With complete combustion of 1.74 g of organic matter, 2.016 l of carbon dioxide and 1.62 g

With complete combustion of 1.74 g of organic matter, 2.016 l of carbon dioxide and 1.62 g of water were obtained. Determine the empirical formula of the substance.

Let’s designate the required formula as CxHyOz. It is impossible to say for sure whether there is oxygen in the composition of this compound, since according to the conditions of the problem this was not said.
So, let’s calculate the Molar masses of carbon dioxide and water:
M (CO2) = 12 + 16 * 2 = 44 g / mol
M (H2O) = 2 * 1 + 16 = 18 g / mol
Now let’s calculate the amount of substance
n (H2O) = mass / molar mass = 1.62 / 18 = 0.09 mol
n (CO2) = volume / molar volume = 2.016 / 22.4 = 0.09 mol
44 grams of CO2 contains 12 g of carbon, respectively, 0.09 mol of CO2 contains 0.09 * 12 = 1.08 carbon
18 water contains 2 g of hydrogen, 0.09 * 2 = 0.18
the mass of the sought-after substance = 1.08 + 0.18 = 1.26 g, and according to the conditions of the problem, the mass of the burnt-out substance = 1.74 g.
Therefore, the remaining mass is the mass of oxygen in the substance
mass (o) = 1.74-1.26 = 0.48 g.
number of islands (o) = 0.48 / 16 = 0.03 mol
n (C): n (H): n (O) = 0.09: 0.09: 0.03 = 2: 2: 1
Hence the simplest empirical formula C2H2O1
Answer С2Н2О1