ChemBench

Empirical Formula Calculator

From raw masses to the simplest formula.

Element 1

Mass % or g

Element 2

Mass % or g

Element 3

Mass % or g

Empirical formula

CH2O

What you entered

C: 40H: 6.7O: 53.3

1
Moles of C
mass ÷ atomic weight= 3.3303
2
Moles of H
mass ÷ atomic weight= 6.6468
3
Moles of O
mass ÷ atomic weight= 3.3315
4
Divide each by the smallest mole value
÷ 3.3303= C: 1, H: 1.9959, O: 1.0004

Result

Empirical formula: CH2O

The simplest whole-number mole ratio of these elements gives the empirical formula CH2O.

Empirical formulas from mass composition

Convert each element's mass to moles, then divide every mole count by the smallest to get the simplest whole-number ratio.

Composition by mass	Empirical formula	A compound that matches
C 40.0 g, H 6.7 g, O 53.3 g	CH2O	Glucose (C6H12O6)
C 92.3 g, H 7.7 g	CH	Benzene (C6H6)
N 30.4 g, O 69.6 g	NO2	Nitrogen dioxide
Na 39.3 g, Cl 60.7 g	NaCl	Sodium chloride

The empirical formula gives the ratio, not the molecule. Glucose, formaldehyde, and ribose all reduce to CH2O; you need the molar mass to recover the molecular formula.

Convert mass to moles, then find the smallest ratio

Each element's mass is divided by its atomic weight to get moles, then every mole value is divided by the smallest one — this converts arbitrary masses into the simplest whole-number ratio of atoms.

Empirical vs. molecular formula

The empirical formula is the simplest ratio (CH₂O), while the molecular formula could be a multiple of it (C₆H₁₂O₆ for glucose) — empirical formula alone can't tell you which multiple without also knowing the molar mass.

Frequently asked questions

My combustion analysis gave 40.0% C, 6.7% H, and 53.3% O — what is the empirical formula?

Convert to moles: C = 40.0/12.01 = 3.33, H = 6.7/1.008 = 6.65, O = 53.3/16.00 = 3.33. Divide by the smallest (3.33): C = 1, H = 2, O = 1. The empirical formula is CH2O.

How do I get the molecular formula from the empirical formula?

Divide the known molar mass by the empirical formula mass. For CH2O (empirical mass 30.03 g/mol) with a measured molar mass of 180.16 g/mol: 180.16/30.03 = 6, so the molecular formula is C6H12O6 (glucose). Use the molar mass calculator to find the exact empirical formula mass.

What if the mole ratios don't come out to whole numbers?

Multiply all ratios by the smallest integer that makes them whole. If you get a ratio of 1:1.5, multiply by 2 to get 2:3. Common multipliers are 2 (for x.5), 3 (for x.33), and 4 (for x.25).

Can two different compounds have the same empirical formula?

Yes. Acetic acid (C2H4O2) and glucose (C6H12O6) both reduce to the same empirical formula CH2O. You need the molar mass to distinguish them.