a) Hợp chất CO có MCO = 12 + 16 = 28 g/mol

    %mO = 100 – 42,86 = 57,14%

Hợp chất CO2: có MCO2 = 12 + 32 = 44 g/mol

    %mO = 100 – 27,3 = 72,7%

b) Hợp chất Fe3O4: MFe3O4 = 3.56 + 4.16 = 232 g/mol

    %mO = 100 – 72,4 = 27,6%

Hợp chất Fe2O3: MFe2O3 = 2.56 + 3.16 = 160 g/mol

    %mO = 100 – 70 = 30%

c) Hợp chất SO2 : MSO2 = 32 + 2.16 = 64 g/mol

    %mO = 100 – 50 = 50%

Hợp chất SO3 : MSO3 = 32 + 16.3 = 80 g/mol

    %mO = 100 – 40 = 60%

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

\(PTK_{CuO}=64+16=80\left(đvC\right)\)

\(\%m_{Cu}=\) \(\dfrac{64}{80}.100=80\%\)

\(\%m_O=100-80=20\%\)

\(PTK_{MgCO_3}=24+12+3.16=84\left(đvC\right)\)

\(\%m_{Mg}=\dfrac{24}{84}.100=28,57\%\)

\(\%m_C=\dfrac{12}{84}.100=14,28\%\)

\(\%m_O=\dfrac{3.16}{84}.100=57,14\%\)

các ý còn lại làm tương tự

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

\(M_{CuSO_4}=64+32+16.4=160\left(\dfrac{g}{mol}\right)\)

\(\%Cu=\dfrac{64}{160}.100\%=40\%\)

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

\(\%O=100\%-20\%-40\%=40\%\)

\(M_{K_2CO_3}=39.2+12+16.3=138\left(\dfrac{g}{mol}\right)\)

\(\%K=\dfrac{39.2}{138}.100\%=56,5\%\)

\(\%C=\dfrac{12}{138}.100\%=8,7\%\)

\(\%O=100\%-56,5\%-8,7\%=34,8\%\)

\(M_{Ná_2SO_4}=23.2+32+16.4=142\left(\dfrac{g}{mol}\right)\)

\(\%Na=\dfrac{23.2}{142}.100\%=32,4\%\)

\(\%S=\dfrac{32}{142}.100\%=22,5\%\)

\(\%O=100\%-32,4\%-22,5\%=45,1\%\)

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

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

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

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

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) 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

a/

+ CO

+CO₂

b/

+Fe₃O₄

+ Fe₂O₃

c/

+SO₂

+ SO₃

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

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

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

a) MK2CO3 = 39.2 + 12 + 16.3 = 138 (g)

b)