An alkene weighing 14 g adds hydrogen bromide, the volume of which is equal to the volume

An alkene weighing 14 g adds hydrogen bromide, the volume of which is equal to the volume of methane weighing 4 g. Determine the alkene formula.

The relative molecular weight of methane is 16 g / mol. Let’s find the amount of substance in methane:
4/16 = 0.25 mol;

Alkenes have the general formula: CnH2n. Let us compose the equation for the reaction of the interaction of alkene with hydrogen bromide:
CnH2n + HBr = CnH2n + 1Br;

The amount of substance for hydrogen bromide is 0.25 mol, since the amount of substance for hydrogen bromide and for methane are equal (the amount of substance for identical volumes of ideal gases always coincides).
The amount of substance in the alkene is 0.25 mol, since the coefficients before the alkene and before the hydrogen bromide are equal. Let’s find the molecular weight of the alkene:
14 / 0.25 = 56 g / mol;

The relative atomic mass of carbon is 12, hydrogen is 1. To determine the alkene formula, we compose the equation:
12n + 2n = 56;

14n = 56;

n = 56/14;

n = 4;

If n = 4, then 2n = 2 * 4 = 8.

Answer: the alkene formula is C4H8.