When burning a hydrocarbon weighing 8.4 grams, 26.4 grams of gas and 10.8 grams

When burning a hydrocarbon weighing 8.4 grams, 26.4 grams of gas and 10.8 grams of water were obtained. The vapor density of this substance in the air is 2.9. Determine the molecular formula.

1.Let’s find the mass of carbon atoms in 26.4 g of carbon dioxide.

44 g – 12 g (C),

26.4 g – m (C).

m = (26.4 × 12): 44,

m = 7.2 g.

2. Let’s find the mass of hydrogen atoms in 10.8 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

10.8 g-m (H).

m = (10.8 × 2): 18,

m = 1.2 g.

m (substance) = 7.2 + 1, 2 = 8.4 g.

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

n = m: M.

M (C) = 12 g / mol.

n (C) = 7.2 g: 12 g / mol = 0.6 mol.

M (H) = 1.2: 1 = 1.2 mol.

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

C: H = 0.6: 1.2 = 1: 2.

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

Relative density is given by air. You need to multiply the molar mass of air (29) by 2.9.

D (air) = 29 x 2.9 = 84.1

Let’s define the formula.

C6H12 – hexene.