Question

The percentage of $C,H$ and $N$ in an organic compound are 40%, 13.3%and 46.7% respectively then the empirical formula is:

• A. $C_3{H}_{13}{N}_3$
• B. $C{H}_{2}N$
• C. $C{H}_{4}N$
• D. $C{H}_{6}N$

Answer 1

The correct option is C $C{H}_{4}N$

To find the molar ratio, we divide the percentage with atomic weight of the element.

1. C, $\frac{40}{12}=3.33$

2. H, $\frac{13.33}{1}=13.33$

3. N, $\frac{46.67}{14}=3.33$

Now to get the simple ratio, we divide the molar ratio with 3.33,

1. C, $\frac{3.33}{3.33}=1$

2. H, $\frac{13.33}{3.33}=4$

3. N, $\frac{3.33}{3.33}=1$

Hence the formula of the compound is $C{H}_{4}N$