Derive the formula for a substance containing 40% carbon, 6.7% hydrogen, 53.3% oxygen.

Derive the formula for a substance containing 40% carbon, 6.7% hydrogen, 53.3% oxygen. The relative hydrogen vapor density of the substance is 30.

Given:

w% (C) = 40%

w% (H) = 6.7%

w% (O) = 53.3%

D (H2) = 30

To find:

Formula -?

Decision:

1) Find the molecular weight of a substance by its hydrogen density:

D (H2) = Mr (substance): Mr (H2)

Mr (substances) = D (H2) * Mr (H2) = 30 * 2 = 60.

2) Knowing the mass fraction of carbon, we find its molecular weight in the composition and the number of atoms:

60 – 100%

x – 40%,

x = 40% * 60: 100% = 24.

N (C) = 24: Mr (C) = 24: 12 = 2 atoms.

3) Knowing the mass fraction of oxygen, we find its molecular weight in the composition and the number of atoms:

60 – 100%

x – 53.3%,

x = 53.3% * 60: 100% = 32

N (O) = 32: Mr (O) = 32: 16 = 2 atoms.

4) Then the molecular weight of hydrogen accounts for 60 – 24 – 32 = 4.

N (H) = 4: Mr (H) = 4: 1 = 4 atoms.

We got a substance with the simplest formula C2O2H4. The presence of 2 oxygen atoms indicates that this substance is a carboxylic acid. A carboxylic acid containing 2 carbon atoms is acetic acid with the formula CH3COOH.

Answer: CH3COOH – acetic acid.