Question

A 1.2516 gram sample of a mixture of CaCO₃ and Na₂SO₄ was analyzed by dissolving the sample and completely precipitating the Ca₂+ as CaC₂O₄. The CaC₂O₄ was dissolved in sulfuric acid and the resulting H₂C₂O₄ was titrated with a standard KMnO₄ solution. The titration required 35.62 milliliters of 0.1092 M KMnO₄. Calculate the number of moles of H₂C₂O₄ that reacted with the KMnO₄. Calculate the number of moles of CaCO₃ in the original sample. Calculate the percentage by weight of CaCO₃ in the original sample.

Answer 1

The number of moles of H₂C₂O₄ reacted with KMnO₄ is calculated to be 0.001555 mol. The number of moles of CaCO₃ in the original sample is the same, 0.001555 mol. The percentage by weight of CaCO₃ in the original sample is found to be 12.46%.

Step-by-step explanation:

To calculate the number of moles of H₂C₂O₄ that reacted with the KMnO₄, we use the volume and molarity of the KMnO₄ solution:

# mol KMnO₄ = (0.03562 L)(0.1092 M) = 0.003887 mol KMnO₄

From the balanced chemical equation for the titration, we know that 2 moles of H₂C₂O₄ react with 5 moles of KMnO₄. Since we have 0.003887 mol of KMnO₄, we can find the moles of H₂C₂O₄:

# mol H₂C₂O₄ = (0.003887 mol KMnO₄) × (2 mol H₂C₂O₄ / 5 mol KMnO₄) = 0.001555 mol H₂C₂O₄

To find the number of moles of CaCO₃ in the original sample, we look at the reaction where CaC₂O₄ dissolves in sulfuric acid to form H₂C₂O₄. There is a 1:1 molar ratio between CaC₂O₄ and H₂C₂O₄, thus the moles of CaCO₃ are also 0.001555 mol.

To calculate the percentage by weight of CaCO₃ in the original sample:

% w/w CaCO₃ = (0.001555 mol CaCO₃ × 100.087 g/mol CaCO₃) / 1.2516 g sample × 100% = 12.46%