Water bottle rocket: Where does the energy go without water? In Portland's OMSI there is a hands-on water bottle rocket station. (https://www.youtube.com/watch?v=cdtmVY76_PQ). The rockets are normal PET bottlers. The visitors fill their bottle with an amount of water and then fill the remaining volume with compressed air at a given pressure.
The challenge is to find the best water-to-air ratio so that the rocket flies highest. Too much water is bad, not only because it makes the rocket heavy. As I explained to my son, the compressed air is also the energy store for this rocket (not the water, since water is almost incompressible). To have less compressed air means to have less energy available.
But then I got stuck because by inversion this means that an "all air" configuration should be best: Most available energy, highest kinetic energy, highest speed of empty bottle. This is obviously wrong. It was clear experimentally that the best ratio is somewhere in the middle. Also it makes intuitive sense that some mass in form of water is needed to produce thrust, since actio = reactio. In order to produce momentum, mass is needed to "push off of".
I'm aware of the fairly complex rocket flight physics. (For example, https://www.ohio.edu/mechanical/programming/rocket/analysis1.html gives an accessible overview.) But because I am not interested in an exact result much of it can be neglected. The basics are fairly simple: Energy stored in the compressed air is transformed into kinetic energy of the expelled water, rocket and earth, plus "losses" through heat from turbulences. 
My question is on a more general, abstract level. Momentum or not, we have a given energy in the air which must go somewhere.
Where does the energy go which is stored in the compressed air in a "compressed air only" configuration? It should be more energy than with a partly water-filled bottle; but the rocket's final velocity (and hence kinetic energy) is much lower. Did we produce that much heat? I don't think so. Did we accelerate the earth? No, the "burn phase" was short.
I am missing something. What is it?
 A: It was already discussed in the comments that a water rocket needs to push "something" out. It is instructive to do the calculation in a little more detail to see where the "energy" goes. For this I will consider the relative share of energy going to the rocket and the "expelled matter" (gas, or water) as a function of the expelled mass. To simplify things, we will assume that all matter is expelled as a single entity with a certain velocity; in reality you might need to integrate, but any inequality that holds for a small amount of expelled matter will hold for the integral over many such amounts.
I will use upper case symbols for quantities relating to the "rest of the" rocket (mass M, velocity V, momentum P - without the expelled mass) and lower case for the expelled matter(m, v, p). From conservation of momentum, $P = -p$ so $M\cdot V = - m\cdot v$.  The energy of the rocket $E_r$ and expelled mass $E_m$ will be respectively:
$$E_{r} = \frac12 M V^2 = \frac{P^2}{2M}\\
E_m = \frac12 m v^2 = \frac{p^2}{2m} = \frac{P^2}{2m}$$
It follows that the ratio of (energy in rocket)/(energy in expelled matter) is
$$\frac{E_r}{E_m} = \frac{m}{M}$$
In other words - the lower the mass of the expelled matter, the greater the relative amount of energy it contains. In the limit of "no water", the little bit of air mass contains virtually all the energy.
