What is the molar mass of Sr3(PO4)2 and how many moles are in 45.8 mg?

The compound $\text{Sr}_3(\text{PO}_4)_2$ contains:

• 3 atoms of Strontium (Sr)
• 2 atoms of Phosphorus (P) (since there are two $\text{PO}_4$ groups)
• 8 atoms of Oxygen (O) (since there are two $\text{PO}_4$ groups, and each has 4 oxygen atoms, $2 \times 4 = 8$)

$$\text{Molar mass} = (3 \times \text{atomic mass of Sr}) + (2 \times \text{atomic mass of P}) + (8 \times \text{atomic mass of O})$$ $$\text{Molar mass} = (3 \times 87.62 \text{ g/mol}) + (2 \times 30.97 \text{ g/mol}) + (8 \times 16.00 \text{ g/mol})$$ $$\text{Molar mass} = 262.86 \text{ g/mol} + 61.94 \text{ g/mol} + 128.00 \text{ g/mol}$$ $$\text{Molar mass} = 452.80 \text{ g/mol}$$

The molar mass of $\text{Sr}_3(\text{PO}_4)_2$ is $\boxed{\text{452.80 g/mol}}$.

II. How many moles are present in 45.8 mg?

$$\text{Moles} = \frac{\text{mass}}{\text{molar mass}}$$ $$\text{Moles} = \frac{0.0458 \text{ g}}{452.80 \text{ g/mol}}$$ $$\text{Moles} \approx 0.0001011 \text{ mol}$$ $$\text{Moles} \approx 1.011 \times 10^{-4} \text{ mol}$$

The number of moles present is $\boxed{1.011 \times 10^{-4} \text{ mol}}$.

III. Determination of the percentage composition of phosphorus atom in this compound.

From Part I, we know there are 2 phosphorus atoms in the compound.

$$\text{Total mass of P} = 2 \times \text{atomic mass of P}$$ $$\text{Total mass of P} = 2 \times 30.97 \text{ g/mol} = 61.94 \text{ g/mol}$$

$$\text{Percentage composition of P} = \frac{\text{total mass of P}}{\text{molar mass of } \text{Sr}_3(\text{PO}_4)_2} \times 100\%$$ $$\text{Percentage composition of P} = \frac{61.94 \text{ g/mol}}{452.80 \text{ g/mol}} \times 100\%$$ $$\text{Percentage composition of P} \approx 0.13679 \times 100\%$$ $$\text{Percentage composition of P} \approx 13.68\%$$

The percentage composition of phosphorus in $\text{Sr}_3(\text{PO}_4)_2$ is $\boxed{13.68\%}$.