Derive the molecular formula of the hydrocarbon, the mass fraction of hydrogen in which is 14.286%
Derive the molecular formula of the hydrocarbon, the mass fraction of hydrogen in which is 14.286% and the density in air is 1.93.
Given:
w% (H) = 14.286%
D (air) = 1.93
To find:
Formula -?
Decision:
1) Let us find the molecular weight of the hydrocarbon 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 = 1.93 * 29 = 56
2) Knowing the mass fraction of hydrogen, we find its molecular weight in the composition and the number of atoms:
56 – 100%
x – 14.286%,
x = 14.286% * 56: 100% = 8.
N (H) = 8: Mr (H) = 8: 1 = 8 atoms.
3) Then the molecular weight of carbon accounts for 56 – 8 = 48.
N (C) = 48: Mr (C) = 48: 12 = 4 atoms.
We got a substance with the formula C4H8 – butene.
Answer: C4H8 is butene.