C₅H₇N
To determine the empirical formula of nicotine, we will follow the following steps:
Determine the moles of each element
[tex]\begin{gathered} \text{moles of C=}\frac{74}{12}=6.17 \\ \text{moles of H=}\frac{8.65}{1}=8.65 \\ \text{moles of N=}\frac{17.35}{14}=1.24 \end{gathered}[/tex]Divide through by the smallest ratio.
[tex]\begin{gathered} \text{For C =}\frac{6.17}{1.24}=4.98\approx5.0 \\ \text{For H =}\frac{8.65}{1.24}=6.98\approx7.0 \\ \text{For }N=\frac{1.24}{1.24}=1.0 \end{gathered}[/tex]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
