$\begin{cases}CH_3COOH:x mol \\ CH_2=CH-COOH:y mol\\CH_3CH_2COOH:z mol \end{cases}\Rightarrow \begin{cases}60x+72y+74z=3,15 \\ y=\frac{3,2}{160}\\x+y+z=0,045 \end{cases}$$\Rightarrow \begin{cases}x=0,01 mol \\ y=0,02 mol\\z=0,015 mol \end{cases}\Rightarrow \begin{cases}\%CH_3COOH=19,05\% \\\%CH_2=CH-COOH=45,71\%\\\%CH_3CH_2COOH=35,24\% \end{cases}$