Question 60

The elemental composition of a compound is $$54.2\%C,\ 9.2\%H$$ and $$36.6\%O.$$ If the molar mass of the compound is $$132\ \text{g mol}^{-1},$$ the molecular formula of the compound is: [Given: Relative atomic masses C:H:O = 12:1:16]

Empirical formula: C: 54.2/12=4.52, H: 9.2/1=9.2, O: 36.6/16=2.29.

Ratio: 4.52/2.29 : 9.2/2.29 : 2.29/2.29 = 1.97 : 4.02 : 1 ≈ 2:4:1

Empirical formula: C₂H₄O. Empirical mass = 44. Molar mass = 132 = 3×44.

Molecular formula: C₆H₁₂O₃.

The correct answer is Option 4: C₆H₁₂O₃.

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