Hỗn hợp E gồm tripeptit X và tetrapeptit Y đều mạch hở. Thủy phân

* Hướng dẫn giải

$$\begin{gathered} \mathit{Đ}\mathit{á}\mathit{p}\mathbf{ }\mathit{á}\mathit{n}\mathbf{ }\mathit{B} \\ \left\{\begin{array}{l}X\\ Y\end{array}\to \left\{\begin{array}{l}CONH:x mol\\ C{H}_2:y mol\\ H_{2}O:0,2 mol\end{array}\right.\right.\left(E\right) \\ E\stackrel{NaOH}{\to }\left\{\begin{array}{l}COONa:x\\ N{H}_2:y\\ C{H}_2:y\end{array}(76,25g)\right. \\ E\stackrel{HCl}{\to }\left\{\begin{array}{l}COOH:x\\ N{H}_{3}Cl:y\\ C{H}_2:y\end{array}(87,125g)\right. \\ \left\{\begin{array}{l}83x+14y=76,25\\ 97,5x+14y=87,125\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}x=0,75\\ y=1\end{array}\right. \\ \left\{\begin{array}{l}{n}_{GlyNa}+n_{AlaNa}=n_{CONH}=0,75\\ 97n_{GlyNa}+111n_{AlaNa}=76,25\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{n}_{GlyNa}=0,5\\ n_{AlaNa}=0,25\end{array}\right. \\ \left\{\begin{array}{l}{n}_X+n_Y=0,2\\ n_X+n_Y=n_{CONH}=0,75\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{n}_X=0,05\\ n_Y=0,15\end{array}\right. \\ \left\{\begin{array}{l}X là Gl{y}_a{\left(Ala\right)}_{3-a}=0,05\\ Y là Gl{y}_b{\left(Ala\right)}_{4-a}=0,15\end{array}\right. \\ \Rightarrow 0,05a+0,15b=0,15 \\ \Rightarrow a=1, b=3 \\ \Rightarrow \left\{\begin{array}{l}X là Gly{\left(Ala\right)}_2\left(M=217\right)\\ Y là Gl{y}_3\left(Ala\right) \left(M=260\right)\end{array}\right. \\ \Rightarrow \%X=\frac{217.0,05}{217.0,05+260.0,15} \\ =21,76\%\approx 22\% \end{gathered}$$