International A LevelChemistryOxford AQARevision NotesPhysical ChemistryAmount of SubstanceEmpirical & Molecular Formula

Empirical & Molecular Formula (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

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

  • The empirical formula 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

fraction numerator 2.72 over denominator 12.0 end fraction = 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

Worked Example

Calculating empirical formula from %

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

Answer:

Elements

Carbon

Hydrogen

Mass of each element
(%)

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

fraction numerator 10.0 over denominator 1.0 end fraction = 10.0

Ratio (divide by smallest value)

fraction numerator 7.5 over denominator 7.5 end fraction = 1

fraction numerator 10.0 over denominator 7.5 end fraction = 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 number 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 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 for each element (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 value)

fraction numerator 2.02 over denominator 2.02 end fraction = 1

fraction numerator 4.1 over denominator 2.02 end fraction = 2.03

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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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

    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.