\(NaNO_3\\ \%m_{Na}=\dfrac{23}{23+14+3.16}.100\approx27,059\%\\ \%m_N=\dfrac{14}{23+14+3.16}.100\approx16,471\%\\ \%m_O=\dfrac{3.16}{23+14+3.16}.100\approx56,471\%\)

Em tương tự làm cho các chất còn lại!

a) Fe₂O₃----> \(\dfrac{56.2}{56.2+16.3}.100\%=70\%\\ =>100\%-70\%=30\%\)

b) Fe₃O₄---->\(\dfrac{56.3}{56.3+16.4}.100\%=72,41\%\\ =>100\%-72,41\%=27,59\%\)

c)SO₃---->\(\dfrac{32}{32+16.3}.100\%=40\%\\ =>100\%-40\%=60\%\)

\(PTK_{NaNO_3}=23+14+3.16=85\left(đvC\right)\)

\(\%m_{Na}=\dfrac{23}{85}.100=27,05\%\)

\(\%m_N=\dfrac{14}{85}.100=16,47\%\)

\(\%m_O=\dfrac{3.16}{85}=56,47\%\)

\(PTK_{K_2CO_3}=2.39+12+3.16=138\left(đvC\right)\)

\(\%m_K=\dfrac{78}{138}.100=56,52\%\)

\(\%m_C=\dfrac{12}{138}.100=8,69\%\)

\(\%m_O=\dfrac{3.16}{138}.100=34,78\%\)

các ý còn lại làm tương tự

\(M_{NaNO_3}=23+14+16.3=85\left(\dfrac{g}{mol}\right)\)

\(\%Na=\dfrac{23}{85}.100\%=27\%\)

\(\%N=\dfrac{14}{85}.100\%=16,5\%\)

\(\%O=\dfrac{16.3}{85}.100\%=56,5\%\)

a)\(Al_2O_3\)

b)\(ZnCl_2\)

c)\(H_2S\)

d)\(Fe\left(OH\right)_3\)

e)\(H_2S\)

f)\(H_2SO_4\)

g)\(SO_3\)

a) Al₂O₃

B) ZnCl₂

C) H₂S

D) Fe(OH)₃

e) H₂S

F) H₂SO₄

G) SO₃

Câu 2:

\(CTHH:X_2O_5\\ M_{X_2O_5}=\dfrac{16}{100\%-43,67\%}=142\left(g\text{/}mol\right)\\ \Rightarrow M_X=\dfrac{142-16.5}{2}=31\left(g\text{/}mol\right)\left(P\right)\\ \Rightarrow CTHH:P_2O_5\)

Câu 3:

Trong 1 mol B: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{342.15,79\%}{27}=2\left(mol\right)\\n_S=\dfrac{342.28,07\%}{32}=3\left(mol\right)\\n_O=\dfrac{342-2.27-3.32}{16}=12\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow CTHH_B:Al_2\left(SO_4\right)_3\)

Câu 4:

\(M_X=8,5.2=17\left(g\text{/}mol\right)\)

Trong 1 mol X: \(\left\{{}\begin{matrix}n_N=\dfrac{17.82,35\%}{14}=1\left(mol\right)\\n_H=\dfrac{17.17,65\%}{1}=3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow CTHH_X:NH_3\)

C1:

\(NaNO3:\)

 \(MNaNO3=23+62=\dfrac{85g}{mol}\)

\(\%Na=\dfrac{23.100}{85}=27\%\)

\(\%N=\dfrac{14.100}{85}=16\%\)

\(\%O=\dfrac{16.3.100}{85}=56\%\) 

\(K2CO3\)

  \(MK2CO3=39.2+60=\dfrac{138g}{mol}\)

\(\%K=\dfrac{39.2.100}{138}=57\%\)

\(\%C=\dfrac{12.100}{138}=9\%\)

\(\%O=\dfrac{16.3.100}{138}=35\%\)

\(Al\left(OH\right)3:\)

\(MAl\left(OH\right)3=27+17.3=\dfrac{78g}{mol}\)

\(\%Al=\dfrac{27.100}{78}=35\%\)

\(\%O=\dfrac{16.3.100}{78}=62\%\)

\(\%H=\dfrac{1.3.100}{78}=4\%\)

\(SO2:\)

