CHEMISTRY 31

Quiz 1 - SOLUTIONS

Feb. 5

 

1.  The percent of oxalate (C2O42-) in a 1.293 g solid sample is determined by a titration with MnO4- that reacts as follows:  2MnO4- + 5C2O42- + 16H+ ↔ 10CO2(g) + 2Mn2+ + 8H2O(l)

If the sample requires 32.17 mL of 0.0921 M MnO4-, what is the % C2O42- in the sample.  The formula weight of oxalate is 88.02 g/mol.  (10 pts)

mass % = (mass C2O42-)*100/(mass sample)

mass C2O42- can be obtained from moles C2O42-, which is obtained through titration:

m(C2O42-) = (0.03217 L)(0.0921 mol L-1)(5 moles C2O42-/2 moles MnO4-)(88.02 g/mol)

m(C2O42-) = 0.6520 g

mass % = (0.6520 g)100/(1.293 g) = 50.4% (3 sig fig due to 3 in 0.0921)