Dear Pete, given an arbitrary field $k$, the scheme $X=SpecA$ where $ A=k[X,Y, Z]/(Z^2-XY)$ seems to be an example of what you wish. The divisor class group is of order 2, generated by  a ruling of the quadratic cone $X$ (this is proved in Hartshorne's Book-whose-title-need-not-be-spelled-out, page 134).The non factoriality essentially comes from the equation $z^2=xy$, which  explicitly tells us that $z^2$ has two factorizations (a short rigorous proof could use a graduation on the relevant ring).See also Hartshorne, page 142 for nullity of Picard group.

 Normality of $A$ is a general fact for factorial rings to which a root of a square-free element is added : Matsumura , Commutative Ring Theory, page 65.

This example is the same as Steve D.'s, with another presentation: if you map $\mathbb A^2$ to $\mathbb A^3$ by $x=u^2, y=uv, z=v^2$ you'll see the correspondence between the two points of view. I don't know Kang's nor Nakajima's results-which are cerainly quite interesting- but for the elementary example discussed, they can be bypassed.