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JAMB Chemistry 2013 Paper

Question : 33

Total: 44

rn(C=12,H =1, O=16)

Solution:

Concept:

The empirical formula is the simplest whole-number ratio of moles of each element in the compound.

Explanation:

Assume a

100\,

g sample, so percentages become masses:

Carbon:

60\,

g, Hydrogen:

13.3\,

g, Oxygen:

26.7\,

Moles of each element:

\text{C} = \frac{60}{12} = 5.0

\text{H} = \frac{13.3}{1} = 13.3

\text{O} = \frac{26.7}{16} = 1.66875

Divide each by the smallest number of moles,

1.66875

\text{C}: \frac{5.0}{1.66875} \approx 3

\text{H}: \frac{13.3}{1.66875} \approx 8

\text{O}: \frac{1.66875}{1.66875} = 1

Thus, the empirical formula is

\mathrm{C}_3\mathrm{H}_8\mathrm{O}

Answer:

\mathrm{C}_3\mathrm{H}_8\mathrm{O}

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