The combustion of 1.85 g of the substance yielded 4.363 g CO2 and 2.25 g H2O. The vapor density of this substance

The combustion of 1.85 g of the substance yielded 4.363 g CO2 and 2.25 g H2O. The vapor density of this substance in the air is 2.55. Output the formula of the substance.

Given:
m (organic matter) = 1.85 g m (CO2) = 4.363 g m (H2O) = 2.25 g D air. = 2.55
Find: formula in-va -?
Decision:
1) Find the amount of CO2: n (CO2) = m (CO2) / Mr (CO2) = 4.363 / 44 = 0.0992 mol;
2) Find the number of C-islands: n (C) = n (CO2) = 0.0992 mol;
3) Find the mass C: m (C) = n (C) * Mr (C) = 0.0992 * 12 = 1.1904 g;
4) Find the number of H2O: n (H2O) = m (H2O) / Mr (H2O) = 2.25 / 18 = 0.125 mol;
5) Find the number of H-islands: n (H) = n (H2O) * 2 = 0.125 * 2 = 0.25 mol;
6) Find the mass H: m (H) = n (H) * Mr (H) = 0.25 * 1 = 0.25 g;
7) Find the mass of O in organic matter: m (O) = m (organic matter) – m (C) – m (H) = 1.85 – 1.1904 – 0.25 = 0.4096 g;
8) Find the number of O: n (O) = m (O) / Mr (O) = 0.4096 / 16 = 0.0256 mol;
9) Find the simplest formula for a substance: C: H: O = 0.0992: 0.25: 0.0256 = 4: 10: 1; C4H10O;
10) Find the molar mass of an unknown substance: Mr (in-va) = D air. * Mr (air) = 2.55 * 29 = 74 g / mol;
11) Find the molar mass of C4H10O: Mr (C4H10O) = 74 g / mol;
12) Because molar masses are equal, then the unknown substance is C4H10O.
Answer: Unknown organic matter – C4H10O.