Derive the molecular formula of the substance if the mass fractions (%) are C-54.55%, H-9.09%

Derive the molecular formula of the substance if the mass fractions (%) are C-54.55%, H-9.09%, O-36.36%. The relative density for hydrogen is 22.

Given:

w% (C) = 54.55%

w% (H) = 9.1%

w% (O) = 36.36%

D (H2) = 22

To find:

Formula -?

Decision:

1) Find the molecular weight of the oxygen-containing compound by its hydrogen density:

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

Mr (substances) = D (H2) * Mr (H2) = 22 * ​​2 = 44

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

44 – 100%

x – 54.55%,

then x = 54.55% * 44: 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:

44 – 100%

x – 36.36%,

then x = 36.36% * 44: 100% = 16

N (O) = 16: Mr (O) = 16: 16 = 1 atom.

4) Then the molecular weight of hydrogen is 44 – 24 – 16 = 4

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

We got a substance with the formula C2OH4, this is acetaldehyde with the formula H3C – CHO.

Answer: Н3С – СНО – acetaldehyde.