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The mass of oxygen in a 20.0 g sample of sodium oxide is 8.2 g. Determine the empirical and molecular formulas of sodium oxide.

Chemistry

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2 parts
a

NaO\text{NaO}

b

Na2O2\text{Na}_2\text{O}_2

Step-by-step solution

7 steps

Here's the solution for question 2:

Part a· 4 steps

Determine the empirical formula of the compound.

  1. Calculate the mass of sodium (Na) in the compound.

    Mass of compound = 20.0 g20.0 \text{ g}
    Mass of oxygen (O) = 8.2 g8.2 \text{ g}
    Mass of sodium (Na) = Mass of compound - Mass of oxygen

    Mass of Na=20.0 g−8.2 g=11.8 g\text{Mass of Na} = 20.0 \text{ g} - 8.2 \text{ g} = 11.8 \text{ g}

  2. Calculate the number of moles for each element.

    Relative atomic mass of Na = 23.0
    Relative atomic mass of O = 16.3
    Moles of Na = Mass of NaRelative atomic mass of Na=11.8 g23.0 g/mol≈0.513 mol\frac{\text{Mass of Na}}{\text{Relative atomic mass of Na}} = \frac{11.8 \text{ g}}{23.0 \text{ g/mol}} \approx 0.513 \text{ mol}
    Moles of O = Mass of ORelative atomic mass of O=8.2 g16.3 g/mol≈0.503 mol\frac{\text{Mass of O}}{\text{Relative atomic mass of O}} = \frac{8.2 \text{ g}}{16.3 \text{ g/mol}} \approx 0.503 \text{ mol}

  3. Determine the simplest mole ratio by dividing by the smallest number of moles.

    Smallest number of moles is 0.503 mol0.503 \text{ mol} (for oxygen).
    Ratio for Na = 0.5130.503≈1.02\frac{0.513}{0.503} \approx 1.02
    Ratio for O = 0.5030.503=1\frac{0.503}{0.503} = 1
    The ratio of Na:O is approximately 1:1.

  4. Write the empirical formula.

    The empirical formula is NaO\boxed{\text{NaO}}.

Part b

Na2O2\text{Na}_2\text{O}_2

Determine the molecular formula of the compound.

  1. Calculate the empirical formula mass.

    Empirical formula mass of NaO = (1 ×\times Ar of Na) + (1 ×\times Ar of O)

    Empirical formula mass=(1×23.0)+(1×16.3)=23.0+16.3=39.3\text{Empirical formula mass} = (1 \times 23.0) + (1 \times 16.3) = 23.0 + 16.3 = 39.3

  2. Determine the factor 'n' by which the empirical formula must be multiplied.

    Relative formula mass of the compound = 78

    n=Relative formula massEmpirical formula mass=7839.3≈1.98≈2n = \frac{\text{Relative formula mass}}{\text{Empirical formula mass}} = \frac{78}{39.3} \approx 1.98 \approx 2

  3. Write the molecular formula.

    Multiply the subscripts in the empirical formula by 'n'.
    Molecular formula = (NaO)2=Na2O2(\text{NaO})_2 = \text{Na}_2\text{O}_2
    The molecular formula of the compound is Na2O2\boxed{\text{Na}_2\text{O}_2}.

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