We have such a claim:
If $A$ is an ideal of C*-algebra $B$, then there is a unique morphism $f\colon B\to M(A)$ such that $f$ is identity on $A$, here $M(A)$ is the multiplier algebra of $A$.
Now if $A$ is only a C*-subalgebra of $B$, does it also true? How about a hereditary C*-subalgebra?
No. Take $B=C[0,1]$, and $A = \mathbb{C}\cdot 1$, where $1$ denotes the constant function on $[0,1]$. Then $A$ is a subalgebra of $B$ and $M(A) = A$. Now, for any $x\in [0,1]$ the evaluation map $$ f\mapsto f(x)\cdot 1 $$ is a map from $B \to M(A)$ that is identity on $A$.
I am not sure about hereditary sub-algebras though.