What is the empirical formula of a compound containing 40.00% carbon, 6.67% hydrogen, and 53.33% oxygen by mass?

Analyze the question: We're given percentage composition and asked for the empirical formula, which is the simplest whole-number ratio of atoms in the compound.

Concept: Assume 100 g of the compound, so the percentages become masses in grams. Divide each mass by the atomic mass to get moles, then divide by the smallest number of moles to get the ratio.

Solution:

Atomic masses: C = 12, H = 1, O = 16

| Element | Mass (g) | Moles | Divide by smallest (3.333) | |---|---|---|---| | C | 40.00 | 40.00 ÷ 12 = 3.333 | 1 | | H | 6.67 | 6.67 ÷ 1 = 6.67 | 2 | | O | 53.33 | 53.33 ÷ 16 = 3.333 | 1 |

The ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.

Checking the options:

• A. C₄H₈O₄ is 4 × (CH₂O). It is a molecular formula, not the simplest ratio.
• B. CH₂O has the simplest ratio 1:2:1. ✔
• C. C₂H₄O₂ is 2 × (CH₂O). It is not the simplest.
• D. C₃H₆O₃ is 3 × (CH₂O). It is not the simplest.

All four options have the same percentage composition, but only B is in the lowest whole-number ratio. That is exactly what "empirical" means.

Answer: B

Exam tip: When all the options are multiples of one another, the empirical formula is always the one that cannot be reduced further. Check whether you can divide all the subscripts by a common factor. Also, if the percentages are given, the quickest route is: % ÷ atomic mass → divide by the smallest → whole numbers.