Copper forms two oxides. One of them contains 79.9%, the other – 88.8% copper. Formulate the two oxides.

Given:
ω1 (Cu) = 79.9%
ω2 (Cu) = 88.8%

To find:
CuxOy -?
CuxOy -?

1) Let m (CuxOy) = 100 g;
2) m (Cu) = ω (Cu) * m (CuxOy) / 100% = 79.9% * 100/100% = 79.9 g;
3) n (Cu) = m / M = 79.9 / 64 = 1.25 mol;
4) m (O) = m (CuxOy) – m (Cu) = 100 – 79.9 = 20.1 g;
5) n (O) = m / M = 20.1 / 16 = 1.26 mol;
6) x: y = n (Cu): n (O) = 1.25: 1.26 = 1: 1;
CuO – copper (II) oxide;
7) m (Cu) = ω (Cu) * m (CuxOy) / 100% = 88.8% * 100/100% = 88.8 g;
3) n (Cu) = m / M = 88.8 / 64 = 1.39 mol;
4) m (O) = m (CuxOy) – m (Cu) = 100 – 88.8 = 11.2 g;
5) n (O) = m / M = 11.2 / 16 = 0.7 mol;
6) x: y = n (Cu): n (O) = 1.39: 0.7 = 2: 1;
Cu2O – copper (I) oxide.

Answer: Unknown substances: CuO – copper (II) oxide; Cu2O – copper (I) oxide.