3.2 Empirical & Molecular Formula

SLT : 6 HRS

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3.2 EMPIRICAL & MOLECULAR FORMULA

At the end of the lesson, you should be able to:
(a) Define the terms empirical and molecular

formulae.
(b) Determine empirical and molecular formulae

from mass composition and combustion data.

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CHEMICAL FORMULA

Is a shorthand notation to represent the ratio of
atoms of elements in a compound.

Type of Chemical Formula

Empirical Molecular Structural
formula formula formula

The chemical The chemical Will be
formula that formula that explained next
shows the
simplest ratio shows the semester
of all elements actual number
in a molecule 3
of atoms in
each element
in a molecule

Relationship Between Empirical Formula and
Molecular Formula

Molecular formula = n (empirical formula) 1
where n = 1, 2, 3,… 2

n = molar mass of molecular formula
molar mass of empirical formula

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Determination of Empirical and Molecular
Formulae of a compound

Empirical and Molecular formula can be determined from:

Mass Combustion
Composition Data

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1. From mass composition

Example 1:

1.08 g of aluminium combines chemically
with 0.96 g of oxygen to form an oxide.
Write the empirical formula of the oxide?

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Solution: Al O
1.08 0.96
Element 1.08 = 0.04 0.96 = 0.06
Mass/g 27 16
Amount/ mol
0.04 = 1 0.06 = 1.5
Simplest 0.04 0.04
ratio 1x2=2 1.5 x 2 = 3

Empirical formula = Al2O3

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Example 2:

Ascorbic acid (vitamin C) is composed of
40.92% carbon, 4.58% hydrogen and 54.50%
oxygen by mass. The molar mass of ascorbic
acid is 176 gmol-1. Find its empirical
formula and molecular formula.

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Solution:

Assume that 100 % of ascorbic acid has a mass of
100 g.
Then, 40.92 % = 40.92 g

4.58 % = 4.58 g
50.54 % = 50.54 g

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Element C HO
Mass/g 40.92 4.58 54.50

Amount 40.92 = 3.41 4.58 = 4.58 54.50 = 3.41
(mol) 12 1 16

Simplest 3.41 = 1 4.58 = 1.33 3.41 = 1
3.41 3.41 3.41
ratio of
relative 1x3 1.33 x 3 1x3
amount
3 4 3

Empirical formula = C3H4O3

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Mass of empirical formula C3H4O3 = (12.0x3)+(1.0x4)+(16.0x3)
= 88 gmol-1

mass of molecular formula
n = mass of empirical formula

= 176 g =2
88 g

Molecular formula = n (empirical formula)

Molecular formula = n (C3H4O3)
= 2 (C3H4O3)
= C6H8O6

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1. From combustion data

Example 1:

A complete combustion of a hydrocarbon
compound yields 1.32 g of carbon dioxide and
0.54 g of water. Write the empirical formula of the
hydrocarbon?

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Solution:

A hydrocarbon compound consist only carbon and
hydrogen atoms.

Chemical equation for combustion of hydrocarbon:

CxHy + O2 CO2 + H2O

1.32 g 0.54 g

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Step 1 : Find the mass of C from CO2
Molar mass CO2 = 44.0 g/mol , Mass CO2 = 1.32 g
Molar mass C = 12.0 g/mol , Mass C = ?

Mass C = Mass CO2 x Molar mass C

Molar mass CO2

= 1.32 x 12.0
44.0

= 0.36 g of carbon

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Step 2 : Find the mass of H from H2O
Molar mass H2O = 18.0 g/mol , Mass H2O = 0.54 g
Molar mass H = 2(1.0) g/mol , Mass H = ?

Mass H = Mass H2O x Molar mass H

Molar mass H2O

= 0.54 x 2.0
18.0

= 0.06 g of hydrogen

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Element C H
Mass/g 0.36 0.06
Amount/ mol 0.36 = 0.03 0.06 = 0.06
12.0 1.0
Simplest
ratio 0.03 = 1 0.06 = 2
0.03 0.03

Empirical formula = CH2

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Example 2:

A compound P that consists of carbon, hydrogen
and oxygen has a mass of 1.470 g. If the complete
combustion of P produced 2.156 g of CO2 and 0.882
g of H2O. Determine the empirical formula of
compound P.

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Solution: CO2 + H2O
2.156 g 0.882 g
CxHyOz + O2
1.470 g

Mass of C from CO2

Mass C = Mass CO2 x Molar mass C

Molar mass CO2

= 2.156 x 12.0
44.0

= 0.588 g of carbon

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Mass of H from H2O

Mass H = Mass H2O x Molar mass H

Molar mass H2O

= 0.882 x 2.0
18.0

= 0.098 g of hydrogen

Mass of O
Mass O = mass compound P – mass C – mass H

= 1.470 – 0.588 – 0.098
= 0.784 g of oxygen

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Element C H O
Mass/g 0.588 0.098 0.784

Amount 0.588 = 0.049 0.098 = 0.098 0.784 = 0.049

(mol) 12 1 16

Simplest 0.049 = 1 0.098 = 2 0.049 = 1
0.049 0.049
ratio 0.049

Empirical formula = CH2O

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END OF SUBTOPIC 3.2

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