This is actually a more complex question than you might think, because the distinction between mass and energy kind of disappears once you start talking about small particles.
So what is mass exactly? There are two common definitions:
- The quantity that determines an object's resistance to a change in motion, the $m$ in $\sum F = ma$
- The quantity that determines an object's response to a gravitational field, the $m$ in $F_g = mg$ (or equivalently, in $F_g = GMm/r^2$)
The thing is, energy actually satisfies both of these definitions. An object that has more energy - of any form - will be harder to accelerate, and will also respond more strongly to a given gravitational field. So technically, when computing the value of $m$ to plug into $\sum F = ma$ or $F_g = mg$ or any other formula that involves mass, you do need to take into account the chemical potential energy, thermal energy, gravitational binding energy, and many other forms of energy. In this sense it turns out that the "mass" we talk about in chemical and nuclear reactions is effectively just a word for the total energy of an object (well, divided by a constant factor: $m_\text{eff} = E/c^2$).
In special relativity, elementary particle physics, and quantum field theory, mass has a completely different definition. That's not relevant here, though.
If mass is just another word for energy, why do we even talk about it? Well, for one thing, people got used to using the word "mass" before anyone knew about all its subtleties ;-) But seriously: if you really look into all the different forms of energy that exist, you'll find that figuring out how much energy an object actually has can be very difficult. For instance, consider a chemical compound - $\mathrm{CO}_2$ for example. You can't just figure out the energy of a $\mathrm{CO}_2$ molecule by adding up the energies of one carbon atom and two oxygen atoms; you also have to take into account the energy required to make the chemical bond, any thermal energy stored in vibrational modes of the molecule or nuclear excitations of the atoms, and even slight adjustments to the molecular structure due to the surrounding environment.
For most applications, though, you can safely ignore all those extra energy contributions because they're extremely small compared to the energies of the atoms. For example, the energy of the chemical bonds in carbon dioxide is one ten-billionth of the total energy of the molecule. Even if adding the energies of the atoms doesn't quite get you the exact energy of the molecule, it's often close enough. When we use the term "mass", it often signifies that we're working in a domain where those small energy corrections don't matter, so adding the masses of the parts gets close enough to the mass of the whole.
Obviously, whether the "extra" energies matter or not depends on what sort of process you're dealing with, and specifically what energies are actually affected by the process. In chemical reactions, the only changes in energy that really take place are those due to breaking and forming of chemical bonds, which as I said are a miniscule contribution to the total energy of the particles involved. But on the other hand, consider a particle accelerator like the LHC, which collides protons with each other. In the process, the chromodynamic "bonds" between the quarks inside the protons are broken, and the quarks then recombine to form different particles. In a sense, this is like a chemical reaction in which the quarks play the role of the atoms, and the protons (and other particles) are the compounds, but in this case the energy involved in the bonds (by this I mean the kinetic energy of the gluons, not what is normally called the "binding energy") is fully half of the energy of the complete system (the protons) - in other words, about half of what we normally consider the "mass" of the proton actually comes from the interactions between the quarks, rather than the quarks themselves. So when the protons "react" with each other, you could definitely say that the mass (of the proton) was converted to energy, even though if you look closely, that "mass" wasn't really mass in the first place.
Nuclear reactions are kind of in the middle between the two extremes of chemical reactions and elementary particle reactions. In an atomic nucleus, the binding energy contributes anywhere from 0.1% up to about 1% of the total energy of the nucleus. This is a lot less than with the color force in the proton, but it's still enough that it needs to be counted as a contribution to the mass of the nucleus. So that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass. That's not the case with chemical reactions; we can just ignore the binding energy when calculating masses, so we say that chemical reactions do not convert mass to energy.
To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.
The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.
When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.
So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.
To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.
EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:
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This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.
The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).
By the way, this image is from a very helpful article which should also be helpful for understanding this issue.
Best Answer
The standard free energy of formation of carbon dioxide from graphite and oxygen is $394\, \rm kJ\, mol^{-1}$ and so in the formation of one molecule of carbon dioxide $6.5 \times 10^{-19} \,\rm J$ of energy are liberated.
$6.5 \times 10^{-19} \,\rm J$ of energy is $4.1 \, \rm eV$ or has the mass equivalent of $4.4 \times 10^{-9}\,\rm u$.
So you will see that even your very accurate and precise value for the mass of an oxygen atom $M(^{16}\rm O)=15.994915 \, u$ is not good enough to find the energy of formation of carbon dioxide.
This is a good example of the difference between Chemistry and Nuclear Physics in that Chemistry deals with energy changes of order $\rm eV$ whereas Nuclear Physics deals with energies of order $\rm MeV$.