Empirical and molecular formulae

An empirical formula is the simplest whole-number ratio of atoms in a compound. You find it from the masses or percentages of the elements: work out the moles of each element, then divide by the smallest to get the ratio. If you also know the Mᵣ, you can scale the empirical formula up to the molecular formula: the actual number of each atom in one molecule.

Key facts

  • Empirical formula: the simplest whole-number ratio of atoms (ethane, C2H6, has the empirical formula CH3).
  • Molecular formula: the actual number of each atom in one molecule.
  • Method: mass of each element → moles (÷ Aᵣ) → divide by the smallest → whole numbers.
  • With percentages, imagine 100 g of the compound: the percentages become masses in grams.
  • If a ratio ends in .5, double every number; if it ends in .33 or .67, multiply every number by 3.
  • Molecular formula = empirical formula × (Mᵣ ÷ the mass of one empirical formula unit).

Worked examples

Example 1

2.4 g of magnesium burns to form 4.0 g of magnesium oxide. Find the empirical formula. (Aᵣ: Mg = 24, O = 16)

  1. Mass of oxygen that joined on. 4.0 − 2.4 = 1.6 g
  2. Moles of each element. Mg: 2.4 ÷ 24 = 0.1; O: 1.6 ÷ 16 = 0.1
  3. Divide by the smallest (0.1). Mg : O = 1 : 1
  4. So the empirical formula is MgO.

Answer: MgO

Example 2

A compound is 82.8% carbon and 17.2% hydrogen by mass, and its Mᵣ is 58. Find its empirical and molecular formulae. (Aᵣ: C = 12, H = 1)

  1. In 100 g there are 82.8 g of carbon and 17.2 g of hydrogen.
  2. Moles of each element. C: 82.8 ÷ 12 = 6.9; H: 17.2 ÷ 1 = 17.2
  3. Divide by the smallest. C : H = 1 : 2.49
  4. 2.49 is close to 2.5, so double both: C : H = 2 : 5. The empirical formula is C2H5.
  5. One C2H5 unit has a mass of 29. 58 ÷ 29 = 2
  6. Double every number: the molecular formula is C4H10 (butane).

Answer: C2H5; C4H10

Example 3

A compound contains 1.2 g of carbon, 0.2 g of hydrogen and 1.6 g of oxygen. Find its empirical formula. (Aᵣ: C = 12, H = 1, O = 16)

  1. Moles of each element. C: 1.2 ÷ 12 = 0.1; H: 0.2 ÷ 1 = 0.2; O: 1.6 ÷ 16 = 0.1
  2. Divide by the smallest (0.1). C : H : O = 1 : 2 : 1
  3. So the empirical formula is CH2O.

Answer: CH2O