Derive the formula of a substance containing 93.75% – carbon and 6.25% – hydrogen, if the vapor density of this substance in the air is 4.41.
w% (C) = 93.75%
w% (H) = 6.25%
D (air) = 4.41
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) * Mr (air) = 4.41 * 29 = 128
2) Knowing the mass fraction of carbon, we find its molecular weight in the composition and the number of atoms:
128 – 100%
x – 93.75%
then x = 93.75% * 128: 100% = 120
N (C) = 120: Mr (C) = 120: 12 = 10 atoms.
3) Then the molecular weight of hydrogen accounts for 128 – 120 = 8.
N (H) = 8: Mr (H) = 8: 1 = 8 atoms.
We got a substance with the formula C10H8 – dekine or decadiene, depending on which class of hydrocarbons was considered.
Answer: С10Н8 – dekin / decadiene.
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