Energy Released in a Fission Reaction I've been told the following is incorrect, but I can't really see how. 
Consider the fission event described by the equation 
$$
\rm ^{235}U+n\rightarrow{}^{93}Rb+{}^{140}Cs+3n
$$ The energy released is given by 
\begin{align}
Q&=\Delta mc^2
\\&=(m_\mathrm U+m_\mathrm n-m_\mathrm{Rb}-m_\mathrm{Cs}-3m_\mathrm n)c^2
\\&=(m_\mathrm U-m_\mathrm {Rb}-m_\mathrm {Cs}-2m_\mathrm n)c^2
\end{align}
 and writing the nuclear masses in terms of the masses of their constituent nucleons less than nuclear binding energy we find 
$$
Q=B_\mathrm{Rb}+B_\mathrm{Cs}-B_\mathrm{U}
$$ 
You could then estimate these binding energies using the SEMF giving a value of roughly $$Q=145\rm\,MeV$$ (in reality it should be something like $200\rm\,MeV$). However I'm just concerned about whether or not the general method is correct or not. Thanks in advance.
 A: Your result for Q in terms of the Bs is okay, but you're not going to find  B's tabulated directly. You will find B/A for each nuclide.  I find it easier to use either atomic masses or mass defects, since both of those are tabulated somewhere.  
The SEMF is a gross approximation and doesn't fit individual nuclides well enough for better than a 10-20% uncertainty in Q calculations. I wouldn't use that.  A good source to look up individual nuclide atomic masses is the NNDC, National Nuclear Data Center, but that can get overwhelming, trying to find exactly what you want.
The Table of Nuclides at   KAERI has things in an easier format, IMO.
Also, if you're familiar with LaTeX, there is a package called nucleardata which allows direct access to nuclide information like masses, half lives, decay modes, etc.  It requires the use of another package LaTeX called pythontex which allows direct typesetting of Python calculations and output.
A: The Nuclear Wallet Cards make this process a little easier by listing mass excesses rather than binding energies.  For your reaction we have
$$
\begin{array}{cr}
\text{nuclide} & \Delta\,\rm (MeV)
\\\hline
\rm n & +8.1 \\
\rm^{140}Cs & -77.1 \\
\rm^{93}Rb & -72.6 \\
\rm^{235}U & +40.9
\end{array}
$$
So your initial state has mass excess $\rm \Delta n + \Delta U = +49.0\,MeV$, your final state has mass excess $\rm \Delta Rb + \Delta Cs + 3 \Delta n = -125.4\,MeV$, and the $Q$-value for this particular fission is apparently about $\rm 175\,MeV$.  I would say that your estimate from the semi-empirical mass formula is not that far off, all things considered.
You can convince yourself that my method and your method are equivalent by using the mass excess to find the binding energy of each nucleus.  For instance, the binding energy per nucleon for U-235 is $7.59\rm\,MeV$, which corresponds to a total binding energy of $\rm 1783\,MeV$.  (This is almost two nucleons' worth of binding energy!)  If I were to break my U-235 nucleus into 92 protons and 143 neutrons my mass excess would go from 40.9 MeV (above) to
$$
\rm 92\Delta p + 143 \Delta n 
= 92 \times 7.29\,MeV + 143 \times 8.07\,MeV = 1824\,MeV
$$
and $\rm 1824\,MeV - 40.9\,MeV = 1784\,MeV$, as we expected.  You're welcome to check the others if you are so inclined.
