As the comment of Andreas Thom indicated here, a separable $C^\star$ algebra $A$ can not contain a $Z^\star$ algebra.(A $Z^\star$ algebra is a $C^\star$ algebra which all elements are zero divisor). So separability is an obstruction for $A$ to contain a $Z^\star$ algebra.
Now we ask:
Is it the only obstruction? What type of other obstructions can be introduced? In particular is it true that every non separable algebra contains a $Z^\star$ algebra?
$A$ contains a $Z^*$ $C^*$-subalgebra iff it contains an uncountable family $\{ a_i \}$ of nonzero positive mutually orthogonal elements. If it has such a family, then $C^*(\{ a_i \})$ is $Z^*$. If it does not have such a family and $B$ is a $C^*$-subalgebra, then any maximal family $\{ a_n \}$ of n.p.m.o. elements in $B$ is at most countable and hence one can define $a := \sum_n 2^{-n}\| a_n \|^{-1} a_n \in B$, which is not a zero divisor. In particular, any $\mathrm{II}_1$ factor does not contain a $Z^*$ $C^*$-subalgebra.
