Chapter Three
Numerical connections –
The Mole
Handouts and Worksheets
o 7 x mole calculations worksheets
The Mole – The Mole Concept
• In Chemistry the Mole Concept assists
in counting atoms and molecules.
• The word ‘Mole’ (mol) represents a
number. It is simply a unit of
measurement (dozen represents 12 ).
• The number of atoms in one mole of
an element is 6 x 1023.
The Mole – The Mole Concept
• A mole is defined as the amount of
substance that contains as many particles
(atoms, ions or molecules or)as there are
atoms in exactly 12g of the C 12 isotope.
• The number of carbon atoms in 12 g of
C12 isotope has been experimentally
estimated to be 6.02 x 1023 atoms –
Avogradro’s number
• The number of atoms in one mole of
an element is 6.02 x 1023.
The Mole – The Mole Concept
• To find out the mass of one mole of
any element, all you have to do is
add ‘g’ to the relative atomic mass.
• What is the mass of one mole of; Na,
Ca, B, Li, F, O, C, Mg
Na
Ca
B
Li
22.990g
40.078g
10.811g
6.941g
F
O
C
Mg
18.998g
12.011g
15.999g
24.305g
 The
Molar Mass (M) of an element is defined
as the mass on 1 mol of the element. The
unit of grams per mole (gmol-1).
 Eg.
Molar mass of carbon atoms = mass of 1 mol of C atoms =
12 g mol-1
Therefore 12g of carbon contains 6.02 x 1023 atoms of
carbon

Molar mass of oxygen atoms = mass of 1 mol of O atoms
= 16 g mol-1
Therefore 16g of oxygen contains 6.02 x 1023 atoms of
oxygen

 The
molar mass of a compound is defined as
the mass of 1 mol of the compound
expressed in grams per mole (g mol-1)
 Whether
a substance is made up of atoms,
molecules or ions, the same principle
applies: one mole of substance contains the
same number of particles. This number is
always 6.02 x 1023.
 Molecular
Compounds

Molar mass of water molecules = mass of 1 mol of H20
molecules = 18 g mol-1
Therefore 18g of water contains 6.02 x 1023 molecules of water

Molar mass of Chlorine gas = mass of 1 mol of Cl2 molecules =
71g mol-1
Therefore 71g of chlorine gas contains 6.02 x 1023 molecules of
chlorine
Calculate the Mr of: N2, C02, CH4, HCl, I2, NH2
 Ionic
Compounds – The Relative Molecular
Mass (Mr) is found by adding together the Ar
of each atom in formula of the compound.
 Eg
Mr of CuSO4 = Ar (Cu) + Ar (S) + 4 x Ar (O)
= 63.5 + 32.1 + 4 x 16.0
= 159.6
Molar mass of CuS04 = 159.6 g mol-1
divide by molar mass
Mass
Moles
multiply by molar mass
n = number of moles, m = mass, M = molar
mass
Practice Problems
n= m
M
mole calculations
multiply by 6.02 x 10 23
number of particles
moles
divide by 6.02 x 1023
number of particles
n=
6.02 x 1023
Work through the Sample problem page 53
 mole calculations


Complete the revision questions page 53
(1 – 3)










2
(a) 2.3 x 6.02 x 1023
(b) 15.8 x 6.02 x 1023 x 2
(c) 3.5 x 6.02 x 1023
(d) 0.5 x 6.02 x 1023 x 2
3
(a) 16.2 / 60
(b) 0.27 X 6.02 X 1023
(c) 1.63 x 1023 / 6.02 x 1023 x 2
(d) 0.540 x 6.02 x 1023

Work through the Sample problem page
54

Complete the revision question page 54
(4)
4
 (a) 5.2 x 1023 / 6.02 x 1023
 (b) 5.2 x 1023 / 6.02 x 1023
 (c) 5.2 x 1023 / 6.02 x 1023 x 4

 Work
54
m
n
=
M
through the Sample Problem page
 Complete
the revision question page 54
(5)
 Work through the Sample Problem
page55
 Complete the revision questions page 55
(6,7)
 Work through the Sample Problem page
55
 Complete the revision questions page 56
(8 – 10)
5
 (a)
n=m/M
46 / 18
 (b) 2.4 / 44
 (c) 67 / 71
 (d) 2.0 / 58.44
 (e) 128 / 249.62
 (f) 38000 / 159.7
6
m=nxM
 (a) 0.41 x 28
 (b) 12.0 x 64
 (c) 3.84 x 342
 (d) 58.2 x 55.85
 (e) 0.0051 x 143.37
 (f) 2.53 x 262.87
7
 (a)
5.05
 (b) 13.076
 (c) 11.56
 8.
 (a) 5.25 x 1024 / 6.02 x 1023 = n

8.721 x 180.12
 (b) 1.83 x 1021 / 6.02 x 1023 = n

0.003 x 46
 (c) 3.56 x 1014 / 6.02 x 1023 = n

5.91x 10-10 x 44
 (d) 4.13 x 1028 / 6.02 x 1023 = n

68604.65 x 76.13
 (e) 3.62 x 1024 / 6.02 x 1023 = n

6.013 x 92.02
9
 (a)
200g
 (b) 5 x 32 = 160g
 (c) 1.2 x 1024 / 6.02 x 1023 = 1.993

