approximately 22,4dm 3 at Standard Temperature
and Pressure (STP). Standard Temperature is 0oC (273K) and Standard Pressure is
101,3kPa.
Equal volumes of all gases at STP contain the same number of molecules.
Therefore: molar volume = volume of substance
Number of moles of substance
n =
V
22,4 dm3
Empirical formula – smallest ratio of atoms in a molecule
True Formula or molecular formula – actual ratio of atoms in a molecule eg. the Empirical
Formula of a substance is COH2 but its molecular formula is C2O2H4
Percentage Composition – shows percentage of each element in a compound compared to
its molecular mass
Page 5 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
Grade 11
SESSION 11
(LEARNER NOTES)
SECTION D: HOMEWORK
Learner Note: Revise steps for each type of calculation and follow the prescribed method.
Attempt all questions and refer to notes when in doubt.
Question 1: 25 minutes
1.1
1.2
1.3
Calculate the relative formula mass of KCℓO3
(3)
Calculate how many times a molecule of methanol (CH3OH) is heavier than a molecule
of water
(5)
Calculate the empirical formula of the substance with the following composition
1.4
45,3 % O; 43 % Na; 11,3 % C
How many potassium atoms are there in 2 g K2SO4
1.5
Fe + S → FeS
(5)
(5)
Which of the two substances will be used up if 10 g Fe and 10 g S are mixed and
heated
(7)
(25)
Page 6 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
Grade 11
SESSION 11
(LEARNER NOTES)
SECTION E: SOLUTIONS TO SESSION 10 HOMEWORK
1.1
When H atoms approach each other, the valence electron of the two atoms attract the
nucleus of the other atom,  these attractive forces are stronger than the repulsive
forces between the atoms.  The protons and electrons of respective atoms attract the
atoms to form the bond. The atoms move closer together and the potential energy
becomes negative. The atoms are most stable at the lowest value of potential energy –
when the orbitals overlap and bonding occurs.  The two hydrogen atoms each share
an electron during bonding; there is a net electrostatic force of attraction between the
atoms. H2 is formed. 
When an H atom approach oxygen atom, the valence electron of the atoms attracts the
nucleus of the other atom,  these attractive forces are stronger than the repulsive
forces between the atoms. The protons and electrons of respective atoms attract the
atoms to form the bond. The atoms move closer together and the potential energy
becomes negative. The atoms are most stable at the lowest value of potential energy –
when the orbitals overlap and bonding occurs. The hydrogen and oxygen atom
share an electron during bonding, there is a net electrostatic force of attraction
between the atoms. H2O is formed. 
1.2
Both H – atoms require an electron to fill the orbital and obtain the noble structure
which is of lower energy. Its valence energy level is not filled.  The H – atoms share
an electron pair, there is a net electrostatic force of attraction, bonding occurs 
1.3
The Helium atom is in the noble state, it has a filled last energy level,  it is stable and
requires a large amount of energy to remove an electron. No bonding occurs. 
1.4
A bond is a net electrostatic force between two atoms.  Atoms bond to obtain a filled
valence orbital – octet rule – 8 electrons in valence orbital increases stability. Except
hydrogen which follows the “rule of two” – its valence orbital can have a maximum of 2
electrons.  When two atoms approach each other, the valence electrons of the two
atoms attract the positive nucleus of the other atom; these attractive forces are
stronger than the repulsive forces between the atoms. The protons and electrons of
respective atoms attract the atoms to form the bond when orbitals overlap to get a full
valence orbital. The atoms move closer together and the potential energy becomes
negative. The atoms are most stable at the lowest value of potential energy – when the
orbitals overlap and bonding occurs. The two atoms each share electrons during
bonding, there is a net electrostatic force of attraction between the atoms. 
1.5a. Different atoms, each with an unpaired valence electron can share these electrons to
form a chemical bond 
Page 7 of 8
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
PHYSICAL SCIENCE
b.
Grade 11
SESSION 11
(LEARNER NOTES)
Different atoms with paired valence electrons called lone pairs of electrons,
cannot share these four electrons and cannot form a chemical bond 
Different atoms, with unpaired valence electrons can share these electrons and
form a chemical bond for each electron pair shared (multiple bond formation) 
Atoms with an incomplete complement of electrons in their valence shell can
share a lone pair of electrons from another atom to form a co-ordinate or dative
covalent bond 
c.
d.
Question 2
2.1
a.

Cℓ
b.
2.2

He
F
a.
F

b.
O
H

H
c.
H
N
H

H
H+
2.3
a.
H
N
H

H
b.
Lewis base 
c.
Lewis acid 
The SSIP is supported by
Page 8 of 8