\(n_C=n_{CO_2}=0,2\left(mol\right);n_H=2n_{H_2O}=2.\dfrac{7,2}{18}=0,8\left(mol\right)\\ M_A=16.M_{H_2}=16.2=32\left(\dfrac{g}{mol}\right)\\ m_C+m_H=0,2.12+0,8.1=3,2\left(g\right)< 6,42\left(g\right)\\ m_O=6,4-3,2=3,2\left(g\right)\\ n_O=\dfrac{3,2}{16}=0,2\left(mol\right)\\ Ta.có:n_C:n_H:n_O=0,2:0,8:0,2=1:4:1\\ \Rightarrow CTTQ.A:\left(CH_4O\right)_m\left(m:nguyên,dương\right)\\ \Leftrightarrow32m=32\\ \Leftrightarrow m=1\\ Vậy.CTPT.A:CH_4O\\ CTCT:CH_3-OH\)