Hỗn hợp X gồm bốn este mạch hở, trong đó có một este đơn chức và...

${{\text{n}}_{{{\text{O}}_{\text{2}}}}}\text{=0,27;} {{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{=0,23;} {{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{=0,18;} {{\text{n}}_{\text{NaOH}}}\text{=0,12;} {{\text{n}}_{\text{hidrocacbon}}}\text{=0,04}$
m gam X + $\underset{\text{0,27} \text{mol}}{\mathop{{{\text{O}}_{\text{2}}}}} $ $\xrightarrow{{{\text{t}}^{\text{o}}}}$ $\underset{\text{0,23} \text{mol}}{\mathop{\text{C}{{\text{O}}_{\text{2}}}}} $ + $\underset{\text{0,18} \text{mol}}{\mathop{{{\text{H}}_{\text{2}}}\text{O}}} $
$\xrightarrow{\text{BTKL}}{{\text{m}}_{\text{X}}}\text{+0,27}\text{.32=0,23}\text{.44+0,18}\text{.18}\Rightarrow {{\text{m}}_{\text{X}}}\text{=4,72}$ gam
$\xrightarrow{\text{BTNT}\left( \text{O} \right)}{{\text{n}}_{\text{O} \left( \text{X} \right)}}\text{+0,27}\text{.2=0,23}\text{.2+0,18}\text{.1}\Rightarrow {{\text{n}}_{\text{O} \left( \text{X} \right)}}\text{=0,1}$ mol
$\Rightarrow {{\text{n}}_{\text{COO} \left( \text{X} \right)}}\text{=}{{\text{n}}_{\text{NaOH} \left( \text{pu} \right)}}\text{=}{{\text{n}}_{\text{OH} \left( \text{Z} \right)}}\text{=}\dfrac{\text{0,1}}{\text{2}}\text{=0,05}$ mol
${{\text{m}}_{\text{Z}}}\text{=}{{\text{m}}_{\text{binh} \text{Na} \text{tang}}}\text{+}{{\text{m}}_{{{\text{H}}_{\text{2}}}}}\text{=2,25+}\dfrac{\text{0,05}}{\text{2}}\text{.2=2,3}$ gam
$\Rightarrow {{\text{M}}_{\text{Z}}}\text{=}\dfrac{\text{2,3}}{\text{0,05}}\text{=46}$ (C2​H5​OH)
Quy đổi: 4,72 gam X gồm $\left\{ \begin{aligned}
& \text{HCOO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \text{a} \text{mol} \\
& {{\left( \text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \right)}_{\text{2}}} \text{b} \text{mol} \\
& \text{C}{{\text{H}}_{\text{2}}} \text{c} \text{mol} \\
& {{\text{H}}_{\text{2}}} \text{d} \text{mol} \\
\end{aligned} \right. $ + $ \underset{\text{0,27} \text{mol}}{\mathop{{{\text{O}}_{\text{2}}}}} $ $ \xrightarrow{{{\text{t}}^{\text{o}}}} $ $ \underset{\text{0,23} \text{mol}}{\mathop{\text{C}{{\text{O}}_{\text{2}}}}} $ + $ \underset{\text{0,18} \text{mol}}{\mathop{{{\text{H}}_{\text{2}}}\text{O}}} $
Chất rắn Y gồm $\left\{ \begin{aligned}
& \text{HCOONa} \text{a} \text{mol} \\
& {{\left( \text{COONa} \right)}_{\text{2}}} \text{b} \text{mol} \\
& \text{C}{{\text{H}}_{\text{2}}} \text{c} \text{mol} \\
& {{\text{H}}_{\text{2}}} \text{d} \text{mol} \\
& \text{NaOH} \text{du} \text{0,12-0,05=0,07} \text{mol} \\
\end{aligned} \right. $ $ \xrightarrow[{{\text{t}}^{\text{o}}}]{\text{CaO} \text{du}}$ 0,04 mol hiđrocacbon duy nhất Cx​Hy​
$\Rightarrow {{\text{n}}_{\text{hidrocacbon}}}\text{=}{{\text{n}}_{\text{muoi}}}\text{=}{{\text{n}}_{\text{X}}}\text{=0,04}$ mol
Ta có hệ: $\left\{ \begin{aligned}
& \text{a+b=0,04} \\
& \text{a+2b=0,05} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{a=0,03} \\
& \text{b=0,01} \\
\end{aligned} \right.$
$\xrightarrow{\text{BTNT}\left( \text{C} \right)}{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{=3a+6b+c=3}\text{.0,03+6}\text{.0,01+c=0,23}\Rightarrow \text{c=0,08}$
$\xrightarrow{\text{BTNT}\left( \text{H} \right)}{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{=3a+5b+c+d=3}\text{.0,03+5}\text{.0,01+0,08+d=0,18}\Rightarrow \text{d=-0,04}$
Ghép liên kết $\pi $ vào este: $0,04=0,03+0,01$
Ghép CH2​ vào este: $0,08=0,03.2+0,01.2$
Vậy X gồm CH2​=CHCOOC2​H5​ (0,03 mol) và C2​H5​OOC – C2​H2​ – COOC2​H5​ (3 đồng phân)
$\to \%{{\text{m}}_{\text{este} \text{don} \text{chuc}}}\text{=}\dfrac{\text{0,03}\text{.100}}{\text{4,72}}\text{.100 }\%\text{ =63,56 }\%$