The empirical formula is the simplest whole-number ratio of atoms. To find it, convert each mass to moles, divide by the smallest, and clear any fractions.
This lesson is part of moles, formulas and equations. It uses the conversions from converting between mass, moles and particles.
What are the steps to an empirical formula?
There are four steps, and they work for both percentages and masses.
- Write the mass of each element, treating percentages as grams in a 100 g sample.
- Divide each mass by its relative atomic mass to get moles.
- Divide every mole value by the smallest one.
- If any value is not close to a whole number, multiply all values by the same small integer.
Worked example: a clean ratio
A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. (C = 12, H = 1, O = 16.)
| Element | Mass (g) | ÷ Ar = moles | ÷ smallest |
|---|---|---|---|
| C | 40.0 | 3.33 | 1 |
| H | 6.7 | 6.7 | 2 |
| O | 53.3 | 3.33 | 1 |
The empirical formula is CH₂O, with mass 12 + 2 + 16 = 30. Dividing 180 by 30 gives 6, so the molecular formula is C₆H₁₂O₆.
Worked example: a ratio that needs doubling
An iron oxide contains 70% iron and 30% oxygen by mass. (Fe = 56, O = 16.)
Moles of Fe = 70 ÷ 56 = 1.25. Moles of O = 30 ÷ 16 = 1.875. Dividing by 1.25 gives Fe 1 and O 1.5.
A ratio of 1 : 1.5 is not whole, and rounding to 1 : 2 would give the wrong compound. Multiply both by 2 to get 2 : 3, so the formula is Fe₂O₃.
The mistake that costs marks
The common slip is rounding a fraction such as 1.5 to the nearest whole number. A second slip is dividing by the proton number instead of the relative atomic mass.
| Ratio found | Wrong move | Right move |
|---|---|---|
| 1 : 1.5 | Round to 1 : 2 | Multiply by 2 to get 2 : 3 |
| 1 : 1.33 | Round to 1 : 1 | Multiply by 3 to get 3 : 4 |
| 1 : 2.02 | Multiply by 2 | Round to 1 : 2 |
Fractions with .5 need doubling, .33 or .67 need tripling, and .25 or .75 need quadrupling. The mole and stoichiometry steps tool gives a template for the moles table if you want to try your own data.
Check yourself
A hydrocarbon is 85.7% carbon and 14.3% hydrogen by mass, with relative molecular mass 56. Find its molecular formula. (C = 12, H = 1.)
Answer
Moles of C = 85.7 ÷ 12 = 7.14. Moles of H = 14.3 ÷ 1 = 14.3.
Divide by 7.14: C = 1, H = 2.0. The empirical formula is CH₂, with mass 14.
56 ÷ 14 = 4, so the molecular formula is C₄H₈.
What to study next
Next, use formulas in reactions with solving reacting-mass and gas-volume calculations. Practise the full chapter with the moles practice set.
If you want a teacher to watch you work through unfamiliar ratios, see online one-to-one Chemistry tuition.