0.672 L of some hydrocarbon was burned. In this case, carbon dioxide with a mass of 3.96 g and water

0.672 L of some hydrocarbon was burned. In this case, carbon dioxide with a mass of 3.96 g and water with a mass of 2.16 g were formed. The relative density of this hydrocarbon for oxygen is 1.375. Determine the molecular formula of this substance.

Given:
V (CxHy) = 0.672 l
m (CO2) = 3.96 g
m (H2O) = 2.16 g
D O2 (CxHy) = 1.375
Find: CxHy -?
Decision:
1) M (CxHy) = D O2 (CxHy) * M (O2) = 1.375 * 32 = 44 g / mol;
2) n (CO2) = m (CO2) / M (CO2) = 3.96 / 44 = 0.09 mol;
3) n (C in CxHy) = n (CO2) = 0.09 mol;
4) n (H2O) = m (H2O) / M (H2O) = 2.16 / 18 = 0.12 mol;
5) n (H in CxHy) = n (H2O) * 2 = 0.12 * 2 = 0.24 mol;
6) x: y = 0.09: 0.24 = 1: 2.66667 = 3: 8;
C3H8;
7) M (C3H8) = 44 g / mol;
Unknown hydrocarbon – C3H8 – propane.
Answer: Unknown hydrocarbon – C3H8 – propane.