Question

How many atoms of phosphorus are in 4.50 mol of copper(ii) phosphate?

Answer 1

Answer: $54.2\times 10^{23}$ atoms

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here copper is having an oxidation state of +2 called as $Cu^{2+}$ cation and phosphate $PO_4^{3-}$ is an anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $Cu_3(PO_4)_2$.

1 mole of $Cu_3(PO_4)_2$ contains = $2\times 6.023\times 10^{23}=12.05\times 10^{23}$ atoms of phosphorous

Thus 4.50 mole of $Cu_3(PO_4)_2$ contain = $\frac{12.05\times 10^{23}}{1}\times 4.50=54.2\times 10^{23} atoms$of phosphorous

Thus there are $54.2\times 10^{23}$ atoms of phosphorous.

Answer 2

Because copper (II) cations have a charge of +2 and phosphate anions have a charge of -3, the formula of anhydrous* copper (II) phosphate is

Cu3(PO4)2

This shows that each formula unit contains 2 atoms of phosphorus. Therefore, the atoms of phosphorus in 4.8 formula units is

4.8 x Avogadro's Number (6.022x10*23), or about 2.9 X 10*24 atoms