Question

A student found that 1.19 g of chromium (`Cr`) formed 1.74 g f chromium oxide. The molar mass of `Cr` is 52.00 g/mol. What is the empirical formula of chromium oxide?

Answer 1

`Cr_2O_3`

Explanation:

As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:

`"Moles of metal"` `=` `(1.19*g)/(52.00*g*mol^-1)=0.0229*mol`

`"Moles of oxygen"` `=` `(1.74*g-1.19*g)/(16.00*g*mol^-1)=0.0343*mol`

And we divide thru by the SMALLEST molar quantity, that of chromium, to give:

`Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50`.

But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by `2` to `Cr_2O_3` as required. Capisce?

How did I know there were `0.55*g` of oxygen present?