What is the formula for copper(II) phosphate? Capitalization and punctuation count. Formula: Cu3(PO4)2. How many atoms of phosphorus are in 3.70 mol of copper(II) phosphate? Number of P atoms: 229.4.
The Correct Answer and Explanation is:
The formula for copper(II) phosphate is Cu₃(PO₄)₂.
To find the number of phosphorus (P) atoms in 3.70 mol of copper(II) phosphate, follow these steps:
- Identify the number of phosphorus atoms in one formula unit of Cu₃(PO₄)₂:
- In Cu₃(PO₄)₂, there are two phosphate ions (PO₄)²⁻.
- Each phosphate ion contains one phosphorus (P) atom.
- Therefore, for each formula unit of Cu₃(PO₄)₂, there are 2 phosphorus (P) atoms.
- Calculate the number of phosphorus atoms in 3.70 mol of copper(II) phosphate:
- 1 mol of Cu₃(PO₄)₂ contains 2 mol of phosphorus atoms because there are two P atoms per formula unit.
- Therefore, 3.70 mol of Cu₃(PO₄)₂ contains 3.70 × 2 = 7.40 mol of phosphorus atoms.
- Convert moles of phosphorus atoms to the number of atoms:
- Avogadro’s number (6.022 × 10²³) tells us how many atoms are in one mole of a substance.
- So, the number of phosphorus atoms in 7.40 mol of phosphorus is: Number of P atoms=7.40 mol×6.022×1023 atoms/mol\text{Number of P atoms} = 7.40 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol}Number of P atoms=7.40 mol×6.022×1023 atoms/mol =4.46×1024 phosphorus atoms.= 4.46 \times 10^{24} \text{ phosphorus atoms}.=4.46×1024 phosphorus atoms.
- Conclusion:
- The number of phosphorus atoms in 3.70 mol of copper(II) phosphate is 4.46 × 10²⁴.
