The semi-empirical formula is used to find the binding energy of a nucleus. But if you know the mass of a nucleus and the number of neutrons and protons that this nucleus consists of (and you know the mass of a neutron and a proton), then you can calculate the binding energy using $E=mc^2$. What is the difference between these two procedures, and do you get the same energy either way?
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2 Answers
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The semi-empirical mass formula is derived from a fit to the measured masses. If you know the numbers of protons and neutrons then the idea is that the SEMF should give you a good estimate of that mass (there are of course small residuals of the order 0.2 MeV to the fit).
$$M(A,Z)c^2 = (A-Z)m_n c^2 + Zm_pc^2 - AE_b,$$ where $A,Z$ are the mass number and atomic number and $E_b$ is the binding energy per nucleon that is described by the SEMF. So of course if you can measure the mass $M$, then you could easily rearrange this formula to give $E_b$. However, the beauty of the SEMF is it gives a very simple way to attempt to understand what is happening to the binding energy as you change the number of nucleons and the neutron/proton ratio. It also allows you to predict the properties of exotic or short-lived nuclei (for example in the crusts of neutron stars), where you may not have the luxury of a laboratory-based measurement.
In direct response to your question. Yes you can do this and the answers should be very similar. But there are small differences because the SEMF is just a model, and a relatively simple one at that. For instance if you were thinking of the Weizsacker liquid-drop version, then this doesn't predict the larger binding energies of the "magic number" nuclei, that are better describe by the shell model.
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The semi-empirical mass formula (SEMF) is much more involved than this, and tries to take into account various other factors that will affect the mass/energy of the nucleus. For any nucleus, of course if you KNOW its mass beforehand then simply subtracting the individual masses of protons and neutrons will give you ($\pm$)the binding energy, but the SEMF at least tries to be more predictive. Some of the things it takes into account are repulsions, nuclear surface tensions, volume effects and asymmetry due to the exclusion principle.
