notam M(OH)2 = hidroxidul unui metal divalent
miu = M+17x2 = M+34 g/mol
100% .......................... 65,3% M
M+34 ......................... M
=> M = 64 => M = Cu => Cu(OH)2
miu.Cu(OH)2 = 64+34 = 98 g/mol
100% .................... a% O
98 g ...................... 32 g O
=> a = 32,65% O