Empirical vs Molecular Formula — How to Get Both from Percentages
Percentage composition questions look intimidating and are actually the most mechanical numericals in the whole mole-concept chapter. There is one procedure, it never changes, and once you have done it four times you will finish these questions in under two minutes. This article gives the procedure, four fully worked determinations — including a combustion-analysis problem — and the three places students consistently lose marks.
The difference in one line
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. The molecular formula is always a whole-number multiple of the empirical formula.
| Compound | Empirical formula | Molecular formula | n |
|---|---|---|---|
| Water | H₂O | H₂O | 1 |
| Hydrogen peroxide | HO | H₂O₂ | 2 |
| Benzene | CH | C₆H₆ | 6 |
| Glucose | CH₂O | C₆H₁₂O₆ | 6 |
| Sodium chloride (ionic) | NaCl | not defined — no discrete molecule | — |
That last row matters. Ionic solids have no molecules, so they have only a formula unit, which is the empirical formula. Asking for the "molecular formula of NaCl" is a trick question.
The five-step method
- Assume 100 g of the compound. Every percentage then becomes a mass in grams directly — 40.00% carbon becomes 40.00 g of carbon.
- Convert each mass to moles: moles = mass ÷ atomic mass.
- Divide every mole value by the smallest one. This gives a ratio starting at 1.
- Clear any fractions by multiplying all the numbers by the same small integer (×2 for .5, ×3 for .33 or .67, ×4 for .25 or .75).
- Find n from the given molar mass and multiply through.
Atomic masses used throughout: H = 1.008, C = 12.011, N = 14.007, O = 15.999, P = 30.974.
Worked example 1 — 40.00% C, 6.71% H, 53.29% O; M = 180.16 g/mol
Step 1–2 (moles in 100 g):
C: 40.00 ÷ 12.011 = 3.3303 mol
H: 6.71 ÷ 1.008 = 6.657 mol
O: 53.29 ÷ 15.999 = 3.3308 mol
Step 3 (divide by the smallest, 3.3303):
C = 1.000 · H = 1.999 · O = 1.000 → ratio 1 : 2 : 1
Empirical formula = CH₂O
Step 5: empirical formula mass = 12.011 + 2(1.008) + 15.999 = 30.026 g/mol.
n = 180.16 ÷ 30.026 = 6.00 → Molecular formula = C₆H₁₂O₆ (glucose).
Worked example 2 — 26.68% C, 2.24% H, 71.08% O; M = 90.03 g/mol
C: 26.68 ÷ 12.011 = 2.2213 mol
H: 2.24 ÷ 1.008 = 2.2222 mol
O: 71.08 ÷ 15.999 = 4.4428 mol
Divide by 2.2213: C = 1.000, H = 1.000, O = 2.000 → empirical formula CHO₂.
Empirical mass = 12.011 + 1.008 + 2(15.999) = 45.017 g/mol.
n = 90.03 ÷ 45.017 = 2.00 → Molecular formula = C₂H₂O₄ (oxalic acid).
Note that CHO₂ is not a real molecule — the empirical formula is only a ratio. That is exactly why the second half of the question exists.
Worked example 3 — a fractional ratio: 43.64% P, 56.36% O; M = 283.9 g/mol
P: 43.64 ÷ 30.974 = 1.4090 mol
O: 56.36 ÷ 15.999 = 3.5227 mol
Divide by 1.4090: P = 1.000, O = 2.500. A .5 appears, so multiply both by 2: P = 2, O = 5 → empirical formula P₂O₅.
Empirical mass = 2(30.974) + 5(15.999) = 61.948 + 79.995 = 141.943 g/mol.
n = 283.9 ÷ 141.943 = 2.00 → Molecular formula = P₄O₁₀.
This one is worth remembering: the oxide everyone calls "phosphorus pentoxide" is really P₄O₁₀. The common name uses the empirical formula.
Worked example 4 — combustion analysis
Boards and JEE often give combustion data instead of percentages. Burning a hydrocarbon sends every carbon into CO₂ and every hydrogen into H₂O, so you work backwards from the masses of those two products.
Problem. 0.5000 g of a hydrocarbon burns completely to give 1.5716 g of CO₂ and 0.6427 g of H₂O. Its molar mass is 56.11 g/mol. Find both formulae.
Carbon. M(CO₂) = 44.009 g/mol.
moles CO₂ = 1.5716 ÷ 44.009 = 0.035711 mol → 0.035711 mol C
mass of C = 0.035711 × 12.011 = 0.4289 g
Hydrogen. M(H₂O) = 18.015 g/mol. Each H₂O carries two H atoms.
moles H₂O = 0.6427 ÷ 18.015 = 0.035676 mol → 0.071352 mol H
mass of H = 0.071352 × 1.008 = 0.0719 g
Check for oxygen. 0.4289 + 0.0719 = 0.5008 g ≈ the 0.5000 g sample, so there is no oxygen in the compound — it really is a hydrocarbon.
Ratio. 0.071352 ÷ 0.035711 = 2.00 → C : H = 1 : 2 →
empirical formula CH₂, empirical mass = 12.011 + 2.016 = 14.027 g/mol.
n = 56.11 ÷ 14.027 = 4.00 → Molecular formula = C₄H₈.
If the masses of C and H had not added up to the sample mass, the shortfall would be the mass of oxygen in the compound, and you would convert that to moles as a third element. Oxygen can never be found from the CO₂ and H₂O directly, because oxygen from the air is mixed into both products.
- Rounding the mole ratio too aggressively. 1.33 is not 1 — it is 4/3, so multiply by 3. Only values within about ±0.02 of a whole number should be rounded. If you get 2.15, re-check your arithmetic before forcing it.
- Forgetting that H₂O has two hydrogens. The single most common combustion error. Moles of H = 2 × moles of H₂O.
- Dividing by the largest mole value instead of the smallest. You then get fractions below 1 and the ratio is much harder to read.
- Giving the empirical formula when the molar mass was supplied. If a molar mass appears anywhere in the question, the examiner wants the molecular formula. Finish the last step.
- Using mass numbers instead of atomic masses. Cl is 35.45, not 35; using 35 shifts the ratio enough to change the answer.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | "Some Basic Concepts of Chemistry" — a standard 3-mark numerical |
| JEE / NEET | Combustion analysis, percentage of an element by mass, formula from vapour density |
| IIT-JAM / CUET-PG | Formula determination combined with degree of unsaturation |
| GATE / CSIR-NET | Elemental analysis (CHN) data in structure-determination problems |
One shortcut worth carrying into the exam hall: for a gas, molar mass = 2 × vapour density. If a question gives vapour density 39, the molar mass is 78 g/mol, and with empirical formula CH that gives n = 78 ÷ 13.019 = 5.99 ≈ 6, so the compound is C₆H₆.
Check every step instantly. The Molar Mass & Composition calculator gives you the empirical formula mass, the molar mass of your final answer, and the element-wise percent composition — so you can verify your molecular formula by working the percentages forwards again.
Open the Molar Mass & Composition Calculator →Need structured help with Class 11–12 chemistry numericals? ABC Chemistry runs Class 11–12 coaching at its Gurugram centre and online across India, plus dedicated IIT-JAM, GATE and CSIR-NET batches — details at abcchemistry.in.