Question

how many atoms of phosphorous are in 4.90 mol of copper(ii) phosphate? the formula for copper(ii) phosphate is Cu3(PO4)2.

Answer 1

`\boxed{5.90 * 10^(24)}`

Step-by-step explanation:

Step 1. Calculate the formula units of Cu₃(PO₄)₂

`\text{4.90 mol Cu$_(3)$(PO$_(4)$)$_(2)$} * \frac{6.022 * 10^(23) \text{ formula units}}{\text{1 mol Cu$_(3)$(PO$_(4)$)$_(2)$}}\\\\=\text{2.951 $*$ 10$^(24)$ formula units Cu$_(3)$(PO$_(4)$)$_(2)$}`

Step 2. Calculate the atoms of P

`\text{Atoms of P}\\\\=\text{2.951 $*$ 10$^(24)$ formula units Cu$_(3)$(PO$_(4)$)$_(2)$} * \frac{ \text{2 atoms P}}{\text{1 formula unit Cu$_(3)$(PO$_(4)$)$_(2)$}}\\\\= \boxed{5.90 * 10^(24) \text{ atoms P}}`