How do you find the empirical formula from percent composition?

Assume a 100 gram sample so each percent becomes grams. Divide each element's mass by its atomic weight to get moles. Divide every mole value by the smallest one to get the mole ratio, then multiply to the nearest whole numbers. Those whole numbers are the subscripts of the empirical formula. For example C 40.0%, H 6.7%, O 53.3% gives CH2O.

What is the difference between empirical and molecular formula?

The empirical formula is the simplest whole-number ratio of atoms, like CH2O. The molecular formula is the actual number of atoms in a molecule, like C6H12O6 for glucose, and is always a whole-number multiple of the empirical formula. To get it, divide the molar mass by the empirical-formula mass and multiply the subscripts by that factor.

How do you calculate percent composition?

For each element, percent by mass = (number of atoms × atomic weight) ÷ molar mass × 100. For water H2O (molar mass 18.02 g/mol): H is 2 × 1.008 ÷ 18.02 = 11.2% and O is 16.00 ÷ 18.02 = 88.8%. The tool parses any formula and shows the molar mass and each element's mass percent.

How does combustion analysis find a formula?

When a hydrocarbon (or CHO compound) burns, all carbon ends up in CO2 and all hydrogen in H2O. Moles of C equal moles of CO2; moles of H equal twice the moles of H2O. The mass of oxygen in the original sample is found by difference: sample mass minus the mass of C and H. Convert to moles, take the ratio, and you have the empirical formula.

Why are my mole ratios not whole numbers?

After dividing by the smallest number of moles you often get values like 1.5, 1.33 or 1.25. Multiply every ratio by a small integer (2, 3 or 4) to clear the fraction — 1.5 becomes 3 when multiplied by 2, and 1.33 becomes 4 when multiplied by 3. This tool does that automatically and shows the multiplier it used.

Is this empirical formula calculator free?

Yes. It is completely free, needs no signup, and uses exact IUPAC atomic weights through the open-source chempy library.

Empirical & Molecular Formula Calculator

Elements — one per line, element: value

Values are

Molar mass (g/mol, optional)

Assumes a 100 g sample (percent = grams). Give the molar mass to also get the molecular formula.

Chemical formula

Parentheses and hydrates supported. Returns molar mass and each element's mass percent.

Sample mass (g)

CO₂ produced (g)

H₂O produced (g)

Molar mass (optional)

All C ends up in CO₂, all H in H₂O; remaining mass is taken as oxygen (CₓHᵧOᵤ). Moles C = mol CO₂, moles H = 2 × mol H₂O.

From percent composition to a formula — the method

Percent → grams. Assume a 100 g sample, so 40.0% C means 40.0 g C.
Grams → moles. Divide each mass by the element's atomic weight.
Moles → ratio. Divide every mole value by the smallest one.
Clear fractions. Multiply by a small integer until the ratios are whole numbers — those are the empirical subscripts.
Empirical → molecular. Divide the molar mass by the empirical-formula mass and multiply the subscripts by that factor.

This tool runs every step on the server with the exact IUPAC atomic weights from chempy, and shows the moles, mole ratios and the multiplier it used so you can follow along.

Frequently asked

Treat each percent as grams (100 g sample), divide by atomic weights to get moles, divide all by the smallest, then multiply to whole numbers. C 40.0%, H 6.7%, O 53.3% → CH₂O.

Empirical is the simplest whole-number ratio (CH₂O); molecular is the real count (C₆H₁₂O₆ = glucose), always a whole multiple. Divide molar mass by empirical mass for the multiplier.

For each element, mass % = (atoms × atomic weight) ÷ molar mass × 100. Water: H = 11.2%, O = 88.8%.

All carbon becomes CO₂ and all hydrogen becomes H₂O. moles C = mol CO₂, moles H = 2 × mol H₂O, and oxygen mass is the sample mass minus C and H. Convert to moles and take the ratio.

Yes — free, no signup, exact atomic weights via the open-source chempy library.

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Empirical & Molecular Formula Calculator

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From percent composition to a formula — the method

Frequently asked

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