Empirical and Molecular Formulas: From Percent Composition to the Real Formula

Find an empirical formula from percent composition or masses, then use molar mass to get the molecular formula. Worked examples with every step.

By the Dayob Editorial Team · Updated · 2 min read

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Analysis of an unknown compound often tells you its percent composition, how much of its mass comes from each element. From that you can find the empirical formula, the simplest whole-number ratio of atoms. Add the molar mass and you get the molecular formula, the actual number of each atom in a molecule.

Empirical vs. molecular formula

CompoundMolecular formulaEmpirical formula
GlucoseC₆H₁₂O₆CH₂O
Hydrogen peroxideH₂O₂HO
BenzeneC₆H₆CH
WaterH₂OH₂O

Step 1: percent to grams

Assume a 100 g sample so each percentage becomes grams. A compound that is 40.0% C, 6.7% H and 53.3% O contains 40.0 g C, 6.7 g H and 53.3 g O.

Step 2: grams to moles

  • C: 40.0 ÷ 12.011 = 3.33 mol
  • H: 6.7 ÷ 1.008 = 6.65 mol
  • O: 53.3 ÷ 15.999 = 3.33 mol

Step 3: divide by the smallest

Divide each by 3.33: C = 1.00, H = 2.00, O = 1.00. The empirical formula is CH₂O.

Step 4: fix non-whole ratios

If a ratio comes out near 1.5, 1.33 or 1.25, multiply all ratios by 2, 3 or 4 respectively. For example, ratios of 1 : 1.5 become 2 : 3. Values like 1.98 or 2.03 are just rounding from experimental data and can be rounded to whole numbers.

Step 5: the molecular formula

Divide the measured molar mass by the empirical formula's mass to get a whole-number multiplier.

  • Empirical formula mass of CH₂O: 12.011 + 2(1.008) + 15.999 = 30.03 g/mol
  • Measured molar mass: 180 g/mol
  • Multiplier: 180 ÷ 30.03 ≈ 6
  • Molecular formula: C₆H₁₂O₆

Check it in the molecular weight calculator: C6H12O6 gives 180.16 g/mol, and the mass percent column matches the original data. Mass percent is explained in percent composition.

Second example: from masses

A 2.50 g sample contains 1.75 g of iron and the rest oxygen.

  • O: 2.50 − 1.75 = 0.75 g
  • Fe: 1.75 ÷ 55.845 = 0.0313 mol; O: 0.75 ÷ 15.999 = 0.0469 mol
  • Divide by 0.0313: Fe = 1, O = 1.50
  • Multiply by 2: Fe₂O₃

Tips

  • Keep at least three significant figures until the division step.
  • The ratios are just like simplifying a ratio, see how to simplify ratios, except experimental data rarely gives exact whole numbers.
  • Ionic compounds are always written as empirical formulas, since they don't form discrete molecules.

Further reading from official sources

More molecular weight guides

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