Derive the molecular formula of a hydrocarbon, the mass fraction of carbon in which is 92.3%

| September 22, 2021 | education

Derive the molecular formula of a hydrocarbon, the mass fraction of carbon in which is 92.3%, and the density of its vapor in air is 2.69.

Given:

w% (C) = 92.3%

D (air) = 2.69

Find:

Formula -?

Solution:

1) Find the molecular weight of a compound by its density in air:

D (air) = Mr (HC): Mr (air), – since air does not have a certain molecular weight, it is conventionally taken as 29.

Mr (HC) = D (air) * 29 = 2.69 * 29 = 78.

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

78 – 100%

x – 92.3,

then x = 92.3% * 78: 100% = 72.

N (C) = 72: Mr (C) = 72: 12 = 6 atoms.

3) Then the molecular weight of hydrogen in the composition is 78 – 72 = 6.

N (H) = 6: Mr (H) = 6: 1 = 6 atoms.

We got a substance with the formula C6H6 – benzene.

Answer: C6H6 is benzene.



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