First concerning your question: most people use $Pic(C)$$\operatorname{Pic}(C)$ for Cartier divisors. I interpret your question as follows: you want to determine whether every divisor on $C$ is linear equivalent with a Cartier divisor (this means factorial).
Now if $C$ is smooth this is true and I think this is also true if $\dim C-\dim C_{sing}>3$.
If $\dim C_{sing}\geq \dim C-3$ things are much more complicated. A necessary condition for being factorial is (roughly said) that the rank of $H^{N-2,N-2}(C,\mathbb{C}) \cap H^{2N-4}(C,\mathbb{Z})$ equals one. (If the MHS on H^{2N-2}$ does not have pure weight you have to be a bit more careful here.)
If $\Sigma=C_{sing}$ then you have an exact sequence $$H^{2N-5}(C)\to H^{2N-5}(C\setminus \Sigma)\to H^{2N-4}_\Sigma(C)\to H^{2N-4}(C).$$
If I remember correctly there should be a copy of $H^2(\Sigma) $ inside $H^{2N-4}_{\Sigma} (C)$. If this is all of $H^{2N-4}_\Sigma$ then you can relatively easily show that $H^{2N-4}(C)$ is one-dimensional and hence each divisor on $C$ is homologically equivalent to a Cartier divisor.
If $H^{2N-4}_\Sigma(C)$ is bigger then $H^2(\Sigma)$ things are getting complicated.
In the case that $\dim \Sigma=0$, i.e., $C$ has isolated singularities then the only interesting case is $N=4$. Now $H^4_\Sigma$ is the part of the cohomology of the Milnor fiber that is invariant under the monodromy. This can be calculated using Singular.
In some case you can actually calculate the cokernel $K$ of $H^3(C\setminus \Sigma)\to H^4_\Sigma(C)$. For this see e.g., Dimca's paper on Betti numbers and defects of linear systems. It turns out that $K$ is the primitive cohomology group $H^4(C,\mathbb{C})$
The formula Francesco mentioned is a special case of Dimca's approach.
Grooten-Steenbrink and Hulek-K. gave similar formula as Dimca for certain classes of nonisolated singularities.
