Calculation of mass from mmol/L Sorry I understand this looks like a "homework question", (I'm actually a 46yr old dentist looking to see what amount of calcium phosphate should be effective in a proprietary toothpaste!). However, if the actual question below can not be worked out can you let me know what the process of the calculation would be to determine mass of a salt to add to gain a mmol/L concentration?

Original Question: 
I wonder if you can help on this? I want to calculate how much calcium phosphate to add to 1g of toothpaste to have between 18mmol/L to 1.5mmol/L of calcium and phosphate on the tooth surface? Also if you are able to work this out can you let me know what assumptions you have made in this calculation?
There are several calcium phosphates.  I'll assume you mean monocalcium posphate, $\ce{Ca(H2PO4)2}$, because the others are not nearly soluble enough in water to allow even a 1.5 mmol/L solution.  (I assume the toothbase's most abundant ingredient is water.)

The formula weight is 234.  (I won't bother with many decimals because your range of molarities, 1.5 to 18 mmol/L, is so wide.)  I got this number from 
http://en.wikipedia.org/wiki/Monocalcium_phosphate although it is easy to compute from scratch by adding the atomic weights for 1 Ca atom, 4 H atoms etc.

To calculate how much  of this compound you need given the molarity, ideally you should be looking at a known volume of paste.  Instead you have a known weight, 1g.  If you know the density of the paste, you can compute the volume by dividing 1g by the density.  Or you can just use an approximation.  I'll assume that 1g = 0.9mL (this will be a good approximation).  From now on I will assume volume = 0.9mL.

Multiplying 1.5mmol/L by 0.9mL gives you 1.35 micro moles (1.35$\times 10^{-6}$ moles).  Since 1 mole is 234 grams (formula weight), this means 316 micrograms or 0.315 mg.  The same calculation with 18mmol/L gives 3.79 mg. 

So your answer (discarding  unwarranted decimals) is: between 0.31 and 3.8 mg .

I hope if your proprietary formulation is commercially successful you will make a generous contribution to the organization behind stackexchange!