$n_{CO_2}=n_{CaCO_3}=\dfrac{5}{100}=0,05(mol)$
Có: $m_{CaCO_3}-m_{CO_2}-m_{H_2O}=2,08g$
$\to n_{H_2O}=\dfrac{5-0,05.44-2,08}{18}=0,04(mol)$
Đặt CTTQ $X$: $C_nH_{2n-2}O_4$
$n_X=n_{H_2O}-n_{CO_2}=0,01(mol)$
$\to n=\dfrac{0,05}{0,01}=5$
$\to $ CTPT: $C_5H_8O_4$
Thuỷ phân $X$ thu muối axit cacboxylic và ancol nên CTCT là:
$HCOO-CH_2-CH_2-CH_2-OOCH$
$HCOO-CH(CH_3)-CH_2-OOCH$
$HCOO-CH_2-CH_2-OOCCH_3$
$CH_3OOC-COOC_2H_5$
$CH_3-OOCCH_2COO-CH_3$