# What is the empirical formula for a compound that is 60.9% arsenic and 39.1% sulfur?

Apr 2, 2016

$A {s}_{2} {S}_{3}$

#### Explanation:

How did we get this formula?

We assume $100.0$ $g$ of compound, and there are $60.9$ $g$ of arsenic, and $39.1$ $g$ of sulfur.

We divide each mass thru by the molar mass of each element:

$A s$ $=$ $\frac{60.9 \cdot g}{74.92 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.813$ $m o l$.

$S$ $=$ $\frac{39.1 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.22$ $m o l$.

Now, we divide thru by the LOWEST molar quantity, $0.813$ $m o l$:

We get $A s {S}_{1.500}$. But we want the whole number ratio, so we get $A {s}_{2} {S}_{3}$ as required.