Question

Nicotine is a poisonous, addictive compound found in tobacco. A sample of nicotine contains 6.16 mmol of C, 8.56 mmol of H, and 1.23 mmol of N [1 mmol (1 millimole) =10^-3 mol]. What is the empirical formula?

Short answer

The final empirical formula of nicotine is written as,${\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{7}}\mathbf{N}$

Step-by-step solution

Determine the Preliminary formula

According to the question,

Given values: Moles of C = 6.16 mmol or$\mathbf{6}\mathbf{\times }\mathbf{16}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{mol}$

Moles of H = 8.56 mmol or$\mathbf{8}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{mol}$

Moles of N = 1.23 mmol or$\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{mol}$

The preliminary formula for any compound is written as:${\mathbf{C}}_{\mathbf{6}\mathbf{.}\mathbf{16}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{H}}_{\mathbf{8}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{N}}_{\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}$

Calculation for the empirical formula of nicotine

So the final preliminary formula can be calculated as follows:

This calculation requires, dividing all the subscripts by the subscript with the smallest value i.e,$\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol}$

${\mathbf{C}}_{\frac{\mathbf{6}\mathbf{.}\mathbf{16}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}}{\mathbf{H}}_{\frac{\mathbf{8}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}}{\mathbf{N}}_{\frac{\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}}$

On further simplification,

${\mathbf{C}}_{\mathbf{5}\mathbf{.}\mathbf{01}}{\mathbf{H}}_{\mathbf{6}\mathbf{.}\mathbf{96}}{\mathbf{N}}_{\mathbf{1}\mathbf{.}\mathbf{00}}\mathbf{\approx }{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{7}}\mathbf{N}$

So the final empirical formula of nicotine is written as,${\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{7}}\mathbf{N}$