# What is the empirical formula of an "aluminum sulfide", for which a mass of 2.03*g contains a 0.73*g mass of "aluminum"?

Jul 6, 2017

We interrogate the empirical formula, and get $A {l}_{2} {S}_{3}$....

#### Explanation:

And so we work out moles of aluminum and moles of sulfur.

$\text{Moles of Al} = \frac{0.73 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1} = 0.0271 \cdot m o l$

$\text{Moles of S} = \frac{2.03 \cdot g - 0.73 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.0406 \cdot m o l$

We divide thru by the smallest molar quantity...........

And get $A l {S}_{1.498}$. But the $\text{empirical formula}$ is specified to be the $\text{simplest WHOLE number ratio}$.........and thus we get an empirical formula of $A {l}_{2} {S}_{3}$.........

How did I know that there were $1.30 \cdot g$ of sulfur?