Establish the molecular formula of nitrogen-containing organic matter, the combustion of which consumed 7.2 g

Establish the molecular formula of nitrogen-containing organic matter, the combustion of which consumed 7.2 g of oxygen; in this case, 4.5 g of H2O and 5.6 dm3 of a gas mixture were obtained, the volume of which, when passed through an excess of alkali, was reduced to 1.12 dm3.

Given:
m (O2) = 7.2 g
m (H2O) = 4.5 g
V1 (gas mixture) = 5.6 dm3 = 5.6 l
V2 (gas mixture) = 1.12 dm3 = 1.12 l

To find:
substance -?

Solution:
1) substance + O2 -> H2O + CO2 + N2;
CO2 + 2KOH => K2CO3 + H2O;
2) n (H2O) = m (H2O) / M (H2O) = 4.5 / 18 = 0.25 mol;
3) n (H) = n (H2O) * 2 = 0.25 * 2 = 0.5 mol;
4) n1 (O) = n (H2O) = 0.25 mol;
5) V (N2) = V2 (gas mixture) = 1.12 l;
6) n (N2) = V (N2) / Vm = 1.12 / 22.4 = 0.05 mol;
7) n (N) = n (N2) * 2 = 0.05 * 2 = 0.1 mol;
8) V (CO2) = V1 (gas mixture) – V (N2) = 5.6 – 1.12 = 4.48 l;
9) n (CO2) = V (CO2) / Vm = 4.48 / 22.4 = 0.2 mol;
10) n (C) = n (CO2) = 0.2 mol;
11) n2 (O) = n (CO2) * 2 = 0.2 * 2 = 0.4 mol;
12) n (O2) = m (O2) / M (O2) = 7.2 / 32 = 0.225 mol;
13) n3 (O) = n (O2) * 2 = 0.225 * 2 = 0.45 mol;
14) n1 (O) + n2 (O)> n3 (O);
0.65> 0.45;
15) The substance consists of H, N, C, O: HxNyCzOa;
16) n (O in the substance) = n1 (O) + n2 (O) – n3 (O) = 0.65 – 0.45 = 0.2 mol;
x: y: z: a = n (H): n (N): n (C): n (O in the substance) = 0.5: 0.1: 0.2: 0.2 = 5: 1: 2: 2;
Substance – H5NC2O2 – C2H5NO2 – aminoacetic acid.

Answer: Substance – C2H5NO2 – aminoacetic acid.