In light of the nitrogen rule mentioned in Problem 12-17, what is themolecular formula of pyridine, M⁺=79?

According to nitrogen rule an organic compound having odd molar mass contain odd number of nitrogen atoms and compound having even molar mass contain even number of nitrogen atoms.

As we have M⁺= 79 is an odd number, pyridine contains an odd number of Nitrogen atom. If pyridine contain one nitrogen atom (atomic weight = 14) the remaining atoms would have formula weight of 65, corresponding to $-{\mathrm{C}}_5{\mathrm{H}}_5$. ${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}$is, in fact the molecular formula of pyridine.

According to rule of 13, divide 79 by 13

$\frac{79}{13}=13\begin{array}{c}6\\ \overline{)\begin{array}{l}79\\ \underset{\_}{-78}\\ 1\\ \end{array}}\end{array}$

The remainder is 1. The basic C-H formula is: ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{(\mathrm{n}+\mathrm{r})}$

Here n=6 and r=1

${\mathrm{C}}_6{\mathrm{H}}_{(6+1)}={\mathrm{C}}_6{\mathrm{H}}_7$

As, molecular formula is odd it also contain odd nitrogen atom in it.

Thus,

As, ${\mathrm{C}}_5{\mathrm{H}}_5=(12\times 5)+(1\times 5)$

= 60 + 5

=65

By adding one odd number of nitrogen atom we have,${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}=60+5+14=79$.