A hydrocarbon contained 14.3% hydrogen. The empirical formula for the hydrocarbon is   [H= 1.0, C= 12.0]

Analyze the question: A hydrocarbon contains only C and H. If 14.3% is hydrogen, then carbon is 100 − 14.3 = 85.7%. We need the simplest whole-number ratio of C to H atoms.

Concept: Empirical formula steps:

1. Assume 100 g of the compound, so percentages become masses.
2. Divide each mass by its atomic mass to get moles.
3. Divide by the smallest number of moles.
4. Round to the simplest whole-number ratio.

Solution:

| Element | Mass (g) | Moles | Ratio (÷ 7.14) | |---|---|---|---| | C | 85.7 | 85.7 ÷ 12 = 7.14 | 1 | | H | 14.3 | 14.3 ÷ 1 = 14.3 | 2.00 |

The ratio of C : H is 1 : 2, so the empirical formula is CH₂.

Check: CH₂ has a mass of 12 + 2 = 14. The H percentage is 2/14 × 100 = 14.3%, which matches the question.

Exam tip: For a hydrocarbon, you only need to find the C:H ratio. A quick shortcut is to test each option by computing %H = (H mass ÷ total mass) × 100. Here, CH₂ gives 14.3% straight away. Also remember that 14.3% H (and about 85.7% C) is the classic signature of alkenes, whose general formula is CₙH₂ₙ.