\(K.L.P.T_{SO_2}=32+16.2=64< amu>.\)

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

\(\%O=100\%-50\%=50\%.\)

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

\(\%_{Al}=\dfrac{27}{78}.100\%=34,6\%\)

\(\%_O=\dfrac{16.3}{78}.100\%=61,5\%\)

\(\%_H=100\%-34,6\%-61,5\%=3,9\%\)

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

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
\(\%M_{\dfrac{Fe}{Fe_2O_3}}=\dfrac{112}{160}.100\%=70\%\)
\(\%M_{\dfrac{O}{Fe_2O_3}}=100\%-70\%=30\%\)
 

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:

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

\(CaO:\%^MCa=\dfrac{40}{40+16}\cdot100\%=71,43\%\\ \%^MO=100\%-71,43\%=28,57\%\)

\(MgCO_3:\%^MMg=\dfrac{24}{24+12+16\cdot3}\cdot100\%=28,57\%\\ \%^MC=\dfrac{12}{24+12+16\cdot3}\cdot100\%=14,29\%\\ \%^MO=100\%-28,57\%-14,29\%=57,14\%\)

\(NaOH:\%^MNa=\dfrac{23}{23+16+1}\cdot100\%=57,5\%\\ \%^MO=\dfrac{16}{23+16+1}\cdot100\%=40\%\\ \%^MH=100\%-57,5\%-40\%=2,5\%\)

a)

\(m_C=\dfrac{52,15.46}{100}=24\left(g\right)=>n_C=\dfrac{24}{12}=2\left(mol\right)\)

\(m_H=\dfrac{13,04.46}{100}=6\left(g\right)=>n_H=\dfrac{6}{1}=6\left(mol\right)\)

\(m_O=46-24-6=16\left(g\right)=>n_O=\dfrac{16}{16}=1\left(mol\right)\)

=> CTHH: C₂H₆O

b) \(n_A=\dfrac{18,4}{46}=0,4\left(mol\right)\)

m_C = 12.0,4.2 = 9,6(g)

m_H = 1.0,4.6 = 2,4 (g)

m_O = 16.0,4.1 = 6,4 (g)

c) \(n_A=\dfrac{13,8}{46}=0,3\left(mol\right)\)

Số nguyên tử C = 2.0,3.6.10²³ = 3,6.10²³

Số nguyên tử H = 6.0,3.6.10²³ = 10,8.10²³

Số nguyên tử O = 1.0,3.6.10²³ = 1,8.10²³

`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%

\(M_{Fe_3O_4}=56.3+16.4=232\left(\dfrac{g}{mol}\right)\)

\(\%Fe=\dfrac{56.3}{232}.100\%=72,4\%\)

\(\%O=\dfrac{16.4}{232}.100\%=27,862\%\)

tìm công thức hóa học của hợp chất có thàn phần các nguyên tố như sau:55,189%K; 14,623%P và còn lại là oxi.Biết 0,05 mol hợp chất có khối lượng là 10,6 gam