On combustion of 33.6 g of organic matter, 53.76 liters of carbon dioxide and 43.2 g of water were formed.

On combustion of 33.6 g of organic matter, 53.76 liters of carbon dioxide and 43.2 g of water were formed. The relative density of vapor of this compound for hydrogen is 42. Determine the molecular formula of the compound.

1. Let’s find the amount of substance carbon dioxide.

n = V: Vn, where Vn is the molar volume of gas, equal to 22.4 l / mol.

n (CO2) = 53.76 L: 22.4 L / mol = 2.4 mol.

Let’s find the mass of carbon dioxide.
m = n M, М (СО2) = 44 g / mol.

m = 2.4 mol × 44 g / mol = 105.6 g.

1. Let’s find the mass of carbon atoms in 105.6 g of carbon dioxide.

44 g – 12 g (C),

105.6 g-m (C).

m = (105.6 × 12): 44,

m = 28.8 g.

2. Let’s find the mass of hydrogen atoms in 43.2 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

43.2 g-m (H).

m = (43.2 × 2): 18,

m = 4.8 g.

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

n = m: M.

M (C) = 12 g / mol.

n (C) = 28.8 g: 12 g / mol = 2.4 mol.

M (H) = 4.8: 1 = 4.8 mol.

m = 4.8 g + 28.8 g = 33.6 g.

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

C: H = 2.4: 4.8 = 1: 2.

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

Relative gravity is given as hydrogen. You need to multiply the molar mass of hydrogen (2) by 21.

D (H2) = 2 × 42 = 84.

The molar mass of the hydrocarbon is 84. The ratio of carbon and hydrogen is 1: 2.

Let’s define the formula.

C 6H12 – hexene.

M (C6H12) = 6 × 12 + 12 = 84 g / mol.

Answer: C6H12 – hexene.