O-Level Chemistry Empirical Formulae: Handle Ratios That Are Not Whole Numbers

Separate a ratio from a rounding problem

In the October 2026 SGExams Chemistry discussion, students described getting a non-integer mole ratio and being unsure whether to round it. The lesson is to inspect the ratio before choosing subscripts. This article uses original examples and does not claim to reproduce the official answers or marking scheme for that examination.

The 2026 Chemistry 6092 syllabus includes calculating empirical and molecular formulae from relevant data. An empirical formula gives the simplest whole-number ratio of atoms of the elements. The mass ratio is not usually the atom ratio because different atoms have different relative masses. Start by identifying exactly what the question gives.

Convert each element to moles

Consider an invented compound sample containing 2.4 g of carbon and 0.6 g of hydrogen. Using relative atomic masses 12 and 1 for this exercise, the amounts are 0.20 mol of carbon atoms and 0.60 mol of hydrogen atoms. Divide both by the smaller amount to obtain C:H = 1:3. The empirical formula is CH3.

If percentages are supplied, a convenient working basis is a 100 g sample, so the percentage values become corresponding masses in grams. Label that assumption. Use the relative atomic masses provided by your question, retain adequate working precision and keep the element labels aligned through every step.

Look for a small common multiplier

Suppose another calculation gives A:B close to 1:1.5. Rounding 1.5 to 2 changes the ratio substantially. Multiplying both entries by 2 gives 2:3 instead. A ratio close to 1:1.333 may suggest multiplying by 3 to obtain 3:4, provided the unrounded values support that interpretation.

Multiply every entry by the same number. For an invented three-element ratio of 1:1.5:2, multiplying by 2 gives 2:3:4. Do not adjust only the awkward entry or independently round each number until the answer looks convenient. If no small multiplier makes sense, check the earlier conversions before forcing a formula.

Check the formula against the given data

For the carbon and hydrogen example, CH3 has a relative formula mass of 15. Carbon contributes 12/15 of that mass, or 80%. The original sample has 2.4/3.0 of its mass as carbon, also 80%. This reverse check helps distinguish a genuine ratio from a convenient-looking set of integers.

A molecular formula requires additional information. If the empirical formula is CH3 and the stated relative molecular mass is 30, the multiplier is 30/15 = 2, giving C2H6. Do not assume the empirical formula and molecular formula are always identical.

Diagnose the first wrong step

Set out columns for element, mass, relative atomic mass, amount in moles and simplified ratio. Mark the earliest point where your answer differs from a worked solution. A wrong mass conversion needs different practice from an incorrect whole-number multiplier. Redo one similar question without the answer open, then check the result by reversing the calculation.

Choose a suitable practice question from our Secondary Pure Chemistry papers, checking the level, year and original instructions.

Sources and further reading

Checked on 6 November 2026. The linked discussion is real; worked examples and practice routines are our original guidance. Older discussions are not evidence of current examination rules.

• 6092/02 O Levels Chemistry Paper 2 Exam Megathread 2026 — 27 October 2026.
• SEAB 2026 syllabus