# A compound was found to have 85.7% carbon and 14.3% hydrogen. Its molecular mass is 84. What is its empirical formula? What is its molecular formula?

Dec 13, 2016

$\text{empirical formula}$ $=$ $C {H}_{2}$.

$\text{molecular formula}$ $=$ ${C}_{6} {H}_{12}$

#### Explanation:

AS with all these problems, we assume (for simplicity) $100 \cdot g$ of unknown compound.

And thus there are (85.7%xx100*g)/(12.011*g*mol^-1) with respect to carbon, i.e. $7.14 \cdot m o l \cdot C$

And (14.3%xx100*g)/(1.00794*g*mol^-1) with respect to hydrogen, i.e. $14.1 \cdot m o l \cdot H$.

And thus the $\text{empirical formula}$, the simplest whole number ratio that defines constituent elements in a species is $C {H}_{2}$.

Now the $\text{molecular formula}$ is alway a whole number multiple of the $\text{empirical formula}$:

$\text{empirical formula} \times n$ $=$ $\text{molecular formula}$

So $\left(12.011 \cdot g \cdot m o {l}^{-} 1 + 2 \times 1.00794 \cdot g \cdot m o {l}^{-} 1\right) \times n = 84 \cdot g \cdot m o {l}^{-} 1.$

So $n = 6$, and $\text{molecular formula}$ $=$ $6 \times C {H}_{2}$ $=$ ${C}_{6} {H}_{12}$