a,
$n_{CO_2}=n_{CaCO_3\downarrow}=\dfrac{2}{100}=0,02(mol)$
$\Rightarrow n_C=n_{CO_2}=0,02(mol)$
$\Delta m_b=1,24g=m_{CO_2}+m_{H_2O}$
$\Rightarrow 44n_{CO_2}+18n_{H_2O}=1,24g$
$\Rightarrow n_{H_2O}=\dfrac{1,24-0,02.44}{18}=0,02(mol)$
$\Rightarrow n_H=2n_{H_2O}=0,04(mol)$
$\Rightarrow n_O=\dfrac{0,6-0,02.12-0,04}{16}=0,02(mol)$
$n_C: n_H: n_O=0,02:0,04:0,02=1:2:1$
$\Rightarrow$ CTĐGN: $CH_2O$
b,
$n_Y=n_{O_2}=\dfrac{1,6}{32}=0,05(mol)$
$\Rightarrow M_Y=\dfrac{3}{0,05}=60$
Đặt CTPT Y: $(CH_2O)_n$
$\Rightarrow n=2$
Vậy CTPT là $C_2H_4O_2$