“A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula.” If a question like that makes you freeze, the good news is that it follows one fixed procedure every single time. This guide teaches that procedure from the mole concept, works through three examples including the awkward “decimal” cases, and shows how to go on to the molecular formula when the exam asks for it.
What an empirical formula is
The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound. It is not always the real molecule: glucose is really C₆H₁₂O₆, but its empirical formula is CH₂O, because 6 : 12 : 6 simplifies to 1 : 2 : 1. Percentage composition questions almost always ask for the empirical formula first, because percentages only tell you the ratio of masses, not the actual number of atoms in one molecule.
The five-step method
The whole calculation is the same five steps in the same order. Set your work out as a table with a row for each element.
- Take the percentages as masses. Assume you have a 100 g sample, so a percentage becomes that many grams. 40.0% carbon becomes 40.0 g of carbon.
- Divide each mass by the relative atomic mass (Ar) of that element. This converts mass into moles, because moles = mass ÷ Ar. Use the Ar values you are given (C = 12, H = 1, O = 16, and so on).
- Find the smallest of your mole values, then divide every mole value by that smallest number. This scales the ratio so the smallest becomes 1.
- Turn the results into whole numbers. If they are already close to whole numbers, round them. If one ends in a clear fraction, like .5, .33 or .25, multiply every value by 2, 3 or 4 to clear it.
- Write the empirical formula using those whole numbers as the subscripts.
Worked example 1: the clean case
A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Find its empirical formula.
| Element | Mass (g) | ÷ Ar | Moles | ÷ smallest (3.33) |
|---|---|---|---|---|
| C | 40.0 | ÷ 12 | 3.33 | 1 |
| H | 6.7 | ÷ 1 | 6.7 | 2.01 ≈ 2 |
| O | 53.3 | ÷ 16 | 3.33 | 1 |
The ratio C : H : O is 1 : 2 : 1, so the empirical formula is CH₂O. Notice the tiny rounding on hydrogen (2.01 to 2), small differences from rounding the data are expected, so do not panic if a value is 1.98 or 2.02.
Worked example 2: clearing a “.5”
An oxide of iron contains 70.0% Fe and 30.0% O by mass. Find its empirical formula. (Ar: Fe = 56, O = 16.)
- Fe: 70.0 ÷ 56 = 1.25
- O: 30.0 ÷ 16 = 1.875
Divide by the smaller value, 1.25:
- Fe: 1.25 ÷ 1.25 = 1
- O: 1.875 ÷ 1.25 = 1.5
A ratio of 1 : 1.5 is not whole, so multiply both by 2 to get 2 : 3. The empirical formula is Fe₂O₃, iron(III) oxide. This is the step students most often get wrong: never round 1.5 to 2. When you see a clear .5, double everything.
Worked example 3: starting from masses, not percentages
The same method works when a question gives you masses directly, you simply skip step 1. 4.8 g of magnesium combines with 3.2 g of oxygen. Find the empirical formula. (Ar: Mg = 24, O = 16.)
- Mg: 4.8 ÷ 24 = 0.2
- O: 3.2 ÷ 16 = 0.2
Divide by 0.2: both give 1, so the ratio is 1 : 1 and the empirical formula is MgO. This is exactly the reasoning behind the classic magnesium-oxide experiment, where you find the formula by burning a known mass of magnesium. You can review the full calculation pattern on our empirical formula page.
Going on to the molecular formula
Sometimes the question also gives the relative molecular mass (Mr) and asks for the molecular formula, the actual number of atoms in one molecule. Two more steps do it:
- Work out the mass of one empirical-formula unit. For CH₂O that is 12 + (2 × 1) + 16 = 30.
- Divide the given Mr by that empirical mass to find n, then multiply every subscript by n.
If CH₂O has an Mr of 180, then n = 180 ÷ 30 = 6, so the molecular formula is (CH₂O)₆ = C₆H₁₂O₆ (glucose). Always find the empirical formula first, then scale up.
Common mistakes to avoid
- Dividing by Mr instead of Ar in step 2. You are converting each element’s mass to moles, so divide by the atomic mass of that element.
- Rounding a .5 or .33 too early. Multiply the whole ratio up to clear the fraction instead.
- Forgetting to divide by the smallest. Raw mole values are rarely a neat ratio until you scale them.
- Confusing empirical and molecular formulae. The empirical formula is the simplest ratio; the molecular formula needs the Mr as well.
Practise until the table is automatic
Every percentage-composition question is the same table: mass, divide by Ar, divide by smallest, clear fractions, write the formula. Drill a few until you can set the table up without thinking, then add the molecular-formula step. While you learn you can check answers with our empirical formula calculator, but aim to do it unaided, since no calculator does the reasoning for you in the exam. If the “.5” and “.33” cases keep tripping you, that is a five-minute fix with a teacher. Our online one-to-one lessons run in English from RM50 an hour, with a paid one-hour trial.
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