BTKL: x + 1,1475.32 = 60,93 + 0,225.28 ⇒ x = 30,51 gam.

E: C_nxH_2nx+2-nN_nO_n+1: a mol

\(\left\{ {\begin{array}{*{20}{c}}
{a\left( {14nx{\rm{ }} + {\rm{ }}29n{\rm{ }} + {\rm{ }}18} \right){\rm{ }} = {\rm{ }}30,51}\\
{a\left( {1,5nx{\rm{ }}--{\rm{ }}0,75n} \right){\rm{ }} = {\rm{ }}1,1475}\\
{na{\rm{ }} = {\rm{ }}0,225.2}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{nxa{\rm{ }} = {\rm{ }}0,99}\\
{na{\rm{ }} = {\rm{ }}0,45}\\
{a{\rm{ }} = {\rm{ }}0,2}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x{\rm{ }} = {\rm{ }}2,2}\\
{n{\rm{ }} = {\rm{ }}2,25}\\
{a{\rm{ }} = {\rm{ }}0,2}
\end{array}} \right.\)

n_X = 4,95 ⇒ peptit 4C chỉ có thể là Gly-Gly (X)

n_X =  0,5.0,2 = 0,1 mol ⇒ muối A là Gly.

\(\overline {{C_{Y,Z}}}  = \frac{{0,99 - 0,14}}{{0,2 - 0,1}} = 5,9 \Rightarrow \exists \) peptit 5C chỉ có thể là Gly-Ala (Y): y (mol)

\(14{\rm{ }} = \underbrace 3_{{O_X}} + \underbrace 3_{{O_Y}} + {\rm{ }}{O_z} \Rightarrow {O_z} = {\rm{ }}8 \Rightarrow Z:heptan{\rm{ }}peptit:{\rm{ }}z{\rm{ }}\left( {mol} \right)\)

\(\left\{ {\begin{array}{*{20}{c}}
{y{\rm{ }} + {\rm{ }}z{\rm{ }} = {\rm{ }}0,1}\\
{BTNT{\rm{ }}N:{\rm{ }}2y{\rm{ }} + {\rm{ }}7z{\rm{ }} + {\rm{ }}0,1.2{\rm{ }} = {\rm{ }}0,225.2}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{y{\rm{ }} = {\rm{ }}0,09}\\
{z{\rm{ }} = {\rm{ }}0,01}
\end{array}} \right.\)

⇒ Do n_Y = n_B = n_Ala ⇒ Z chỉ chứa Gly ⇒ Z là Gly₇

\(\% Z = \frac{{0,01.(75.7 - 18.6).100\% }}{{30,51}} = 13,67\% \)