Question

A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Fe2+ in acid and then titrating the Fe2+ with MnO4-. A 1.0373 g sample was dissolved in acid and then titrated with 25.0 mL of 0.01183 M KMnO4. The balanced equation is given below. 8H+(aq) + 5Fe2+(aq) + MnO4-(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) Calculate the mass percent of iron in the ore.

Answer 1

Given :

Mass of sample = 1.0373 g

Volume of KMnO4 = 25.0 mL =0.025 L

Molarity of KMnO4 = 0.01183 M

Balanced equation :

8H+(aq) + 5Fe2+(aq) + MnO4-(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Calculation of number of moles of KMnO4

Mole = Molarity x volume in L

= 0.01183 M x 0.025 L

=0.000296 mol

Moles of KMnO4 = moles of MnO4

Calculation of moles of iron:

Mole ratio of Fe to MnO4- is 5 : 1

Moles of iron = moles of MnO4- x 5 mol Fe / 1 mol MnO4-

= 0.000293 mol MnO4- x 5 mol Fe / 1 mol MnO4-

= 0.001479 mol

Calculation of mass of Fe

Mass of Fe= moles x molar mass

= 0.001479 mol x 55.845 g/mol Fe

= 0.082581 g

Mass percent of Fe = Mass of Fe/mass of sample x 100

=0.082581 g / 1.0373 g   x 100

= 8.0 %

Mass percent of Fe in the ore is 8.0 %