# What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

Apr 25, 2016

${C}_{3} {H}_{5} {O}_{2}$

#### Explanation:

We assume $100$ $g$ of unknown compound.

There are thus $\frac{50.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $=$ $4.163$ $m o l \cdot C$.

And $\frac{6.7 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $6.65$ $m o l \cdot H$.

And $\frac{43.3 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.71$ $m o l \cdot O$.

We divide thru by the lowest molar quantity, that of oxygen to get a formula of ${C}_{1.54} {H}_{2.45} {O}_{1}$, which we double to give a whole number EMPIRICAL formula of ${C}_{3} {H}_{5} {O}_{2}$.