Giả sử $a=44g$
$\Rightarrow n_C=n_{CO_2}=\dfrac{44}{44}=1(mol)$
$\Rightarrow b=\dfrac{3a}{11}=12g$
$\Rightarrow n_H=2n_{H_2O}=\dfrac{12.2}{18}=\dfrac{4}{3}(mol)$
$x=\dfrac{3(a+b)}{7}=24g$
Ta có $m_O=x-m_C-m_H=24-1.12-\dfrac{4}{3}=\dfrac{32}{3}g$
$\Rightarrow n_O=\dfrac{\frac{32}{3}}{16}=\dfrac{2}{3}(mol)$
$n_C : n_H : n_O=1:\dfrac{4}{3}:\dfrac{2}{3}=3:4:2$
$\Rightarrow$ CTĐGN $(C_3H_4O_2)_n$
$M_A<3.29=87\Rightarrow 72n<87$
$\Rightarrow n<1,2$
$\Rightarrow n=1$
Vậy CTPT A là $C_3H_4O_2$