# What is the empirical formula of an aluminum fluoride that is 32% by mass with respect to the metal?

Feb 16, 2017

We find an empirical formula of $A l {F}_{3}$..........

#### Explanation:

We assume $100 \cdot g$ of $\text{aluminum fluoride}$. And we work out the molar quantities of each constituent, i.e.

$\text{Moles of metal} = \frac{32.0 \cdot g}{27.0 \cdot g \cdot m o {l}^{-} 1} = 1.19 \cdot m o l .$

$\text{Moles of fluorine} = \frac{68.0 \cdot g}{19.0 \cdot g \cdot m o {l}^{-} 1} = 3.58 \cdot m o l .$

We divide thru each molar quantity thru by the smaller molar quantity, that of aluminum to give:

$A l : \frac{1.19 \cdot m o l}{1.19 \cdot m o l} = 1$; $F : \frac{3.58 \cdot m o l}{1.19 \cdot m o l} = 3$, and thus we get an empirical formula of $A l {F}_{3}$.