Establish the molecular formula of one of the dibromoalkaces containing 85.11% bromine.

| July 25, 2021 | education

Establish the molecular formula of one of the dibromoalkaces containing 85.11% bromine. Which isomers correspond to this molecular formula. Name them.

Let the formula of an unknown dibromoalkane be RBr2, where R is an unknown alkyl radical.

Determine the molecular weight of the unknown dibromoalkane:

M (RBr2) = M (Br2) * 100 / w (Br2) = 160 * 100 / 85.11 = 188 (amu).

Determine the molecular weight of the unknown alkyl radical:

M (R) = M (RBr2) – M (Br2) = 188 – 160 = 28 (amu).

This molecular weight of the radical corresponds to two CH2 groups:

M (CH2CH2) = 12 * 2 + 1 * 4 = 28 (amu)

or one CH3 group and one CH group:

M (CH3CH) = 12 * 2 + 1 * 4 = 28 (amu).

Thus, the formula for these isomeric dibromoalkanes: C2H4Br2

and it corresponds to BrCH2CH2Br (1,2-dibromoethane) and CH3CHBr2 (1,1-dibromoethane).



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.