Đặt $n_{HCOOH}=n_{CH_3COOH}=x$
$\Rightarrow 46x+60x=16,96$
$\Leftrightarrow x=0,16$
$\Rightarrow n_X=0,16.2=0,32 mol$
$\overline{M_X}=\dfrac{16,96}{0,32}=53=M_{RCOOH}$
$\Rightarrow M_R=8$
Đặt $n_{CH_3OH}=2y\Rightarrow n_{C_2H_5OH}=3y$
$\Rightarrow 32.2y+46.3y=8,08$
$\Leftrightarrow y=0,04$
$\Rightarrow n_Y=5y=0,2 mol$
$\overline{M_Y}=\dfrac{8,08}{0,2}=40,4=M_{R'OH}$
$\Rightarrow M_{R'}=23,4$
$RCOOH+R'OH\rightleftharpoons RCOOR'+H_2O$
Theo lí thuyết tạo 0,2 mol este.
$\Rightarrow m=0,2.(8+44+23,4).80\%=12,064g$