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.