How many atoms of phosphorus are in 8.90 mole of copper (ii) phosphate?

Question

anonymous · Answer

Well to start, you have to get the molecular formula of Copper (II) Phosphate, which is $$Cu _{3}(PO _{4})_{2}$$
By looking at this formula we can see that there are 3 moles of Cu, 2 moles of P, and 8 moles of O. (You distribute the 2 to both the phosphorus and oxygen atoms.) Now from here, we need to use dimensional analysis.
So, we know that for every 1 mole of Cu3(PO4)2, there are 2 moles of phosphorus. We also know that 1 mole of Cu3(PO4)2 has 6.022 x 10^23 atoms (Avogadro's number).
Thus with this information, we can use dimensional analysis to convert moles into atoms.
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First you multiply all the top values, then multiply all the bottom values, and divide those two values.
You should get (53.5958 x 10^23)/2 = 26.7979 x 10^23 atoms of P