Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra and let $\pi : \mathfrak{g} \to \mathfrak{gl}(V)$ be a faithful and irreducible representation of $\mathfrak{g}$. Let $\{X_i\}$ be a basis for $\mathfrak{g}$.
Then given any nondegenerate symmetric bilinear form $B$ on $\mathfrak{g}$ we can find elements $\{X^i\}$ of $\mathfrak{g}$ so that $B(X^i, X_j) = \delta_{i,j}$, and it is a theorem that the associated Casimir element $$ C = \sum_{i=1}^n X_i X^i $$ in the universal enveloping algebra $U\mathfrak{g}$ of $\mathfrak{g}$ is independent of the choice of basis and acts as a nonzero scalar on $V$ (for this we need that $V$ is irreducible, and just apply Shur's lemma using the fact that the action of $C$ on $V$ is an intertwiner).
However, to my eyes there are two natural choices for the bilinear form $B$; we could choose the Killing form $K$ on $\mathfrak{g}$, which is nondegenerate by the semisimplicity hypothesis, or we could choose the bilinear form $T(X,Y) = \operatorname{tr}(\pi(X) \pi(Y))$ for $\pi : \mathfrak{g} \to \mathfrak{gl}(V)$ our representation. Since we assumed that $\pi$ is faithful, the same proof that shows the Killing form is nondegenerate shows that this trace pairing is nondegenerate too.
Hence, it seems like I get two different "Casimir elements" $C_K$ and $C_T$ out of this construction, since these two bilinear forms give two different dual bases $\{X^i_K\}$ and $\{X^i_T\}$. They can't be the same, since by construction the latter element has $$ \operatorname{tr}(\pi(C_T)) = \sum_{i = 1}^n \operatorname{tr}(\pi(X_i) \pi(X^i_T)) = \sum_{i = 1}^n T(X_i, X^i_T) = n, $$ while it is well known that the standard Casimir element $C_K$ coming from the Killing form doesn't act this way (there is a formula for this in Wikipedia).
How are these two "Casimir elements" related?