Question

nicotine contains 74.0 % carbon, 8.72 % hydrogen, and 17.3 % nitrogen. what is the empirical formula of nicotine?
o c4h10n2
o c3h4n2
o c2h4n2
o c5h7n
o c10h14n2

Explanation:

Step1: Assume 100g of nicotine

We assume a 100 - g sample of nicotine. So, we have 74.0 g of C, 8.72 g of H, and 17.3 g of N.

Step2: Calculate the moles of each element

The molar mass of C is 12.01 g/mol, of H is 1.01 g/mol, and of N is 14.01 g/mol.
Moles of C: $n_{C}=\frac{74.0\ g}{12.01\ g/mol}\approx6.16\ mol$
Moles of H: $n_{H}=\frac{8.72\ g}{1.01\ g/mol}\approx8.63\ mol$
Moles of N: $n_{N}=\frac{17.3\ g}{14.01\ g/mol}\approx1.23\ mol$

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (1.23 mol in this case).
For C: $\frac{6.16\ mol}{1.23\ mol}\approx5$
For H: $\frac{8.63\ mol}{1.23\ mol}\approx7$
For N: $\frac{1.23\ mol}{1.23\ mol}=1$

Answer:

E. $C_{5}H_{7}N$