Energy from 1 gram of fuel in a nuclear power plant? It might go without saying, but I am asking about a fission reactor. I assume the fuel still has mass after it is used. If I am right about that, I can't use $E=m*c^2$ to answer my question. Hence my question.
 A: The total energy from the fission of one atom of U-235 is 202.5 MeV according to Kaye and Laby.
Typical nuclear fuel is enriched to about 5% U-235, but never more than about 18% - higher than that and you are talking about material for bombs. It is very hard to define the "energy content" of nuclear fuel without knowing what kind of reactor you are using, and what the U-235 level is that this design has for "fresh" fuel, and for "spent" fuel (note that the fuel rods are moved around the reactor - IIRC, the most enriched rods start on the periphery where the neutron flux is lower, and over time they are moved towards the center where higher neutron flux means you can squeeze the last few fission reactions out of them).
If, for example, you went from 10% enrichment to 2% enrichment, you would have "burnt" 0.08 gram of U-235. The energy released would be the number of atoms in a gram of fuel, multiplied by the energy that fission of one atom releases (I hope you can follow where these numbers came from without me having to devise symbols and explain...):
$$E = \frac{0.08}{235}\cdot N_A \cdot 202.5\cdot 10^6 \cdot 1.6\cdot 10^{-19} J=6.6\mathrm{\; GJ}$$
Comparing this to coal, with roughly 24 kJ/g source - but note there is LOTS of variability, you see that Uranium has the advantage by about a factor of 275,000 - or 1 gram of uranium equivalent to 275 kg of coal.
Typically, you will see numbers like "3000 kg of coal". But that is comparing pure uranium - here we were looking at fuel in which just 8% of the weight is converted to energy. The same calculation would give around 3500 kg of coal equivalent to 1 gram of 100% U-235. That is close enough to the commonly quoted value. Of course the amount of heat given off by coal varies a lot with the type - sources vary by more than 2x between the lowest and highest energy density coal.
A: You have to use the change in mass. First, figure out to mass of the element/fuel when it is whole. Then, find the mass of the result of the reaction: the two new elements and the neutrons. Take the result on step 1 and subtract the result on step 2. This is the mass that you plug into the equation.
