PREPWALK

1. CHEMISTRY
CHEMICAL COMBINATIONS:
moles, molecular and empirical
formulae
Short revision series

2. Relative atomic mass
 The relative atomic mass A of an element is
the number of times the average mass of one
atom of that element is heavier than 1/12th
the mass of one carbon-12 atom
 To calculate for eg the relative atomic mass
of say Oxygen
= average mass of 1 atom of oxygen
1/12th mass of 1 atom of carbon-12

3. Relative molecular mass
 Relative molecular mass Mr of an element or
a compound is the number of times the
average mass of one molecule of it is heavier
than one-twelfth the mass of one atom of
carbon-12
Eg calculate the relative molecular mass of
H2SO4
(H = 1, S = 32, O = 16)

4.  Note that one molecule of H2SO4 has 2 H, 1
S, and 4 O atoms
 RMM = 2 x 1 + 1 x 32 + 4 x 16
= 2 + 32 + 64
= 98

5. Mole
 One mole of a substance is the amount
containing as many elementary entities as
the the number of atoms in exactly 12 grams
of carbon-12.
 1 mole or 12 grams of carbon-12 contains
6.02 x 1023 atoms.
 6.02 x 1023 = Avogadro number.
 Therefore 1 mole of oxygen atoms contains
6.02 x 1023 oxygen atoms.

6.  Number of moles of a substance =
number of particles
6.02 x 1023
 Eg 1: How many moles are in 10g of H2SO4?

7.  Molecular mass of H2SO4 = 98g
 98g of H2SO4 = 1 mole
 10g = Xmoles
 X = 10 x 1 = 0.1 mol
98
EG 2 how many atoms are in 16 g of sulphur?
(S = 32)

8.  1 mole = 6x1023, S = 32
 32 g of S = 1 mole
 16 g = 16/32 = 0.5 mole
 1 mole = 6.02 x 1023 atoms
 Therefore 0.5 or ½ mole = 6.02 x 1023
2
= 3 x 1023 atoms

9. Calculations: percentage by
mass
 Calculate the percentage by mass of
component elements in H2SO4
(H = 1, S = 32, O = 16)

10. % by mass = mass of element x 100%
mass of compound
Mass of H2SO4 = 98
% by mass of hydrogen = 2H x 100 = 2 x 100
98 98
= 2.04%
 % by mass of S = 32/98 x 100 = 32.65%
 % by mass of O = 4 x 16 x 100 = 65.31%
98
Check your answer: the sum of all percentages must be
= 100

11. Calculations: Empirical and
molecular formulae
 Find the empirical and molecular formula of
a compound which contains 2g of C, 0.34g of
H, and 2.67g of O.
(Relative molecular mass = 60; C =12, H =1, O
=16)

12.  C  2/12 = 0.17
 H  0.34/1 = 0.34
 O  2.67/16 = 0.17
Divide by the smallest number
C = 0.17/0.17 = 1
H = 0.34/0.17 = 2
O = 0.17/0.17 = 1
 The empirical formular is CH2O

13.  The molecular formular is
(CH2O)x = 60
30x = 60
x = 60/30 = 2
the molecular formular is C2 H4 O2
== CH3 COOH

14. Empirical and molecular formulae
 Eg A compound of Cl contains 26.17%
nitrogen and 7.48% of hydrogen. Find the
empirical formulae.
(Cl = 35.5, H = 1, N = 14; Relative molecular
mass of the compound is

15.  A Compound of Cl therefore the percentage
of Cl is 100 – (26.17 + 7.48) = 66.35%
 For Cl = 66.35/35.5 = 1.87
 For N = 26.17/14 = 1.87
 For H = 7.48/1 = 7.48
Divide by the smallest number
Cl = 1.87/1.87 = 1
N = 1.87/1,.87 = 1
H = 7.48/1.87 = 4
 The empirical formula is NH4Cl