How many calcium ions are in a sample of 7.36 grams of calcium phosphate?

5.72 x 10^22. There are 5.72 x 10^22 Ca^(2+) ions in 7.36 g Ca3(PO4)2.

a)Moles of Ca3(PO4)2

MM of Ca3(PO4)2 = [3x40.08 + 2(30.97 + 4x16.00)] g/mol

= (120.24 + 2x94.97) g/mol = (120.24 + 189.94) g/mol = 310.18 g/mol

Moles of Ca3(PO4)2

= 7.36 g Ca3(PO4)2 x [1 mol Ca3(PO4)2]/[310.18 g Ca3(PO4)2]

= 0.023 73 mol Ca3(PO4)2

b)Formula units (FU) of Ca3(PO4)2

Formula units of Ca3(PO4)2

= 0.023 73 mol Ca3(PO4)2 x [6.022 x 10^23 FU Ca3(PO4)2]/[1 mol Ca3(PO4)2] = 1.429 x 10^22 FU Ca3(PO4)2

c)Ions of Ca

No. of Ca ions

= 1.429 x 10^22 FU Ca3(PO4)2 x [4 Ca^(2+) ions]/[1 FU Ca3(PO4)2]

= 5.72 x 10^22 Ca^(2+) ions

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