How to calculate empirical and molecular formulae

Empirical formula questions look harder than they are. Every one of them follows the same short recipe, and once you have drilled it a few times you can answer them almost on autopilot. The trick is to separate two ideas that students often blur together, the empirical formula and the molecular formula, and to know exactly which piece of information gives you each. This guide sets out the method step by step with worked numbers you can follow along.

Two formulae, two meanings

The empirical formula is the simplest whole-number ratio of the atoms in a compound. The molecular formula is the actual number of each atom in one molecule. They are often different. Glucose has the molecular formula C₆H₁₂O₆, but its empirical formula is CH₂O, because 6 : 12 : 6 simplifies to 1 : 2 : 1. Ethanoic acid, CH₃COOH, also has the empirical formula CH₂O even though its molecular formula is C₂H₄O₂. So the empirical formula tells you the ratio; the molecular formula tells you the real count. You always find the empirical formula first, then scale it up to the molecular formula if the question gives you a relative molecular mass.

The four-step method for the empirical formula

Use this exact routine every time:

1. Write the mass of each element. If you are given percentages, assume 100 g of compound so each percentage becomes a mass in grams. If the data comes from an experiment, use the measured masses directly.
2. Divide each mass by the element’s relative atomic mass to get moles of atoms.
3. Divide every answer by the smallest of them. This scales the ratio down so the smallest becomes 1.
4. Turn the ratio into whole numbers. If a value is still not whole, for example 1.5, multiply every part by a small factor (here, 2) until they all become integers.

That is the whole method. You can practise it in full at empirical formula, and check any answer with our empirical formula calculator.

A worked example from percentages

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

• Step 1: assume 100 g, so 40.0 g C, 6.7 g H, 53.3 g O.
• Step 2: divide by relative atomic masses (C = 12, H = 1, O = 16). Moles of C = 40.0 ÷ 12 = 3.33; moles of H = 6.7 ÷ 1 = 6.7; moles of O = 53.3 ÷ 16 = 3.33.
• Step 3: divide by the smallest, 3.33. That gives C : H : O = 1 : 2.01 : 1.
• Step 4: these are already whole numbers, so the empirical formula is CH₂O.

Every empirical question, whether from percentages or masses, is this same four-step walk.

From empirical to molecular formula

To go further you need one extra fact: the relative molecular mass, Mr. The molecular formula is always a whole-number multiple of the empirical formula, and you find that multiple, n, like this:

n = relative molecular mass ÷ empirical formula mass

Suppose the compound above has an Mr of 180. The empirical formula mass of CH₂O is 12 + (2 × 1) + 16 = 30. So n = 180 ÷ 30 = 6, and the molecular formula is (CH₂O)₆ = C₆H₁₂O₆, glucose. If instead the Mr had been 60, then n = 60 ÷ 30 = 2, giving C₂H₄O₂, ethanoic acid. Same empirical formula, different molecule, decided entirely by the Mr. You can drill this second stage at molecular formula.

When the data comes from an experiment

SPM often gives the numbers from a real experiment rather than a percentage. A classic is burning magnesium in oxygen and weighing the magnesium oxide formed. If 2.4 g of magnesium combines with 1.6 g of oxygen, then moles of Mg = 2.4 ÷ 24 = 0.1 and moles of O = 1.6 ÷ 16 = 0.1. The ratio is 1 : 1, so the empirical formula is MgO. The method is identical; only the source of the masses has changed.

The slips to avoid

Three mistakes cause most lost marks. First, dividing by the wrong relative atomic mass, always take the values from the Periodic Table you are given, and double-check O = 16, not 8. Second, forgetting step 3 and trying to read a ratio straight from the moles. Third, mishandling a non-whole ratio: a value like 1.33 means multiply everything by 3, and 1.5 means multiply by 2, never just round it off, because rounding changes the compound. Round only genuine experimental noise, such as 2.01 to 2.

Master this and empirical and molecular formulae become some of the most predictable marks in the paper. If you would like a teacher to check your ratios and catch a rounding habit before it costs you in the exam, our online one-to-one lessons with our experienced SPM Chemistry teachers give you exactly that kind of feedback, from RM50 an hour with a paid one-hour trial.