Empirical and Molecular Formulae

This lesson covers how to calculate empirical formulae and molecular formulae from experimental data. These are Higher tier topics required by the Edexcel GCSE Chemistry specification (1CH0).

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What Is an Empirical Formula?

The empirical formula of a compound shows the simplest whole number ratio of atoms of each element in that compound.

For example:

Compound	Molecular formula	Empirical formula
Water	H₂O	H₂O
Glucose	C₆H₁₂O₆	CH₂O
Ethane	C₂H₆	CH₃
Butene	C₄H₈	CH₂

Notice that the empirical formula for water is the same as its molecular formula, because H₂O is already in its simplest ratio.

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What Is a Molecular Formula?

The molecular formula shows the actual number of atoms of each element in one molecule of the compound.

The molecular formula is always a whole number multiple of the empirical formula.

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Calculating Empirical Formula from Masses

The Step-by-Step Method

flowchart TD
    A["Step 1: Write the mass of each element"] --> B["Step 2: Divide each mass by the Ar of that element"]
    B --> C["Step 3: Divide each result by the smallest value"]
    C --> D["Step 4: If needed, multiply to get whole numbers"]
    D --> E["Step 5: Write the empirical formula"]
    style A fill:#e6f3ff,stroke:#333
    style B fill:#e6f3ff,stroke:#333
    style C fill:#e6f3ff,stroke:#333
    style D fill:#e6f3ff,stroke:#333
    style E fill:#e6f3ff,stroke:#333

Worked Example 1: From Masses

Question: A compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula.

Step	Carbon	Hydrogen
Mass (g)	2.4	0.8
÷ Ar	2.4 ÷ 12 = 0.2	0.8 ÷ 1 = 0.8
÷ smallest (0.2)	0.2 ÷ 0.2 = 1	0.8 ÷ 0.2 = 4

Ratio → C : H = 1 : 4

Empirical formula: CH₄

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Worked Example 2: From Masses (Three Elements)

Question: A compound contains 12.0 g of carbon, 2.0 g of hydrogen and 16.0 g of oxygen. Find its empirical formula.

Step	Carbon	Hydrogen	Oxygen
Mass (g)	12.0	2.0	16.0
÷ Ar	12.0 ÷ 12 = 1.0	2.0 ÷ 1 = 2.0	16.0 ÷ 16 = 1.0
÷ smallest (1.0)	1.0	2.0	1.0

Ratio → C : H : O = 1 : 2 : 1

Empirical formula: CH₂O

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Calculating Empirical Formula from Percentages

If given percentage composition instead of masses, the method is identical — just treat the percentages as masses (since percentages are equivalent to the mass in a 100 g sample).

Worked Example 3: From Percentage Composition

Question: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula.

Step	Carbon	Hydrogen	Oxygen
"Mass" (from %)	40.0	6.7	53.3
÷ Ar	40.0 ÷ 12 = 3.33	6.7 ÷ 1 = 6.7	53.3 ÷ 16 = 3.33
÷ smallest (3.33)	1.0	2.01 ≈ 2	1.0

Ratio → C : H : O = 1 : 2 : 1

Empirical formula: CH₂O

Exam Tip: If you only have the percentage of some elements, find the missing percentage by subtracting from 100%. For example, if a compound contains 52.2% carbon and 13.0% hydrogen, then oxygen = 100 - 52.2 - 13.0 = 34.8%.

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Worked Example 4: Needing to Multiply

Question: A compound contains 43.7% phosphorus and 56.3% oxygen. Find its empirical formula.

Step	Phosphorus	Oxygen
"Mass" (from %)	43.7	56.3
÷ Ar	43.7 ÷ 31 = 1.41	56.3 ÷ 16 = 3.52
÷ smallest (1.41)	1.0	2.50

The ratio 1 : 2.5 is not whole numbers. Multiply both by 2:

Ratio → P : O = 2 : 5

Empirical formula: P₂O₅

Exam Tip: If dividing by the smallest gives you a ratio ending in .5, multiply everything by 2. If it ends in .33, multiply by 3. If it ends in .25, multiply by 4.

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Finding the Molecular Formula

Once you have the empirical formula, you can find the molecular formula if you know the relative formula mass (Mr) of the compound.

Method

1. Calculate the Mr of the empirical formula.
2. Divide the actual Mr by the empirical formula Mr to find the multiplier (n).
3. Multiply all subscripts in the empirical formula by n.

$n = \frac{M_r \text{ of compound}}{M_r \text{ of empirical formula}}$n=Mr​ of empirical formulaMr​ of compound​

Worked Example 5

Question: A compound has the empirical formula CH₂O and an Mr of 180. Find its molecular formula.

Solution:

1. Mr of CH₂O = 12 + 2 + 16 = 30
2. n = 180 ÷ 30 = 6
3. Molecular formula = C₆H₁₂O₆ (multiply each subscript by 6)

Molecular formula: C₆H₁₂O₆ (this is glucose)

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Worked Example 6

Question: A compound has the empirical formula CH₃ and an Mr of 30. Find its molecular formula.

Solution:

1. Mr of CH₃ = 12 + 3 = 15
2. n = 30 ÷ 15 = 2
3. Molecular formula = C₂H₆

Molecular formula: C₂H₆ (this is ethane)

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Worked Example 7: Complete Problem

Question: Analysis of a hydrocarbon shows it contains 85.7% carbon and 14.3% hydrogen. Its Mr is 56. Find its empirical and molecular formulae.

Step	Carbon	Hydrogen
"Mass"	85.7	14.3
÷ Ar	85.7 ÷ 12 = 7.14	14.3 ÷ 1 = 14.3
÷ smallest (7.14)	1.0	2.0

Empirical formula: CH₂

Mr of CH₂ = 14

n = 56 ÷ 14 = 4

Molecular formula: C₄H₈ (this is butene)

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Using Experimental Data

In experiments, you may be given:

• The mass of a metal and the mass of the oxide formed (subtract to find mass of oxygen).
• The mass of a compound and the masses of individual elements produced by decomposition.

Worked Example 8

Question: 4.6 g of sodium reacts with oxygen to form 6.2 g of sodium oxide. Find the empirical formula of sodium oxide.