\(a)n_{CO_2}=\dfrac{8,8}{44}=0,2mol\\
n_C=n_{CO_2}=0,2mol\\
n_{H_2O}=\dfrac{7,2}{18}=0,4mol\\
n_H=2n_{H_2O}=0,8mol\\
m_H=0,8.1=0,8g\\
m_C=0,2.12=2,4g\\
m_{C\&H}=0,8+2,4=3,2g< m_A\)
\(\Rightarrow\)A có chứa \(C,H,O\)
\(b)CTPT\left(A\right):C_xH_yO_z\\
m_O=6,4-3,2=3,2g\)
Ta có tỉ lệ:
\(\dfrac{12x}{2,4}=\dfrac{y}{0,8}=\dfrac{16z}{3,2}=\dfrac{32}{6,4}\\ \Rightarrow\left\{{}\begin{matrix}x=1\\y=4\\z=1\end{matrix}\right.\\ \Rightarrow CTPT\left(A\right)CH_4O\)