Thủy phân hoàn toàn m gam hỗn hợp M gồm tetrapeptit X và pentapeptit Y đều mạch hở bằng dung dịch KOH vừa đủ

\(m(g)\left\{\begin{matrix} X: C_xH_yO_5N_4: a\\ Y: C_nH_mO_6N_5: b \end{matrix}\right. \xrightarrow[]{+KOH}(M+11,42) \left\{\begin{matrix} C_3H_6O_2NK\\ C_5H_{10}O_2NK \end{matrix}\right.+H_2O\)
Áp dụng qui tắc bảo toàn khối lượng ta có:
\(\left\{\begin{matrix} M+ \underbrace{ 56(4a+5b) }_{M_{HOK}}=\underbrace{ m+11,42 }_{M_{muoi}}+\underbrace{ 18(a+b) }_{M_{H_2O}} \Rightarrow 206a+262b=11,42\\ n_N = 4a+5b=0,11 \times 2 \hspace{7,3cm} \end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} a=0,03\\ b=0,02 \end{matrix}\right.\)
\(\Rightarrow n_{KOH\ (pu)}=0,22 \ mol \xrightarrow[]{BTNT:\ K}n_{K_2CO_3}=0,11\ mol\)
\((m+11,42)Y\left\{\begin{matrix} C_3H_6O_2NK\\ C_5H_{10}O_2NK \end{matrix}\right. \xrightarrow[]{+O_2,\ t^0}\underbrace{K_2CO_3}_{0,11\ mol}+\underbrace{CO_2+H_2O}_{50,96\ g}+\underbrace{N_2}_{0,11\ mol}\)
Áp dụng bảo toàn khối lượng cho hỗn hợp Y, ta có:
\(Y\left\{\begin{matrix} n_C=0,11+n_{CO_2} \hspace{0,5cm}\\ n_H=2n_{H_2O}\hspace{1,6cm}\\ n_{O/muoi}\hspace{2,5cm}\\ n_N=n_K=0,22\ mol \end{matrix}\right. \rightarrow \left\{\begin{matrix} 12(0,11+n_{CO_2})+2n_{H_2O}+\underbrace{18,7}_{m_O+m_N+m_K}=m+11,42\\ 44n_{CO_2}+ 18n_{H_2O}=50,96\ g \hspace{4,4cm}\\ 2n_{H_2O}=2(0,11+ )n_{CO_2} \hspace{5,2cm} \end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} m=19,88\ g \ \ \ \ \ \\ n_{CO_2}=0,79\ mol\\ n_{H_2O}=0,9\ mol \ \end{matrix}\right. \left\{\begin{matrix} \xrightarrow[]{BTNT:\ C}3n_{C_3H_6O_2NK}+ 5n_{C_5H_{10}O_2NK}=(0,79+0,11)\ mol\\ n_{C_3H_6O_2NK}+ n_{C_5H_{10}O_2NK}=0,22\ mol \hspace{3,8cm} \end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} n_{C_3H_6O_2NK}=0,1\ mol\\ n_{C_5H_{10}O_2NK} = 0,12\ mol \end{matrix}\right.\)
Gọi \(\Rightarrow \left\{\begin{matrix} tetrapeptit.X:0,03\ mol \left\{\begin{matrix} x\ Ala \ \ \ \ \ \ \\ (4-x)Val \end{matrix}\right.\\ pentapeptit.Y:0,02\ mol \left\{\begin{matrix} y\ Ala \ \ \ \ \ \ \\ (5-y)Val \end{matrix}\right. \end{matrix}\right.\)   \(\xrightarrow[]{BT:\ Ala}0,03x+0,02y=0,1\ mol \xrightarrow[]{x\geq 1\rightarrow x<3,5}\left\{\begin{matrix} x=2\\ y=2 \end{matrix}\right.\)
⇒ X: Val – Val – Ala – Ala
Và Y: Val – Val – Val – Ala – Ala
⇒ %m_Y = 45,98%