Determine the molecular formula of an organic substance if, upon combustion of 30.6 g of this substance

Determine the molecular formula of an organic substance if, upon combustion of 30.6 g of this substance, 79.2 g of carbon monoxide (IV) and 37.8 g of water are formed. The relative density of its vapors for hydrogen is 51. Make up its structural formula.

1.Let’s find the mass of carbon atoms in 79.2 g of carbon dioxide.

Мr (СО2) = 44 g / mol.

44 g – 12 g (C),

79.2 g – m (C).

m = (79.2 × 12): 44,

m = 21.6 g.

2. Let’s find the mass of hydrogen atoms in 37.8 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

37.8 g-m (H).

m = (37.8 × 2): 18,

m = 4.2 g.

m (substance) = 4.2 g + 21.6 g = 25.8 g.

30.6 – 25.8 = 4.8 (oxygen atoms).

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

n = m: M.

M (C) = 12 g / mol.

n (C) = 21.6 g: 12 g / mol = 1.8 mol.

M (H) = 4.2: 1 = 4.2 mol.

n (O) = 4.8 g: 16 g / mol = 0.3 mol.

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

C: H: O = 1.8: 4.2: 0.3 = 6: 14: 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 hydrogen (2) by 21.

D (H2) = 2 × 51 = 102.

The molar mass of the hydrocarbon is 102. Let’s define the formula.

C6H14O.

M (C6H14O) = 12 × 6 + 14 + 16 = 102 g / mol.




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