Hydrocarbon of the acetylene series weighing 136 g is added 89.6 liters of chlorine to complete saturation.

Hydrocarbon of the acetylene series weighing 136 g is added 89.6 liters of chlorine to complete saturation. Determine the molecular formula of the starting substance and write down the possible isomers of this composition, name the substances according to the systematic nomenclature.

Given:
m (CnH2n-2) = 136 g
V (Cl2) = 89.6 l

To find:
CnH2n-2 -?

Decision:
1) CnH2n-2 + 2Cl2 => CnH2n-2Cl4;
2) n (Cl2) = V (Cl2) / Vm = 89.6 / 22.4 = 4 mol;
3) n (CnH2n-2) = n (Cl2) / 2 = 4/2 = 2 mol;
4) M (CnH2n-2) = m (CnH2n-2) / n (CnH2n-2) = 136/2 = 68 g / mol;
5) M (CnH2n-2) = Mr (CnH2n-2) = Ar (C) * N (C) + Ar (H) * N (H);
68 = 12 * n + 1 * (2n – 2);
n = 5;
6) Unknown substance – C5H8;
7) Isomers: pentin-1; pentin-2; methylbutin; pentadiene-1,2; pentadiene 1,3; pentadiene-1,4; pentadiene-2,3; cyclopentene; methylcyclobutene-1; methylcyclobutene-2; methylcyclobutene-3; 1,2-dimethylcyclopropene; 1,3-dimethylcyclopropene; 1-ethylcyclopropene; 2-ethylcyclopropene; 3-ethylcyclopropene.

Answer: Unknown substance – C5H8.