When burning 1.05 gas, 3.3 g of carbon dioxide and 1.35 g of water were released, the density

When burning 1.05 gas, 3.3 g of carbon dioxide and 1.35 g of water were released, the density for argan was 1.05 g, determine the formula of the substance.

1.Let’s find the mass of carbon atoms in 3.3 g of carbon dioxide.

M (CO2) = 44 g / mol.

44 g – 12 g (C),

3.3 g-m (C).

m = (3.3 × 12): 44,

m = 0.9 g.

2. Let’s find the mass of hydrogen atoms in 1.35 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

1.35 g – m (H).

m = (1.35 × 2): 18,

m = 0.15 g.

m (substance) = m (C) + m (H) = 0.15 g + 0.9 g = 1.05 g.

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

n = m: M.

M (C) = 12 g / mol.

n (C) = 0.9 g: 12 g / mol = 0.075 mol.

M (H) = 0.15: 1 = 0.15 mol.

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

C: H = 0.075: 0.15 = 1: 2.

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

Relative gravity is given for argon. You need to multiply the molar mass of argon (40) by 1.05.

D (Ar) = 40 × 1.05 = 42

The molar mass of the hydrocarbon is 42. Let’s define the formula.

C3H6 – propene.

M (C3H6) = 42 g / mol.