In Tourlakis' Mathematical Logic, he claims that $\models A $ if and only if $\emptyset \models A$. This question is on page 36. The first statement implies the second is correct but the converse is incorrect.
A counterexample would be any contradiction say $A \land \neg A$.
Am I missing something ?
I think I see your error. I'm not quite sure what you mean by state, but let me try to informally translate the definitions you gave to symbols.
$\models A$ means that $$\forall_{states} A,$$ whereas $\Gamma\models A$ for a collection of statements $\Gamma$ means $$\forall_{states}\left(\left(\forall_{\psi\in\Gamma}\psi\right)\implies A\right).$$
Note the careful parentheses in this second definition, since I'm fairly sure the error is one of misinterpreting the grouping of the quantifiers and symbols here. If I now put $\Gamma=\varnothing$, then I have $\forall_{\psi\in\Gamma}\psi$ becomes vacuously true, or in other words $$\varnothing\models A$$ means that $$\forall_{states} \mathrm{True}\implies A,$$ or $$\forall_{states} A,$$ since $$\mathrm{True}\implies A\text{ if and only if }A.$$