Question

Find the empirical formula for a compound which contains 32.8%chromium and 67.2% chlorine?

Answer 1

Well, I make it `CrCl_3`...

Explanation:

For empirical formula calculations it is useful to assume that you gots a `100*g` of reagent....and then we interrogate its atomic composition on that basis...

And so for a `100*g` mass of the stuff...

`"Moles of metal"=(32.8*g)/(52.00*g*mol^-1)=0.631*mol`

`"Moles of chlorine"=(67.2*g)/(35.45*g*mol^-1)=1.90*mol`

And as usual we divide thru by the element in LEAST molar quantity to give an empirical formula of...

`Cr_((0.631*mol)/(0.631*mol))Cl_((1.90*mol)/(0.631*mol))=CrCl_3`

And `Cr(+III)` is a highly reasonably oxidation state for chromium compounds...