Question

A chromium chloride salt contains `32.8%` metal content by mass. What is its empirical formula?

Answer 1

We assume a `100*g` mass of compound, and get an empirical formula of `CrCl_3`.

Explanation:

In all problems of this sort, it is convenient to assume a `100*g` mass of compound, and then work out the individual molar quantities of each element:

`"Moles of chromium"=(32.8*g)/(52.0*g*mol^-1)=0.631*mol`

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

We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of `CrCl_3`.