My textbook definition says that operator spaces are the closed subspaces of a $C^*$ algebra. I was wondering if we can give an operator space structure to a Hilbert space? My idea was the following mapping.
$f:H\to B(H)$ such that $h\mapsto \left<.,e\right>h$, where $e$ is any normal element. Am I doing this right?
Yes, this is an isometric embedding if $e$ has norm $1$ (I guess that's what you mean by normal) and therefore defines an operator space structure on $H$.
However, note that every Banach space $E$ can be embedded isometrically into the $C^\ast$-algebra $C(B_{E^\ast})$, where $B_{E^\ast}$ is the unit ball of $E^\ast$ endowed with the weak$^\ast$ topology. So having some operator space structure is nothing that distinguishes Hilbert spaces from other Banach spaces.