A sample of 2.12g copper was heated in oxygen. The resulting oxide weighed 2.65g. Calculate the empirical formula of the compound.

Please help me, I know how to figure out empirical formula(another normal) but not this question.
I need help.

First you need to know the reaction -- at least half of it

Cu + O_2 = Cu_xO_y

2.12 \, g\, of\, Cu \,\div\, \frac {63.546 \, g\, of\, Cu}{mole\,of\,Cu} = 0.03336\,moles\,of\,Cu

The weight of the oxide is 2.65 grams. The difference in weight is 2.65 g-2.12 g = 0.53 g. Since you know that this is an oxide (the problem said so), you can safely assume that the 0.53 grams is oxygen. Note that we are not using MOLES of oxygen (O2), but gram-atoms of O.

0.53 \, g\, O \,\div\, \frac {15.9994\,g}{gram-atom\,of\, O} = 0.0331\,gram-atoms\,of\,O

The copper-to-oxygen ratio is, therefore

\frac {Cu}{O} = \frac {0.03336}{0.0331} \approx 1

Therefore, the copper-to-oxygen ratio is approximately 1. The empirical formula will, therefore, be CuO.