# Question #0969e

May 23, 2017

The differences in physical properties become smaller because the increases in molar mass are smaller percentages of the total mass.

#### Explanation:

Consider the melting and boiling points of the alkanes.

The melting points are plotted as blue squares and the boiling points are represented by orange diamonds.

Melting Points

(a) Compare propane and pentane

The molar mass increases from 44 g/mol to 72 g/mol (an increase of 86 %), and the melting point increases from -187.7 °C to -129.8 °C (an increase of 57.9 °C).

(b) Compare triacontane ${\text{C"_30"H}}_{62}$ and dotriacontane ${\text{C"_32"H}}_{66}$.

The molar mass increases from 423 g/mol to 451 g/mol, (an increase of 5.4 %), and the melting point increases from 66 °C to 69°C (an increase of only 3 °C).

The increased molecular size is so small that it has little effect on increasing the London dispersion forces between adjacent molecules.

Boiling points

(a) Compare propane and pentane.

The molar mass increases from 44 g/mol to 72 g/mol, (an increase of 86 %), and the boiling point increases from -42.1 °C to 36.1 °C (an increase of 78.2 °C).

(b) Compare triacontane ${\text{C"_30"H}}_{62}$ and dotriacontane ${\text{C"_32"H}}_{66}$.

The molar mass increases from 423 g/mol to 451 g/mol, (an increase of 5.4 %), and the boiling point increases from 450 °C to 467°C (an increase of only 17 °C).

Again, the increased molecular size is so small that it has little effect on increasing the London dispersion forces between adjacent molecules.