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0.2 g of an organic compound containing C, H and O on combustion yielded 0.147 g CO, and 0.12 g water. The percentage of oxygen in it is:

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78.45 %

83.23 %

73.34 %

89.50 %

answer is A.

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Detailed Solution

$% of C = \frac{12 \times 0.0147}{44 \times 0.2} \times 100 = 20 % % of H = \frac{2 \times 0.12}{18 \times 0.2} \times 100 = 6.66 % ∴ % of O = 100 - 20 - 6.66 = 73.34 %$

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