Hỗn hợp E chứa các chất hữu cơ mạch hở gồm tetrapeptit X; pentapeptit Y và este

Sơ đồ hóa bài toán:
$36,86(g)E\,\left\{ \begin{array}{l} tetrapeptit\,X\\ pentapeptit\,Y\\ Z:\,{H_2}NC{H_2}{\rm{COOC}}{{\rm{H}}_3} \end{array} \right.\,\, + \,KOH{\,_{vua\,du}}\, \to \left| \begin{array}{l} Muoi\,\left\{ \begin{array}{l} {H_2}NC{H_2}{\rm{COOK:}}\,{\rm{a}}\,{\rm{mol}}\\ {\rm{C}}{{\rm{H}}_3}CH(N{H_2})C{\rm{OOK:}}\,b\,mol \end{array} \right.\, + \,{O_2}\, \to \,\left\{ \begin{array}{l} C{O_2}\\ {H_2}O\\ {N_2}\\ {K_2}C{O_3} \end{array} \right.\\ C{H_3}OH:\,\frac{{3,84}}{{32}} = 0,12;\,{H_2}O \end{array} \right.$
Bảo toàn nguyên tố K → ${n_{KOH}} = \,{n_{muoi}}\, = \,2{n_{{K_2}C{O_3}}}\, = \,2.0,25 = \,0,5\,mol$
→ a + b = 0,5 (1)
Bảo toàn nguyên tố O: $2{n_{muoi}}\, + \,2{n_{{O_2}}}\, = \,2{n_{C{O_2}}}\, + {n_{{H_2}O}}\, + \,3{n_{{K_2}C{O_3}}}$
→ 2. (a +b) + 2.1,455 = 2. (2a+3b-0,25)+(2a+3b)+3.0,25
Từ (1) và (2) → $\left\{ \begin{array}{l} a = 0,28 = \,{n_{{H_2}NC{H_2}{\rm{COOK}}}} = \,{n_{goc\,Gly}}\\ b = 0,22 = \,{n_{C{H_3}CH(N{H_2})C{\rm{OOK}}}} = \,{n_{goc\,Ala}} \end{array} \right.$
→ ${m_{muoi}} = \,$ 0,28.113+0,22.127=59,58(g)
Bảo toàn khối lượng: $\underbrace {{m_{hh\,E}}}_{36,86}\, + \,\underbrace {{m_{KOH}}}_{0,5.56}\, = \underbrace {\,{m_{muoi}}}_{59,58}\, + \,\underbrace {{m_{C{H_3}OH}}}_{3,84}\, + \,{m_{{H_2}O}}$
→ ${m_{{H_2}O}} = \,0,08\,mol$
Các quá trình phản ứng thủy phân như sau
$\underbrace {X\,}_x + \underbrace {4KOH}_{4x} \to \,4\,muoi\, + \,\underbrace {{H_2}O}_x$
$\underbrace Y_y\, + \,\underbrace {5KOH}_{5y}\, \to 5\,Muoi\, + \underbrace {\,{{\rm H}_2}O}_y$
${H_2}NC{H_2}{\rm{COOC}}{{\rm{H}}_3}\, + \,\underbrace {KOH\,}_{0,12} \to \,{H_2}NC{H_2}{\rm{COOK}}\,{\rm{ + }}\,\underbrace {{\rm{C}}{{\rm{H}}_3}OH}_{0,12}$
Ta có: ${n_{{H_2}O}} = \,x + y = 0,08$, ${n_{KOH}} = \,4x + 5y + 0,12 = 0,5$
→ x = 0,02 = ${n_X}$ và y = 0,06 = ${n_Y}$
Gọi công thức $\left\{ \begin{array}{l} X:\,{(Ala)_n}{(Gly)_{4 - n}} = \,0,02\,(n\, \le 4)\\ Y:\,{(Ala)_m}{(Gly)_{5 - m}} = \,0,06\,(m\, \le 5) \end{array} \right.$
Bảo toàn Na: 0,02n + 0,06m = 0,22
→ $\left\{ \begin{array}{l} n = 2\\ m = 3 \end{array} \right.(t/m) \to \,\left\{ \begin{array}{l} X:{(Ala)_2}{(Gly)_2}\\ Y:\,{(Ala)_3}{(Gly)_2} \end{array} \right.\, \to \,{M_X}\, = \,89.2 + 75.3 - 18.3 = 274$
%${m_{X(E)}} = \,\frac{{274.0,02}}{{36,86}}.100\% = 14,87\%$