$n_{C_2H_5OH}<n_{CH_2=C(CH_3)-COOH}\Rightarrow $Ta có sơ đồ chuyển hóa sau :$C_2H_5OH\overset{60\%}{\rightarrow}CH_2=C(CH_3)COOC_2H_5$
$\overset{60\%}{\rightarrow} [-CH_2-C(CH_3)(COOC_2H_5)-]_n$
$\Rightarrow m=\frac{114.128.60.60}{46.100.100}\approx 114,2 g$