Question

Given the following reaction

8 H+ + 5 Cu+ + MnO4- =====> 4 H2O + Mn+2 + 5 Cu 2+

Determine the weight percent Cu of a 1.84 gram sample if it takes 19.6 ml of 0.111 Molar MnO4- to titrate it. Find the moles of Cu first, then grams.

Answer 1

The balanced equation is 8H+ + 5 Cut + MnO4 =====> 4 H2O + Mn+2 + 5 Cu2+ Number of moles of MnO4- = M*V = 0.111 M * 19.6 mL = 2.1756 mmol According to balanced equation, 1 mole of MnO4- reacts with 5 mole of Cu+ So, 2.1756 mmol of MnO4- would react with 2.1756 mmol * 5/1 = 10.878 mol of Cu+ Number of moles of Cu+ = 10.878 mmol = 0.010878 mol Mass of Cu+ = moles * mol.wt. = 0.010878 mol * 63.55 g/mol = 0.691297 g Mass of Copper present = 0.691 g Mass percent of copper = mass of Cu/mass of the sample * 100 Mass percent of copper = 0.691 / 1.84 * 100 = 37.57 %