A sample of organic matter with a mass of 4.3 g was burned in oxygen. The reaction products were CO

A sample of organic matter with a mass of 4.3 g was burned in oxygen. The reaction products were CO with a volume of 6.72 liters and water with a mass of 6.3 g. The hydrogen vapor density of the substance is 43. Determine the formula of the substance?

The reaction equation is:
X + O2 → CO2 + H2O
Substance X contains C, H and possibly O atoms.
Let’s find the amount of C, H and O, for this we calculate the amount of CO2 and H2O:
ν (CO2) = V (CO2) / Vm (CO2) = 6.72 / 22.4 = 0.3 mol
ν (H2O) = m (H2O) / M (H2O) = 7.2 / 18 = 0.35 mol

In 1 molecule of CO2 there is 1 C atom (ratio 1: 1) ⇒ In 0.3 mol of CO2 0.3 mol of C atoms
In 1 H2O molecule there are 2 H atoms (ratio 1: 2) ⇒ B 0.35 mol H2O 0.35 * 2 = 0.7 mol H atoms

Let’s find the amount of O in the original substance:
m (substances) = m (C) + m (H) + m (O) ⇒ m (O) = m (substances) – m (C) – m (H)
m (C) = ν * M = 0.3 * 12 = 3.6 g
m (H) = ν * M = 0.7 * 1 = 0.7 g
m (O) = m (X) – m (C) – m (H)
m (O) = 4.3 – 3.6 – 0.7 = 0

This means that there are no oxygen atoms in the formula of a substance.

The ratio of C, H atoms in the formula of a substance: n (C): n (H) = 0.3: 0.7: 0.4 = 3: 7

Let’s get simple. f-lu C3H7.

D (X / H2) = M (X) / M (H2)
M (C2H6) = 1 * 2 = 2 g / mol
M (X) = D (X / C2H6) * M (H2) = 43 * 2 = 86 g / mol
M (C3H7) = 12 * 3 + 7 * 1 = 43 g / mol
The ratio M (X) / M (C3H7) = 86/43 = 2
those. molecular formula X is doubled simple. f-la = (C3H7) 2 = C6H14
Answer: the formula of the substance C6H14 is hexane.