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Difficulty: Very hardPercentage Composition and Percentage Purity Calculations

A 10.00 g10.00\text{ g} sample of impure hydrated iron(II) tetraoxosulfate(VI), FeSO4⋅7H2O\text{FeSO}_4 \cdot 7\text{H}_2\text{O}, was dissolved in acidic medium and titrated against 0.050 mol dm−30.050\text{ mol dm}^{-3} potassium tetraoxomanganate(VII) solution. If exactly 40.00 cm340.00\text{ cm}^3 of the KMnO4\text{KMnO}_4 solution was required for complete oxidation of the Fe2+\text{Fe}^{2+} ions, what is the percentage purity of the hydrated salt sample? ([H=1, O=16, S=32, Fe=56][\text{H}=1,\text{ O}=16,\text{ S}=32,\text{ Fe}=56])

  1. $27.80\%Answer
  2. B
    $15.20\%
  3. C
    $5.56\%
  4. D
    $5.60\%

Answer

The percentage purity of the hydrated salt sample is 27.80%27.80\%.
The correct answer is 27.80%27.80\%. Calculating the moles of potassium tetraoxomanganate(VII) used (0.050×0.04000=0.0020 mol0.050 \times 0.04000 = 0.0020\text{ mol}) and applying the redox stoichiometry ratio (5 Fe2+:1 MnO4−5\text{ Fe}^{2+} : 1\text{ MnO}_4^-) gives 0.010 mol0.010\text{ mol} of pure FeSO4⋅7H2O\text{FeSO}_4 \cdot 7\text{H}_2\text{O}. Multiplying by its molar mass (278 g mol−1278\text{ g mol}^{-1}) gives 2.78 g2.78\text{ g} of pure salt, which represents 27.80%27.80\% of the 10.00 g10.00\text{ g} impure sample.

Step-by-Step Solution

1
Calculate the moles of KMnO4\text{KMnO}_4 used in the titration.
Moles of KMnO4=Molarity×Volume in dm3=0.050 mol dm−3×40.001000 dm3=0.0020 mol\text{Moles of } \text{KMnO}_4 = \text{Molarity} \times \text{Volume in dm}^3 = 0.050\text{ mol dm}^{-3} \times \frac{40.00}{1000}\text{ dm}^3 = 0.0020\text{ mol}.
Molarity and volume yield the amount of oxidizing agent delivered at the endpoint.
2
Determine the moles of Fe2+\text{Fe}^{2+} (and thus pure FeSO4⋅7H2O\text{FeSO}_4 \cdot 7\text{H}_2\text{O}) present.
From the redox ionic equation 5Fe2++MnO4−+8H+→5Fe3++Mn2++4H2O5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}, the mole ratio is 5 mol Fe2+:1 mol MnO4−5\text{ mol Fe}^{2+} : 1\text{ mol MnO}_4^-. Therefore, moles of Fe2+=5×0.0020 mol=0.010 mol\text{moles of Fe}^{2+} = 5 \times 0.0020\text{ mol} = 0.010\text{ mol}.
Stoichiometry of the redox reaction establishes the mole relationship between reactant species.
3
Calculate the molar mass of hydrated iron(II) tetraoxosulfate(VI), FeSO4⋅7H2O\text{FeSO}_4 \cdot 7\text{H}_2\text{O}.
Molar mass=56+32+(4×16)+7×(2×1+16)=56+32+64+126=278 g mol−1\text{Molar mass} = 56 + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = 56 + 32 + 64 + 126 = 278\text{ g mol}^{-1}.
The entire hydrated formula mass must be used to convert moles of pure salt to mass.
4
Calculate mass of pure hydrated salt and the percentage purity.
Mass of pure salt=0.010 mol×278 g mol−1=2.78 g\text{Mass of pure salt} = 0.010\text{ mol} \times 278\text{ g mol}^{-1} = 2.78\text{ g}.
Percentage Purity=(Mass of Pure SaltTotal Impure Sample Mass)×100%=(2.78 g10.00 g)×100%=27.80%\text{Percentage Purity} = \left( \frac{\text{Mass of Pure Salt}}{\text{Total Impure Sample Mass}} \right) \times 100\% = \left( \frac{2.78\text{ g}}{10.00\text{ g}} \right) \times 100\% = 27.80\%.
Percentage purity expresses the mass fraction of pure compound relative to total sample mass.

Key Concept

Redox Titration Stoichiometry and Percentage Purity
Estimated Time:2m 30s
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Topics

Percentage Composition and Percentage Purity Calculations

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