Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?

Why do molecules like alkanes with higher relative molecular mass ($M_\mathrm r$) have stronger intermolecular forces? For example, methane ($\ce{CH4}$) has a weaker intermolecular force than pentane ($\ce{C5H12}$). Pentane has a higher $M_\mathrm r$ than methane and therefore pentane possesses stronger intermolecular force than methane. These are members of the same family in the homologous series, known as alkanes.

My issue is why does pentane have stronger intermolecular force than methane? Why is there a relationship between relative molecular mass ($M_\mathrm r$) and inter-molecular forces. The stronger the intermolecular force the higher the boiling point and the melting point?

3 Answers

There are a lot of reasons for why some molecules have stronger IMF's than others, but the trend of increasing IMF for increasing relative molecular mass ($M_\mathrm r$) is due to an increase in London Dispersion Forces, part of the greater set of van der Waals forces. As you may or may not know, these alkanes are non-polar molecules (that is, they don't have a real internal dipole--all $\ce{C-C}$ bonds are non-polar and the electronegativity of the $\ce{C-H}$ bond is close enough that those bonds are also considered non-polar). This matters because polar interactions are a relatively strong IMF.

Let's take a detour into e- distributions. Typically, electrons have the highest probability of being evenly distributed about the atoms in a molecule, proportional to the electronegativity of those atoms. That being said, that is only the most probable state, and so, of course, there are times when the e- are NOT evenly distributed. Remember how alkanes normally aren't polar? Well, when the electrons get unbalanced in the molecule, surprise! An Instantaneous Dipole is temporarily introduced.

What does this mean? Well, when an alkane is next to another alkane that is subject to an instantaneous dipole, the electrostatic forces from the quasi-polarized alkane cause the other alkane to also attain a lopsided distribution of electrons, causing it to gain a temporary polarity, and this effect both attracts these molecules while also propagating throughout the surrounding molecules.

So now that you have a little understanding of what LDF's or instantaneous dipoles are, why the trend with molecular weight? The short answer is that larger molecules have more e-, and are affected by LDF's more than smaller molecules with less e-. With more electrons and more atoms for those electrons to be distributed over, there is a greater chance for frequent or large instantaneous dipoles, increasing the overall Van der Waals forces.

In the case of alkanes, with increasing molecular mass, the number of atoms in a molecule also increase, which means an increase in the chain length of molecules. A molecule with a large chain length experiences stronger London Dispersion Forces. The reason is that longer molecules have more places where they can be attracted to other molecules. This is the reason why pentane (longer chain molecule) experiences stronger intermolecular forces of attraction than methane. As alkanes are non-polar, therefore, they will only exhibit London Dispersion Forces.

Larger molecules have larger electron clouds so therefore stronger instantaneous-induced dipole bonds.