Magnesium phosphate has the chemical formula Mg3(PO4)2 and has a formula mass of 262.9 g/mol. How many moles of the element phosphorus are in 117.4 g Mg3(PO4)2? Provide a numerical answer reported to three significant figures. Do NOT include units.
The Correct Answer and Explanation is :
To determine how many moles of phosphorus are present in 117.4 grams of magnesium phosphate (Mg₃(PO₄)₂), we need to follow these steps:
Step 1: Find the molar mass of Mg₃(PO₄)₂
The formula of magnesium phosphate is Mg₃(PO₄)₂, which consists of:
- 3 magnesium (Mg) atoms
- 2 phosphate (PO₄) groups
We can break it down into the molar masses of the elements:
- Magnesium (Mg) has a molar mass of approximately 24.31 g/mol.
- Phosphorus (P) has a molar mass of approximately 30.97 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
Now calculate the molar mass of Mg₃(PO₄)₂:
- Molar mass of Mg₃ = 3 × 24.31 = 72.93 g/mol
- Molar mass of PO₄ = (1 × 30.97) + (4 × 16.00) = 30.97 + 64.00 = 94.97 g/mol
- Molar mass of Mg₃(PO₄)₂ = (3 × 24.31) + (2 × 94.97) = 72.93 + 189.94 = 262.87 g/mol
(Note: The slight discrepancy between the calculated and given molar mass of 262.9 g/mol is due to rounding.)
Step 2: Determine the number of moles of Mg₃(PO₄)₂
To find the number of moles of Mg₃(PO₄)₂ in 117.4 grams, use the formula:
[
\text{moles of Mg₃(PO₄)₂} = \frac{\text{mass of Mg₃(PO₄)₂}}{\text{molar mass of Mg₃(PO₄)₂}}
]
[
\text{moles of Mg₃(PO₄)₂} = \frac{117.4 \, \text{g}}{262.9 \, \text{g/mol}} = 0.447 \, \text{mol}
]
Step 3: Determine the moles of phosphorus (P)
In the formula Mg₃(PO₄)₂, there are 2 phosphorus atoms per formula unit of Mg₃(PO₄)₂. Therefore, the moles of phosphorus is twice the moles of Mg₃(PO₄)₂:
[
\text{moles of phosphorus} = 2 \times \text{moles of Mg₃(PO₄)₂} = 2 \times 0.447 \, \text{mol} = 0.894 \, \text{mol}
]
Thus, there are 0.894 moles of phosphorus in 117.4 grams of magnesium phosphate.
Conclusion:
The number of moles of phosphorus in 117.4 grams of magnesium phosphate is 0.894.