Question

A 4.979 g sample containing the mineral tellurite was dissolved and then treated with 50.00 mL of 0.03187 M K₂Cr₂O₇. Upon completion of the reaction, the excess Cr₂O₇^2- required 37.58 mL of 0.1197 M Fe²⁺ solution. Calculate the percentage by mass of TeO₂ (FM 159.60) in the sample. The reactions are

3 TeO₂ + Cr₂O₇^2- + 8 H⁺ → 3 H₂TeO₄ + 2 Cr³⁺ + H₂O
Cr₂O₇^2- + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O

% TeO₂

Answer 1

Moles of Fe²⁺ required = molarity * volume / 1000 = 0.1197 * 37.58 / 1000 = 0.004498 mol

From the second balanced equation,

0.004498 mol of Fe²⁺ requires 1 * 0.004498 / 6 = 0.0007497 mol of Cr₂O₇^2-

Initial moles of Cr₂O₇^2- = 0.03187 * 50.00 / 1000 = 0.001594 mol

Now, moles of Cr₂O₇^2- consumed in the reaction with TeO₂ = 0.001594 - 0.0007497 = 0.0008443 mol

From the first balanced equation,

1 mol Cr₂O₇^2- requires 3 mol TeO₂

then, 0.0008443 mol Cr₂O₇^2- requires 3 * 0.0008443 = 0.002533 mol TeO₂

Therefore, mass of TeO₂ = moles * molar mass = 0.002533 * 159.60 = 0.4043 g.

% by mass of TeO₂ = mass of TeO₂ * 100 / mass of sample = 0.4043 * 100 / 4.979 = 8.119 %