\(M_A=8.2=16\left(\dfrac{g}{mol}\right)\\ n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\Rightarrow n_H=0,4\left(mol\right)\Rightarrow m_H=0,4.1=0,4\left(g\right);\\ m_C=1,6-0,4=1,2\left(g\right)\Rightarrow n_C=\dfrac{1,2}{12}=0,1\left(mol\right)\\ Đặt.CTTQ.A:C_xH_y\left(x,y:nguyên,dương\right)\\ x:y=0,1:0,4=1:4\\ \Rightarrow CTĐGN:\left(CH_4\right)_m\left(m:nguyên,dương\right)\\ M_A=16m=16\\ \Leftrightarrow m=1\\ Vậy.CTPT.A:CH_4\)