• # question_answer A compound has C = 40%, H =13.33% and N = 46.67. The empirical formula is A) $C{{H}_{4}}N$                  B) $C{{H}_{4}}{{H}_{4}}$C) $C{{H}_{2}}N$                  D) ${{C}_{2}}{{H}_{5}}N$

 Element Percent Atomic Mass Relative Number of Moles Simple Ratio C 40 12 $\frac{40}{12}=3.3$ $\frac{3.3}{3.3}=1$ H 13.3 1 $\frac{13.3}{1}=13.3$ $\frac{13.3}{3.3}=4.0$ N 46.67 14 $\frac{46.67}{14}=3.3$ $\frac{3.3}{3.3}=1$
Empirical formula$={{C}_{1}}{{H}_{4}}{{N}_{1}}$ Hence, the correct option is [a].