Question

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

Answer 1

We find an empirical formula of `AlF_3`..........

Explanation:

We assume `100*g` of `"aluminum fluoride"`. And we work out the molar quantities of each constituent, i.e.

`"Moles of metal"=(32.0*g)/(27.0*g*mol^-1)=1.19*mol.`

`"Moles of fluorine"=(68.0*g)/(19.0*g*mol^-1)=3.58*mol.`

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

`Al:(1.19*mol)/(1.19*mol)=1`; `F:(3.58*mol)/(1.19*mol)=3`, and thus we get an empirical formula of `AlF_3`.