How many atoms (total) are in one formula unit of Ca

_{3}(PO

_{4})

_{2}? (I know that the answer is 13 b/c of the answer key, but I am having trouble doing the actual work and I come no where close to 13.)

My attempt:

The total grams of the above is 310.18 grams. I found the % of each of the elements by dividing the total grams of each element by the total grams (310.18) x 100 and got the following: (Ca - 39%), (P - 20%) and (0 - 41%).

I then found the amount of moles of each element by dividing its percent by its atomic mass and I got the following: (Ca - .9730538922), (P - .6457862448) and (O - 2.5625).

I then divided the amount of moles that I got for each element by the smallest one (the moles of phosphorus) to get the empirical formula and came up with Ca

_{2}PO

_{4}.

For my conversion from grams to atoms I calculated the number of grams in the empirical formula and came up with 175.13 g. I converted this to atoms by dividing this number by 310.18g (the original number of grams in the original molecular formula given) and multiplied it by 6.02x10

^{23} and I am getting 3.4x10

^{23}.

I have looked it over a few times and re-done the calculations and not sure if I am missing a step, doing a step completely wrong ...