IB Chemistry SL
Topic 1 — Counting Particles by Mass
Paper 1 & 2
Core skill
~9 min read
Empirical Formulae
The empirical formula is a compound stripped down to its simplest atom ratio. From a set of masses or percentages you can find it in three steps — and with one extra piece of information, work all the way up to the molecular formula.
📘 What you need to know
- The molecular formula shows the actual number of atoms of each element in a molecule (e.g. ethanoic acid is C2H4O2).
- The empirical formula shows the simplest whole-number ratio of atoms (ethanoic acid is CH2O).
- Find the empirical formula by dividing each element’s mass (or %) by its Ar, then dividing by the smallest result.
- Ionic compounds are always written as empirical formulae.
- To get the molecular formula: divide the Mr by the empirical formula mass, then multiply the empirical formula by that whole number.
Empirical vs molecular formula
The two describe the same compound at different levels of detail.
- The molecular formula is the real count — ethanoic acid genuinely has 2 C, 4 H and 2 O atoms per molecule (C2H4O2).
- The empirical formula is that ratio reduced to its simplest form — dividing through by 2 gives CH2O.
Organic compounds often have different empirical and molecular formulae, while the formula of an ionic compound is always the empirical (simplest-ratio) form.
Finding the empirical formula
The method is always the same three steps, whether you’re given masses or percentages.
🧪 How to do it
- Write down the mass (or % — treat % as grams) of each element.
- Divide each by that element’s Ar to get moles.
- Divide every answer by the smallest of them to get the simplest whole-number ratio.
WORKED EXAMPLEA compound contains 10 g of hydrogen and 80 g of oxygen. Find its empirical formula. (Ar: H = 1.01, O = 16.00)
Divide mass by A_r
H: 10 ÷ 1.01 = 10 mol
O: 80 ÷ 16.00 = 5 mol
Divide by the smallest (5)
H: 10 ÷ 5 = 2 O: 5 ÷ 5 = 1
Empirical formula = H₂O
WORKED EXAMPLEA compound is 85.7% carbon and 14.3% hydrogen by mass. Find its empirical formula. (Ar: C = 12.01, H = 1.01)
Treat % as grams, divide by A_r
C: 85.7 ÷ 12.01 = 7.14 mol
H: 14.3 ÷ 1.01 = 14.2 mol
Divide by the smallest (7.14)
C: 1 H: 2
Empirical formula = CH₂
From empirical to molecular formula
The empirical formula only gives the ratio — to find the real molecular formula you need the compound’s relative molecular mass (Mr).
🧪 How to do it
- Work out the mass of the empirical formula (add up the Ar values).
- Divide the molecular mass (Mr) by that empirical mass — this gives a whole number.
- Multiply the empirical formula by that number.
WORKED EXAMPLEThe empirical formula of X is C4H10S and its Mr is 180.42. Find the molecular formula. (Ar: C = 12.01, H = 1.01, S = 32.07)
Empirical formula mass
(4 × 12.01) + (10 × 1.01) + 32.07 = 90.21
Divide M_r by empirical mass
180.42 ÷ 90.21 = 2
Multiply empirical formula by 2
Molecular formula = C₈H₂₀S₂
💡 Exam tip
- If a ratio comes out close to a whole number (like 2.01 or 0.99), round it — small errors come from rounding Ar values.
- If a ratio lands on something like 1.5, multiply everything by 2 to clear the fraction rather than rounding.
Up next: Concentration of Solutions — measuring how much solute is dissolved in a solution, in mol dm⁻³, g dm⁻³ and ppm.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
Book a Free Session →