Let $X$ be a variety (irreducible, by Hartshorne's definition).
I know that when $X$ is non-singular, the Picard group $\text{Pic}(X)$ coincides with the divisor class group $\text{Cl}(X)$. This can be seen by considering the following group homomorphism: \begin{align*} \gamma:\text{CaDiv}(X)&\to \text{Div}(X)\\ \{(U_i,f_i)\}_i&\mapsto \sum_{Y}n_YY \end{align*} where $Y$ runs through all codimension $1$ subvarieties of $X$ and $n_Y$ is the order of $f_i$ with respect to the codimension $1$ subvariety $U_i\cap Y\subset U_i$ (for some $i$ with $U_i\cap Y\neq \emptyset$, of course).
Since $X$ is non-singular, every Weil divisor can be represented as a Cartier divisor, therefore $\gamma$ is surjective. Furthermore, principal divisors are preserved under $\gamma$, so $\gamma$ induces an isomorphism $\text{Pic}(X)\simeq \text{Cl}(X)$.
But what happens in non-singular varieties?
Take for example Hartshothe's example $6.11.3$, where $X=Z(xy-z^2)\subset \mathbb{A}^3$, where $Y=Z(x,z)$ defines a Weil divisor but not a Cartier divisor (because it is not locally principle around the origin).
In this case, clearly $\gamma$ is not surjective. But we still have an inclusion $\text{Pic}(X)\hookrightarrow \text{Cl}(X)$, don't we? Can we say in general that $\text{Pic}(X)$ is a subgroup of $\text{Cl}(X)$?