how many atoms are in 8.70 mol of copper (II) Phosphate

Question

anonymous · Answer

The formula of copper (II) phosphate is $\sf Cu_3(PO_4)_2$.

As you know, 1 mole of any substance consists of $\sf 6.022×10^{23}$ molecules. From here, you can calculate number of $\sf Cu_3(PO_4)_2$ molecules in 8.70 moles.
              $8.70\times 6.022 \times10^{23}\approx 5.24\times 10^{24}$ molecules

Next, from the formula $\sf Cu_3(PO_4)_2$, you can easily find the number of atoms:
$3+2\times (1+4)=13$.

Therefore, the number of atoms is  $\sf 13\times 5.24\times 10^{24}\approx 6.81\times 10^{25}$ atoms