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

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

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

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

a) \(PTK_{CaCO_3}=40+12+3.16=100\left(đvC\right)\)

\(\%m_{Ca}=\dfrac{40}{100}.100=40\%\)

\(\%m_C=\dfrac{12}{100}.100=12\%\)

\(\%m_O=\dfrac{3.16}{100}.100=48\%\)

b) \(PTK_{Fe_2O_3}=2.56+3.16=160\left(đvC\right)\)

\(\%m_{Fe}=\dfrac{2.56}{160}.100=70\%\)

\(\%m_O=\dfrac{3.16}{160}.100=30\%\)

\(\%Fe=\dfrac{56}{56.2+16.3}.100\%=35\%\\ \%O=\dfrac{16}{56.2+16.3}.100\%=10\%\)

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)

\(M_{\left(NH_4\right)_2SO_4}=\left(14+1.4\right).2+32+16.4=132\left(\dfrac{g}{mol}\right)\)

\(\%N=\dfrac{14.2}{132}.100\%=21,2\%\)

\(\%H=\dfrac{1.4.2}{132}.100\%=6\%\)

\(\%S=\dfrac{32}{132}.100\%=24,2\%\)

\(\%O=\dfrac{16.4}{132}.100\%=48,5\%\)

Khối lượng mol của \(Fe2O3\)  là :

\(M_{Fe_2}_{O_3}=56,2+16,3=160\left(g.mol\right)\)

\(\%Fe=\dfrac{56.2}{100}.100\%=70\%\)

\(\%O=100\%-70\%=30\%\)

M_{Fe\(_2O_3\)} = 56.2+16.3 = 160 ( g/mol )

\(\%Fe=\dfrac{56.2}{160}.100\%=70\%\)

\(\Rightarrow\%O=100\%-70\%=30\%\)

\(M_{Fe_2O_3}=56.2+16.3=160(g/mol)\\ \%m_{Fe}=\dfrac{56.2}{160}.100\%=70\%\\ \%m_O=100\%-70\%=30\%\)

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

\(\%m_{Ca}=\dfrac{40}{164}.100=24,39\left(\%\right)\\ \%m_N=\dfrac{28}{164}.100=17,07\left(\%\right)\\ \%m_O=\dfrac{48.2}{164}.100=58,54\left(\%\right)\)

$M_{Ca(NO_3)_2} = 164$

$\%Ca = \dfrac{40}{164}.100\% = 24,39\%$
$\%N = \dfrac{14.2}{164}.100\% = 17,07\%$

$\%O = 100% -24,39\% -17,07\% = 58,54%%

Câu 2:

Trong 1 mol X: \(\left\{{}\begin{matrix}n_{Ag}=\dfrac{170.63,53\%}{108}=1\left(mol\right)\\n_N=\dfrac{170.8,23\%}{14}=1\left(mol\right)\\n_O=\dfrac{170\left(100\%-63,53\%-8,23\%\right)}{16}=3\left(mol\right)\end{matrix}\right.\)

Vậy CTHH của X là \(AgNO_3\)

Câu 1:

\(a,\%_{Fe}=\dfrac{56}{180}\cdot100\%=31,11\%\\ \%_N=\dfrac{14\cdot2}{180}\cdot10\%=15,56\%\\ \%_O=100\%-31,11\%-15,56\%=53,33\%\\ b,\%_{N\left(N_2O\right)}=\dfrac{14\cdot2}{44}\cdot100\%=63,63\%\\ \%_{O\left(N_2O\right)}=100\%-63,63\%=36,37\%\\ \%_{N\left(NO\right)}=\dfrac{14}{30}\cdot100\%=46,67\%\\ \%_{O\left(NO\right)}=100\%-46,67\%=53,33\%\\ \%_{O\left(NO_2\right)}=\dfrac{16\cdot2}{46}\cdot100\%=69,57\%\\ \%_{N\left(NO_2\right)}=100\%-69,57\%=30,43\%\)