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

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Answer 1 · 2017-02-16T15:55:42

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

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 $AlF_3$ .

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