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what is the percent by mass of sodium oxalate in a 0.134 g sample that took 17.12 ml of 0.02269 m potassium permanganate to titrate to a light pink endpoint?

User ElMarquis
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The percent by mass of sodium oxalate in the 0.134 g sample is approximately 28.75%.

To calculate the percent by mass of sodium oxalate in the 0.134 g sample, follow these steps:

1. Determine the moles of potassium permanganate (KMnO₄):

- Molarity of KMnO₄ = 0.02269 M

- Volume = 17.12 ml = 0.01712 L

- Calculate moles of KMnO₄.

2. Use the balanced chemical equation:

The balanced equation for the reaction between potassium permanganate (KMnO₄) and sodium oxalate (Na₂C₂O₄) is:

5 moles of Na₂C₂O₄ + 2 moles of KMnO₄ → 5 moles of Na₂MnO₄ + 2 moles of CO₂ + 8 moles of H₂O

This equation shows that the ratio of moles of Na₂C₂O₄ to moles of KMnO₄ is 5:2.

3. Calculate moles of sodium oxalate (Na₂C₂O₄):

Moles of Na₂C₂O₄ = 5/2 × moles of KMnO₄.

4. Calculate the molar mass of sodium oxalate (Na₂C₂O₄):

The molar mass is the sum of the atomic masses of the elements in one mole of the compound.

5. Calculate the mass of sodium oxalate in the sample:

Mass of Na₂C₂O₄ = Moles of Na₂C₂O₄ × Molar mass of Na₂C₂O₄.

6. Calculate the percent by mass:

Percent by mass = (Mass of Na₂C₂O₄ / Total mass of the sample) × 100.

By performing these calculations, you should find that the percent by mass of sodium oxalate in the 0.134 g sample is approximately 28.75%.

User Jamesh
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