$n_{CO_2}=\dfrac{33,44}{44}=0,76(mol)$
$\to n_C=n_{CO_2}=0,76(mol)$
$n_{H_2O}=\dfrac{8,208}{18}=0,456(mol)$
$\to n_{H}=2n_{H_2O}=0,912(mol)$
$\to n_O=\dfrac{14,896-0,76.12-0,912.1}{16}=0,304(mol)$
$n_C: n_H: n_O=0,76:0,912:0,304=5:6 :2$
$\to$ CTPT: $C_5H_6O_2$
$k=3$, gốc axit và ancol giống nhau nên CTCT là:
$CH_2=CHCOOCH=CH_2$
$CH_2=CHCOOCH=CH_2+NaOH\xrightarrow{{t^o}} CH_2=CHCOONa+CH_3CHO$