IGCSEChemistryOxford AQARevision NotesQuantitative ChemistryUse of Amount of Substance In Relation to Masses of Pure SubstancesEmpirical Formula

Empirical Formula (Oxford AQA IGCSE Chemistry)

Revision Note

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Oxford AQA ChemistryOxford AQA Combined Science Double Award: 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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    Author
    Reviewer
    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.