# Calculation of loss of mass during fission – how does the math add up?

The following is a question and sample answer related to the calculation of loss of mass from fission and, in turn, the calculation of energy released.

In a fission reaction a neutron is absorbed by a uranium-235 nucleus. Barium-139 and krypton-94 nuclei are released as well as some neutrons.

Write a nuclear equation for this reaction. $$^{235}_{92}{\rm U} + {}^1_0{\rm n} \rightarrow {}^{139}_{56}{\rm Ba} + {}^{94}_{36}{\rm Kr} + 3\,{}^1_0{\rm n}$$ Calculate the energy released, in MeV, in this reaction. \text{loss in mass}=3.0\times10^{-28}\,\mathrm{kg}\\\begin{align}E&=mc^2\\E&=2.47\times10^{-11}\mathrm J\\E&=171\,\mathrm{MeV.}\end{align}

How does the math add up for the calculation of "loss in mass"? The mass numbers of the various particles/atoms are equal on both sides of the equation and thus there doesn't appear to be any loss in mass.

What am I missing?

• Note that $171\,\mathrm{MeV}/c^2$ corresponds to a little under $0.2\rm\,u$.