- Khi gộp X và Y với tỉ lệ mol tương ứng là 1 : 3 có $2X+Y\xrightarrow{{}}{{X}_{2}}Y+2{{H}_{2}}O\,\,(1)$

+ Từ: $\frac{{{n}_{Gly}}}{{{n}_{Tyr}}}=\frac{0,075}{0,06}=\frac{5}{4}$ $\Rightarrow {{X}_{2}}Y$ là ${{(Gly)}_{5k}}{{(Tyr)}_{4k}}$

mà $\sum{\underbrace{s\grave{e}\text{ m }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ c x }\!\!\acute{\mathrm{Y}}\!\!\text{ c}{{\text{h}}_{(\min )}}}_{(5+2).{{n}_{X}}}\,\,\text{}\,\,\,\sum{\underbrace{s\grave{e}\text{ m }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ c x }\!\!\acute{\mathrm{Y}}\!\!\text{ ch c }\!\!\tilde{\mathrm{n}}\!\!\text{ a }X{{Y}_{3}}}_{5k+4k}\,}\,\,\,\text{}\,\,\,\sum{\underbrace{s\grave{e}\text{ m }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ c x }\!\!\acute{\mathrm{Y}}\!\!\text{ c}{{\text{h}}_{(max)}}}_{(5+2).{{n}_{Z}}}}\text{ }}\to 7.1<9k<7.2\Rightarrow k=1$

+ Với k = 1 Þ ${{n}_{{{(Gly)}_{5}}{{(Tyr)}_{4}}}}={{n}_{{{X}_{2}}Y}}=\frac{{{n}_{Gly}}}{5}=\frac{{{n}_{Tyr}}}{4}=0,015\,mol$

 (1) $\xrightarrow{BTKL}{{m}_{X}}+{{m}_{Y}}={{m}_{{{X}_{2}}Y}}+18{{n}_{{{H}_{2}}O}}=$14,865(g)