Can you pre-determine the result of a coin toss? The question is fairly simple. Consider the following parameters which are known to you:
1.) Mass of the coin
2.) The force applied on the coin
3.) The point where the force is applied on the coin
4.) $g$ of the place where the coin is tossed (acceleration due to gravity)
5.) Radius of the Coin
Based on this data, how would you calculate the result of the coin toss?
 A: At the very least you would also need the dimensions of the coin (mass alone is not enough) and the elastic properties of both the coin and of the surface on which you land... that has a big impact (since the coin can bounce or "stick", depending on the surface). It may be possible to predict for certain sizes of coin and height of drop (a carefully executed half-flip of a huge coin) - but as the number of revolutions completed increases, and particularly if the calculated position at the time of landing approaches the metastable position (landing on edge in such a way that it is equally likely to fall to either side) the uncertainty in the result also increases, until you end up with a resounding "no".
In general, then, the answer is "no". For specific cases the answer can be "yes". It all depends...
But if I were asked to do the calculation - which was your question - I would just solve the equations of motion. Assuming that you have the following information
m = mass of coin
I = moment of inertia about length axis
g = acceleration of gravity
h = initial height
d = distance from center of mass where impulse is applied
p = impulse applied to coin

you can then compute the initial vertical velocity from
$$v = \frac{p}{m}$$
and thus the time to landing by first computing the maximum height (which will be reached after $t=v/g$ seconds, and be at $h_{max}=h+\frac12 g t^2$), and then the time from this height to the ground, given by $t_2=\sqrt{2 h_{max}/g}$. You would probably want the time until you reach "ground minus radius of coin" so you can compute the time when you actually strike the surface - this depends on the angular position of the coin.
Next you would compute the angular velocity of the coin from
$$\omega = \frac{p d}{I}$$
With these two numbers you can calculate how many revolutions the coin has completed when it gets close to the ground, and then you would have to figure out when the coin first hits the surface - this is not when the center of mass hits h=0, but rather when a rim of the coin does - and that depends on the angular position.
Finally, you would have to take into account the properties of the surface to determine how the coin behaves when it hits the ground. This can be the trickiest part of the calculation - which is why I started out by saying that the surface properties absolutely need to be part of the input.
A: There's different levels on which you can answer this question:
First, because the coin is a large enough macroscopic system that quantum-mechanical effects are negligible (i.e. we don't have to worry about the wave nature of the coin).  The variables that will affect the final outcome of the coin toss are all classical in nature, and so can be measured and calculated classically without a problem.  Therefore, if you know the properties of the system that Floris mentioned in his answer (moment of inertia $I$ of coin, height $h$ of coin, etc) you can in principle calculate the outcome of the coin toss.
The real problem here, then, is that a coin toss can quickly become chaotic (in the physical/mathematical sense of the word).  Very small changes in initial conditions can produce different results.  Very small changes in the distance the flick is applied from the axis of rotation can produce a different result.  Because coins also do not have uniform mass distribution, the coin will experience precession and wobble as it rotates through the air.  And although I'm not certain about this, I would not be surprised if we even need to consider the fluid dynamics of the air around the coin.  Due to the sensitivity to initial conditions that exceeds the amount of precision in our measurements, it's easy to see that the task becomes unmanageable in practice.
Please note, however, then when other answers say this task is impossible, I think this gives the wrong impression.  Determining when a single atom of a radioactive isotope will decay is impossible.  Due to quantum mechanical effects, there is indeterminancy built into the system on the quantum level that absolutely cannot be avoided.  This is very different than a coin toss, where we could always say "If only I could measure ($x$, $I$, $p$, $h$ or whatever) more precisely..."
A: To do it completely without error would be impossible. There are simply too many parameters to be determined to exact precision in order for there to be no error. 
One of the answers in John's comment links to this research that provides further details. 
