What would happen if I mixed a single water molecule in a beaker of hexane?
This is a clever question because the molecular weight of water is only 18.01528 g/mol while the molecular weight of hexane is 86.17536 g/mol. Naturally, you would expect the lower mass to rest on top, but water has a higher density... and we all know that water will sink! But there are other forces at work.

So, supposing that you neglect the fact that one molecule of water would be in a vapor-state, Google says the surface tension of Hexane @ 20 °C is 18.43 mN/m... so placing the water molecule ontop of the hexane would probably cause the molecule to float (like a feather on water). 

However, if you suberged the molecule it should dissolve (Google says that the solubility of water in hexane is 0.01% at 20°C). Now, once the saturation point of hexane has been met, such that one more single molecule of water would cause a change to occur, that change would probably result in a condensation and precipitation of the dissolved water, because the reality is, that the hexane will become super-saturated, such that the extra water molecule will cause many water molecules to fall out of solution.

Understanding that you are transferring energy is key to understanding this. Consider pushing a ball over a hill, once the ball reaches the top and begins to fall, it falls into the valley. Once you push a boat to the tipping point, it falls to the bottom.

So the number of molecules necessary for hydrogen bonding together to become more dense than hexane would be the number of water molecules dissolved at the maximum supersaturation point (maybe 0.0105%, +1 molecule) minus the number of molecules at the saturation point (0.01%)

Now having a saturated solution, with water on the bottom of the container, adding another molecule of water would generally cause one of the other dissolved molecules to fall out solution (because there is an equilibrium... some of the water molecules at the bottom are dissolving, while others are falling out of solution).