\(m_C=\dfrac{60.40}{100}=24\left(g\right)\)
\(m_O=\dfrac{60.53,33}{100}\approx32\left(g\right)\)
\(m_H=\dfrac{60.6,56}{100}\approx4\left(g\right)\)
\(\Rightarrow\) \(n_C=\dfrac{24}{12}=2\left(mol\right)\)
\(n_O=\dfrac{32}{16}=2\left(mol\right)\)
\(n_H=\dfrac{4}{1}=4\left(mol\right)\)
Vậy CTHH là \(C_2O_2H_4\)