\(MSO2=32+16.2=\dfrac{64g}{mol}\)

\(\%S=\dfrac{32.100}{64}=50\%\)

\(\%O=\dfrac{16.2.100}{64}=50\%\)

\(SO3:\)

\(MSO3=32+16.3=\dfrac{80g}{mol}\)

\(\%S=\dfrac{32.100}{80}=40\%\)

\(\%O=\dfrac{16.3.100}{80}=60\%\)

\(Fe2O3:\)

\(MFe2O3=56.2+16.3=\dfrac{160g}{mol}\)

\(\%Fe=\dfrac{56.2.100}{160}=70\%\)

\(\%O=\dfrac{16.3.100}{160}=30\%\)

C5:

a,MX=2,207.29=64đvC

b, gọi cthh của hợp  chất này là SxOy

Ta có: 32x:16y=50:50

=>x:y=\(\dfrac{50}{32}:\dfrac{50}{16}\)

         = 1,5625:3,125

         =     1      :  2

Vậy CTHH của hợp chất này là SO2

 C2,3,4 lm r nên t bổ sung thim:>

a) \(M_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(DvC\right)\)

\(\%Ca=\dfrac{40.1}{74}.100\%=54\%\)

\(\%O=\dfrac{16.2}{74}.100\%=43\%\)

\(\%H=100\%-54\%-43\%=3\%\)

a) \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{40.1}{74}.100\%=54,054\%\\\%m_O=\dfrac{16.2}{74}.100\%=43,243\%\\\%m_H=\dfrac{2.1}{74}.100\%=2,703\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{137.1}{208}.100\%=65,865\%\\\%Cl=\dfrac{35,5.2}{208}.100\%=34,135\%\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\%m_K=\dfrac{39.1}{56}.100\%=69,643\%\\\%m_O=\dfrac{16.1}{56}.100\%=28,571\%\\\%m_H=\dfrac{1.1}{56}.100\%=1,786\%\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.2}{102}.100\%=52,94\%\\\%m_O=\dfrac{16.3}{102}.100\%=47,06\%\end{matrix}\right.\)

e) \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{23.2}{106}.100\%=43,396\%\\\%m_C=\dfrac{12}{106}.100\%=11,321\%\\\%m_O=\dfrac{16.3}{106}.100\%=45,283\%\end{matrix}\right.\)

g) \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56.1}{72}.100\%=77,78\%\\\%m_O=\dfrac{16.1}{72}.100\%=22,22\%\end{matrix}\right.\)

h) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{65.1}{161}.100\%=40,373\%\\\%m_S=\dfrac{32.1}{161}.100\%=19,876\%\\\%m_O=\dfrac{16.4}{161}.100\%=39,751\%\end{matrix}\right.\)

i) \(\left\{{}\begin{matrix}\%m_{Hg}=\dfrac{201.1}{217}.100\%=92,627\%\\\%m_O=\dfrac{16}{217}.100\%=7,373\%\end{matrix}\right.\)

k) \(\%m_{Na}=\dfrac{23.1}{85}.100\%=27,06\%;\%m_N=\dfrac{14.1}{85}.100\%=16,47\%\%;\%m_O=\dfrac{16.3}{85}.100\%=56,47\%\)

`a,` \(K.L.P.T_{Fe_2O_3}=56.2+16.3=160< amu>.\) 

\(\%Fe=\dfrac{56.2.100}{160}=70\%\)

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

`b,`\(K.L.P.T_{CaCO_3}=40+12+16.3=100< amu>.\)

\(\%Ca=\dfrac{40.100}{100}=40\%\)

\(\%C=\dfrac{12.100}{100}=12\%\) 

\(\%O=100\%-40\%-12\%=48\%\)

`c,` \(K.L.P.T_{HCl}=1+35,5=36,5< amu>.\)

\(\%H=\dfrac{1.100}{36,5}\approx2,74\%\)

\(\%Cl=100\%-2,74\%=97,26\%\)

a: \(\%Fe=\dfrac{56\cdot2}{56\cdot2+16\cdot3}=70\%\)

=>%O=30%

b: \(\%Ca=\dfrac{40}{40+12+16\cdot3}=40\%\)

\(\%C=\dfrac{12}{100}=12\%\)

%O=100%-12%-40%=48%

c: %H=1/36,5=2,74%

=>%Cl=97,26%