X, Y là hai peptit mạch hở, có tổng số nguyên tử oxi là 9 và đều...

$\left\{ \begin{aligned}
& {{C}_{x}}{{H}_{2x-1}}NO:a mol \\
& {{n}_{peptit}}:b mol \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 0,6.1,25.56-8,52=0,6.\left( \left( 14x+29 \right)a+18b \right)+18b \\
& 1,5ax-0,75a=2,43 \\
& 44ax-a.18\left( x-0,5 \right)-b.18=51 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& ax=1,92 \\
& a=0,6 \\
& b=0,24 \\
\end{aligned} \right.$
$\Rightarrow \overline{x}=\dfrac{xa}{a}=\dfrac{1,92}{0,6}=3,2$. Đặt $\left\{ \begin{aligned}
& Gly\left( 2C \right):{a}' \\
& Val\left( 5C \right):{b}' \\
\end{aligned} \right.\Rightarrow $ Đ. chéo: $ \dfrac{{{a}'}}{{{b}'}}=\dfrac{5-3,2}{3,2-2}=\dfrac{3}{2}=\dfrac{0,36}{0,24}$
$\begin{aligned}
& E\left\{ \begin{aligned}
& {{X}_{m}}:t mol \\
& {{Y}_{n}}:k mol \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& m+1+n+1=9 \\
& m<\dfrac{a}{b}<n \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& m+n=7 \\
& m<2,5<n \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& m=2 \\
& n=5 \\
\end{aligned} \right. \\
& \Rightarrow \left\{ \begin{aligned}
& t+k=0,24 \\
& \dfrac{t}{k}=\dfrac{5-2,5}{2,5-2}=\dfrac{5}{1} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=0,2 \\
& k=0,04 \\
\end{aligned} \right. \\
& E\left\{ \begin{aligned}
& X{{\left( Gly \right)}_{{{a}''}}}{{\left( Val \right)}_{2-{a}''}}:0,2 \\
& Y{{\left( Gly \right)}_{{{b}''}}}{{\left( Val \right)}_{5-{b}''}}:0,04 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& BT.Gly:0,2{a}''+0,04{b}''=0,36 \\
& {a}''\le 2;{b}''\le 5 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {a}''=1 \\
& {b}''=4 \\
\end{aligned} \right. \\
& \Rightarrow \%{{m}_{Y}}=\dfrac{\left( 75.4+117-18.4 \right).0,04}{14.1,92+29.0,6+18.0,24}.100\%=28,40\% \\
\end{aligned}$