Bảo toàn khối lượng \(\Rightarrow m_X=1.44+2.18-2.32=16\left(g\right)\)
\(n_{CO_2}=1\left(mol\right)\Rightarrow n_C=1\left(mol\right)\Rightarrow m_C=12\left(g\right)\)
\(n_{H_2O}=2\left(mol\right)\Rightarrow n_H=4\left(mol\right)\Rightarrow m_H=4\left(g\right)\)
Ta có: \(m_X=m_C+m_H=16\left(g\right)\)
\(\Rightarrow CTDC:C_xH_y\)
\(x:y=1:4\)
\(\Rightarrow CT:CH_4\)