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Complete combustion of 1.80 g of an oxygen containing compound $(C_{x} H_{y} O_{z})$ gave 2.64 g of ${CO}_{2}$ and 1.08 g of $H_{2} O$ . The percentage of oxygen in the organic compound is :

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51.63

53.33

63.53

50.33

answer is B.

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Detailed Solution

$\begin{array}{l} % C = \frac{12}{44} \times \frac{2.64}{1.8} \times 100 = 40 \\ % H = \frac{2}{18} \times \frac{1.08}{1.80} \times 100 = 6.66 % O = 100 - [40 + 6.66] = 53.34 \end{array}$

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