# Question #05d76

Dec 16, 2016

The $\text{empirical formula}$ is $A {s}_{2} {O}_{5}$.

#### Explanation:

As with all these problems, we ASSUME that there are $100 \cdot g$ of unknown compound.

Given the elemental percentages, thus there are:

$\text{Moles of oxygen}$ $=$ $\frac{34.8 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 2.18 \cdot m o l$.

$\text{Moles of arsenick}$ $=$ $\frac{65.2 \cdot g}{74.9 \cdot g \cdot m o {l}^{-} 1} = 0.871 \cdot m o l$.

We divide thru by the smallest molar quantity to get, $O : A s$, $2.5 : 1.0$.

But by definition, the $\text{empirical formula}$ is the simplest WHOLE number ratio that defines constituent atoms in a species, so we take this preliminary determination, and DOUBLE it to give WHOLE numbers:

$A {s}_{2} {O}_{5}$. Capisce?