# Question 61897

Mar 3, 2017

Use the compound's molar mass.

#### Explanation:

The idea here is that the molecular formula is always a multiple of the empirical formula

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{molecular formula" = "empirical formula} \times n}}}$

Here $n$ must be a whole number.

In other words, the mass of $1$ mole of molecular formulas of nicotine must be equal to the mass of $1$ mole of empirical formulas of nicotine multiplied by a constant $n$.

You know that the empirical formula for nicotine is

$\text{C"_5"H"_7"N}$

use the molar masses of the elements that make up the empirical formula to find the molar mass of the empirical formula

$5 \times \text{12.01 g mol"^(-1) " } +$
$7 \times {\text{1.008 g mol}}^{- 1}$
$1 \times {\text{14.07 g mol}}^{- 1}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
${\text{ 81.18 g mol}}^{- 1}$

Now look up the molar mass of nicotine, which you'll find listed as

${M}_{\text{M nicotine" = "162.23 g mol}}^{- 1}$

You can thus say that

$162.23 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))) = 81.18 color(red)(cancel(color(black)("g mol}}^{- 1}}}} \times \textcolor{b l u e}{n}$

This will get you

$\textcolor{b l u e}{n} = \frac{162.23}{81.18} = 1.998 \approx 2$

Therefore, you can say that the molecular formula for nicotine is

color(darkgreen)(ul(color(black)(("C"_5"H"_7"N")_color(blue)(2) = "C"_10"H"_14"N"_2)))#