Question

A hydrocarbon has the empirical formula CH2. When 1.595 grams of the hydrocarbon are dissolved in 15 grams of benzene, the boiling point of the solution is 81.84°
c. What is the molecular formula for the hydrocarbon?

Answer 1

We know that ?T b = Kbx m
Where
?T b= elevation in boiling point
= boiling point of solution - boilinging point of pure solvent
= 81.84 - 80.1

= 1.74 °C
Kb = elevation in boiling point constant of benzene= 2.65 °C/m
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
= (1.595 g/ M) / 0.015 kg
= 106.3/M
Plug the values we get M = 162 g/ mol

empirical formula is CH₂ it's mass is 12+2=14

N= molarmass/ empirical mass

N= 162/14

N~12

So molecular formula is 14(CH₂)= C₁₄H₂₈