Calculate the percentage of phosphorus in:
Calcium phosphate $C a_{3} (P O_{4})_{2}$

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Solution

Relative molecular mass of $C a_{3} (P O_{4})_{2}$
$= (40.07 \times 3) + (30.9 \times 2) + (16 \times 8)$
$= 120.21 + 61.8 + 128$
$= 310.01$
$310.01 g C a_{3} (P O_{4})_{2}$ contains $61.8 g P$
So, $100 g C a (H_{2} P O_{4})_{2}$ contains
$% o f p h o s p h o r o u s = \frac{2 \times 31}{310.17} \times 100$
$% o f P h o s p h o r o u s = 19.98 %$

The $%$ of $P$ in $C a (H_{2} P O_{4})_{2}$ is $19.98 %$

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Calculate the percentage of phosphorus in:Calcium phosphate Ca3(PO4)2

Relative molecular mass of Ca3(PO4)2=(40.07×3)+(30.9×2)+(16×8)=120.21+61.8+128=310.01310.01gCa3(PO4)2 contains 61.8gPSo, 100gCa(H2PO4)2 contains%ofphosphorous=2×31310.17×100%ofPhosphorous=19.98% The % of P in Ca(H2PO4)2 is 19.98%

Calculate the percentage of phosphorus in:
Calcium phosphate $C a_{3} (P O_{4})_{2}$