IGCSEScienceOxford AQACombined Science Double AwardChemistryRevision NotesQuantitative ChemistryUse of Amount of Substance in Relation to Masses of Pure SubstancesEmpirical Formula

Empirical Formula (Oxford AQA IGCSE Combined Science Double Award)

Revision Note

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Oxford AQA Combined Science Double Award: ChemistryOxford AQA Chemistry

How to Calculate Empirical Formula

  • The simplest formula is a whole number ratio of the atoms of each element present in one molecule or formula unit of the compound

    • The simplest formula is often called the empirical formula 

  • The empirical formula of an organic molecule is often different to its chemical / molecular formulae

    • For example, ethanoic acid has the chemical formula CH3COOH or C2H4O2 but its empirical formula is CH2O

  • The chemical formula of an ionic compound is always its empirical formula 

    • For example, sodium chloride has the chemical formula NaCl, which is also its empirical formula

Worked Example

A sample of a compound was found to contain 10 g of hydrogen and 80 g of oxygen.

Calculate the empirical formula of this compound.

Ar(H) = 1   Ar(O) = 16

Answer:

 

hydrogen

oxygen

Write the mass of each element 

10 g

80 g

Calculate the number of moles

(Divide each mass by the Ar)

open parentheses 10 over 1 close parentheses = 10

open parentheses 80 over 16 close parentheses = 5

Find the simplest whole number molar ratio

(Divide by the smallest number)

open parentheses 10 over 5 close parentheses = 2 

open parentheses 5 over 5 close parentheses = 1 

So, the empirical formula = H2O

Worked Example

Carbohydrate X was analysed and found to contain 31.58% carbon and 5.26% hydrogen by mass.

Find the empirical formula of carbohydrate X.

Ar (H) = 1   Ar (C) = 12   Ar (O) = 16

Answer:

A carbohydrate contains carbon, hydrogen and oxygen

The percentages do not add up to 100%, which means that you need to calculate the percentage of oxygen needs to be calculated

Percentage of oxygen = 100 - 31.58 - 5.26 = 63.16%

 

carbon

hydrogen

oxygen

Convert % to g

(Assume 100 g of substance is present)

31.58 g

5.26 g

63.16 g

Calculate the number of moles

(Divide each mass by the Ar)

open parentheses fraction numerator 31.58 over denominator 12 end fraction close parentheses = 2.63

open parentheses fraction numerator 5.26 over denominator 1 end fraction close parentheses = 5.26

open parentheses fraction numerator 63.16 over denominator 16 end fraction close parentheses = 3.95

Find the simplest molar ratio

(Divide by the smallest number)

open parentheses fraction numerator 2.63 over denominator 2.63 end fraction close parentheses = 1

open parentheses fraction numerator 5.26 over denominator 2.63 end fraction close parentheses = 2 

begin mathsize 14px style open parentheses fraction numerator 3.95 over denominator 2.63 end fraction close parentheses end style = 1.5

Obtain a whole number ratio

(Multiply all by 2)

1 x 2 = 2

2 x 2 = 4

1.5 x 2 = 3

So, the empirical formula = C2H4O3

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    Alexandra Brennan

    Author: Alexandra Brennan

    Expertise: Chemistry

    Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom 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.