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

1 Answer
anor277
Feb 16, 2017

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#.

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