Ask Me Help Desk - Calculate the empirical formula of the compound.

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.