MCu2O=144g/mol
MCuO=80g/mol
raportul molar este 1/2
consideram un mol Cu2O (144g) in care exista 16g O
consideram 2moli CuO (2*80g) in care exista 32g O
masa amestec oxizi=144+160=304g
in 304 g amestec...........(16+32)g O
100g amestec................x=100*48/304=15,78% O