TITLE: What's the deal with empty models in first-order logic?
QUESTION [32 upvotes]: Asaf's answer here reminded me of something that should have been bothering me ever since I learned about it, but which I had more or less forgotten about. In first-order logic, there is a convention to only work with non-empty models of a theory $T$. The reason usually given is that the sentences $(\forall x)(x = x)$ and $(\forall x)(x \neq x)$ both hold in the "empty model" of $T$, so if we want the set of sentences satisfied by a model to be consistent, we need to disallow the empty model.
This smells fishy to me. I can't imagine that a sufficiently categorical setup of first-order logic (in terms of functors $C_T \to \text{Set}$ preserving some structure, where $C_T$ is the "free model of $T$" in an appropriate sense) would have this defect, or if it did it would have it for a reason. So something is incomplete about the standard setup of first-order logic, but I don't know what it could be. 
The above looks like an example of too simple to be simple, except that I can't explain it to myself in the same way that I can explain other examples.

REPLY [2 votes]: Since nobody else has mentioned the phrase, I think what you are looking for is called free logic by those who adopt the convention about what "first-order logic" means that would imply all models are nonempty.
We can construct a $C_T$ that is defective in the same way; we include a bunch of arrows $1 \to T$ to represent indeterminate elements of $T$. Then any lex functor $C_T \to \mathbf{Set}$ must send $T$ to a non-empty set.
Of course, the usual categorical approach of interpreting free variables as generalized elements doesn't have this defect.