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On complete combustion, 0.3 g of an organic compound gave 0.2 g of $CO_2$ and 0.1 g of $H_2O$. The percentage composition of carbon and hydrogen in the compound, respectively is:
A
4.07% and 15.02%
B
18.18% and 3.70%
C
15.02% and 4.07%
D
3.70% and 18.18%
Detailed Solution
%C $=\dfrac{12\times m_{CO_2}\times100}{44\times m}=\dfrac{12\times0.2\times100}{44\times0.3}=18.18\%$. %H $=\dfrac{2\times m_{H_2O}\times100}{18\times m}=\dfrac{2\times0.1\times100}{18\times0.3}=3.70\%$.