The relative density of hydrocarbon vapors with respect to nitrogen is 4. upon combustion of 11.2 g of this compound

The relative density of hydrocarbon vapors with respect to nitrogen is 4. upon combustion of 11.2 g of this compound, 35.2 g of carbon monoxide (4) and 14.4 g of water were formed. establish the molecular formula of this compound

1.Let’s find the mass of carbon atoms in 35.2 g of carbon dioxide.

Мr (СО2) = 44 g / mol.

44 g – 12 g (C),

35.2 g m (C).

m = (35.2 × 12): 44,

m = 9.6 g.

2.Let’s find the mass of hydrogen atoms in 14.4 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

14.4 g-m (H).

m = (14.4 × 2): 18,

m = 1.6 g.

m (substance) = 1.6 g + 9.6 g = 11.2 g.

Let’s find the amount of substance of carbon atoms, hydrogen.

n = m: M.

M (C) = 12 g / mol.

n (C) = 9.6 g: 12 g / mol = 0.8 mol.

n (H) = 1.6: 1 = 1.6 mol.

Let’s find the ratio of the amounts of the substance carbon and hydrogen.

C: H = 0.8: 1.6 = 1: 2.

Let’s find the molar mass of the hydrocarbon by the relative density.

Relative gravity is given as nitrogen. You need to multiply the molar mass of nitrogen by 4.

M (N2) = 14 × 2 = 28 g / mol.

D (nitrogen) = 4 × 28 = 112.

The molar mass of the hydrocarbon is 112. The ratio of carbon and hydrogen is 1: 2.

Let’s define the formula.

С8Н16 – octene.

M (C8H16) = 12 × 8 + 16 = 112 g / mol.