# What would happen if I mixed a single water molecule in a beaker of hexane?

What would happen if I mixed a single water molecule in a beaker of hexane? Would it sink or float?

How about 2 water molecules? 3? Etc?

In other words, how many water molecules does it take to be hydrogen bonding together to become more dense than 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).

• Need we consider Brownian motion here? – a-cyclohexane-molecule Jul 22 '16 at 17:52
• I think supersaturation is a moot point because it is usually achieved by raising the temperature of a solvent to allow more solute to dissolve before allowing the solvent to cool slowly. For hexane at 20 °C, it would be as you pointed out 0.1%. Then if you add a single water molecule above that limit it would probably float to the top. However, once enough molecules are added (around 6 I think) to create a solvation shell around a single water molecule, then the density of that 7H2O complex would be higher than hexane and sink to the bottom. – Nova Jul 22 '16 at 17:57
• @nova Supersaturation is practically done like in the way that you described, but the solution would also supersaturate by delicately adding one molecule at a time. And again, they would all float due to surface tension... they would need to be physically pushed to submerge. Then they would be dissolved; then the hexane would force a bunch of the other molecules out of solution... like rain. – Ben Welborn Jul 22 '16 at 18:06
• Ok, good point. But my argument still stands for what happens once the supersaturated point is reached. Any addition molecules would begin to float up until enough amalgamate to create a sphere of hydration that would then sink. – Nova Jul 22 '16 at 18:21
• @nova They would not float. They are dissolved throughout the solution. At supersaturation, they would continue to be dissolved in solution. Adding one more would cause a cascade of many water molecules to condense and precipitate. – Ben Welborn Jul 22 '16 at 18:30

It seems like you assume that water is not soluble in hexane at all. This is incorrect. It is not miscible, but solubility of water in hexane 0.01% at 20°C. That is 100 mg/L or 5.5 mM which is actually a decent amount of molecules. Distribution in liquid in a beaker will be the same at all heights.

A good question would be "what is the distribution of water molecules on the walls of the beaker and in the volume?" Beaker walls can accomodate small amount of molecules. The extent of absorption is following Langmuir adsorption model. It can be ignored at high concentrations (mM of dissolved material), but starts playing a significant role when you go into nanomolar or picomolar range. At that concentrations the actual amount of material in solution is significantly smaller than calculated because of absorption.

Suppose that you have a sealed beaker with some hexane liquid in it at room temperature. Now add one water molecule. The answer is obviously that the single water molecule will spend some time in the vapour phase and some dissolved in the hexane. Once the water molecule hits the liquid surface there is a chance that it will enter the liquid. If there is a strong interaction between the water and the liquid then the tendency will be to spend longer in the liquid than vapour. Hexane and water have a poor interaction, the hexane cannot effectively solvate a water molecule's dipole as it has no appreciable dipole itself and hence a small dielectric constant (relative permittivity)$^*$. The low dielectric solvent means that the electric filed of the water's dipole can spread out over many hexane molecules causing a positive interaction energy. Exactly how big is difficult to say.

The water molecule can occasionally overcome any intermolecular interactions in the liquid phase as there is an exponential distribution of energies in all molecules at a finite temperature. This is given by the Boltzmann distribution. Thus a few molecules have far greater energy than the average ($3RT/2$) and eventually these molecules will impart some of their excess energy to the water molecule and it will be ejected into the vapour phase. These processes will continually repeat themselves.

Now if we consider this from a thermodynamic viewpoint, we will have to assume numerous water molecules in the presence of the hexane. We need to calculate the free energy which consists of the enthalpy change $\Delta H$ and entropy change $\Delta S$ as in the formula $\Delta G=\Delta H -T\Delta S$.

The enthalpy of water entering the hexane will probably be small and positive for reasons outlined above, but there is also an entropic factor. When two liquids are mixed the (ideal) entropy of mixing$^{**}$ is $\Delta S= RT(n_1ln(x_1)+n_2ln(x_2) )$ where $x_1$ is the mole fraction of water and $x_2$ that of hexane and $n_1$ and $n_2$ are the respective number of moles. As the solubility of water in hexane is low approx $10^{-2}$ molar, if not smaller, then the mole fraction of hexane can be approximated as 1 and $\Delta S= RTn_1ln(x_1)$ which evaluates to approx -1 kJ/mol.

The enthalpic (heat) term can be shown to be $(n_1+n_2)x_1x_2w$ where $w$ is an energy term allowing for the interaction between solute-solute $\epsilon_{11}$, solvent-solvent and solute-solvent molecules, $w \approx 2\epsilon_{1,2}-\epsilon_{1,1}-\epsilon_{2,2}$ where $\epsilon$ represents an interaction energy. As the interaction between water and hexane is smaller than hexane-hexane interaction($\epsilon_{1,1}$) and water-water interactions, the interaction energy is positive. This means that the free energy for dissolving $\Delta G$ is going to be small and probably negative overall, and this means that the solubility is going to be small also.

As the amount of water increases approaching the solubility limit there will be microscopic regions of pure water and pure hexane until the water separates and being the most dense of the two liquid forms the lower layer.In this situation water is dissolved in the hexane, to its solubility limit, and hexane also in water. The extent of solubility is given by Henry's law. The vapour pressure P of a solute above a solution is given by $P= k_Hx$ where x is the solute mole fraction and $k_H$ the Henry Law constant which is only determined by experimental and depends on solute/solvent and temperature. Values of $k_H$ (in units of Bar) are several thousands for relatively insoluble solutes such as water in hexane or vice versa.

$^*$ A low dielectric constant $\epsilon _0$ means that an electric field spreads a considerable distance r from its source compared to a solvent with a high constant. The interaction energy scales as $\approx 1/(\epsilon _0 r)$, low dielectric (alkane) solvents have $\epsilon _0$ in the range 2-4, whereas water is 78 so the effect is considerable.

$**$ Entropy is defined from statistical thermodynamics as $S=k_Bln(\Omega)$ , where $k_B$ is the Boltzmann constant and $\Omega$ the number of configurations available to the system. The calculation of the entropy starts by working out the number of ways of placing molecules into sites. Suppose that there are N1 and N2 molecules and that they can occupy in total N1+N2 sites. There is a chance of putting the first molecule into N1+N2 sites , the second in N1+N2-1 sites etc. The total number of possibilities is (N1+N2)!. However, molecules of one type cannot be distinguished from one another thus we must divide by N1! and similarly for N2 by N2!. This produces $\Omega=\frac{(N1+N2)!}{N1!N2!}$. Using Sterling's approximation for factorials the equation for $\Delta S$ can be obtained.

This is the answer to the previous version of the question with ethane instead of hexane.

Ethane is a gas, did you mean ethane? Assume that your beaker is sealed but in thermal contact with the outside. i.e. sitting on the bench. On adding a single water molecule the thermal energy of the ethane will impart its energy to the water and it will be found in the gas phase. If you keep adding water the same thing happens but eventually liquid water will form at the bottom/sides of the container in equilibrium with water vapour; rate of evaporation equal to rate of condensation. Adding more water will allow more ethane to dissolve as the water volume and gas pressure increases in the container (as it is sealed). Some small amount of the ethane will also dissolve in the water. The amount is determined by Henry's law.

• Changed ethane to hexane. Typo. – Nova Jul 22 '16 at 16:26