Empirical and Molecular Formulae
Two formulae describing the same compound in different ways. The calculation follows a fixed four-step routine — and the errors are almost all in step 3.
1. The difference
EMPIRICAL formula — the SIMPLEST WHOLE-NUMBER RATIO of atoms in a compound. MOLECULAR formula — the ACTUAL number of each atom in one molecule.
| Compound | Molecular | Empirical |
|---|---|---|
| Ethene | C₂H₄ | CH₂ |
| Butene | C₄H₈ | CH₂ |
| Hydrogen peroxide | H₂O₂ | HO |
| Glucose | C₆H₁₂O₆ | CH₂O |
| Water | H₂O | H₂O (already simplest) |
| Benzene | C₆H₆ | CH |
Confusing the two was recorded three separate times. Read the question carefully — “empirical” means simplify; “molecular” means the real formula.
The molecular formula is always a WHOLE-NUMBER MULTIPLE of the empirical formula.
Note that ethene and butene share the same empirical formula — the empirical formula alone does not identify a compound.
2. Calculating an empirical formula
1. Write down the MASS (or percentage) of each element. 2. DIVIDE each by its RELATIVE ATOMIC MASS → moles. 3. DIVIDE all the answers by the SMALLEST one. 4. Round to whole numbers → the ratio.
Worked example. A compound contains 2.4 g carbon and 0.8 g hydrogen.
| Carbon | Hydrogen | |
|---|---|---|
| Mass | 2.4 | 0.8 |
| ÷ Ar | 2.4 ÷ 12 = 0.2 | 0.8 ÷ 1 = 0.8 |
| ÷ smallest (0.2) | 1 | 4 |
Empirical formula = CH₄
With percentages: treat the percentages as grams out of 100 g and proceed identically.
Percentages convert straight to grams — a recorded confusion. If a compound is 40% carbon, take 40 g of carbon in a 100 g sample.
You MUST divide by the Ar to get moles. Not doing so was recorded — the percentages are masses, and the formula is a ratio of atoms, so you have to convert.
Divide by the SMALLEST, not by anything else. Confusion about why we divide by the smallest was recorded: it is simply to make the smallest value equal 1, which gives the simplest ratio.
Use the ATOMIC mass, not the diatomic molecular mass. A recorded error used 2 for hydrogen and 32 for oxygen. In a compound you have atoms, so use H = 1 and O = 16.
3. Handling awkward ratios
Sometimes step 3 gives numbers that are not close to whole.
If a value ends in .5, MULTIPLY EVERYTHING by 2. If it ends in .33 or .67, multiply everything by 3. If it ends in .25 or .75, multiply by 4.
Example: a ratio of 1 : 1.5 → multiply both by 2 → 2 : 3
Do NOT round 1.5 to 2. Incorrectly rounding empirical formula ratios, and struggling to convert decimals to whole numbers, were both recorded. Only round when the value is very close to a whole number (within about 0.1) — anything else needs multiplying up.
Values like 2.98 or 1.02 are rounding artefacts from the data — round those to 3 and 1.
4. From empirical to molecular formula
1. Find the Mr of the EMPIRICAL formula. 2. DIVIDE the compound’s actual Mr by that number → n. 3. MULTIPLY every subscript in the empirical formula by n.
Worked example. Empirical formula CH₂, and the compound’s Mr is 56.
- Mr of CH₂ = 12 + 2 = 14
- n = 56 ÷ 14 = 4
- Molecular formula = C₄H₈
Not knowing how to get from the empirical formula and Mr to the molecular formula was recorded — it is just this division and multiplication.
n must be a whole number. If it isn’t, check your empirical formula.
5. Empirical formula from a reaction
Some questions give the mass of a metal and the mass of its oxide.
Mass of oxygen = mass of oxide − mass of metal.
Then proceed as normal.
Example: 4.8 g of magnesium forms 8.0 g of magnesium oxide.
- Oxygen = 8.0 − 4.8 = 3.2 g
- Mg: 4.8 ÷ 24 = 0.2; O: 3.2 ÷ 16 = 0.2
- Ratio 1 : 1 → MgO
6. Mistakes that cost marks
Confusing empirical with molecular formula.
Not dividing by the Ar to get moles.
Using diatomic masses (2 for H, 32 for O) instead of atomic.
Dividing by the wrong value instead of the smallest.
Rounding 1.5 up to 2 instead of doubling everything.
Forgetting to multiply ALL the numbers when clearing a fraction.
Not simplifying the final ratio.
Multiplying the empirical formula by the wrong n.
Forgetting to subtract to find the oxygen mass.
Frequently asked questions
What is an empirical formula? The simplest whole-number ratio of atoms in a compound.
What is a molecular formula? The actual number of each atom in one molecule.
What is the empirical formula of C₆H₁₂O₆? CH₂O.
How do I calculate an empirical formula? Mass ÷ Ar for each element, then divide all by the smallest, then round or scale to whole numbers.
What do I do with percentages? Treat them as grams in 100 g and proceed the same way.
Why divide by the smallest? To make the smallest value 1, giving the simplest ratio.
What if I get 1.5? Multiply everything by 2 — don’t round.
How do I find the molecular formula? Divide the compound’s Mr by the empirical formula mass, then multiply all subscripts by that number.
Can two compounds share an empirical formula? Yes — ethene and butene are both CH₂.
How do I find the mass of oxygen in an oxide? Mass of oxide − mass of metal.
Quick revision checklist
- I know the difference between empirical and molecular
- I can give the empirical formula of common compounds
- I follow the four steps in order
- I always divide by Ar to get moles
- I use atomic, not diatomic, masses
- I divide by the smallest value
- I multiply up rather than rounding .5, .33 or .25
- I can find the molecular formula from Mr ÷ empirical mass
- I check n is a whole number
- I can find an oxygen mass by subtraction
These notes cover empirical and molecular formulae in the Cambridge IGCSE Chemistry (0620) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Chemistry lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and data booklet for your own exam series.
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