Empirical formula calculator with working

Magnesium oxide: 2.4 g of Mg combines with 1.6 g of O.

Empirical formula: MgO.

A hydrocarbon: 80% carbon, 20% hydrogen. C: $80 \div 12 = 6.67$; H: $20 \div 1 = 20$. Dividing by 6.67 gives 1 : 3, so CH₃.

Iron oxide: 70% Fe, 30% O.

• Fe: $70 \div 56 = 1.25$; O: $30 \div 16 = 1.875$.
• Divide by 1.25: Fe 1, O 1.5.
• 1.5 isn't a whole number, so multiply both by 2: Fe 2, O 3.

Empirical formula: Fe₂O₃. Ratios ending in .5 need ×2; ending in .33 or .67 need ×3; ending in .25 or .75 need ×4. Don't round 1.5 to 2.

The molecular formula is a whole-number multiple of the empirical formula. Divide the compound's $M_r$ by the empirical formula's mass. CH₃ has a mass of 15; if the compound's $M_r$ is 30, the multiplier is 2, so the molecular formula is C₂H₆ (ethane). Our relative formula mass calculator gives the $M_r$ of any formula.

A classic practical heats magnesium ribbon in a crucible to form magnesium oxide. Weigh the crucible empty, with magnesium, and with the oxide after heating; the gain in mass is the oxygen. Lift the lid now and then to let air in, without letting the white smoke escape, or the oxygen mass will come out too low.

• Dividing by the atomic number instead of the relative atomic mass.
• Rounding too early (keep at least 2 decimal places in the ratio).
• Using a capital letter wrongly in symbols (Co is cobalt; CO is carbon monoxide).

Practise with our empirical formula questions.

• Lay out your working as a table with a column for each element.
• Ratios within about 0.1 of a whole number can be rounded; anything ending in .5, .33 or .25 needs multiplying.
• If the masses don't add up to the compound's mass, the missing mass may be oxygen: subtract to find it.