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Chemistry

Calculate percentage composition of the elements in calcium phosphate Ca3(PO4)2.

Stoichiometry

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Answer

Relative molecular mass of Calcium phosphate, Ca3(PO4)2
     = 3 x 40 + 2(31 + 4 x 16)
     = 120 + 2(31 + 64)
     = 120 + 190
     = 310 amu

310 g of Ca3(PO4)2 contains 120 g of Calcium

∴ 100 g of Ca3(PO4)2 contains 120×100310\dfrac{120 \times 100}{310} g of Calcium

= 12000310\dfrac{12000}{310} = 38.70 % of Calcium.

310 g of Ca3(PO4)2 contains 62 g of Phosphorus

∴ 100 g of Ca3(PO4)2 contains 62×100310\dfrac{62 \times 100}{310} g of Phosphorus

= 6200310\dfrac{6200}{310} = 20 % of Phosphorus.

310 g of Ca3(PO4)2 contains 128 g of Oxygen.

∴ 100 g of Ca3(PO4)2 contains 128×100310\dfrac{128 \times 100}{310} g of Oxygen.

= 12800310\dfrac{12800}{310} = 41.3 % of Oxygen.

∴ In Ca3(PO4)2 : Ca = 38.70%, P = 20% and O = 41.3%

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