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

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Solution

Relative molecular mass of $C a (H_{2} P O_{4})_{2}$
$= 40.07 + (1 \times 4) + (30.9 \times 2) + (16 \times 8)$
$= 40.07 + 4 + 61.8 + 128$
$= 233.87$
$233.87 g C a (H_{2} P O_{4})_{2}$ contains $61.8 g P$
So, $100 g C a (H_{2} P O_{4})_{2}$ contains
$\frac{100 \times 61.8}{233.87} = 26.42 g$
The $%$ of P in $C a (H_{2} P O_{4})_{2}$ is $26.42 %$

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Calculate the percentage of phosphorus in:Calcium hydrogen phosphate Ca(H2PO4)2

Relative molecular mass of Ca(H2PO4)2=40.07+(1×4)+(30.9×2)+(16×8)=40.07+4+61.8+128=233.87233.87gCa(H2PO4)2 contains 61.8gPSo, 100gCa(H2PO4)2 contains100×61.8233.87=26.42gThe % of P in Ca(H2PO4)2 is 26.42%

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