Question

A molecule of nicotine contains 74.0% Carbon, 8.65% of Hydrogen, and 17.35% Nitrogen. Determine the empirical formula of nicotine. Please write down all work ! :)

Answer 1

C₅H₇N

Explanations:

To determine the empirical formula of nicotine, we will follow the following steps:

Determine the moles of each element

`\begin{gathered} \text{moles of C=}(74)/(12)=6.17 \\ \text{moles of H=}(8.65)/(1)=8.65 \\ \text{moles of N=}(17.35)/(14)=1.24 \end{gathered}`

Divide through by the smallest ratio.

`\begin{gathered} \text{For C =}(6.17)/(1.24)=4.98\approx5.0 \\ \text{For H =}(8.65)/(1.24)=6.98\approx7.0 \\ \text{For }N=(1.24)/(1.24)=1.0 \end{gathered}`

Determine the empirical formula.

The ratios show that there are 5 atoms of Carbon, 7 atoms of hydrogen, and 1 mole of Nitrogen in the empirical formula of nicotine. Hence the empirical formula of nicotine will be C₅H₇N