The pitching moment probably has little (if any) relevance in the case of an airfoil attached to a race car.
In particular, the pitching moment is related to the fact that with a cambered airfoil, the center of pressure will forward and backward depending on the angle of attack at which the airfoil is operating.
In the case of a race car, the angle of attack won't normally vary much, so the pitching coefficient probably isn't very relevant.
I'm not sure it's really relevant, but the aerodynamic center is normally taken as being at 25% of MAC. The Cm basically then defines a torque that's exhibited at the aerodynamic center.
One other point: I'd at least assume that on a race car, the airfoil is mounted upside down compared to an aircraft--that is, it's intended to produce pressure downward, not upward. To the (I think minimal) extent that pitching moment is relevant at all, it'll be positive (whereas it's normally negative for the main wing of an aircraft). That is to say, on an aircraft the pitching moment leads to a force pushing the nose downward, but in your case it'll be a force tending to push the nose up (and the tail down).
Okay, so based on the comment, what's really desired in this case is to compute the location of the center of pressure. To do this, we really need two things: the Cm and the CL.
For the moment, I'm going to assume a Reynolds number around 500,000 or so. For an E423, your quoted Cm of -0.22 corresponds to an Alpha (angle of attack) of around 5 degrees. With Re = 500K and alpha=5, it looks like the CL is about 1.6.
The formula we need is:
hcp = hac - Cmac / CL.
Plugging in the numbers, we get: 0.25 - (-0.22)/1.6.
Running that through the calculator, we get 0.3875. So your center of pressure should be approximately 38.75% of MAC. Of course, you'll need to adjust if you're actually basing your number for Cm on a different combination of angle of attack and Reynolds number.
As noted above, in your case the airfoil is (presumably) upside-down, so that should really be a Cm of +0.22 and a lift of -1.6, but the signs cancel so we get the same result either way (exactly as you'd expect).
- Airfoil data