Calculate the weight percent %(w/w).

Question

The amount of oxalic acid (𝐻₂𝐶₂𝑂4) in a sample of rhubarb was determined by

reacting with 𝐹𝑒³⁺. After extracting a 10.62 g of rhubarb with a solvent, oxidation of the oxalic acid required 36.44 mL of 0.0130 𝑀 𝐹𝑒³⁺. Calculate the weight percent %(w/w) of oxalic acid in the sample of rhubarb.

2 Fe³⁺(aq) + H₂C₂O₄ (aq) + H₂O(l) ------> 2 Fe²⁺(aq) + 2CO₂(g) + 2 H₃O+(aq)

Answer 1

2Fe³⁺ + OA + H2O ==> 2Fe²⁺ + CO2 + H₃O⁺

mols Fe³⁺ needed = 36.44 ml x 1 L /1000 ml x 0.0130 mol/L = 0.000474 moles Fe³⁺

mols OA present = 0.000474 mols Fe³⁺ x 1 mol OA / 2 mols Fe³⁺ = 0.000237 mols OA

grams OA present = 0.000237 mols OA x 90 g/mol = 0.02133 g OA

% by mass (w/w) = 0.02133 g OA / 10.62 g (x100%) = 0.2008% oxalic acid