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Empirical formula from percent composition

Enter each element and its percentage by mass; get the simplest whole-number formula.

Work it out

Total 100.0%

CH2O

  • Take 100 g, so each percentage is a mass in grams.
  • C: divided by its atomic weight gives 3.33028 mol, which is 1 relative to the smallest.
  • H: divided by its atomic weight gives 6.64683 mol, which is 1.99588 relative to the smallest.
  • O: divided by its atomic weight gives 3.33146 mol, which is 1.00035 relative to the smallest.
  • Those are already whole numbers.

A worked example

40.0% carbon, 6.7% hydrogen, 53.3% oxygen.

CH₂O

  1. 1In 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
  2. 2Moles: 40.0/12.011 = 3.33; 6.7/1.008 = 6.65; 53.3/15.999 = 3.33.
  3. 3Divide by the smallest (3.33): 1.00, 2.00, 1.00.
  4. 4That gives CH₂O — the empirical formula of glucose, ribose and formaldehyde alike.

What this assumes

  • An empirical formula is the simplest ratio, not the molecular formula. CH₂O is the empirical formula of both formaldehyde (CH₂O) and glucose (C₆H₁₂O₆); you need the molar mass to tell them apart.
  • Percentages should total 100%. A shortfall usually means an element — most often oxygen — was measured by difference and left out.

The terms used here

Empirical formula
The simplest whole-number ratio of atoms in a compound.
Molecular formula
The actual count of atoms per molecule. Always a whole-number multiple of the empirical formula.
Percent composition
What fraction of a compound's mass each element accounts for.

This calculator runs in your browser. Nothing you type is sent anywhere, and no account is needed. If you want to go further than arithmetic — look a structure up, see what a reaction has been recorded as giving, or follow a mechanism — the library is open too.