$M_{hợp\ chất} = 98 + 63 = 161(đvC)$
$\%H = \dfrac{2 + 1}{161}.100\% = 1,86\%$
$\%S = \dfrac{32}{161}.100\% = 19,88\%$

$\%N = \dfrac{14}{161}.100\% = 8,7\%$
$\%O = \dfrac{16.4 + 16.3}{161}.100\% = 69,56\%$

\(PTK=12.12+1.22+16.11=342< mol>.\)

\(\%C=\dfrac{12.12.100}{342}\approx42,11\%\)

\(\%H=\dfrac{1.22.100}{142}=15,49\%\)

\(\%O=100\%-42,11\%-15,49\%=42,4\%\)

+) 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

\(M=242+132=374(đvc)\\ \%_{Fe}=\frac{56}{374}.100=14,97\%\\ \%_N=\frac{14.3+14.2}{374}.100=18,72\%\\ \%_H=\frac{1.4.2}{374}=2,14\%\\ \%_S=\frac{32}{374}.100=8,57\%\\ \%_O=55,6\%\)

\(M=242+132=374(đvc)\\ \%_{Fe}=\frac{56}{374}.100=14,97\%\\ \%_{N}=\frac{14.3+14.2}{374}.100=18,72\%\\ \%_H=\frac{1.4.2}{374}.100=2,14\%\\ \%_S=\frac{32}{374}.100=8,57\%\\ \%_O=55,6\%\)

\(Fe(NO_3)_3\\ \%_{Fe}=\frac{56}{242}.100\%=23,14\%\\ \%_N=\frac{14.3}{242}.100\%=17,36\%\\ \%_O=595,5\%\\ (NH_4)_2SO_4\\ \%N=\frac{14.2}{132}=21,21\% \\ \%H=\frac{(1.4).2.}{132}=6,06\% \\ \%S=\frac{32}{132}=24,24\% \\ \%O=\frac{16.4}{132}=48,49\%\\ \)

`@` `\text {MgO}`

\(\text{PTK = 24 + 16 = 40 < amu>}\)

\(\%\text{O}=\dfrac{16\cdot100}{40}=40\%\)

Vậy, khối lượng `%` của `\text {O}` trong `\text {MgO}` là `40%`

`@` `\text {Fe}_2 \text {O}_3`

\(\text{PTK = }56\cdot2+16\cdot3=160\text{ }< \text{amu}\text{ }>\)

\(\%\text{Fe}=\dfrac{56\cdot2\cdot100}{160}=70\%\)

Vậy, khối lượng `%` của `\text {Fe}` trong `\text {Fe}_2 \text {O}_3` là `70%`