"Evaporation energy" is not the same as chemical potential, because you forgot about entropy. For example, at $100\,\rm^\circ C$, the chemical potential of liquid water and water vapor is exactly the same, because they are in "diffusive equilibrium" (the number of molecules going out of the vapor into the gas is, on average, equal to the number moving the opposite direction). But the binding energy is much greater for liquid – it takes a lot of energy to rip a liquid molecule away from the surrounding liquid molecules (to which it is attracted), but it takes virtually no energy to pull a water vapor molecule away from the other water vapor molecules.
Evaporation energy sounds like the same as enthalpy per particle, whereas chemical potential is Gibbs free energy per particle.
UPDATE WITH MORE DETAILS: How is it, microscopically, that the liquid and vapor at $100\,\rm^\circ C$ are in diffusive equilibrium, even though the liquid molecules have a huge binding energy sucking themselves together, and the vapor molecules have almost none? Well, at any given moment, there is an entire layer of liquid molecules at the topmost surface, and any one of them could randomly get enough energy to fly off.
On the other hand, at any given moment, there are only a tiny handful of vapor molecules that happen to be close enough to the liquid surface to potentially be sucked in and join the liquid. In other words, if you show me a random molecule at the liquid-vapor interface, I would say it is much more likely to choose to stay in the liquid rather than the vapor, but this is balanced by the fact that if you show me a random molecule at the liquid-vapor interface, you are much more likely to be showing me a molecule coming from liquid than from vapor. (This is a common way that an entropic effect can manifest itself microscopically.)
In a microscopic simulation at $100\,\rm^\circ C$, you would indeed see that just as many liquid molecules go into vapor as the other way around. That means by definition liquid and vapor have the same chemical potential at $100\,\rm^\circ C$.
How about at $50\,\rm^\circ C$? What is the microscopic definition of chemical potential difference between liquid and vapor at $50\,\rm^\circ C$? Certainly, the liquid chemical potential is lower, because more vapor molecules turn into liquid than vice-versa. But if you bias the system in favor of creating vapor molecules, you can rig the simulation to get diffusive equilibrium. For example, you add some artificial force field into the simulation that covers the liquid surface and has the effect that a liquid molecule gains $1\,\rm eV$ of kinetic energy when it goes into vapor, and conversely, a vapor molecule has to lose $1\,\rm eV$ of kinetic energy in order to reach the liquid surface.
Maybe now, it turns out that the liquid and vapor are in diffusive equilibrium. If so, then you have proven through this simulation that the chemical potential difference between liquid and vapor at $50\,\rm^\circ C$ is exactly $1\,\rm eV$. (Chemical potentials are not calculated like this in practice but it is a nice conceptual way to imagine it.)