How to find the percent by mass

Question

I need to find the percent by mass of H2O2 in a solution of KMnO4. To do so, I need to know how many moles of H2O2 reacted in the titration to do my calculation (mass of H2O2 reacting/ mass of H2O2 weighed X 100). The problem is that I don't know how to find the moles reacting.

I have access to these values: [KMnO4] = 0.02 M , Volume of KMnO4 used to titrate = 10.60 ml , Mass of H2O2 weighed = 0.6041g

Answer 1

The very FIRST thing you need is a correctly balanced equation showing the reaction between the KMnO4 and the H2O2. Assuming the titration was done in acid (H2SO4, eg) you would have something like this...

2KMnO₄ + 3H₂SO₄ + 5H₂O₂ ==> 2MnSO₄ + K₂SO₄ + 5O₂ + 8H₂O

Next, you use the stoichiometry of this balanced equation along with the information you have, and you determine the moles of H₂O₂ present:

moles KMnO₄ used = 10.60 ml x 1 L / 1000 ml x 0.02 mol / L = 0.000212 moles KMnO₄

moles H₂O₂ present = 0.000212 mols KMnO₄ x 5 mol H₂O₂ / 2 mol KMnO₄ = 0.000530 mols H₂O₂

From here we can easily calculate mass H₂O₂ reacting:

mass H₂O₂ = 0.000530 mols x 34.01 g / mol H₂O₂ = 0.0180 g H₂O₂ present

Percent (by mass) of H₂O₂ in the original sample = 0.0180 g / 0.6041 g (x100%) = 2.93%

This should probably be reported as 3% (1 sig.fig.) since 0.02 M has only 1 sig.fig.