\(a,\%Cu=\dfrac{m_{Cu}}{M_{CuSO_4}}=\dfrac{64}{160}=40\%\\ \%S=\dfrac{m_S}{M_{CuSO_4}}=\dfrac{32}{160}=20\%\\ \%O=100\%-\%Cu-\%S=100\%-40\%-20\%=40\%\)
\(b,\%Fe=\dfrac{m_{Fe}}{M_{Fe_3O_4}}=\dfrac{168}{232}=72,41\%\\ \%O=100\%-\%Fe=100\%-72,41\%=27,59\%\)
\(c,\%Fe=\dfrac{m_{Fe}}{M_{FeO}}=\dfrac{56}{72}=77,77\%\\ \%O=100\%-\%Fe=100\%-77,77\%=22,23\%\)
\(d,\%K=\dfrac{m_K}{M_{K_2SO_3}}=\dfrac{78}{138}=56,52\%\\ \%C=\dfrac{m_C}{M_{K_2SO_4}}=\dfrac{12}{138}=8,69\%\\ \%O=100\%-\%K-\%C=100\%-56,52\%-8,69\%=34,79\%\)
a) \(M_{CúSO_4}=64+32+16.4=160\left(DvC\right)\)
\(m_{Cú}=\dfrac{64}{160}.100\%=40\%\)
\(m_S=\dfrac{32}{160}.100\%=20\%\)
\(m_O=\dfrac{16.4}{160}.100\%=40\%\)
b) \(M_{Fe_3O_4}=56.3+16.4=232\left(DvC\right)\\ \%Fe=\dfrac{56.3}{232}.100\%=72\%\\ \%O=100\%-74\%=26\%\)
c) \(M_{FeO}=56+16=72\left(DvC\right)\)
\(\%Fe=\dfrac{56}{72}.100\%=77\%\\ \%O=100\%-77\%=33\%\)
