Empirical Formula (DP IB Chemistry): Revision Note

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Philippa Platt

Updated on

Empirical Formula

  • The molecular formula shows the actual number and type of atoms in a molecule:

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

  • The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound

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

    • It can be determined from percentage composition by mass data

  • Organic compounds often have different empirical and molecular formulae

  • The formula of an ionic compound is always written as an empirical formula

Worked Example

Determine the empirical formula of a compound that contains 10 g of hydrogen and 80 g of oxygen.

Answer:

 

Hydrogen

Oxygen

Note the mass of each element

10 g

80 g

Divide the masses by atomic masses

10
1.01

= 10 mol

80
16.00

= 5 mol

Divide by the lowest figure to obtain nearest whole number ratio

10
5.0 

=2

 5.0

 5.0 

=1

Empirical formula

H2O

Worked Example

Determine the empirical formula of a compound that contains 85.7% carbon and 14.3% hydrogen.

Answer:

 

Carbon

Hydrogen

Note the X by mass of each element

85.6

14.3

Divide the X by atomic masses

85.7
12.01

=7.14 mol

14.3
1.01

= 14.2 mol

Divide by the lowest figure to obtain nearest whole number ratio

7.14
7.14

= 1

14.2
7.14

= 2

Empirical formula

CH2

Molecular formula

  • The molecular formula shows the actual number of atoms of each element in a compound

  • To determine the molecular formula:

    1. Divide the compound’s relative molecular mass (Mr) by the relative mass of the empirical formula

    2. This gives a whole number multiplier

    3. Multiply the empirical formula by this number to get the molecular formula

Worked Example

The empirical formula of X is C4H10S and the relative molecular mass of X is 180.42.

What is the molecular formula of X?

Relative Atomic Mass: Carbon: 12.01, hydrogen: 1.01, sulfur: 32.07

Answer:

  • Step 1: Calculate the relative mass of empirical formula

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

Relative empirical mass = (12.01 x 4) + (1.01 x 10) + (32.07 x 1)

Relative formula mass = 90.21

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

fraction numerator 180.42 over denominator 90.21 end fraction equals 2

  • Step 3: Multiply the empirical formula by 2

2 x C4H10S = C8H20S2

The molecular formula of X is C8H20S2 

👀 You've read 1 of your 5 free revision notes this week
An illustration of students holding their exam resultsUnlock more revision notes. It's free!

By signing up you agree to our Terms and Privacy Policy.

Already have an account? Log in

Did this page help you?

Alexandra Brennan

Author: Alexandra Brennan

Expertise: Chemistry Content Creator

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.

Philippa Platt

Reviewer: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Download notes on Empirical Formula