From percent to formula

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

The empirical formula shows the simplest whole-number ratio of atoms in a compound; the molecular formula shows the actual atom count per molecule.
Convert percent (or mass) composition to moles by dividing each element’s mass by its atomic mass, then divide every mole value by the smallest to obtain the mole ratio.
Apply n = M(compound) ÷ M(empirical), then write the molecular formula as n × empirical formula.
In combustion analysis, C is recovered from CO₂ (mass × 12/44) and H from H₂O (mass × 2/18); any oxygen is found by difference.
Two compounds (C₂H₄ and C₃H₆) can share the same empirical formula (CH₂) yet differ in molar mass.
Examiner tip: show every division step; a missing unit on the final molar mass forfeits a mark.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

From percent to formula

Start every empirical-formula problem by assuming a 100 g sample, so each percentage becomes a mass in grams. Convert each mass to moles using the element’s atomic mass. The number of moles is proportional to the atom count, so dividing every mole value by the smallest one gives a clean ratio.

Worked relationship

For a compound containing only C, H, and O with percent values 40.0 % C, 6.7 % H, and 53.3 % O:

C: 40.0 / 12.0 = 3.33 mol
H: 6.7 / 1.0 = 6.70 mol
O: 53.3 / 16.0 = 3.33 mol

Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O (mass = 30 g mol⁻¹).

Combustion analysis

In the standard Liebig train, a known mass of an organic compound is burned in excess oxygen. Water is absorbed by anhydrous CaCl₂ and carbon dioxide by KOH (potash). The mass gain of each tube directly yields the H and C content; oxygen is the balance if no N, S, or halogen is reported.

Common ratio traps

Mole ratio obtained	Action
1.00 : 2.00 : 1.00	Write directly
1.00 : 1.33 : 1.00	Multiply all by 3 → 3 : 4 : 3
1.00 : 1.50 : 1.00	Multiply by 2 → 2 : 3 : 2
1.00 : 1.98 : 1.00	Round to 1 : 2 : 1 only if within ±0.05

A ratio like 1 : 1.33 must be multiplied by 3, never rounded to 1 : 1, because the multiplier n in the final molecular formula depends on the true empirical unit.

Up-scaling to molecular formula

If the compound’s molar mass (from a separate measurement, e.g. vapour density) is 180 g mol⁻¹, then n = 180 / 30 = 6, and the molecular formula is C₆H₁₂O₆. Always state units (g mol⁻¹) when writing molar mass.

Exam pattern

Makerere University Chemistry typically frames this topic as a single Section-B calculation worth 5–8 marks: a printed table of percentage data plus a supplied molar mass, with stepwise marks allocated for each conversion.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge cases and analytical nuances

When the smallest mole value is itself a small fraction (for example, 0.25 mol), it is cleaner to invert the divisor before multiplication — i.e. divide 1.00 mol by 0.25 to get 4, then express every other value relative to that 4. This avoids decimal-heavy ratios that examiners find hard to mark.

For compounds containing nitrogen, halogens, or sulfur, combustion alone cannot locate the heteroatom; the element is detected separately (Dumas test for N, Beilstein for halogens, sodium fusion for S and halogens). Its mass in the original sample is then computed as % element × sample mass ÷ 100, before the mole table is built.

Mechanisms behind the algebra

The percent-to-mole conversion works because moles = mass ÷ atomic mass, and Avogadro’s number cancels when ratios are taken. The empirical-to-molecular step uses the fact that n must be a positive integer — if the division gives 6.0, the empirical unit must appear six whole times per molecule.

Connections to adjacent topics

This skill is a prerequisite for stoichiometry, limiting-reactant calculations, and organic structural identification (where the empirical formula narrows the molecular formula, which is then matched against degree-of-unsaturation data). It also feeds into volumetric analysis when percent purity is requested.

Frequent mistakes to avoid

Confusing molar mass of the empirical unit (e.g. 30 g mol⁻¹ for CH₂O) with the molecular molar mass supplied in the question.
Forgetting to subtract C and H masses from the original sample before reporting “oxygen by difference.”
Writing a molecular formula without checking that n is an integer; if it is not, re-examine the ratio rounding.
Misreading 30 as 3.0 g mol⁻¹, which produces an erroneous n = 60 and a non-existent molecular formula.

Practice prompts

A compound is 52.2 % C, 13.0 % H, and 34.8 % O, with molar mass 46 g mol⁻¹. Determine both formulae and state the multiplier.
Combustion of 0.46 g of a C–H–O compound gives 0.88 g CO₂ and 0.54 g H₂O; its molar mass is 60 g mol⁻¹. Find the molecular formula.

Time budget on exam day: ≈ 4 minutes for an empirical step and 2 minutes for the molecular up-scaling; total ≈ 6 minutes per question, well within Section-B pacing.

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From percent to formula