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. In this compound, copper exists in a +2 oxidation state, and phosphate is the anion with the formula PO₄³⁻. The structure shows that there are 3 copper ions (Cu²⁺) and 2 phosphate ions (PO₄³⁻) per formula unit.
To determine how many phosphorus (P) atoms are present in 3.70 mol of copper(II) phosphate (Cu₃(PO₄)₂), follow these steps:
Step 1: Understanding the formula
In the formula Cu₃(PO₄)₂, there are 2 phosphate ions (PO₄) per formula unit. Each phosphate ion contains 1 phosphorus (P) atom. Therefore, for every formula unit of Cu₃(PO₄)₂, there are 2 phosphorus atoms.
Step 2: Find the number of phosphorus atoms per mole of copper(II) phosphate
Since there are 2 phosphorus atoms in 1 formula unit of Cu₃(PO₄)₂, there are also 2 phosphorus atoms in 1 mole of Cu₃(PO₄)₂.
Step 3: Use Avogadro’s number to calculate the number of atoms
Avogadro’s number tells us that 1 mole of a substance contains 6.022 × 10²³ entities (atoms, molecules, etc.). Therefore, 1 mole of Cu₃(PO₄)₂ contains 2 × 6.022 × 10²³ phosphorus atoms.
Step 4: Calculate the number of phosphorus atoms in 3.70 mol of copper(II) phosphate
The total number of phosphorus atoms in 3.70 mol of Cu₃(PO₄)₂ is:3.70 mol×(2×6.022×1023 atoms/mol)=4.45×1024 P atoms3.70 \, \text{mol} \times \left(2 \times 6.022 \times 10^{23} \, \text{atoms/mol}\right) = 4.45 \times 10^{24} \, \text{P atoms}3.70mol×(2×6.022×1023atoms/mol)=4.45×1024P atoms
Thus, the number of phosphorus atoms in 3.70 mol of copper(II) phosphate is 4.45 × 10²⁴ phosphorus atoms.
Final Answer:
The number of phosphorus atoms in 3.70 mol of copper(II) phosphate is 4.45 × 10²⁴ P atoms.
