Question

With the completion of the determinations of % potassium, % iron, and % oxalate in the crystals, you may calculate the % water. The percentage compositionof the crystals, K_xFe(C₂O₄)_y · zH₂O, has then been completely determined experimentally. The simplest formula (x,y,z) can now be calculated from the the percentage composition. Once the formula is know it is then possible to calculate the percent yield of product that was obtained in the preparation and purification of the crystals.

From Part A:

Mass of K_xFe(C₂O₄)_y · zH₂O prepared : 4.800 g
Mass of FeCl₃ : 1.60 g

From Part B:

% Potassium in compound : 19.43 %
% Iron (from ion exchange & titration vs. NaOH) : 10.95 %

From Part C:

% Oxlate : 43.45 %

Unsure about my first two answers so if you could check those it would be greatly appreciated!

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 5 Fe(C204) 6 .6 H20 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/3 Previous Tries Now that the formula of the complex salt is known, the percent yield can be determined. Calculate the moles of FeCl3 used in preparation: 0.00986 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/3 Previous Tries Calculate the theoretical moles of KyFe(C204)y. 2H20: Submit Answer Tries 0/3 Calculate the actual moles of KyFe(C204)y : 2H20 synthesized: Submit Answer Tries 0/3 Calculate the percent yield: Submit Answer Tries 0/3

Answer 1

Q1. % water of hydration = 100 - (% Potassium + % Iron + % Oxalate)

                                 = 100 - (19.43 + 10.95 + 43.45)

                                = 26.17 %

Assuming 100 g of sample is taken:

Elememt/molecule	Mass (g) in 100 g of compound	molar mass (in g/mol)	mol present (=mass/molar mass)	relative ratio of mole	relative ratio in whole number
Potassium (K)	19.43	39.10	0.4969	(0.4969/0.1961) = 2.52	3 = x
Iron (Fe)	10.95	55.84	0.1961	(0.1961/0.1961) = 1.00	1
Oxalate (C2O4)2-	43.45	88.02	0.4936	(0.4936/0.1961) =2.52	3 = y
Water (H2O)	26.17	18.01	1.453	(1.453/0.1961) = 7.41	7 = z

So, the empirical formula : K3Fe(C2O4)3 · 7H2O

Q2. Mass of FeCl3 : 1.60 g = (1.60/162.2) = 0.00986 = mole of Fe

(As, molar mass of FeCl3 = 162.2 g/mol).

For Q3. And Q4. Molar mass of K3Fe(C2O4)3 · 7H2O= 3 x 39.10 + 55.84 + 3 x88.01 + 7 x 18.01

                                          = 563.11 g/mol   

Q 3. 1 mole of FeCl3 contain 1 mole Fe , And

as 1 mole of Fe is present in the 1 mole of complex,

Therefore, 0.00986 mole of Fe should give 0.00986 mole of complex = theoretical moles

Q 4. Actual mole synthesized = (4.800 g) /(563.11 g/mol) = 0.00852

Q 5. Percent yield = (actual moles) /theoretical moles

                              = (0.00852)/ (0.00986) = 0.8645

So, in % = 86.45