Đáp án:

\(\begin{array}{l}
7)\\
CTHH:{C_6}{H_{12}}{O_6}\\
8)\\
CTHH:N{a_2}S\\
{M_{N{a_2}S}} = 78(g/mol)\\
9)\\
CTHH:F{e_2}{O_3}\\
{M_{F{e_2}{O_3}}} = 160(g/mol)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
7)\\
CTHH\,:\,{X_6}{H_{12}}{O_6}\\
{M_{{X_6}{H_{12}}{O_6}}} = \dfrac{{45}}{{16}} \times 2 \times {M_{{O_2}}} = \dfrac{{45}}{{16}} \times 2 \times 32 = 180(g/mol)\\
 \Rightarrow 6{M_X} + 12 \times {M_H} + 6 \times {M_O} = 180\\
 \Rightarrow {M_X} = 12(g/mol)\\
 \Rightarrow X:Cacbon(C)\\
 \Rightarrow CTHH:{C_6}{H_{12}}{O_6}\\
8)\\
CTHH:\,N{a_x}{S_y}\\
\% {m_S} = 100 - 59 = 41\% \\
x:y = \dfrac{{59}}{{23}}:\dfrac{{41}}{{32}} = 2,57:1,28 \approx 2:1\\
 \Rightarrow CTHH:N{a_2}S\\
{M_{N{a_2}S}} = 23 \times 2 + 32 = 78(g/mol)\\
9)\\
CTHH:F{e_x}{O_y}\\
{m_{Fe}}:{m_O} = 7:3 \Rightarrow 56x:16y = 7:3\\
 \Rightarrow x:y = 2:3\\
 \Rightarrow CTHH:F{e_2}{O_3}
\end{array}\)

\({M_{F{e_2}{O_3}}} = 56 \times 2 + 16 \times 3 = 160(g/mol)\)