Your Name Date lab done
LAB #4 : Preparation and Purification of an Iron-Containing Salt
DATA
Mass of potassium oxalate _____2.253______g
Mass of ferric nitrate monohydrate _____1.500______g
Mass of empty glass beaker _____51.125_____g
Mass of beaker + K3Fe(C2O4)3 _____52.444_____g
Mass of K3Fe(C2O4)3 _____1.319______g
CALCULATIONS
1. Theoretical Yield:
Fe(NO3)3 • 9H2O + 3K2C2O4 •H2O → K3Fe (C2O4)3•3H2O + 3KNO3 + 9H2O
1 mol Fe(NO3)3 • 9H2O 1 mol K3Fe (C2O4)3•3H2O
a) 1.500 g Fe(NO3)3 • 9H2O x 404.0 g
x 1 mol Fe(NO3)3 • 9H2O
= 0.003713 mol
491.2 g K3Fe (C2O4)3•3H2O
b) 0.03713 mol K3Fe (C2O4)3 x = 1.824 g
1 mol
Since Fe(NO3)3 • 9H2O is the limiting reagent, the theoretical yield is 1.824 g of K3Fe (C2O4)3•3H2O.
2. Percent Yield :
actual yield from experiment 1.319 g
% = x 100% = x 100% = 72.31%
theoretical yield 1.824 g
LAB #4 : Preparation and Purification of an Iron-Containing Salt
DATA
Mass of potassium oxalate _____2.253______g
Mass of ferric nitrate monohydrate _____1.500______g
Mass of empty glass beaker _____51.125_____g
Mass of beaker + K3Fe(C2O4)3 _____52.444_____g
Mass of K3Fe(C2O4)3 _____1.319______g
CALCULATIONS
1. Theoretical Yield:
Fe(NO3)3 • 9H2O + 3K2C2O4 •H2O → K3Fe (C2O4)3•3H2O + 3KNO3 + 9H2O
1 mol Fe(NO3)3 • 9H2O 1 mol K3Fe (C2O4)3•3H2O
a) 1.500 g Fe(NO3)3 • 9H2O x 404.0 g
x 1 mol Fe(NO3)3 • 9H2O
= 0.003713 mol
491.2 g K3Fe (C2O4)3•3H2O
b) 0.03713 mol K3Fe (C2O4)3 x = 1.824 g
1 mol
Since Fe(NO3)3 • 9H2O is the limiting reagent, the theoretical yield is 1.824 g of K3Fe (C2O4)3•3H2O.
2. Percent Yield :
actual yield from experiment 1.319 g
% = x 100% = x 100% = 72.31%
theoretical yield 1.824 g