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

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

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

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

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

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

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

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

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

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

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

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

a, %C = 12n/(12n + 2n + 2) = 75%

=> n = 1

CTHH: CH4

b, CaHo ???

\(\%Ca=\dfrac{40.1}{56}.100\%=71,4\%\\ \%O=\dfrac{16.1}{56}.100\%=28,6\%\)

\(M_{Na_2SO_4}=23.2+32+16.4=142\left(DvC\right)\\ \%Na=\dfrac{23.2}{142}.100\%=32,4\%\\ \%S=\dfrac{32}{142}.100\%=22,5\%\\ \%O=100\%-32,4\%-22,5\%=45,1\%\)

\(-Na_2SO_4\)

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

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

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

\(-Fe_3O_4\)

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

\(\%_O=100\%-72,4\%=27,6\%\)

\(-BaCl_2\)

\(\%_{Ba}=\dfrac{137}{208}.100\%=65,9\%\)

\(\%_{Cl}=100\%-65,9\%=34,1\%\)

* Na₂SO₄:

\(\%m_{Na}=\dfrac{NTK_{Na}.2}{NTK_{Na}.2+NTK_S+4.NTK_O}.100\%=\dfrac{23.2}{23.2+32+4.16}.100=\dfrac{46}{142}.100\approx32,394\%\\ \%m_S=\dfrac{NTK_S}{NTK_{Na}.2+NTK_S+4.NTK_O}=\dfrac{32}{142}.100\approx22,535\%\\ \Rightarrow\%m_O\approx100\%-\left(32,394\%+22,535\%\right)\approx45,071\%\)

Với các chất còn lại cách làm tương tự nha em!

a) Hợp chất với hidro là XH₄ 

Oxit xao nhất của X là XO₂ 

Ta lập được phương trình: \(\dfrac{X}{X+4}=1,875\cdot\dfrac{X}{X+32}\) 

\(\Rightarrow X=28\)  (Silic)

b) \(\%Si_{\left(SiO_2\right)}=\dfrac{28}{28+16\cdot2}\cdot100\%\approx46,67\%\)

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

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

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

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

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

+) Trong H₂SO₄ có: \(\left\{{}\begin{matrix}\%m_H=\dfrac{2}{98}\cdot100\%\approx2,04\%\\\%m_S=\dfrac{32}{98}\cdot100\%\approx32,65\%\\\%m_O=65,31\%\end{matrix}\right.\)

+) Trong HNO₃ có: \(\left\{{}\begin{matrix}\%m_H=\dfrac{1}{63}\cdot100\%\approx1,59\%\\\%m_N=\dfrac{14}{63}\cdot100\%\approx22,22\%\\\%m_O=76,19\%\end{matrix}\right.\)

a)Ta có:\(m\%_H=\dfrac{2.100\%}{98}=2,04\%\)

        \(m\%_S=\dfrac{32.100\%}{98}=32,65\%\)

        \(m\%_O=100-2,04-32,65=65,31\%\)

b) tương tự

Câu 30. Phần trăm về khối lượng của nguyên tố N trong (NH₂)₂CO là :

A. 46,67%                   B. 31,81%                               C. 32,33%                   D. 63,64%