# A sample of a compound is 80% carbon and 20% hydrogen by mass. Its formula mass is 30 amu. What is the molecular formula?

Mar 10, 2017

We have $\text{ethane}$.

#### Explanation:

As with all these problems, it is usually assumed that we have a $100 \cdot g$ mass of unknown compound, and we work out the molar quantities:

And thus $\text{moles of carbon} \equiv \frac{80 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 6.66 \cdot m o l$.

And thus $\text{moles of hydrogen} \equiv \frac{20 \cdot g}{1.008 \cdot g \cdot m o {l}^{-} 1} = 19.8 \cdot m o l$.

We divide the molar quantities thru by the SMALLER molar quantity:

$C : \frac{6.66 \cdot m o l}{6.66 \cdot m o l} = 1$

$H : \frac{19.8 \cdot m o l}{6.66 \cdot m o l} = 2.97$

And thus the $\text{empirical formula}$ $=$ $C {H}_{3}$

And we know that the molecular formula is always a multiple of the empirical formula, and thus is terms of mass:

i.e. $\text{molecular formula}$ $=$ $n \times \text{empirical formula}$

But we have a molecular mass of $30 \cdot \text{amu}$

So $30 \cdot \text{amu}$ $=$ $n \times \left(12.01 + 3 \times 1.01\right) \cdot \text{amu}$

Clearly, $n = 2$, and the MOLECULAR FORMULA is ${C}_{2} {H}_{6}$.

This is not a realistic problem, as few analysts would perform combustion on a liquid, and no analyst could perform combustion on a gas.