Can the encodings set of a non-trivial class of languages which contains the empty set be recursively enumerable?
to complete Raphael's answer, there is an extension of Rice's theorem that says the following:


  Generalized Rice's Theorem
  
  Let $S \subseteq RE$ be some property, and let $L_S$ be all the TMs that satisfy the property $S$, that is,
  $$L_S = \{ \langle M \rangle \mid L(M) \in S \}.$$
  Then, $L_S \in RE$ if and only if all the following conditions hold:
  
  
  for any $L_1,L_2 \in RE$, if $L_1 \in S$ and $L_1 \subseteq L_2$ then $L_2 \in S$.
  if $L_1\in S$ then there exists a finite $L_2 \subseteq L_1$ such that $L_2 \in S$.
  The language of 'all finite languages in $S$' is in RE.  (in other words, there exists a TM $M_S$ that, if $L$ is a finite language $L=\{w_1, w_2, \ldots w_k)$, and $(w_1, w_2, \ldots, w_k)$ is given to $M_S$ as an input, $M$ accepts only if $L\in S$.
  


Now back to the original question. We now that $\langle M_{loopy}\rangle\in L$ so $L(\langle M_{loopy}\rangle)\in C$. But $L(\langle M_{loopy}\rangle)=\emptyset$ since this TM never halts. This means that $\emptyset \in C$.

Now lets look on the first condition of the above theorem. ANY language $L$ satisfies $\emptyset \subseteq L$. Thus in order to satisfy condition 1, it must be that $C=RE$. However, the question states that $C\subsetneq RE$ and therefore, by the theorem, $L\notin RE$.