Question

An organic compound contains $C=40\%,H=13.33\%$ and $N=46.67\%$. Its emperical formula is

• A. $C_2{H}_{2}N$
• B. $C_3{H}_{7}N$
• C. $C{H}_{4}N$
• D. $CHN$

Answer 1

Answer: (C)

Element	Percent by mass	Atomic mass	Relative number of moles of elements	Simplest molar ratio
C	40	12	$\frac{40}{12}=3.33$	$\frac{3.33}{3.33}=1$
H	13.33	1	$\frac{13.33}{1}=13.33$	$\frac{13.33}{3.33}=4$
N	46.67	14	$\frac{46.67}{14}=3.33$	$\frac{3.33}{3.33}=1$

Hence, emperical ratio is $C{H}_{4}N$