Đáp án:
`a) M_(H_2SO_4)=1.2+32+16.4=98` $(g/mol)$
`%m_H=2/98 . 100%=2,04%`
`%m_S=32/98 .100%=32,65%`
`%m_O=100%-2,04%-32,65%=65,31%`
`b) M_(CuSO_4)=64+32+16.4= 160` $(g/mol)$
`%m_(Cu)=64/160 . 100%=40%`
`%m_S=32/160 .100%=20%`
`%m_O=100%-40%-20%=40%`