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Empirical & Molecular Formula (AQA A Level Chemistry): Revision Note

Exam code: 7405

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on 3 July 2026

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, often have similar empirical and molecular formulae
Ionic compounds always have similar empirical and molecular formulae

Empirical & Molecular Formulae Calculations

Empirical formula

The 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 the 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 / A_r	$\frac{2.72}{12.0}$ = 0.227	$\frac{7.28}{16.0}$ = 0.455
Ratio (divide by the smallest value)	$\frac{0.227}{0.227}$ = 1	$\frac{0.455}{0.227}$ = 2

So, the empirical formula of the compound is CO₂

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

Worked Example

Calculating the empirical formula from the percentages

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 / A_r	$\frac{90.0}{12.0}$ = 7.5	$\frac{10.0}{1.0}$ = 10.0
Ratio (divide by the smallest value)	$\frac{7.5}{7.5}$ = 1	$\frac{10.0}{7.5}$ = 1.33
Convert to a 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

Worked Example

Finding the empirical formula of an oxide from mass data.

Determine the empirical formula of an oxide of molybdenum if 5.0 g of the oxide contains 4.0 g of molybdenum.

Answer

Elements	Molybdenum	Oxygen
Mass of each element (g)	4.0	5.0-4.0= 1.0
Atomic mass	96.0	16.0
Moles = mass / A_r	$\frac{4.0}{96.0}$ = 0.0417	$\frac{1.0}{16.0}$ = 0.0625
Ratio (divide by the smallest value)	$\frac{0.0417}{0.0417}$ = 1	$\frac{0.0625}{0.0417}$ = 1.5
Convert to a whole number ratio (x3 for this example)	1 x 2 = 2	1.5 x 2 = 3

So, the empirical formula of the compound is Mo₂O₃

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?

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

Answer:

Step 1: Calculate the 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 the relative formula mass of X by the relative empirical mass

Ratio between M_r of X and the M_rof the empirical formula = 180/90
Ratio between M_r of X and the M_rof 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)
The 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 (M_r = 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 / A_r	$\frac{24.2}{12.0}$ = 2.02	$\frac{4.1}{1.0}$ = 4.1	$\frac{71.7}{35.5}$ = 2.02
Ratio (divide by smallest)	$\frac{2.02}{2.02}$ = 1	$\frac{4.1}{2.02}$ = 2	$\frac{2.02}{2.02}$ = 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) + (1 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 M_r of X and the M_rof the empirical formula = 99.0/49.5
- Ratio between M_r of X and the M_rof the empirical formula = 2
Multiply each number of elements by 2
- (C_{1 x 2}) + (H_{2 x 2}) + (Cl_{1 x 2}) = (C₂) + (H₄) + (Cl₂)
- The molecular formula of X is C₂H₄Cl₂

Related topics

Examiner Tips and Tricks

AQA mark schemes award no credit for a final formula given without working shown — students who jump straight to the molecular formula from the relative molecular mass can lose most of the marks even if the answer is correct.

When percentages are given for only some elements, check that they sum to 100% first. A recent AQA examiner report noted a significant number of students missed finding a hidden element by subtraction before starting the empirical formula calculation.

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Empirical & Molecular Formula (AQA A Level Chemistry): Revision Note

Empirical & Molecular Formulae

Empirical & Molecular Formulae Calculations

Empirical formula

Worked Example

Worked Example

Worked Example

Molecular formula

Worked Example

Worked Example

Examiner Tips and Tricks

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Author: Stewart Hird

Reviewer: Caroline Carroll