Nuclear fission problem

If during a nuclear fission U(235) decomposes into Xe(140), Sr(94) and n(1), how is it possible that the original U(235) has bigger mass than the three resulting nuclei together? Should it not be the case that the U(235) has a deeper bonding energy so the lower mass; this way its binding energy is released during the fission and the resulting elements have smaller mass defect?

• No, binding energy is inherently a negative quantity. A "deeper binding energy" means that the nucleus is more tightly bound together, just as Mercury is more "tightly bound" to the Sun than the Earth is. So if U-235 was more tightly bound (i.e. had a greater mass defect) than the proposed daughter nuclei, we would need to provide energy for this decay to occur (just as we'd need to give Mercury a lot of energy to raise it up into Earth's orbit.) – Michael Seifert May 18 '15 at 19:50
$$M_\text Uc^2 =\frac{m_\text{Xe}c^2}{\sqrt{1-\frac{v_\text{Xe}^2}{c^2}}} +\frac{m_\text{Sr}c^2}{\sqrt{1-\frac{v_\text{Sr}^2}{c^2}}} +\frac{m_\text{n}c^2}{\sqrt{1-\frac{v_\text{n}^2}{c^2}}}$$
where the M's correspond to the rest mass of each object. Note that (In case you don't know) $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ is the total energy in relativistic dynamics. (I haven't written it this way for uranium since it is assumed that its stationary and $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=1$ for an object with zero velocity.)
So if you analyse the equation I wrote above, you'll notice that it is necessary that $M_U>m_{Xe}+m_{n}+m_{Sr}$ since the $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is always greater than one.