$M_{NH_4NO_3} = 80(đvC)$
$\%N = \dfrac{14.2}{80}.100\% = 35\%$

$\%H = \dfrac{4}{80}.100\% = 5\%$
$\%O = \dfrac{16.3}{80}.100\% = 60\%$

\(\%K=\dfrac{m_K}{M_{K_2SO_3}}=\dfrac{78}{158}=49,36\%\\ \%S=\dfrac{m_S}{M_{K_2SO_3}}=\dfrac{32}{158}=20,25\%\\ \%O=100\%-\%K-\%S=100\%-49,36\%-20,25\%=30,39\%\)

+) Trong H₂SO₄ có: \(\left\{{}\begin{matrix}\%m_H=\dfrac{2}{98}\cdot100\%\approx2,04\%\\\%m_S=\dfrac{32}{98}\cdot100\%\approx32,65\%\\\%m_O=65,31\%\end{matrix}\right.\)

+) Trong HNO₃ có: \(\left\{{}\begin{matrix}\%m_H=\dfrac{1}{63}\cdot100\%\approx1,59\%\\\%m_N=\dfrac{14}{63}\cdot100\%\approx22,22\%\\\%m_O=76,19\%\end{matrix}\right.\)

a)Ta có:\(m\%_H=\dfrac{2.100\%}{98}=2,04\%\)

        \(m\%_S=\dfrac{32.100\%}{98}=32,65\%\)

        \(m\%_O=100-2,04-32,65=65,31\%\)

b) tương tự

Ta có: M_K2SO4 = 39.2 + 32 + 16.4 = 174 (g/mol)

\(\Rightarrow\left\{{}\begin{matrix}\%K=\dfrac{39.2}{174}.100\%\approx44,8\%\\\%S=\dfrac{32}{174}.100\%\approx18,4\%\\\%O\approx36,8\%\end{matrix}\right.\)

Bạn tham khảo nhé!

Ta có: M_K2SO4 = 39.2 + 32 + 16.4 = 174 (g/mol)

\(\%C=\dfrac{12\cdot2}{28}\cdot100\%=85.71\%\)

\(\%H=100-85.71=14.29\%\)

`%C=(12.2)/(12.2+4.1) .100%=85,71%`

`%H=(4.1)/(12.2+4.1) .100%=14,29%`

\(M_{KMnO_4}=158(g/mol)\\ \%_{K}=\dfrac{39}{158}.100\%=24,68\%\\ \%_{Mn}=\dfrac{55}{158}.100\%=34,81\%\\ \%_O=100\%-24,68\%-34,81\%=40,51\%\)

\(M_{CH_3COOH}=60\left(\dfrac{g}{mol}\right)\)

\(\%C=\dfrac{24}{60}\cdot100\%=40\%\)

\(\%H=\dfrac{4}{60}\cdot100\%=6.67\%\)

\(\%O=100-40-6.67=53.33\%\)

\(CaCO_3\\ \%m_{Ca}=\dfrac{40}{40+12+3.16}.100=40\%\\ \%m_C=\dfrac{12}{40+12+16.3}.100=12\%\\ \Rightarrow\%m_O=100\%-\left(40\%+12\%\right)=48\%\\ H_2SO_4\\ \%m_H=\dfrac{2.1}{2.1+32+4.16}.100\approx2,041\%\\ \%m_S=\dfrac{32}{2.1+32+4.16}.100\approx32,653\%\\ \%m_O=\dfrac{4.16}{2.1+32+4.16}.100\approx65,306\%\\ Fe_2O_3\\ \%m_{Fe}=\dfrac{56.2}{56.2+16.3}.100=70\%\\ \Rightarrow\%m_O=100\%-70\%=30\%\)

CaCO₃
\(\%M_{\dfrac{Ca}{CaCO_3}}=\dfrac{40}{100}.100\%=40\%\)
\(\%M_{\dfrac{C}{CaCO_3}}=\dfrac{12}{100}.100\%=12\%\)
\(\%M_{\dfrac{O}{CaCO_3}}=100\%-\left(40\%+12\%\right)=48\%\)
H₂SO₄
\(\%M_{\dfrac{H_2}{H_2SO_4}}=\dfrac{2}{98}.100\%=2,04\%\)
\(\%M_{\dfrac{S}{H_2SO_4}}=\dfrac{32}{98}.100\%=32,65\%\)
\(\%M_{\dfrac{O}{H_2SO_4}}=100\%-\left(2,04\%+32,65\%\right)=65,31\%\)
Fe₂O3
\(\%M_{\dfrac{Fe}{Fe_2O_3}}=\dfrac{112}{160}.100\%=70\%\)
\(\%M_{\dfrac{O}{Fe_2O_3}}=100\%-70\%=30\%\)