Question

Nicotine, a component of tobacco, is composed of c, h, and n. A 2.625-mg sample of nicotine was combusted, producing 7.1210 mg of co2 and 2.042 mg of h2o. What is the empirical formula for nicotine?

Answer 1

The mass sample of nicotine combusted = 2.625 mg  (given)

Mass of $CO_2$ produced = 7.1210 mg  (given)

Mass of $H_2O$ produced = 2.042 mg  (given)

Molar mass of $CO_2$ = 44 g/mol

Molar mass of $H_2O$ = 18 g/mol

Percentage of Carbon = $\frac{12 g}{44 g/mol}\times \frac{7.120 mg of CO_2}{2.625 mg sample}\times 100 = 74.04$%

Percentage of hydrogen = $\frac{2 g}{18 g/mol}\times \frac{2.04 mg of H_2O}{2.625 mg sample}\times 100 = 8.62$%

Now, for percentage of nitrogen = $100 - (74.04+8.62) = 17.34$%

Calculating the moles of each element:

$Number of moles = \frac{given mass}{Molar mass}$

• For $C$

$\frac{74.04 g}{12 g/mol} = 6.17 mol$

• For $H$

$\frac{8.62 g}{1 g/mol} = 8.62 mol$

• For $N$

$\frac{17.34 g}{14 g/mol} = 1.24 mol$

Dividing with the smallest mole value to calculate the molar ratio of each element:

$C_{\frac{6.17}{1.24}} = C_{4.9} = C_5$

$H_{\frac{8.62}{1.24}} = H_{6.9} = H_7$

$N_{\frac{1.24}{1.24}} = N_1$

Hence, the empirical formula for nicotine is $C_5H_7N$.