[jeremyfischer76]

As a student physician, I want to determine the equivalent dosage of Strontium citrate needed to obtain the same dose of elemental strontium as is obtained in the studied form of strontium, which is strontium ranelate. What I did so far was look up the molecular mass of strontium ranelate, 513.491g/mol, and calculate the percentage of that as Strontium (34.127%), which gives 0.68g elemental strontium with 2g/day of strontium ranelate, the studied dosing strategy.

Then I planned to determine a. the molecular mass of strontium citrate, b. the percentage of that as elemental strontium, and c. the amount of strontium citrate required to yield 0.68g elemental strontium. However, I was confused by the information I found about the molecular mass on Strontium citrate. One site ( http://www.chemblink.com/products/813-97-8.htm ) lists the chemical formula as C6H6O7Sr, gives a molecular weight of 277.73, and yet also calls it tristrontium dicitrate. When I add up the atomic weights of the individual atoms in the C6H6O7Sr, compared with that of tristrontium dicitrate, they are obviously very different. Furthermore, although Wikipedia doesn't have strontium citrate as a listing, they do have Calcium citrate, and list the formula as Ca3(C6H5O7)2, so there is one less hydrogen per citrate. It's been awhile since chemistry classes, and my memory fails me here, but I'm wondering if aqueous form verse dry form has something to do with the variance in numbers of hydrogens listed. I'm also guessing that the C6H6O7Sr versioin of the formula for Strontium citrate couldn't possibly give proper ratios, since there would need to be 3 strontiums per citrate ion to balance the polarities.

Anyone skilled in this area who could help me solve the problem?

Thanks,

Jeremy

[Borek]

Simplest approach calls for (C₆H₅O₇)₂Sr₃.

Although I have no idea if it contains (or not) crystalline water, which will change the molar mass slightly.