Empirical vs Molecular Formula
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 — always a whole-number multiple of the empirical formula. Tell them apart and the calculations follow a fixed recipe.
The difference in one line
Glucose has the molecular formula C₆H₁₂O₆ but the empirical formula CH₂O — the same 1 : 2 : 1 ratio, scaled down. Many compounds share an empirical formula but differ in molecular formula: formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) all reduce to CH₂O, yet they are completely different substances.
| Compound | Molecular formula | Empirical formula |
|---|---|---|
| Hydrogen peroxide | H₂O₂ | HO |
| Benzene | C₆H₆ | CH |
| Glucose | C₆H₁₂O₆ | CH₂O |
| Water | H₂O | H₂O (already simplest) |
Finding the empirical formula
- Take the mass (or percent) of each element. Percent works directly — assume a 100 g sample so percentages become grams.
- Convert each to moles by dividing by its atomic mass.
- Divide every mole value by the smallest one to get a ratio.
- If the ratio is not whole numbers, multiply all of them by a small integer until it is.
Finding the molecular formula
You also need the molar mass of the whole compound, usually given or found from a separate experiment. Divide it by the empirical-formula mass to get the multiplier n, then multiply every subscript by n:
n is always a whole number (within rounding). If it comes out near 2.0, the molecular formula is double the empirical one, and so on.
Worked example: glucose
A compound is 40.0% C, 6.71% H, 53.3% O, with a molar mass of 180 g/mol.
- Moles in 100 g: C 40.0 ÷ 12.01 = 3.33; H 6.71 ÷ 1.008 = 6.66; O 53.3 ÷ 16.00 = 3.33.
- Divide by the smallest (3.33): C 1, H 2, O 1 → empirical CH₂O (mass 30.03 g/mol).
- n = 180 ÷ 30.03 ≈ 6 → molecular formula C₆H₁₂O₆.
Worked example: combustion analysis
Combustion analysis is the classic way these problems are dressed up. A hydrocarbon is burned, and all its carbon ends up as CO₂ while all its hydrogen ends up as H₂O. From the masses of those products you back-calculate the original element masses.
Burning a sample gives 0.660 g CO₂ and 0.270 g H₂O:
H: 0.270 × (2.016 ÷ 18.02) = 0.0302 g → 0.0300 mol
ratio C : H = 0.0150 : 0.0300 = 1 : 2 → empirical formula CH₂
If a separate measurement gave a molar mass of 28 g/mol, then n = 28 ÷ 14.03 = 2 and the molecular formula is C₂H₄ (ethene).
Common mistakes
- Rounding the mole ratio too aggressively, turning 1.33 into 1 instead of multiplying by 3.
- Forgetting oxygen in combustion analysis when the compound contains it — find O by subtracting the C and H masses from the sample mass.
- Confusing the two formulas: the molar mass is needed only for the molecular formula, never for the empirical one.
Check empirical-formula masses quickly with the Molar Mass Calculator, and see how moles drive everything in How to Approach Stoichiometry.
The General Chemistry Workbook covers empirical and molecular formulas, including combustion analysis, with worked solutions.
Frequently Asked Questions
The empirical formula is the simplest whole-number ratio of atoms, while the molecular formula is the actual count of atoms in one molecule. The molecular formula is always a whole-number multiple of the empirical formula.
Assume a 100 g sample so percentages become grams, convert each element to moles, divide every value by the smallest, and scale up to whole numbers if needed. The result is the empirical formula.
Divide the compound's molar mass by the empirical-formula mass to get a whole-number multiplier n, then multiply every subscript in the empirical formula by n.
A technique where a compound is burned so all its carbon becomes CO₂ and all its hydrogen becomes H₂O. The masses of those products are used to back-calculate the carbon and hydrogen in the original sample.
Yes. Formaldehyde, acetic acid and glucose all reduce to CH₂O despite being different substances. That is exactly why the molar mass is needed to pin down the molecular formula.