a) \(\left\{{}\begin{matrix}\%Fe=\dfrac{56.2}{160}.100\%=70\%\\\%O=100\%-70\%=30\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\%Al=\dfrac{27.2}{342}.100\%=15,79\%\\\%S=\dfrac{32.3}{342}.100\%=28,07\%\\\%O=\dfrac{16.12}{342}.100\%=56,14\%\end{matrix}\right.\)

\(a,\%Cu=\dfrac{m_{Cu}}{M_{CuSO_4}}=\dfrac{64}{160}=40\%\\ \%S=\dfrac{m_S}{M_{CuSO_4}}=\dfrac{32}{160}=20\%\\ \%O=100\%-\%Cu-\%S=100\%-40\%-20\%=40\%\)

\(b,\%Fe=\dfrac{m_{Fe}}{M_{Fe_3O_4}}=\dfrac{168}{232}=72,41\%\\ \%O=100\%-\%Fe=100\%-72,41\%=27,59\%\)

\(c,\%Fe=\dfrac{m_{Fe}}{M_{FeO}}=\dfrac{56}{72}=77,77\%\\ \%O=100\%-\%Fe=100\%-77,77\%=22,23\%\)

\(d,\%K=\dfrac{m_K}{M_{K_2SO_3}}=\dfrac{78}{138}=56,52\%\\ \%C=\dfrac{m_C}{M_{K_2SO_4}}=\dfrac{12}{138}=8,69\%\\ \%O=100\%-\%K-\%C=100\%-56,52\%-8,69\%=34,79\%\)

a) \(M_{CúSO_4}=64+32+16.4=160\left(DvC\right)\)

\(m_{Cú}=\dfrac{64}{160}.100\%=40\%\)

\(m_S=\dfrac{32}{160}.100\%=20\%\)

\(m_O=\dfrac{16.4}{160}.100\%=40\%\)

b) \(M_{Fe_3O_4}=56.3+16.4=232\left(DvC\right)\\ \%Fe=\dfrac{56.3}{232}.100\%=72\%\\ \%O=100\%-74\%=26\%\)

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

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

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

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

\(M_{Fe_2O_3}=56\cdot2+16\cdot3=160\left(đvc\right)\)

\(\%m_{Fe}=\dfrac{112}{160}\cdot100\%=70\%\)

\(\%m_O=\dfrac{48}{160}\cdot100\%=30\%\)

\(\left\{{}\begin{matrix}\%Fe=\dfrac{56.2}{56.2+16.3}.100\%=70\%\\\%O=100\%-70\%=30\%\end{matrix}\right.\)

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

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

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

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

a) Với Fe₃O₄ thì Fe là 72,4% và O là 27,6%;
    Với Fe₂O₃ thì Fe là 70% và O là 30%
b) Với SO₂ thì S là 50% và O là 50%
    Với SO₃ thì S là 40% và O là 60%
c) m_Cu= \(\dfrac{80.80}{100}\)=64(g) ; m_O=\(\dfrac{80.20}{100}\)=16(g)
    n_Cu=\(\dfrac{64}{64}\)=1(mol) ; n_O=\(\dfrac{16}{16}\)=1(mol)
    Vậy CTHH của oxit đồng màu đen là: CuO
d) d_A/H2=\(\dfrac{Ma}{2}\)=17 => M_A=2.17=34(đvC)
   H =\(\dfrac{5,88.34}{100}\)\(\approx\)2(đvC) ; S =\(\dfrac{94,12.34}{100}\)\(\approx\)32
=> CTHH của chất khí A là SH₂

Câu 1 :

\(M_{K_2CO_3}=39.2+12+16.3=138\left(dvC\right)\)

\(\%K=\dfrac{39.2}{138}.100\%=56,52\%\)

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

\(\%O=100\%-56,52\%-8,69\%=34,79\%\)

Còn lại cậu làm tương tự nhá

Bài 2 :

\(M_S=\dfrac{64.50\%}{100\%}=32\left(g\right)\)

\(n_S=\dfrac{32}{32}=1\left(mol\right)\)

\(M_O=64-32=32\left(g\right)\\ n_O=\dfrac{32}{16}=2\left(mol\right)\)

\(=>CTHH:SO_2\)

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

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