Question

Calcium phosphate is added to some foods, such as yogurt, to boost the calcium content and is also used as an anticaking agent.

A) How many formula units are in 75.5g of calcium phosphate?

B) HOw many phosphate ions are present in this sample?

Answer 1

A)

Molar mass of Ca3(PO4)2,

MM = 3*MM(Ca) + 2*MM(P) + 8*MM(O)

= 3*40.08 + 2*30.97 + 8*16.0

= 310.18 g/mol

mass(Ca3(PO4)2)= 75.5 g

use:

number of mol of Ca3(PO4)2,

n = mass of Ca3(PO4)2/molar mass of Ca3(PO4)2

=(75.5 g)/(3.102*10^2 g/mol)

= 0.2434 mol

use:

number of molecules = number of mol * Avogadro’s number

number of molecules = 0.2434 * 6.022*10^23 molecules

number of molecules = 1.466*10^23 molecules

Answer: 1.47*10^23

B)

1 Ca3(PO4)2 has 2 PO43- ions

So,

number of PO43- = 2*number of Ca3(PO4)2

= 2*1.47*10^23

= 2.94*10^23

Answer: 2.94*10^23