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Empirical & Molecular Formula (AQA A Level Chemistry)

Revision Note

Stewart Hird

Last updated

Empirical & Molecular Formulae

  • The molecular formula is the formula that shows the number and type of each atom in a molecule

    • E.g. the molecular formula of ethanoic acid is C2H4O2

  • The empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound

    • E.g. the empirical formula of ethanoic acid is CH2O

  • Organic molecules often have different empirical and molecular formulae

  • Simple inorganic molecules however have often similar empirical and molecular formulae

  • Ionic compounds always have similar empirical and molecular formulae

Empirical & Molecular Formulae Calculations

Empirical formula

  • Empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound

  • It is calculated from knowledge of the ratio of masses of each element in the compound

  • The empirical formula can be found by determining the mass of each element present in a sample of the compound

  • It can also be deduced from data that gives the percentage compositions by mass of the elements in a compound

Worked Example

Calculating empirical formula from mass

Determine the empirical formula of a compound that contains 2.72 g of carbon and 7.28 g of oxygen.

Answer:

Elements

Carbon

Oxygen

Mass of each element
(g)

2.72

7.28

Atomic mass

12.0

16.0

Moles = mass / Ar

begin mathsize 14px style fraction numerator 2.72 over denominator 12.0 end fraction end style = 0.227

fraction numerator 7.28 over denominator 16.0 end fraction = 0.455

Ratio (divide by smallest value)

fraction numerator 0.227 over denominator 0.227 end fraction = 1

fraction numerator 0.455 over denominator 0.227 end fraction = 2

  • So, the empirical formula of the compound is CO2

  • The above example shows how to calculate empirical formula from the mass of each element present in the compound

  • The example below shows how to calculate the empirical formula from percentage composition

Worked Example

Calculating empirical formula from percentage

Determine the empirical formula of a hydrocarbon that contains 90.0% carbon and 10.0% hydrogen.

Answer:

Elements

Carbon

Hydrogen

Mass of each element
(g)

90.0

10.0

Atomic mass

12.0

1.0

Moles = mass / Ar

fraction numerator 90.0 over denominator 12.0 end fraction = 7.5

begin mathsize 14px style fraction numerator 10.0 over denominator 1.0 end fraction end style = 10.0

Ratio (divide by smallest value)

fraction numerator 7.5 over denominator 7.5 end fraction = 1

begin mathsize 14px style fraction numerator 10.0 over denominator 7.5 end fraction end style = 1.33

Convert to whole number ratio
(x3 for this example)

1 x 3 = 3

1.33 x 3 = 4

  • So, the empirical formula of the compound is  C3H4 

Molecular formula

  • The molecular formula gives the exact numbers of atoms of each element present in the formula of the compound

  • The molecular formula can be found by dividing the relative formula mass of the molecular formula by the relative formula mass of the empirical formula

  • Multiply the number of each element present in the empirical formula by this number to find the molecular formula

Worked Example

Calculating molecular formula

The empirical formula of X is C4H10S and the relative molecular mass of X is 180

What is the molecular formula of X?

(Ar data: C = 12, H = 1, S = 32)

Answer:

Step 1: Calculate relative mass of the empirical formula

  • Relative empirical mass = (C x 4) + (H x 10) + (S x 1)

  • Relative empirical mass = (12 x 4) + (1 x 10) + (32 x 1)

  • Relative empirical mass = 90

Step 2: Divide relative formula mass of X by relative empirical mass

  • Ratio between Mr of X and the Mr of the empirical formula = 180/90

  • Ratio between Mr of X and the Mr of the empirical formula = 2

Step 3: Multiply each number of elements by 2

  • (C4 x 2) + (H10 x 2) + (S x 2)     =    (C8) + (H20) + (S2)

  • Molecular Formula of X is C8H20S2

Worked Example

Calculating empirical formula and molecular formula

Analysis of a compound X shows that it contains 24.2 % by mass of carbon, 4.1 % by mass of hydrogen and 71.7% by mass of chlorine.

Calculate the empirical formula of X.

Use this empirical formula and the relative molecular mass of X (Mr = 99.0) to calculate the molecular formula of X.

Answer:

Elements

Carbon

Hydrogen

Chlorine

Value (g or %)

24.2

4.1

71.7

Atomic mass

12.0

1.0

35.5

Moles = mass / Ar

fraction numerator 24.2 over denominator 12.0 end fraction = 2.02

fraction numerator 4.1 over denominator 1.0 end fraction = 4.1

fraction numerator 71.7 over denominator 35.5 end fraction = 2.02

Ratio (divide by smallest)

fraction numerator 2.02 over denominator 2.02 end fraction = 1

fraction numerator 4.1 over denominator 2.02 end fraction = 2

fraction numerator 2.02 over denominator 2.02 end fraction = 1

  • So, the empirical formula of compound X is CH2Cl

  • The relative formula mass of the empirical formula is:

    • Relative formula mass = (1 x C) + (2 x H) + (1 x Cl)

    • Relative formula mass = (1 x 12.0) + (2 x 1.0) + (2 x 35.5)

    • Relative formula mass = 49.5

  • Divide the relative formula mass of X by the relative formula mass of the empirical formula

    • Ratio between Mr of X and the Mr of the empirical formula = 99.0/45.9

    • Ratio between Mr of X and the Mr of the empirical formula = 2

  • Multiply each number of elements by 2

    • (C1 x 2) + (H2 x 2) + (Cl1 x 2) = (C2) + (H4) + (Cl2)

    • The molecular formula of X is C2H4Cl2

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    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.