В порядке уменьшения W (Cu)

CuSO4, Cu (OH) 2, CuO, Cu2O

Mr (CuSO4) = 64+32+4*16 = 160

W (Cu) = 64/160=0.4=40%

Mr (Cu (OH) 2) = 64+2*16+2*1 = 98

W (Cu) = 64/98=0.65=65%

Mr (CuO) = 64+16=80

W (CuO) = 64/80=0.8=80%

Mr (Cu2O) = 2*64+16=144

W (Cu) = 128/144=0.89=89%

Ответ: Cu2O, CuO, Cu (OH) 2, CuSO4