Empirical and molecular formula

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of atoms in one molecule. For glucose the molecular formula is C₆H₁₂O₆ and the empirical formula is CH₂O: the ratio 6 : 12 : 6 cancels down to 1 : 2 : 1.

Sometimes the two are the same, as in water and carbon dioxide. Ionic compounds such as sodium chloride have only an empirical formula, because they are not made of separate molecules.

Finding the empirical formula from percentages

Assume 100 g of the compound, so each percentage becomes a mass in grams.
Divide each mass by the atomic weight of that element to get moles.
Divide every answer by the smallest of them.
If the results are not whole numbers, multiply all of them by 2, 3 or 4 until they are.

Example. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen.

Element	Grams in 100 g	÷ atomic weight	Moles	÷ smallest
Carbon	40.0	12.011	3.33	1
Hydrogen	6.7	1.008	6.65	2
Oxygen	53.3	15.999	3.33	1

The ratio is 1 : 2 : 1, so the empirical formula is CH₂O.

From empirical formula to molecular formula

You need one more fact: the molar mass of the compound, found by experiment.

Work out the mass of the empirical formula. For CH₂O it is 30.03.
Divide the molar mass by it. The answer is a whole number, n.
Multiply every subscript in the empirical formula by n.

If the compound above has a molar mass of about 180 g/mol, then n = 180 ÷ 30.03 = 6 and the molecular formula is C₆H₁₂O₆, glucose. If its molar mass is about 60, n = 2 and it is C₂H₄O₂, acetic acid. If about 30, it is formaldehyde itself. Three very different substances share one empirical formula.

When the ratio is not whole

Suppose step 3 gives 1 : 1.5. Do not round 1.5 to 2. Multiply both by 2 to get 2 : 3. A result near 1.33 or 1.67 needs multiplying by 3; near 1.25 or 1.75, by 4. An oxide of iron with 72.4% iron works out to Fe 1 : O 1.33, which becomes 3 : 4, the formula Fe₃O₄.

From an experiment

In the classic school experiment, a weighed strip of magnesium is burned in a crucible and the white ash is weighed. If 0.24 g of magnesium gives 0.40 g of oxide, the oxygen gained is 0.16 g. Moles of magnesium: 0.24 ÷ 24.30 = 0.0099. Moles of oxygen: 0.16 ÷ 16.00 = 0.0100. The ratio is 1 : 1, so the formula is MgO.

Pairs worth knowing

Benzene C₆H₆ and ethyne C₂H₂: both CH.
Hydrogen peroxide H₂O₂: empirical formula HO.
Phosphorus(V) oxide: molecule P₄O₁₀, empirical formula P₂O₅.
Butane C₄H₁₀: empirical formula C₂H₅.

Which values? This site uses the abridged standard atomic weights published by IUPAC (CIAAW), read on 12 November 2026. Your textbook or question paper may round them differently, for example H = 1 and O = 16, so its answer can differ slightly. In an exam, use the values the paper gives.

Hydrates and water of crystallisation All guides