Coi X gồm : $CH_3COOH(a\ mol) ; C_2H_3COOCH_3(b\ mol)$
Ta có :
$60a + 86b = 4,02$
$a + b = \dfrac{1,344}{22,4} = 0,06$
Suy ra $a = \dfrac{57}{1300} ; b = \dfrac{21}{1300}$
Bảo toàn H : $n_{H_2O} = 2a + 3b = \dfrac{177}{1300}$
$m = 18.\dfrac{177}{1300} = 2,45(gam)$