Determine the molecular formula of a monobromine substituted alkane if the mass fraction of bromine in it is 65%.

Let’s designate the molecular formula of an unknown monobromalkane as CnH2n + 1Br, where CnH2n + 1 is an unknown alkyl radical.

Determine the mass fraction of the alkyl radical:

w (CnH2n + 1) = 100 – w (Br) = 100 – 65 = 35 (%).

Determine the molecular weight of the alkyl radical:

M (CnH2n + 1) = M (Br) * w (CnH2n + 1) / w (Br) = 80 * 35/65 = 43 (amu).

Let’s compose and solve the equation:

12 * n + 1 * (2n + 1) = 43,

n = 3, whence the molecular formula of monobromalkane: C3H7Br.

Answer: C3H7Br.