Let's take the simple system of a deuterium nucleus, that is the bound state of a proton and neutron.
It's tempting to think of the binding energy as something that has to be added to a proton and neutron to glue them together into a deuteron, and therefore that the deuteron must weigh more than the proton and neutron because it has had something extra added to it. However this is the exact opposite of what actually happens.
Suppose we start with a proton and a neutron at a large separation, and we let them go and allow them to attract each other. As they move towards each other their potential energy decreases, so because of energy conservation their kinetic energy increases. They accelerate towards each other in just the same way you accelerate towards the Earth if you jump out of a window. This means that when the proton and neutron meet they are moving very rapidly towards each other, and they just flash past each other and coast back out to a large separation. This is not a bound state. To form a deuteron you have to take energy out of the system so that when the proton and neutron meet they are stationary with respect to each other. Then they can bond to each other to form a deuteron.
And this is the key point. To form a bound state you need to take an amount of energy out that is equal to the kinetic energy gained as the two particles collide. This energy is the binding energy, which is 2.2 MeV for a deuteron. If you take the energy 2.2 MeV and convert it to a mass using Einstein's equation $E = mc^2$ you get a mass of $3.97 \times 10^{-30}$ kg, and if you compare the mass of a deuteron with the mass of a proton + neutron you find the deuteron is indeed $3.97 \times 10^{-30}$ kg lighter.
Just for completeness let's look at this the other way round. Suppose we start with a deuteron and we want to separate it into a proton and neutron. Because the two particles in a deuteron attract each other we need to do work to pull them apart, which means we need to add energy to the system. To pull them completely apart we need to add an amount of energy equal to the binding energy of 2.2 MeV. That means our separated proton and neutron have 2.2 MeV more energy than the deuteron, so the total mass of the separated particles has increased by 2.2 MeV and they are $3.97 \times 10^{-30}$ kg heavier than the deuteron.
Best Answer
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.