1.993 x 4 = 7.97g
 (d) 3.5 x 1022 / 6.02 x 1023 = 0.058

0.058 x 73 = 4.234g
Formulae summary check
1. Calculating number of particles (atoms,
molecules, ions)
– Number of particles = n x 6.02 x 1023 (avogadro’s number)
2. Calculating number of moles given number
of particles
– n = number of particles ÷ 6.02 x 1023
3. Calculating moles from mass
– n = mass ÷ M
4. Calculating mass for numbers of particles
– First calculate n (using formula 2)
– Then apply n = mass ÷ M
Percentage Composition
• The composition of a compound is often
expressed in terms of the percentage that each
element contributes to its mass – percentage
composition.
• Work through the Sample Problem page 57
Review
• Complete the revision question page 57 (11)
Review answers
• 11
• (a) Mr = (12 x 3) + (1 x 8) = 44
% mass C = (12 x 3) / 44 x (100 / 1)
= 81.8%
% mass H = (1 x 8) / 44 x (110 / 1)
= 18.2%
Review answers
• (b) Na – 22.99 , H – 1 , S – 32.06 , O – (4 x 16)
Percentage composition of hydrated
compounds
• Calculate the Mr of the compound including
the water
• Calculate the mass of the water
• Calculate the % of water in the compound
• Work through the Sample Problem pages
58,59
Review
• Complete the revision question page 59 (12)
Empirical Formulae
• The Empirical Formulae of a compound
gives the simplest whole number ratio of
the atoms or ions present in the compound
and can be found only by experiment.
Empirical Formulae
1. Write down the symbols of the elements present
2. Assume that the mass of the sample is 100g and
all the percentages become grams
3. Convert masses to moles
4. Find the simplest whole number ratio of the
atoms by dividing all numbers of moles by the
smallest number of moles
5. If necessary, multiply by a factor to convert all
numbers to whole numbers
•
Work through the Sample Problem page 60
Review
• Complete the revision questions page 60
(13,14)
Empirical Formula of a hydrated compound
1. Calculate the mass of the water and the
compound
2. Calculate number of moles of each
3. Calculate the mole ratio of compound to
water
• Work through the Sample Problem page
60
Review
• Complete the revision questions page 61
(15,16)
The Molecular Formula of a compound
represents the actual composition of a
compound that is made up of molecules.
 The Molecular Formula is either the same as the
empirical formula or a whole number multiple of
it
 Compounds with different molecular formulae
may have very different properties.
 The Molecular Formula of a compound may be
determined from its empirical formula only if its
molar mass is also known

Since the empirical formula is the lowest ratio
the actual molecule would weigh more.
 By a whole number multiple.
 Divide the actual molar mass by the mass of
one mole of the empirical formula.
 Caffeine has a molar mass of 194 g. what is its
molecular formula?


Find x if
molar mass
x
empirical formula mass
•194 g
•97 g
2X
C4H5N2O1
C8H10N4O2.
=2

A compound is known to be composed of
71.65 % Cl, 24.27% C and 4.07% H. Its molar
mass is known (from gas density) is known to
be 98.96 g. What is its molecular formula?

71.65 Cl
24.27C
4.07 H.
1mol
35.5 g
1mol
12 g
1mol
1g
= 2.0mol
= 2.0mol
= 4.0mol

Cl2C2H4
We divide by
lowest (2mol )
•Cl1C1H2
would give an empirical wt of 48.5g/mol
Its molar mass is known (from gas density)
is known to be 98.96 g. What is its molecular
formula?
•
would give an empirical wt of 48.5g/mol
Its molar mass is known (from gas density)
is known to be 98.96 g. What is its molecular
formula?
molar mass
x
=
empirical formula mass
98.96 g
=2
48.5 g
2 X Cl1C1H2
= Cl2C2H4

Balancing equations on line multiple choice

Complete the revision questions page 62 (17 –
19)
Work through the Sample Problem page 62 Process Calculate the empirical formula
Write down the symbols of the elements present
Assume that the mass of the sample is 100g and all
the percentages become grams
3. Convert masses to moles
4. Find the simplest whole number ratio of the atoms by
dividing all numbers of moles by the smallest number
of moles
5. If necessary, multiply by a factor to convert all
numbers to whole numbers


1.
2.


Calculate the empirical formula mass
ratio = molar mass ÷ empirical formula mass
Review
•
The Mole on line questions
•
The Mole on line multiple choice
•
Complete the revision questions page 63 (20,21)
•
Complete the multiple choice questions. Check and review your
answers.
•
Complete ; Mole Calculations, Percentage Composition, Empirical
Formulae, Molecular Formulae. Check and review your answers.
•
Complete the Exam Practice Questions (1, 2). Ensure ALL your
working is shown.
•
Ensure ALL of the Chapter 3 practice worksheets are complete