What is the best way to determine if an organic molecule is really aromatic?

More often than not, knowing how to check via resonance structures for the full #pi#-electron conjugation throughout the ring will show aromaticity, if you know what you are doing.

The conditions for aromaticity can be summarized like so:

A planar ring with #\mathbf(4n + 2)# #pi# electrons within the #pi# system, where every atom in the ring can participate in the delocalization of the #pi# electrons, is aromatic.

On the other hand, something antiaromatic follows the #\mathbf(4n)# rule for the number of #pi# electrons, but would otherwise be aromatic.

Therefore, you have to prove aromaticity before you can assert antiaromaticity.

There is no clear-cut, "best", "most efficient" test for aromaticity, but in general it usually works to check the resonance structure for whether or not the #pi# electrons are fully conjugated around the ring. I wouldn't trust Hückel's Rule necessarily.

CHECKING #\mathbf(pi)# ELECTRON LOCALIZATION

Proper resonance structures are usually, but not always, a good indication of aromaticity.

You have to understand, from examining the electron geometry, which electrons are truly delocalized within the ring, and which ones are not. That matters.

For example, consider pyridine, which I'm going to tell you ahead of time is aromatic.

MISTAKEN USAGE OF HÜCKEL'S RULE DOES NOT DISPROVE AROMATICITY

Count how many #pi# electrons it has. Two per double bond gives #6#. That means #4n + 2 = \mathbf(6)# and #n = \mathbf(1)# (Hückel's Rule). That satisfies one potentially difficult condition for aromaticity.

If you counted #8#, then you must have counted nitrogen's #sp^2# lone pair of electrons, but wait a minute---those don't count.

Why? Because they are perpendicular to the rest of the ring, and thus are localized outside of the ring. They do not participate in the resonance structure and are NOT going to count in the #4n + 2# rule. Counting #8# would mean they follow the #4n# rule, which would falsely indicate that pyridine is antiaromatic. It's not!

PYRIDINE'S #\mathbf(pi)# ELECTRONS ARE FULLY DELOCALIZED WITHIN THE RING

If you proceed to check the resonance structures, you can see that the #pi# electrons are indeed conjugated throughout the ring, but the #sp^2# lone pair on nitrogen does NOT actually participate:

That satisfies probably the most difficult condition for aromaticity.

Finally, it is obvious that pyridine is cyclic (it's a ring). Pyridine is also planar due to its depiction having alternating double bonds (#3# electron groups around each atom in the ring). Therefore, it satisfies all conditions for aromaticity.

SOMETIMES RESONANCE STRUCTURES FAIL

Resonance structures don't always work, so you should always consider all of the conditions for aromaticity.

For instance, sometimes you may think "hey, this is fully conjugated around the ring. It's aromatic, right?" However, there are exceptions.

One such exception is 1,3-cyclobutadiene.

I won't go too much into detail on this, but this molecule is neither aromatic nor antiaromatic, not only because of its #4n# electrons where #n = 1#, but because its electrons aren't even delocalized. As such, you shouldn't even be able to draw resonance structures for it at all.

I've written a more in-depth discussion of this here:
http://socratic.org/questions/why-is-1-3-cyclobutadiene-not-aromatic

CHALLENGE: Try drawing the resonance structures for the 1,3-cyclopentadiene anion (deprotonated on its only #CH_2#). Is it aromatic? How do you know?