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Q.

0.15 gm of an organic compound on combustion gave 0.44 gm of $C O_{2}$ and 0.27gm of $H_{2} O$ . The molecular formula of organic compound can be

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a

$C_{4} H_{8}$

b

$C_{2} H_{6}$

c

$C_{2} H_{2}$

d

$C_{2} H_{4}$

answer is B.

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Detailed Solution

$% o f C = \frac{12}{44} \times \frac{0.44}{0.15} \times 100$

$= 80$
$% o f H = \frac{2}{18} \times \frac{0.27}{0.15} \times 100$
= 20
Question Image

E.F. = $C H_{3}$
$M . F = C_{2} H_{6}$

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