Question

Determine the molecular formula and calculate the percent composition of each element present in nicotine, which has a molecular weight of 162 g/mol and an empirical formula of $C_5{H}_{7}N$.

• A. $C_{10}{H}_{14}{N}_2;$ 74.1% C, 8.6% H, 17.3% N
• B. $C_{15}{H}_{21}{N}_3;$ 84.1% C, 8.6% H, 17.3% N
• C. $C_{10}{H}_{14}{N}_2;$ 54.1% C, 3.6% H, 17.3% N
• D. $C_{10}{H}_{21}{N}_2;$ 84.1% C, 18.6% H, 17.3% N

Answer 1

The correct option is A $C_{10}{H}_{14}{N}_2;$ 74.1% C, 8.6% H, 17.3% N

The molecular formula of of nicotine is $C_{10}{H}_{14}{N}_2$.
C mass % = $\frac{10\times 12}{162}\times 100=74.1$%
H mass % = $\frac{14\times 1}{162}\times 100=8.6$%
N mass % = $\frac{2\times 14}{162}\times 100=17.3$%