\(n_{CO_2}=\dfrac{17,6}{44}=0,4mol\Rightarrow n_C=0,4mol\Rightarrow m_C=4,8g\)
\(n_{H_2O}=\dfrac{7,2}{18}=0,4mol\Rightarrow n_H=0,8mol\Rightarrow m_H=0,8g\)
Nhận thấy \(m_C+m_H=5,6< m_A=12g\Rightarrow\)có chứa oxi
\(m_O=12-5,6=6,4mol\Rightarrow n_O=0,4mol\)
Gọi CTHH là \(C_xH_yO_z\)
\(xy:z=n_C:n_H:n_O=0,4:0,8:0,4=1:2:1\)
CTHH: \(CH_2O\)
CTCT: \(HCHO\)