# An organic compound containing 78.5% carbon, 13.1% nitrogen, 8.4% hydrogen. If the molar mass of the compound is 107 g/mol, what is the empirical and molecular formulas?

Jan 10, 2017

$\text{Empirical formula}$ $=$ $\text{Molecular formula}$ $=$ ${C}_{7} {H}_{9} N$.

#### Explanation:

With all these problems, we assume a mass of $100 \cdot g$ and divide the individual elemental masses thru by the atomic mass, i.e.

$C :$ $\frac{78.5 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 6.54 \cdot m o l .$

$H :$ $\frac{8.4 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 8.33 \cdot m o l .$

$N :$ $\frac{13.1 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 0.935 \cdot m o l .$

We divide thru by the smallest molar quantity, that of nitrogen, to get an empirical formula of ${C}_{7} {H}_{9} N$. The hydrogen ratio took a bit of rounding up.

Now the $\text{molecular formula}$ is always a multiple of the $\text{empirical formula}$. So we solve for $n$ in the equation $n \times \left(7 \times 12.011 + 9 \times 1.00794 + 1 \times 14.01\right) \cdot g \cdot m o {l}^{-} 1 = 107 \cdot g \cdot m o {l}^{-} 1.$

Clearly, $n = 1$, and the $\text{molecular formula}$ $\equiv$ ${C}_{7} {H}_{9} N$.