\(\begin{array}{l}
Bao\,toan\,C:\,n_C=n_{CO_2}=\frac{4,4}{44}=0,1(mol)\\
Bao\,toan\,H:\,n_H=2n_{H_2O}=2.\frac{1,8}{18}=0,2(mol)\\
\to \%m_C=\frac{0,1.12}{2,2}.100\%≈54,55\%\\
\to \%m_H=\frac{0,2}{2,2}.100\%≈9,09\%\\
\to \%m_O=100-9,09-54,55=36,36\%
\end{array}\)