During the combustion of organic matter weighing 8.8 g, carbon dioxide with a volume of 8.96 liters

During the combustion of organic matter weighing 8.8 g, carbon dioxide with a volume of 8.96 liters and water weighing 7.2 g were formed. The vapor of this substance is 2 times heavier than carbon dioxide. As a result of hydrolysis of this substance, alcohol and carboxylic acid are formed. Derive the molecular formula of a substance and make up the structural formulas of the isomers corresponding to the condition of the problem. Give them names.

Given:
m (in-va) = 8.8 g
V (CO2) = 8.96 l
m (H2O) = 7.2 g
M (in-va) = M (CO2) * 2

To find:
in -?

1) n (CO2) = V / Vm = 8.96 / 22.4 = 0.4 mol;
2) n (C) = n (CO2) = 0.4 mol;
3) m (C) = n * M = 0.4 * 12 = 4.8 g;
4) n (H2O) = m / M = 7.2 / 18 = 0.4 mol;
5) n (H) = n (H2O) * 2 = 0.4 * 2 = 0.8 mol;
6) m (H) = n * M = 0.8 * 1 = 0.8 g;
7) m (O) = m (in-va) – m (C) – m (H) = 8.8 – 4.8 – 0.8 = 3.2 g;
8) n (O) = m / M = 3.2 / 16 = 0.2 mol;
9) CxHyOz;
10) x: y: z = 0.4: 0.8: 0.2 = 2: 4: 1;
11) C2H4O;
12) M (CxHyOz) = M (CO2) * 2 = 44 * 2 = 88 g / mol;
13) M (C2H4O) = Mr (C2H4O) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 2 + 1 * 4 + 16 * 1 = 44 g / mol;
14) M (CxHyOz) / M (C2H4O) = 2;
15) (C2H4O) 2 -C4H8O2;
16) Isomers: HCOOCH2CH2CH3 – propyl formate; HCOOCH (CH3) 2 isopropyl formate; CH3COOCH2CH3 – ethyl acetate; CH3CH2COOCH3 – methyl propionate.

Answer: Unknown substance – C4H8O2.