Bài 1 

Coi khối lượng Fe:7 gam

      Khối lượng O:3 gam 

-> n_Fe=7:56=0,125 mol

    n_o=3:16=0,1875 mol

Tỉ lệ n_Fe:n_o=o,125:0,1875=2:3

->công thức:Fe₂O₃

b khối lượng mol là:

56* 2+16*3=160

a.

\(\%_{Fe_{\left(Fe_2O_3\right)}}=\dfrac{56.2}{160}.100\%=70\%\)

\(\%_{O_{\left(Fe_2O_3\right)}}=100\%-70\%=30\%\)

b.

\(\%_{C_{\left(C_6H_{12}O_6\right)}}=\dfrac{12.6}{180}.100\%=7\%\)

\(\%_{H_{\left(C_6H_{12}O_6\right)}}=\dfrac{1.12}{180}.100\%=6,7\%\)

\(\%_{O_{\left(C_6H_{12}O_6\right)}}=100\%-7\%-6,7\%=86,3\%\)

c.

\(\%_{C_{\left(\left(C_6H_{10}O_5\right)_n\right)}}=\dfrac{12.6}{162n}.100\%=44,4n\%\)

\(\%_{H_{\left(\left(C_6H_{10}O_5\right)_n\right)}}=\dfrac{1.10}{162n}.100\%=6,2n\%\)

\(\%_{O_{\left(C_6H_{1o}O_5\right)}}=\dfrac{16.5}{162n}.100\%=49,4n\%\)

\(\Rightarrow49,4n\%=100\%-44,4n\%-6,2n\%\)

\(\Leftrightarrow n=1\)

\(\Rightarrow\left\{{}\begin{matrix}\%_C=44,4\%\\\%_H=6,2\%\\\%_O=49,4\%\end{matrix}\right.\)

d. 

\(\%_{Na_{\left(NaCl\right)}}=\dfrac{23}{58,5}.100\%=39,3\%\)

\(\%_{Cl_{\left(NaCl\right)}}=100\%-39,3\%=60,7\%\)

B4:

\(CTTQ:Na_xS_yO_z\left(x,y,z:nguy\text{ê}n,d\text{ươ}ng\right)\\ n_{Na}=\dfrac{4.6}{23}=0,2\left(mol\right);n_S=\dfrac{3,2}{32}=0,1\left(mol\right);n_O=\dfrac{4,8}{16}=0,3\left(mol\right)\\ x:y:z=0,2:0,1:0,3=2:1:3\\ \Rightarrow x=2;y=1;z=3\\ \Rightarrow CTHH:Na_2SO_3\)

Mình cần phải có tư duy, chớ nên phụ thuộc vào người khác quá nhiều nhé!

Bài 3:

\(CTTQ:Al_xS_yO_z\left(x,y,z:nguy\text{ên},d\text{ươ}ng\right)\\ x:y:z=\dfrac{15,8\%}{27}:\dfrac{28,1\%}{32}:\dfrac{56,1\%}{16}\approx0,0059:0,0087:0,0351\approx2:3:12\\ \Rightarrow CT\text{Đ}GN:Al_2S_3O_{12}\)

Nó cứ sai sai

Gọi CTHH là \(S_xO_y\)

ta có M S : M O = \(\dfrac{32x}{16y}=\dfrac{2}{3}\)

=> \(\dfrac{x}{y}=\dfrac{1}{3}\)

vậy CTHH là \(SO_3\)

\(\dfrac{32x}{16y}=\dfrac{2}{3}=>\dfrac{2}{1}.\dfrac{x}{y}=\dfrac{2}{3}=>\dfrac{x}{y}=\dfrac{2}{3}:2=\dfrac{1}{3}\)