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Empirical formula practice problems
An empirical formula gives the simplest whole-number atom ratio. Convert masses to moles before comparing elements, because equal masses do not represent equal numbers of atoms.
By ExamCat · Original study lesson · Published September 24, 2026
Understand the method
For percentage composition, imagine a 100 g sample so each percentage becomes grams. Divide by each element’s molar mass, then divide all mole amounts by the smallest. If ratios are near simple fractions, multiply every ratio by a common small integer.
Worked example, step by step
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Use rounded molar masses C = 12.0, H = 1.0 and O = 16.0 g/mol.
- In a 100 g sample, amounts are 40.0 g C, 6.7 g H and 53.3 g O.
- Convert to approximately 3.33 mol C, 6.7 mol H and 3.33 mol O.
- Divide all by 3.33 to get approximately 1:2:1.
- The empirical formula is CH₂O. Percentage rounding explains the slightly imperfect hydrogen ratio.
A mistake to watch for
The empirical formula does not establish molecular size. CH₂O and C₆H₁₂O₆ share the same simplest ratio, so a molecular formula needs additional information such as molar mass.
Try it yourself: three questions
Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.
1. Mole amounts are 0.20 mol A and 0.30 mol B. Find the simplest ratio.
Show answer and explanation
Divide by 0.20 to obtain 1:1.5, then multiply both by two: A₂B₃.
2. Why not round a ratio of 1.5 directly to 2?
Show answer and explanation
That changes the composition. Multiply all ratios to remove the half while preserving proportions.
3. An empirical formula has mass 30 g/mol and molecular molar mass is 180 g/mol. What multiplier applies?
Show answer and explanation
180/30 = 6. Multiply every empirical subscript by six, rather than adding six to each one.
Build your next practice session
Mark the first step where your reasoning differed from the explanation. Turn that step into a question, answer it without looking, then change one condition in the example and explain what would change. A correct guess is a reason to revisit the method.
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Further reading
OpenStax Chemistry 2e provides textbook background. The worked scenario and practice questions on this page were written independently for ExamCat.
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