[KingofSpades]

I've spent a long time looking at this and cannot do it. My teacher explained it, but went too fast. Could somebody explain to me how to do this calculation, and why it is done the way it is.

The bones of an adult human skeleton have a mass of approximately 9kg. Of this, 1kg is calcium.

The calcium in bones can be assumed to be present as calcium phosphate. A phosphate ion has the formula PO4 (3-)

a.) What is the formula of calcium phosphate

- Ca3PO4(2)

b.) Estimate the percentage by mass of calcium phosphate in an adult human skeleton.




I'm not sure how to do this. I did 1/9 times 100, and this is wrong.

Could somebody explain how to do this please.

[Borek]

1 kg of calcium is present in how many kilograms of calcium phosphate?

[sci994]

m_Ca=1kg=1000g
M_Ca=40g/mol

M_Ca3(PO4)2=(3*40+2*(31+64))=215 g/mol

120g : 215g =1000g : m(calcium phosphate)

m(calcium phosphate)=1791.67g=1.79167 kg

(1.79167/9)*100%=19.91 %

[DrCMS]

Correct approach but the wrong answer; look again at the formula weight of calcium phosphate.

[sci994]

m_Ca=1kg=1000g
M_Ca=40g/mol

M_Ca3(PO4)2=(3*40+2*(31+64))=120+2*95=310g/mol

120g : 310g =1000g : m(calcium phosphate)

m(calcium phosphate)=2583.33g
(2.58333/9)*100%=28.70 %

DrCMS tnx I made a mistake XD

[vmelkon]

If by 1 Kg of Calcium, they really mean the calcium atoms only, then :

(3*40+2*(31+64)) = 310 g/mol for Ca3(PO4)2  (<---HERE IS YOUR mistake)

120g/mol for the 3 calciums in Ca3(PO4)2
190g/mol for the 2 PO4 in Ca3(PO4)2

Let's assume 1 mol.
In other words, 120g/mol * 1 mol = 120g of 3 calciums in Ca3(PO4)2
310 g/mol * 1 mol = 310 g of Ca3(PO4)2

(310 g * 1000 g) / 120g = 2583 g of Ca3(PO4)2

2.583 Kg/9.000 Kg = 28.7 %

[Borek]

If by 1 Kg of Calcium, they really mean the calcium atoms only, then :

(3*40+2*(31+64)) = 310 g/mol for Ca3(PO4)2  (<---HERE IS YOUR mistake)

120g/mol for the 3 calciums in Ca3(PO4)2
190g/mol for the 2 PO4 in Ca3(PO4)2

Let's assume 1 mol.
In other words, 120g/mol * 1 mol = 120g of 3 calciums in Ca3(PO4)2
310 g/mol * 1 mol = 310 g of Ca3(PO4)2

(310 g * 1000 g) / 120g = 2583 g of Ca3(PO4)2

2.583 Kg/9.000 Kg = 28.7 %

Not only OP already corrected the mistake, but you also gave the final answer, which is against forum rules.

[vmelkon]

Sorry about that.