%O trong hợp chất X: 100 - 43,66 = 56,34 %
\(m_P=\dfrac{\%A.M_{hc}}{100\%}=\dfrac{43,66.142}{100}\approx62\left(g\right)\)
\(m_O=\dfrac{\%A.M_{hc}}{100\%}=\dfrac{56,34.142}{100}\approx80\left(g\right)\)
\(n_P=\dfrac{m}{M}=\dfrac{62}{31}=2\left(mol\right)\)
\(n_O=\dfrac{m}{M}=\dfrac{80}{16}=5\left(mol\right)\)
\(\Rightarrow CTHH:P_2O_5\)