Empirical Formula Calculator

Quick Answer

Last updated: 3 November 2026 ·How we calculate this: the mole ratio method

How This Empirical Formula Calculator Works

Type the symbol of each element in the compound and how much of it there is. You can use percent by mass or grams from an experiment. The calculator divides each amount by the element’s atomic weight to get moles, divides every result by the smallest, and then finds the smallest whole numbers that fit. It shows the empirical formula, its formula mass and a table of every ratio.

If you also know the compound’s molar mass, enter it. The calculator divides the molar mass by the empirical formula mass and multiplies the subscripts to give the molecular formula.

How to Use the Empirical Formula Calculator

1. Enter the symbol of each element, such as C, H or O.
2. Enter each amount as a percent or in grams.
3. Optionally enter the molar mass to get the molecular formula.
4. Press Calculate.
5. Read the empirical formula, the mole ratios and the steps.

The Steps

1. Moles: moles of each element = mass ÷ atomic weight.
2. Ratio: divide every mole value by the smallest one.
3. Whole numbers: if a ratio ends in about .5, multiply all by 2; about .33 or .67, by 3; about .25 or .75, by 4.
4. Molecular formula: n = molar mass ÷ empirical formula mass, then multiply each subscript by n.

Key Terms

• Empirical formula: the simplest whole-number ratio of atoms in a compound.
• Molecular formula: the actual number of atoms of each element in a molecule.
• Mole ratio: the ratio of moles of each element in a compound.
• Percent composition: each element’s share of a compound’s mass.
• Atomic weight: the average mass of an element’s atoms, in g/mol.

Worked Example

An iron oxide is 69.94% iron and 30.06% oxygen by mass. Take a 100 g sample.

• Fe: 69.94 ÷ 55.845 = 1.2524 mol
• O: 30.06 ÷ 15.999 = 1.8789 mol
• Divide by the smaller: Fe = 1, O = 1.5
• Multiply by 2 to clear the half: Fe = 2, O = 3, so the formula is Fe₂O₃

Real-Life Examples

• Homework: 40.0% C, 6.71% H and 53.3% O gives CH₂O.
• Metal oxides: 69.94% Fe and 30.06% O gives Fe₂O₃.
• Phosphorus oxide: 43.64% P and 56.36% O gives P₂O₅.
• Molecular formula: CH₂O with a molar mass of 180.16 g/mol is C₆H₁₂O₆.

Empirical Formula vs Molecular Formula

• Empirical formula: the simplest whole-number ratio of atoms.
• Molecular formula: the actual number of each atom in one molecule.
• Link: the molecular formula is a whole-number multiple of the empirical formula.

Empirical and Molecular Formulas

Tips for Good Results

• Use enough digits: rounding the data too early can turn 1.5 into 1.4 and give the wrong formula.
• Check the total: percentages should add to about 100%. If oxygen is missing, it is often the remainder.
• Watch the case: Co is cobalt, but CO is carbon and oxygen.
• Expect small errors: real lab data rarely gives perfect whole numbers.

Empirical Formulas From Combustion Analysis

Chemists often find the make-up of an organic compound by burning a sample and weighing the carbon dioxide and water it gives off. Every carbon atom ends up in CO₂ and every hydrogen atom ends up in H₂O, so the masses of those products tell you how much carbon and hydrogen were in the sample.

• Carbon: grams of C = grams of CO₂ × 12.011 ÷ 44.009.
• Hydrogen: grams of H = grams of H₂O × 2.016 ÷ 18.015.
• Oxygen: the sample mass minus the carbon and hydrogen, if the compound has only C, H and O.

Enter those three masses in grams here to get the empirical formula.

Common Mistakes to Avoid

• Using percentages without moles. Divide by atomic weights first; percent ratios are not atom ratios.
• Rounding 1.5 to 2. A ratio of 1.5 means multiply everything by 2.
• Rounding too early. Keep at least three or four significant figures.
• Wrong capital letters. Element symbols are case sensitive.
• Confusing empirical and molecular formulas. You need the molar mass to get the molecular one.

Frequently Asked Questions

References & Sources