Bài 1:
a) \(m_{Fe}=\dfrac{160.70}{100}=112\left(g\right)=>n_{Fe}=\dfrac{112}{56}=2\left(mol\right)\)
\(m_O=\dfrac{160.30}{100}=48\left(g\right)>n_O=\dfrac{48}{16}=3\left(mol\right)\)
=> CTHH: Fe2O3
b) MA = 23.2 = 46 (g/mol)
\(m_C=\dfrac{52,17.46}{100}=24\left(g\right)=>n_C=\dfrac{24}{12}=2\left(mol\right)\)
\(m_H=\dfrac{13,04.46}{100}=6\left(g\right)=>n_H=\dfrac{6}{1}=6\left(mol\right)\)
\(m_O=\dfrac{34,79.46}{100}=16\left(g\right)=>n_O=\dfrac{16}{16}=1\left(mol\right)\)
=> CTHH: C2H6O
Bài 2
a) \(\left\{{}\begin{matrix}\%Mg=\dfrac{24.1}{58}.100\%=41,38\%\\\%O=\dfrac{2.16}{58}.100\%=55,17\%\\\%H=\dfrac{2.1}{58}.100\%=3,45\%\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\%Cu=\dfrac{64.1}{80}.100\%=80\%\\\%O=\dfrac{16.1}{80}.100\%=20\%\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\%Fe=\dfrac{56.1}{72}.100\%=77,78\%\\\%O=\dfrac{16.1}{72}.100\%=22,22\%\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\%Fe=\dfrac{56.2}{160}.100\%=70\%\\\%O=\dfrac{16.3}{160}.100\%=30\%\end{matrix}\right.\)
