Ta có: \(dX/O_2=\dfrac{M_X}{M_{O_2}}=1,3125\Rightarrow M_X=1,3125.32=42\left(g/mol\right)\)
\(m_H=14,29\%.42=6\left(g\right)\Rightarrow n_H=\dfrac{6}{1}=6\left(mol\right)\)
\(\Rightarrow\%_C=100\%-14,29\%=85,71\%\)
\(\Rightarrow m_C=85,71\%.42\approx36\left(g\right)\Rightarrow n_C=\dfrac{36}{12}=3\left(mol\right)\)
Vậy CTHH của hợp chất X là : \(C_3H_6\)