Percentage of elements in a compound are C = 40%, H = 6.6%, O = 53.4%. Find the molecular formula of the compound if its molecular weight is 60.

C_{2} H_{4} O_{2}

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C H_{4} O_{2}

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C_{2} H_{4} O

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C_{2} H_{2} O_{2}

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Solution

The correct option is A $C_{2} H_{4} O_{2}$
Moles of C $= \frac{60 \times 0.4}{12} = 2$

Moles of H $= \frac{60 \times 0.066}{1} \approx 4$

Moles of O $= \frac{60 \times 0.534}{16} \approx 2$

Therefore, molecular formula $C_{2} H_{4} O_{2}$

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