Hỗn hợp E chứa ba peptit đều mạch hở gồm peptit X (C4H8O3N2), peptit Y (C7HxOyNz)

* Hướng dẫn giải

$$\begin{gathered} E \to \left\{\begin{array}{l}CONH: a\\ C{H}_2: b\\ H_{2}O: c\end{array}\underset{\left(1\right)}{\overset{NaOH}{\to }}\left\{\begin{array}{l}COONa: a\\ N{H}_2: a\\ C{H}_2: b\end{array}\right.\right. \\ \underset{\left(2\right)}{\overset{1,155 mol,t^o}{\to }}N{a}_{2}C{O}_3 + ... \\ \left\{\begin{array}{l}a=n_{CONH/E}=2.n_{N{a}_{2}C{O}_3}=0,44\\ m_E=43a + 14b + 18c=28,42\\ BT e \left(2\right): 3a + 6b = 1,155.4\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}a = 0,44\\ b = 0,55\\ c = 0,1\end{array} \right. \\ {\overline{M}}_{CONH/E} = \frac{0,44}{0,1} =4,4. \\ \Rightarrow Z phải chứa 4 nhóm CONH vì \\ \left\{\begin{array}{l}X chứa 2 nhóm CONH\\ Y chứa 3 nhóm CONH\end{array}\right. \\ \left\{\begin{array}{l}X: C_4{H}_8{O}_3{N}_2\\ Z: C_4{H}_8{O}_3{N}_2\\ Y: C_7{H}_x{O}_y{N}_z\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}4=2+2\\ 11=2.4+3\\ 7=2+5=2.2+3\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}X là Gl{y}_2 (g mol)\\ Z là Gl{y}_{4}Ala (h mol)\\ Y là Gly Val (k mol) hoặc Gl{y}_{2}Ala\end{array}\right. \\ \left\{\begin{array}{l}{n}_{CONH}=2g+5h+2k=0,44\\ n_E=g+h+k=0,1\\ m_E=132g+317h+174k=28,42\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}g=0,01\\ h=0,08\\ z=0,01\end{array}\right. \\ \Rightarrow \%X=\frac{0,01.132}{28,42} = 4,64\% \end{gathered}$$