Empirical Formula Calculator
Turn percent composition into the empirical formula, and into the molecular formula if you know the molar mass.
Supply this to get the molecular formula as well as the empirical one.
Molecular formula
C₆H₁₂O₆
Empirical formula CH₂O, empirical mass 30.026 g/mol, multiplier 6
How the ratio was found
| Element | % | ÷ atomic mass | ÷ smallest | Subscript |
|---|---|---|---|---|
| Carbon (C) | 40 | 3.3303 | 1.000 | 1 |
| Hydrogen (H) | 6.7 | 6.6468 | 1.996 | 2 |
| Oxygen (O) | 53.3 | 3.3315 | 1.000 | 1 |
The method, in order
- Treat each percentage as grams in a 100 g sample.
- Divide by the atomic weight to convert each mass to moles.
- Divide every mole figure by the smallest of them.
- If the ratios are not whole numbers, multiply them all by the smallest factor that makes them whole.
The table under the answer shows each of those steps for your own numbers, which is what makes it checkable against a worked example in a textbook.
Empirical against molecular
An empirical formula alone cannot identify a compound. CH is the empirical formula of both acetylene, C₂H₂, and benzene, C₆H₆ — same ratio, very different substances. That is why the molar mass is needed to get from one to the other.
Related: molar mass calculator, chemical equation balancer, periodic table.
How it works
An empirical formula is the simplest whole-number ratio of atoms in a compound. You get there by turning each element's mass share into moles, then scaling the smallest to 1 — because a formula counts atoms, and only moles count atoms.
Empirical formula from percent composition
moles = percent ÷ atomic weight ratio = moles ÷ smallest moles molecular formula = empirical formula × n, where n = molar mass ÷ empirical mass
- percent
- the element's share of the compound's mass
- atomic weight
- from the periodic table, g/mol
- n
- whole-number multiplier from empirical to molecular formula
A compound that is 40.0% C, 6.7% H, 53.3% O, molar mass 180 g/mol
- C 40.0%, H 6.7%, O 53.3%
- Molar mass 180 g/mol
- C: 40.0 ÷ 12.011 = 3.33 mol
- H: 6.7 ÷ 1.008 = 6.65 mol
- O: 53.3 ÷ 15.999 = 3.33 mol
- Divide by the smallest, 3.33 → C 1, H 2, O 1
- Empirical formula CH₂O, empirical mass 30.03
- n = 180 ÷ 30.03 = 6
Empirical formula CH₂O; molecular formula C₆H₁₂O₆ — glucose.
Good to know
- Assume 100 g of compound and each percentage becomes a mass in grams. That is why percentages can be divided by atomic weights directly.
- A ratio near .5, .33 or .25 is a real fraction, not rounding error. 1 : 1.5 is 2 : 3 — multiply through rather than rounding down.
- An empirical formula alone cannot identify a compound: CH is the empirical formula of both acetylene, C₂H₂, and benzene, C₆H₆. The molar mass is what separates them.
- If your percentages fall well short of 100%, an element is usually missing — oxygen most often, because it is frequently found by difference rather than measured.
Related Calculators
Frequently Asked Questions
What is an empirical formula?
The empirical formula is the simplest whole-number ratio of the elements in a compound. Glucose is C₆H₁₂O₆, but its empirical formula is CH₂O — the same 1:2:1 ratio, reduced as far as it will go.
How do you find an empirical formula from percentages?
Assume 100 g of the compound, so each percentage becomes a mass in grams. Divide each mass by that element’s atomic weight to get moles, divide every result by the smallest one, and round the ratios to whole numbers — multiplying through if you land on a clean half or third.
When do I multiply the ratio instead of rounding?
When a ratio lands near .5, .33 or .25 it is not rounding error, it is a genuine fraction. A ratio of 1 : 1.5 is really 2 : 3, so multiply everything by 2. Rounding 1.5 down to 1 would give the wrong compound.
How do I get the molecular formula?
Divide the compound’s actual molar mass by the mass of the empirical formula. The result is a whole number, and multiplying every subscript by it gives the molecular formula. For glucose: 180 ÷ 30.03 = 6, and CH₂O becomes C₆H₁₂O₆.
Why do my percentages not add to 100?
Usually an element has been left out — oxygen is the common omission, because it is often found by difference rather than measured. A gap of more than a percent or two is worth chasing before trusting the formula.