The relative density of the vapors of the organic compound in the air is 4.69. When 95.2 g of this substance

The relative density of the vapors of the organic compound in the air is 4.69. When 95.2 g of this substance is burned, 154 g of carbon dioxide and 75.6 g of water are formed. Print the formula for this substance.

carbon atoms in 154 g of carbon dioxide.

Мr (СО2) = 44 g / mol.

44 g – 12 g (C),

154 g – m (C).

m = (154 × 12): 44,

m = 42 g.

2. Let’s find the mass of hydrogen atoms in 75.6 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

75.6 g-m (H).

m = (75.6 × 2): 18,

m = 8.4 g.

m (substance) = 8.4 g + 42 g = 50.4 g.

95.2 g – 50.4 g = 44.8 g (oxygen).

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

n = m: M.

M (C) = 12 g / mol.

n (C) = 42 g: 12 g / mol = 3.5 mol.

n (H) = 8.4: 1 = 8.4 mol.

n (O) = 44.8: 16 = 2.8 mol.

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

C: H: O = 3.5: 8.4: 2.8 = 1.25: 3: 1.

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 air (29) by 4.69.

D (air) = 4.69 × 29 = 136.

Molar mass of hydrocarbon 136. Ratio of carbon and hydrogen

1: 3: 1.

Let’s define the formula.

C5H12O4.

M (C5H12O4) = 12 × 5 + 12 + 16 × 4 = 136 g / mol.