X, Y là hai peptit mạch hở hơn kém nhau một liên kết peptit, Z là este của

Đáp án B

$E\left\{\begin{array}{l}N{H}_{2}C{H}_{2}COO{C}_2{H}_5:x\\ C_2{H}_{3}NO:y\\ C{H}_2:z\\ H_{2}O:t\end{array}\right.\to \left\{\begin{array}{l}103+57y+14z+18t=43,75\left(1\right)\\ 9x+4y+2z+t=3,5\left(2\right)\end{array}\right.$

$\begin{array}{l}{n}_E{n}_{NaOH}\\ \left(x+t\right)\to \left(x+y\right)\\ 0,3\to 0,55\end{array}$

$$\begin{gathered} \to 0,3.\left(x+y\right)=0,55.\left(x+t\right)\left(3\right) \\ E\left\{\begin{array}{l}N{H}_{2}C{H}_{2}COO{C}_2{H}_5:x\\ C_2{H}_{3}NO:y\\ C{H}_2:z\\ H_{2}O:t\end{array}\right.\to \left\{\begin{array}{l}{C}_2{H}_{4}N{O}_{2}Na:x+y\\ C{H}_2:z\end{array}\right. \\ \begin{array}{l}{n}_E{n}_{NaOH}\\ \left(x+t\right)\to 97\left(x+y\right)+14z\\ 0,3\to 55,03\end{array} \\ \to 0,3.\left(97x+97y+14z\right)=55,03.\left(x+t\right)\left(4\right) \end{gathered}$$

Từ (1), (2), (3), (4) $\to \left\{\begin{array}{l}x=0,19\\ y=0,36\\ z=0,12\\ t=0,11\end{array}\right.$

$n_{C{H}_2}<n_{este}$ nên este Z có công thức $N{H}_{2}C{H}_{2}COO{C}_2{H}_5$

 $\frac{n_{C{H}_2}}{n_{H_{2}O}}=3,27$ nên X, Y là tripeptit và tetrapeptit

$$\begin{gathered} \to \left\{\begin{array}{l}{X}_3:a\\ Y_4:b\end{array}\right.\to \left\{\begin{array}{l}a+b=0,11\\ 3a+4b=0,36\end{array}\right.\to \left\{\begin{array}{l}a=0,08\\ b=0,03\end{array}\right. \\ \left\{\begin{array}{l}Gl{y}_3{\left(C{H}_2\right)}_n:0,08\\ Gl{y}_4{\left(C{H}_2\right)}_m:0,03\end{array}\right.\to 0,08.n+0,03.m=0,12\to \left\{\begin{array}{l}n=0\\ m=4\end{array}\right. \\ \to \left\{\begin{array}{l}X:Gl{y}_3:0,08\\ Y:Gl{y}_{2}AlaVal:0,03\\ Z:N{H}_{2}C{H}_{2}COO{C}_2{H}_5:0,19\end{array}\right. \\ \to \%m_X=34,56\% \end{gathered}$$