Chemistry, asked by Anonymous, 9 months ago

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Please help me!!!!!

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Answers

Answered by Anonymous

11

Given :

$\sf{Calcium\:hydrogen\:phosphate}\:\bf{Ca(H_2PO_4)_2}$

$\sf{Calcium\:phosphate}\:\bf{Ca_3(PO_4)_2}$

To Find :

⟶ Percentage composition of P in both componds.

Solution :

➢ For determining the % composition of each element in a compound :

$\bigstar\:\underline{\boxed{\bf{Mass\:\%\:of\:an\:element=\dfrac{mass\:of\:the\:element\:in\:the\:compound\times 100}{molar\:mass\:of\:the\:compound}}}}$

✴ Atomic weight :

Ca = 40 g/mol
H = 1 g/mol
P = 31 g/mol
O = 16 g/mol

✴ Molecular weight :

(I) Calcium hydrogen phosphate :

$\leadsto\sf\:M_w=Ca+4(H)+2(P)+8(O)$

$\leadsto\sf\:M_w=40+4(1)+2(31)+8(16)$

$\leadsto\sf\:M_w=40+4+62+128$

$\leadsto\bf\:M_w=234\:gmol^{-1}$

(II) Calcium phosphate :

$\leadsto\sf\:M_w=3(Ca)+2(P)+8(O)$

$\leadsto\sf\:M_w=3(40)+2(31)+8(16)$

$\leadsto\sf\:M_w=120+62+128$

$\leadsto\bf\:M_w=310\:gmol^{-1}$

✴ % composition of Phosphorus :

(I) Calcium hydrogen phosphate :

$\mapsto\sf\:\% P=\dfrac{2(P)\times 100}{M_w}$

$\mapsto\sf\:\% P=\dfrac{2(31)\times 100}{234}$

$\mapsto\sf\:\% P=\dfrac{6200}{234}$

$\mapsto\underline{\boxed{\bf{\% P=26.5\%}}}$

(II) Calcium phosphate :

$\mapsto\sf\:\% P=\dfrac{2(P)\times 100}{M_w}$

$\mapsto\sf\:\% P=\dfrac{2(31)\times 100}{310}$

$\mapsto\sf\:\% P=\dfrac{6200}{310}$

$\mapsto\underline{\boxed{\bf{\% P=20\%}}}$

Anonymous: Great!

mddilshad11ab: Wonderful explaination ✔️ bhai

Anonymous: Thank you :)

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