M(OH)2
16+8+0,5=24,5
MM(OH)2= M+ 2•16+2•1= (M+34)g/mol
24,5 g M(OH)2..........16 g M........8 g O.........0,5 g H
M+34 g M(OH)2.........M g M.......32 g O.........2 g H
16(M+34)=24,5M
16M+544=24,5M
544=8,5M
M=64=> Cu(cupru)
formula bazei/hidroxidului: Cu(OH)2