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An organic compound containing C, H and N have the percentage 40, 13.33 and 46.67 respectively. Its empirical formula may be:

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$C_{2} H_{7} N$

$C_{2} H_{7} N_{2}$

${CH}_{4} N$

${CH}_{5} N$

answer is C.

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Detailed Solution

% Relative no. of atoms Simple ratio

C	40	$\frac{40}{12} = 3.33$	$\frac{3.33}{3.33} = 1$
H	13.33	$\frac{13.33}{1} = 13.33$	$\frac{13.33}{3.33} = 4$
N	46.67	$\frac{46.67}{14} = 3.33$	$\frac{3.33}{3.33} = 1$

$∴ Empirical formula is {CH}_{4} N .$

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