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I understand that the name derives from the empirical formula; however, I do not understand why it does not exist as $\ce{P2O5}$. Is there a compound that exists with this as its molecular formula?

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    $\begingroup$ See Wikipedia's articlePhosphorus pentoxide. Essentially $\ce{P4O10}$ is a molecule and $\ce{P2O5}$ is not. $\endgroup$ – MaxW Mar 11 at 20:26
  • $\begingroup$ Just out of curiosity why do you think isolated P2O5 molecules would be stable? As far as I know only 1 element (Iodine) forms a pentoxide for which the most stable form in the solid state is an isolated X2O5 molecule, all other pentoxides have more extended structures. Why should phosphorous be different from the norm? $\endgroup$ – Ian Bush Mar 11 at 20:44
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    $\begingroup$ @Karl not really, it could look like N2O5 does. $\endgroup$ – Mithoron Mar 11 at 20:54
  • $\begingroup$ Unfortunately, "exists" is not well-defined enough. $\endgroup$ – Zhe Mar 11 at 22:30
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There is no clear rule about how to name a compound which is actually existing as a dimer, a trimer... or an octamer, as long as it does not make a difference in a reaction equation, which is usually the case.

It does not matter in an equation if you count sulfur as a single atom entity (S) or as its true form (S8). You will find the same quantity in both cases.

The same stands for P2O5/P4O10: the true form is usually P4O10 (depending on the solvent), but the 2 forms give the same result at the end.

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The short answer is probably that in any case of stoichiometry it doesn't matter. Much as SteffX explained it above. But while we have it already. There are cases where I think it has at least some benefit if we use the not shortened formula. This would probably be one of these cases. As you may know, the $\ce{P4O10}$ is formed when white Phosphorus $\ce{P4}$ reacts with air. The reason for this is that in the $\ce{P4}$-tetrahedron we have smaller bond angles (60°) than we would expect from a head-on bond formation of three p-type orbitals (90°). Therefore the orbitals cannot overlap perfectly and there is some considerable bond strain.

My former Professor for Inorganic Chemistry is a famous phosphorus chemist and he explained it in much more details back then. But if it's about angles in $\ce{P-P}$-bonds, triangular shapes are some of the worst things that can happen. And for the $\ce{P4}$-tetrahedron (white phosphorus) we have a polyhedron that consists of triangular faces only.

So if we introduce oxygen into the system it could enter between every $\ce{P-P}$-bond to increase the angle and therefore reduce the strain. If you draw this on some paper you will see that a $\ce{P4O6}$ results (sometimes called $\ce{P2O3}$). And what you should be able to see is that you can still connect all $\ce{P}$-atoms to get back the original tetrahedron. So the overall shape did not change. We call this a 'topotactic oxidation', oxidation, where the original shape remains, we simply add something inbetween. And in the final step, the phosphorus is at $\ce{P^3+}$ now, we can even oxidize it up to it's highest possible oxidation state $\ce{P^5+}$ by adding more oxygen. It will attack the terminal positions, so we get four additional oxygens and a final $\ce{P4O10}$. This is a relationship many people often forget about. So it's quite nice to remind them of the original $\ce{P4}$-tetrahedron by not shortening the formula down to $\ce{P2O5}$. And as many suggested above it's just the unit you will see.

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