a) \(M_{CuSO_4}=64+32+16\times4=160\left(g\right)\)
\(\%Cu=\dfrac{64}{160}\times100\%=40\%\)
\(\%S=\dfrac{32}{160}\times100\%=20\%\)
\(\Rightarrow\%O=100\%-40\%-20\%=40\%\)
b) \(n_{CuSO_4}=\dfrac{8}{160}=0,05\left(mol\right)\)
Ta có: \(n_{Cu}=n_{CuSO_4}=0,05\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,05\times64=3,2\left(g\right)\)