Empirical and Molecular Formulae

This lesson covers how to determine the empirical formula and molecular formula of a compound from experimental data including percentage composition by mass and combustion analysis.

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Definitions

• Empirical formula: The simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula: 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.

Molecular formula = (empirical formula) × n

where n = Mᵣ of compound / Mᵣ of empirical formula

Examples

Compound	Molecular formula	Empirical formula	n
Water	H₂O	H₂O	1
Ethane	C₂H₆	CH₃	2
Glucose	C₆H₁₂O₆	CH₂O	6
Benzene	C₆H₆	CH	6
Ethanoic acid	C₂H₄O₂	CH₂O	2

Key Point: Ionic compounds do not have molecular formulae — their formula is always an empirical formula (e.g., NaCl, MgO, CaCl₂).

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Method: From Percentage Composition by Mass

To determine the empirical formula from percentage composition:

Step 1: Write down the percentage of each element (if one is missing, find it by subtracting the others from 100%).

Step 2: Divide each percentage by the Aᵣ of that element to get the number of moles.

Step 3: Divide all the mole values by the smallest to get the simplest ratio.

Step 4: If the ratio is not a whole number, multiply through to get whole numbers.

Worked Example 1: Percentage Composition

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

Element	% by mass	÷ Aᵣ	Moles	÷ smallest	Ratio
C	40.0	÷ 12.0	3.33	÷ 3.33	1
H	6.7	÷ 1.0	6.7	÷ 3.33	2
O	53.3	÷ 16.0	3.33	÷ 3.33	1

Empirical formula = CH₂O

Worked Example 2: With Rounding

A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen. Determine its empirical formula.

Element	% by mass	÷ Aᵣ	Moles	÷ smallest	Ratio
C	52.2	÷ 12.0	4.35	÷ 2.175	2
H	13.0	÷ 1.0	13.0	÷ 2.175	5.98 ≈ 6
O	34.8	÷ 16.0	2.175	÷ 2.175	1

Empirical formula = C₂H₆O

Worked Example 3: Non-integer Ratios

A compound contains 36.8% nitrogen and 63.2% oxygen. Determine its empirical formula.

Element	% by mass	÷ Aᵣ	Moles	÷ smallest	Ratio
N	36.8	÷ 14.0	2.629	÷ 2.629	1
O	63.2	÷ 16.0	3.950	÷ 2.629	1.50

The ratio 1 : 1.50 is not whole numbers. Multiply both by 2:

Ratio = 2 : 3

Empirical formula = N₂O₃

Exam Tip: If you get ratios like 1.5, multiply by 2. If you get ratios like 1.33, multiply by 3. If you get ratios like 1.25, multiply by 4. Recognise common fractions and find the appropriate multiplier.

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Method: From Masses (Experimental Data)

If given actual masses instead of percentages, the method is the same but start from masses directly.

Worked Example 4: From Experimental Masses

1.35 g of aluminium reacts with 1.20 g of oxygen to form an oxide. Determine the empirical formula.

Element	Mass (g)	÷ Aᵣ	Moles	÷ smallest	Ratio
Al	1.35	÷ 27.0	0.0500	÷ 0.0375	1.33
O	1.20	÷ 16.0	0.0750	÷ 0.0375	2

Ratio = 1.33 : 2. Multiply by 3: ratio = 4 : 6 = 2 : 3.

Empirical formula = Al₂O₃

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Method: From Combustion Analysis