Print the molecular formula of a hydrocarbon whose air density is 1.93. Mass fraction of carbon 85,7%, hydrogen 14.3%.

Given:

w% (C) = 85.7%

w% (H) = 14.3%

D (air) = 1.93

To find:

Formula -?

Decision:

1) Find the molecular weight of a 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 carbon, we find its molecular weight in the composition and the number of atoms:

56 – 100%

x – 85.7%,

then x = 85.7% * 56: 100% = 48.

N (C) = 48: Mr (C) = 48: 12 = 4 atoms.

3) Then the molecular weight of hydrogen is 56 – 48 = 8.

N (H) = 8: Mr (H) = 8: 1 = 8 atoms.

We got a substance with the formula C4H8 – butene or cyclobutane, depending on which class of hydrocarbons was considered.

Answer: C4H8 – butene or cyclobutane.