# Determine the molecular formula of an organic substance if it is known that during the combustion of 6.4 g

Determine the molecular formula of an organic substance if it is known that during the combustion of 6.4 g of this substance 4.48 liters of carbon dioxide were released and 7.2 g of water were formed. The relative density of this substance for hydrogen is 16.

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

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

n (CO2) = 4.48 L: 22.4 L / mol = 0.2 mol.

2.Let’s find the mass of carbon dioxide.

m = n M, М (СО2) = 44 g / mol.

m = 0.2 mol × 44 g / mol = 8.8 g.

3.Let’s find the mass of carbon atoms in 8.8 g of carbon dioxide.

44 g – 12 g (C),

8.8 g – m (C).

m = (8.8 × 12): 44,

m = 2.4 g.

4. Let’s find the mass of hydrogen atoms in 7.2 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

7.2 g-m (H).

m = (7.2 × 2): 18,

m = 0.8 g.

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

n = m: M.

M (C) = 12 g / mol.

n (C) = 2.4 g: 12 g / mol = 0.2 mol.

M (H) = 0.8: 1 = 0.8 mol.

m = 0.8 g + 2.4 g = 3.2 g.

m (O) = 6.4 – 3.2 = 3.2.

n (O) = 3.2 g: 16 g / mol = 0.2 mol.

6.Let’s find the ratio of the amounts of matter carbon and hydrogen, oxygen.

C: H: O = 0.2: 0.8: 0.2 = 1: 2: 1.

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

Relative density is given as hydrogen. You need to multiply the molar mass of hydrogen (2) by 16.

D (H2) = 2 × 16 = 32.

The molar mass of hydrocarbon is 32. The ratio of carbon, hydrogen, oxygen is 1: 2: 1.

8. Let’s define the formula.

CH3OH – methanol.

M (CH3OH) = 12 + 1 × 3 + 16 + 1 = 32 g / mol.