Empirical and Molecular Formulae

Hand an examiner 85.7% carbon and 14.3% hydrogen and they expect C, H and a clean CH2 within four lines. This is the most algorithmic question type on Paper 4, worth 2-4 marks, and Supplement-only. Candidates lose marks here in two ways: rounding 1.33 to 1, and stopping at the empirical formula when the question asked for the molecular one. Both are procedure errors, and procedure is exactly what this page installs on top of the Ar values you already use.

The two formulae defined (Supplement)

The empirical formula gives the simplest whole-number ratio of the atoms (or ions) of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule, and is always a whole-number multiple of the empirical formula. Glucose: molecular C6H12O6, empirical CH2O. They can coincide: water’s empirical and molecular formulae are both H2O. For ionic and giant structures, the formula quoted (NaCl, SiO2) is inherently empirical: it states a ratio, because no molecule exists.

From composition to empirical formula (Supplement)

The algorithm, which works identically for masses and for percentages (treat % as grams per 100 g):

Step	Action	Example: 80% C, 20% H
1	Divide each mass/% by the element’s Ar	C: 80 ÷ 12 = 6.67; H: 20 ÷ 1 = 20
2	Divide all results by the smallest	C: 6.67 ÷ 6.67 = 1; H: 20 ÷ 6.67 = 3
3	Clear non-integers by multiplying (×2 for .5, ×3 for .33/.67)	Already whole: ratio 1 : 3
4	Write the formula	CH3

The step-3 rule matters: a ratio of 1 : 1.5 is not “about 1 : 2”; multiply both by 2 to get 2 : 3. A ratio ending in .33 or .67 multiplies by 3. Rounding these to the nearest integer is the costliest error in the topic.

Water-of-crystallisation questions run the same algorithm with formula units instead of atoms: given the masses of anhydrous salt and of water lost on heating, divide each by its Mr and take the ratio to find x in, say, MgSO4·xH2O.

From empirical to molecular formula (Supplement)

The molecular formula needs one extra datum: the Mr, usually given (or derivable from a mole calculation such as M = m/n, using the methods in the mole subtopic). Compute the empirical formula mass, then multiplier = Mr ÷ empirical mass, then scale every subscript. CH3 has mass 15; with Mr = 30 the multiplier is 2 and the molecular formula is C2H6, ethane, which is how this skill resurfaces inside organic chemistry questions. More mixed practice sits in our mole calculations technique guide.

Worked exam question

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56. (a) Calculate the empirical formula of the hydrocarbon. (3) (b) Deduce its molecular formula. (2) [Ar: C 12, H 1]

Model answer: (a) C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3 (1). Divide by smallest: C = 7.14/7.14 = 1, H = 14.3/7.14 = 2 (1). Empirical formula CH2 (1). (b) Empirical formula mass = 12 + 2 = 14; 56 ÷ 14 = 4 (1). Molecular formula = C4H8 (1).

Mark-by-mark: (a) the divide-by-Ar line, the divide-by-smallest line and the formula are three separate awards. A bare “CH2” with no working risks 1 of 3. (b) the multiplier calculation earns its own mark before the final formula; writing CH2 again, or “C4H8 = butene” without the formula, leaves marks behind. Hydrocarbon percentages must sum to 100. If they don’t, re-read the question for a third element.

The mistakes that cost marks

1. Rounding ratios prematurely: 1 : 1.33 forced to 1 : 1 instead of multiplied to 3 : 4. Anything within about 0.1 of a whole number rounds; thirds and halves multiply.
2. Dividing by the wrong quantity: percentages divided by 100 first, or by the compound’s Mr instead of each element’s Ar. Each element gets its own Ar.
3. Stopping at the empirical formula when the question says molecular (or vice versa). The Mr in the question is the signal that a second stage exists.
4. Untabulated working. Numbers scattered around the page cannot be matched to marking points; two labelled rows, one per element, make every mark visible.

How examiners want it phrased

Typical student wording	Accepted mark-scheme wording
”I divided and got CH2-ish"	"C: 85.7/12 = 7.14; H: 14.3/1 = 14.3; ratio 1 : 2, empirical formula CH2"
"Then I doubled it because it felt small"	"Mr ÷ empirical formula mass = 56 ÷ 14 = 4, so molecular formula = C4H8"
"Empirical is the small version"	"The empirical formula is the simplest whole-number ratio of atoms in the compound"
"1.5 rounds to 2"	"Ratio 1 : 1.5 = 2 : 3 (multiply both by 2)“

The Malaysia note

This calculation rewards exam-week freshness more than understanding. Malaysian students who learned it comfortably in class still botch the .33 multiplication under May/June pressure if they have not drilled it recently. Keep one empirical-formula question in every revision session from March onward; the algorithm decays fast and reloads fast. Want your child’s calculation routine pressure-tested before the real paper? That is a standing offer in the free 1-hour trial lesson.

Test yourself

All three are Supplement level. Set the working out in a table, then click to check.

Q1 (3 marks). A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Calculate its empirical formula. [Ar: C 12, H 1, O 16]

Show answer

• Divide by Ar: C 40.0/12 = 3.33; H 6.7/1 = 6.7; O 53.3/16 = 3.33 [1] • Divide by smallest: C = 1, H = 2, O = 1 [1] • Empirical formula CH2O [1]

Q2 (2 marks). A compound has the empirical formula C2H5 and a relative molecular mass of 58. Deduce its molecular formula.

Show answer

• Empirical formula mass = 2(12) + 5(1) = 29; multiplier = 58 ÷ 29 = 2 [1] • Molecular formula = C4H10 [1]

Q3 (3 marks). 4.99 g of hydrated copper(II) sulfate, CuSO4·xH2O, is heated until all the water of crystallisation is driven off, leaving 3.19 g of anhydrous CuSO4. Calculate the value of x. [Mr: CuSO4 160, H2O 18]

Show answer

• n(CuSO4) = 3.19 ÷ 160 = 0.0200 mol [1] • Mass of water lost = 4.99 − 3.19 = 1.80 g; n(H2O) = 1.80 ÷ 18 = 0.100 mol [1] • x = 0.100 ÷ 0.0200 = 5 [1]