Conservation of angular momentum and spin I have a point mass particle r from a axis of rotation 

however the molecule is two atoms spinning about it's own center of mass 

If I wanted to calculate total angular momentum of my system should I not include the spin momentum as well 
While I understand that they cancel to zero can't they interact with translational momentum to alter the balance? 
 A: Yes, total angular momentum is the sum of the orbital angular momentum and the spin angular momentum. However, spin is not the rotation of the molecule around its own axis. It is a little confusing in that you first talk about a point mass particle and then about a molecule consisting of two atoms. To avoid confusion, I'll try to give you a general understanding.
In any system one can compute the orbital angular momentum by computing
$$ \mathbf{L} = \sum_n \mathbf{r}_n \times \mathbf{p}_n $$
for all the particles/point masses in the system. Here $\mathbf{p}_n$ represents their momentum vectors and $\mathbf{r}_n$ are their position vectors relative to some common origin. This means that in general the orbital angular momentum depends on the definition of the origin.
Spin is associated with an internal degree of freedom of the particles. For fermions, for instance, one can have $\mathbf{S}=\pm \hbar/2 \hat{z}$ (assuming we measure it along the $z$-direction). It is important to note that spin does not correspond to a physical rotation of the particle or system of particles. It is an intrinsic property of the particles.
The total angular momentum is now the sum of the orbital angular momentum and the spin angular momentum
$$ \mathbf{J} = \mathbf{L} + \mathbf{S} . $$
Conservation of angular momentum applies to the total angular momentum and not to the orbital angular momentum or the spin angular momentum separately. Hope this clarifies the matter and answers you question.
