Question

compare the boiling points of propane (C3H8) with that of (C4H10) .. explain your answer ...plzzz guys help me with this​

Answer 1

i may be wrong but Use the normal boiling points: propane, C3H8, –42.1˚C; butane, C4H10, –0.5˚C; pentane, C5H12, 36.1˚C; hexane, C6H14, 68.7˚C; heptane, C7H16, 98.4˚C; to estimate the normal boiling point of octane, C8H18.

Answer 2

Answer:

C₄H₁₀ > C₃H₈

Explanation:

Propane (C₃H₈) is symmetrical and non-polar, so the only inter-molecular forces are London dispersion forces.

Butane (C₄H₁₀) is symmetrical and non-polar, so the only inter-molecular forces are also London dispersion forces.

The only difference is the size.  Butane has more atoms, more surface area, and therefore more inter-molecular forces.  So it has a higher boiling point.

C₄H₁₀ > C₃H₈