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Identify the atomic masses of each element.

Chemistry

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3 parts
a

452.80 g/mol452.80 \text{ g/mol}

b

1.01×10−4 mol1.01 \times 10^{-4} \text{ mol}

c

13.68%13.68\%

Step-by-step solution

6 steps

Here's how to solve question 11:

Part a· 2 steps

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

  1. Identify the atomic masses of each element.

    • Strontium (Sr): 87.62 g/mol87.62 \text{ g/mol}
    • Phosphorus (P): 30.97 g/mol30.97 \text{ g/mol}
    • Oxygen (O): 16.00 g/mol16.00 \text{ g/mol}

  2. Calculate the molar mass of Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2.

    The formula indicates 3 Sr atoms, 2 P atoms (from 2×P2 \times \text{P} in (PO4)2(\text{PO}_4)_2), and 8 O atoms (from 2×4 O2 \times 4 \text{ O} in (PO4)2(\text{PO}_4)_2).

    Molar mass=(3×87.62)+(2×30.97)+(8×16.00)\text{Molar mass} = (3 \times 87.62) + (2 \times 30.97) + (8 \times 16.00) Molar mass=262.86+61.94+128.00\text{Molar mass} = 262.86 + 61.94 + 128.00 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{452.80 \text{ g/mol}}.

Part b

1.01×10−4 mol1.01 \times 10^{-4} \text{ mol}

How many moles are present in there?

  1. 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}

  2. Calculate the number of moles using the mass and the molar mass from part a.

    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}

    Rounding to three significant figures:
    The number of moles present is 1.01×10−4 mol\boxed{1.01 \times 10^{-4} \text{ mol}}.

Part c

13.68%13.68\%

Determination the percentage composition of phosphorus atom in this compound.

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

    There are 2 phosphorus atoms in the formula Sr3(PO4)2\text{Sr}_3(\text{PO}_4)_2.

    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}

  2. Calculate the percentage composition of phosphorus.

    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 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 is 13.68%\boxed{13.68\%}.

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