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Calculate the molar mass of Sr3(PO4)2 and the number of moles in 45.8 mg.

Chemistry

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To solve these problems, we will use the following atomic masses:

  • Strontium (Sr): 87.62 g/mol
  • Phosphorus (P): 30.97 g/mol
  • Oxygen (O): 16.00 g/mol

I. What is the molar mass (g/mol) of Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2?

  1. Determine the number of atoms of each element in the compound Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2.

    • Strontium (Sr): 3 atoms
    • Phosphorus (P): 2×1=22 \times 1 = 2 atoms (from the (PO4)2(\text{PO}_4)_2 group)
    • Oxygen (O): 2×4=82 \times 4 = 8 atoms (from the (PO4)2(\text{PO}_4)_2 group)

  2. Calculate the total mass contributed by each element.

    • Mass from Sr: 3×87.62 g/mol=262.86 g/mol3 \times 87.62 \text{ g/mol} = 262.86 \text{ g/mol}
    • Mass from P: 2×30.97 g/mol=61.94 g/mol2 \times 30.97 \text{ g/mol} = 61.94 \text{ g/mol}
    • Mass from O: 8×16.00 g/mol=128.00 g/mol8 \times 16.00 \text{ g/mol} = 128.00 \text{ g/mol}

  3. Sum the masses to find the molar mass of Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2.

    Molar mass=262.86 g/mol+61.94 g/mol+128.00 g/mol\text{Molar mass} = 262.86 \text{ g/mol} + 61.94 \text{ g/mol} + 128.00 \text{ g/mol} Molar mass=452.80 g/mol\text{Molar mass} = 452.80 \text{ g/mol}

    The molar mass of Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2 is 452.80 g/mol\boxed{\text{452.80 g/mol}}.

    II. How many moles are present in 45.8 mg?

  4. Convert the given mass from milligrams (mg) to grams (g).

    45.8 mg=45.8×10−3 g=0.0458 g45.8 \text{ mg} = 45.8 \times 10^{-3} \text{ g} = 0.0458 \text{ g}

  5. Use the molar mass calculated in Part I to find the number of moles.

    Moles=massmolar mass\text{Moles} = \frac{\text{mass}}{\text{molar mass}} Moles=0.0458 g452.80 g/mol\text{Moles} = \frac{0.0458 \text{ g}}{452.80 \text{ g/mol}} Moles≈0.000101148 mol\text{Moles} \approx 0.000101148 \text{ mol} Moles≈1.01×10−4 mol\text{Moles} \approx 1.01 \times 10^{-4} \text{ mol}

    The number of moles present is 1.01×10−4 mol\boxed{1.01 \times 10^{-4} \text{ mol}}.

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

  6. Calculate the total mass of phosphorus in one mole of Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2.

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

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

  7. Calculate the percentage composition of phosphorus using the molar mass from Part I.

    Percentage composition of P=total mass of Pmolar mass of Sr3(PO4)2×100%\text{Percentage composition of P} = \frac{\text{total mass of P}}{\text{molar mass of } \text{Sr}_3(\text{PO}_4)_2} \times 100\% Percentage composition of P=61.94 g/mol452.80 g/mol×100%\text{Percentage composition of P} = \frac{61.94 \text{ g/mol}}{452.80 \text{ g/mol}} \times 100\% Percentage composition of P≈0.136793×100%\text{Percentage composition of P} \approx 0.136793 \times 100\% Percentage composition of P≈13.68%\text{Percentage composition of P} \approx 13.68\%

    The percentage composition of phosphorus in Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2 is 13.68%\boxed{13.68\%}.

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