The hydrocarbon contains 88.89% carbon. Its air density is 1.862.

The hydrocarbon contains 88.89% carbon. Its air density is 1.862. Find the molecular formula of this hydrocarbon, write the structural formulas of possible isomers and name them.

Let’s write down given:

W (C) = 88.89% or 0.8889

D (substances by air) = 1.862

Find the molecular formula of a hydrocarbon -?

Write down and name isomers -?

Decision.

1) According to the condition of the problem, the desired substance is hydrocarbon (CxHy)

Find the mass fraction of hydrogen in the substance

W (H) = 100% – 88.89% = 11.11% or 0.1111

2) Calculate the relative molecular weight of the hydrocarbon

D (air) = Mr (CxHy) / Mr (air)

Mr (CxHy) = D (air) * Mr (air)

Mr (air) = 29

Mr (CxHy) = 1.862 * 29 = 54

3) Calculate the number of carbon and hydrogen atoms in the substance by the formula

W (element) = n * Ar (element) / Mr (substance), hence

n (element) = W (element) * Mr (substance) / Ar (element)

Ar (C) = 12, Ar (H) = 1

n (C) = 0.8889 * 54/12 = 4, therefore the hydrocarbon contains 4 carbon atoms

n (H) = 0.1111 * 54/1 = 6, therefore the hydrocarbon contains 6 hydrogen atoms.

The formula of the desired substance C4 H6 is butin. The hydrocarbon has a triple bond.

4) Let’s write down the structural formulas of butyne and its isomers:

CH Ξ C – C H2 – CH3 butyne – 1

CH3 – C Ξ C – CH3 butyne -2

CH2 = CH – CH = CH2 butadiene 1, 3

CH3 – CH = C = CH2 butadiene 1, 2

Answer: Hydrocarbon C4H6 has four isomers.