When 6.25grams of pure iron are allowed to react with oxygen, a black oxide forms. If the products weighs 8.15g, what is empirical formula of the oxide?

Answer 1

Fe + O2 ==> FexOy
6.25g + ? = 8.15g

Therefore, the amount of oxygen added must be 8.15-6.25 = 1.90.
%Fe = (6.25/8.15)*100 = about 77% but you can (and must) do that more accurately.
%O = (1.90/8.15)*100 = about 233%. Do all of these plus all that follows more accurately.

Take 100 g sample which gives you
approx 77 g Fe
approx 23 g O

Convert to mols.
approx 77/55.85 = approx 1.4
approx 23/16 = approx 1.4

So the ratio is 1:1 and the formula si FeO.

Remember all of my numbers above are estimates.
Convert to mols.

Answer 2

Fe1.4O1.4 = FeO