0.765g of an organic compound on combustion gave 0.535 g $C O_{2}$ and 0.138g of $H_{2} O$ . Then the
ratio of the percentage of carbon and hydrogen is 19 : x. Then the value of x is ___ .

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Detailed Solution

$% C = \frac{12}{44} \times \frac{M a s s o f C O_{2}}{M a s s o f O . C} \times 100$

$= \frac{12}{44} \times \frac{0.535}{.765} \times 100$

$= \frac{12}{44} \times 0.6993$

$= 19 %$

$% H = \frac{2}{18} \times \frac{M a s s o f H_{2} O}{M a s s o f O . C} \times 100$

$= \frac{2}{18} \times \frac{0.138}{0.765} \times 100$

$= 0 11 111 \times 0.18 \times 100$

$= 2 %$

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