Question

From Part A:

Mass of K_xFe(C₂O₄)_y · zH₂O prepared : 5.100 g
Mass of FeCl₃ : 1.60 g

From Part B:

% Potassium in compound : 17.95 %
% Iron (from ion exchange & titration vs. NaOH) : 8.35 %

From Part C:

% Oxlate : 43.97 %

Calculate the % water of hydration :

Calculate the following for Fe³⁺:

g in 100 g sample	mol in 100 g sample	mol/mol Fe (3 sig figs)	mol/mol Fe (whole number)

Calculate the following for K⁺:

g in 100 g sample	mol in 100 g sample	mol/mol Fe (3 sig figs)	mol/mol Fe (whole number)

Calculate the following for C₂O₄^2-:

g in 100 g sample	mol in 100 g sample	mol/mol Fe (3 sig figs)	mol/mol Fe (whole number)

Calculate the following for H₂O

g in 100 g sample	mol in 100 g sample	mol/mol Fe (3 sig figs)	mol/mol Fe (whole number)

Using your chemical knowledge and literature references (don’t forget to include the references in your lab report and discuss possible sources of errors) answer the questions below:

Enter the simplest formula of the Iron Oxalate Complex Salt:

K

Fe(C2O4)

·

H2O

Now that the formula of the complex salt is known, the percent yield can be determined.

Calculate the moles of FeCl₃ used in preparation:

Calculate the theoretical moles of K_xFe(C₂O₄)_y · zH₂O:

Calculate the actual moles of K_xFe(C₂O₄)_y · zH₂O synthesized:

Calculate the percent yield:

Answer 1

Given

mass of FeCl3 taken = 1.6 g

mass of complex formed = 5.1 g

% Fe = 8.35 %

% K = 17.95 %

% C2O4 = 43.97%

Total = 70.27%

% H2O = 100 - 70.27 = 29.73%

Therefore % water of hydration = 29.73%

For Fe3+

g / 100 g of sample = 8.35 g

mole / 100g of sample = 8.35 / 55.85 = 0.15 mole

mole of Fe / mole of Fe = 0.15 / 0.15 = 1.00

                                                     = 1 ( whole number)

For K:

g / 100 g of sample = 17.95 g

mole / 100g sample = 17.95 / 39 = 0.46 mole

mole of K / mole of Fe = 0.46 / 0.15 = 3.07

mole of K / mole of Fe = 3 ( whole number)

For C2O4

g / 100 g of sample = 43.97 g

mole / 100 g of sample = 43.97 / 88 = 0.50 mole

mole of C2O4 / mole of Fe = 0.50 / 0.15 = 3.33

                                       = 3 ( whole number)

For H2O

g / 100 g of sample = 29.73 g

mole / 100 g of sample = 29.73 / 18 = 1.65

mole of H2O / mole of Fe = 1.65 / 0.15 = 11.0

                                                       = 11 (whole number)

The simplest formula for the iron oxalate complex salt is

K3Fe(C2O4)3 . 11 H2O

since 1 mole of Fe combined with 3 mole of K, 3 mole of oxalate and 11 mole of water.

moles of FeCl3 in the preparation = 1.6 g / 162.2 = 0.00986 moles

Therefore theoretical moles of the complex formed = 0.00986 moles [ since the complex contains 1 mole of Fe]

Actual moles of the product formed = mass / molecular mass

                                                  = 5.1 g / 634.85 g / mol [ molar mass of the complex is 634.85]

                                                  = 0.00803 moles

Therefore % yield = [0.00803 / 0.00983] x 100 %

                         = 81.7 %