Question

Organic matter in a 0.95 g sample of burn ointment was removed by gravimetric burning method. A sparingly soluble ZnC2O4 precipitate was formed when the remaining solid ZnO residue was treated with (NH4)2C2O4 after dissolving in acid. When the solid was filtered, washed and dissolved again in dilute acid, 38.00 mL was spent as a result of titration of the released H2C2O4 with 0.02MKMnO4. Calculate the percentage of ZnO in the sample and write the necessary reactions

Answer 1

The percentage of ZnO in the burn ointment sample was calculated to be 25.9% after performing a series of chemical reactions and titration to determine the amount of ZnC2O4, which corresponds stoichiometrically to the original amount of ZnO.

Step-by-step explanation:

To calculate the percentage of ZnO present in the burn ointment sample following gravimetric analysis and titration, we need first to establish the stoichiometry of the reaction between oxalic acid and potassium permanganate, as it will be directly related to the amount of oxalic acid (and hence the ZnC2O4) present in our sample. The balanced chemical equation for the titration reaction is:

5 H2C2O4 + 2 KMnO4 + 6 H2SO4 → 10 CO2 + 2 MnSO4 + 8 H2O + K2SO4

From the equation, we see that 2 moles of KMnO4 react with 5 moles of H2C2O4. Using the molarity and volume of KMnO4 used in the titration, we can find the moles of H2C2O4, and subsequently, the moles of ZnC2O4 and ZnO.

Moles of KMnO4 = 0.02 M * 0.038 L = 7.6 x 10^-4 moles
Based on stoichiometry, Moles of H2C2O4 = (5/2) * Moles of KMnO4 = 1.9 x 10^-3 moles

To find the mass of ZnC2O4:

Moles of ZnC2O4 = Moles of H2C2O4 (they are in 1:1 ratio because ZnC2O4 decomposes to give H2C2O4)
Mass of ZnC2O4 = Moles of ZnC2O4 * Molar mass of ZnC2O4
Molar mass of ZnC2O4 = 65.4 + 2(12.01) + 4(16.00) = 129.42 g/mol
Mass of ZnC2O4 = 1.9 x 10^-3 moles * 129.42 g/mol = 0.246 g

Since ZnO converts to ZnC2O4 on a 1:1 molar basis, the mass of ZnO in the original sample is the same as the mass of ZnC2O4 determined.

Percentage of ZnO = (Mass of ZnO / Mass of the sample) * 100
Percentage of ZnO = (0.246 g / 0.95 g) * 100 = 25.9%

Answer 2

To find the percentage of ZnO in the sample, we use the gravimetric method and the subsequent titration results to calculate the mass of ZnC₂O₄ formed, which represents the mass of ZnO. By comparing this to the original mass of the sample, we determine that the percentage of ZnO is 23.88%.

Step-by-step explanation:

To calculate the percentage of ZnO in the burn ointment sample, we first need to understand that the reaction of ZnO with (NH₄)₂C₂O₄ results in the formation of ZnC₂O₄, which is then dissolved in acid to release H₂C₂O₄. The amount of H₂C₂O₄ is then determined by titration with KMnO₄, which reacts with H₂C₂O₄ according to the reaction:

5 H₂C₂O₄ + 2 KMnO₄ + 6 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 10 CO₂ + 18 H₂O

The moles of KMnO₄ used in the titration are calculated by multiplying the volume used (in liters) by the molarity:

Moles of KMnO₄ = 0.038 L × 0.02 mol/L = 0.00076 mol

The stoichiometry of the titration reaction shows that 2 moles of KMnO₄ react with 5 moles of H₂C₂O₄, so we determine the moles of H₂C₂O₄:

Moles of H₂C₂O₄ = (5/2) × 0.00076 mol = 0.0019 mol

Each mole of ZnC₂O₄ produced one mole of H₂C₂O₄. Therefore, moles of ZnC₂O₄ are also 0.0019. Using the molar mass of ZnC₂O₄ (which is 119.4 g/mol since Zn has an atomic weight of 65.4 g/mol and C₂O₄ has a formula mass of 54 g/mol), we calculate the mass of ZnC₂O₄:

Mass of ZnC₂O₄ = 0.0019 mol × 119.4 g/mol = 0.22686 g

Since all Zn in the original sample was converted to ZnC₂O₄, the mass of ZnC₂O₄ solely represents the mass of ZnO in the original sample. Therefore, the percentage of ZnO in the original sample is:

Percentage of ZnO = (Mass of ZnC₂O₄ / Mass of sample) × 100

Percentage of ZnO = (0.22686 g / 0.95 g) × 100 = 23.88%