Đáp án:
$C_2H_4O_2$
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CTHH:{C_x}{H_y}{O_z}\\
x:y:z = \dfrac{{40}}{{12}}:\dfrac{{6,67}}{1}:\dfrac{{53,33}}{{16}} = 3,33:6,67:3,33\\
\Rightarrow x:y:z = 1:2:1\\
\Rightarrow CTDGN:C{H_2}O\\
CTPT\,A:{(C{H_2}O)_n}\\
{M_A} = 60g/mol \Rightarrow 30n = 60 \Rightarrow n = 2\\
CTPT:{C_2}{H_4}{O_2}\\
b)\\
C{H_3}COOH:{\rm{ax}}it\\
HCOOC{H_3}{\rm{:es}}te
\end{array}\)