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.

You have entered the following values :

From Part A:

Mass of K_xFe(C₂O₄)_y · zH₂O prepared : 6.000 g
Mass of FeCl₃ : 1.60 g

From Part B:

% Potassium in compound : 11.50 %
% Iron (from ion exchange & titration vs. NaOH) : 14.70 %

From Part C:

% Oxlate : 40.56 %

The questions below are part of the final analysis, they are due at the beginning of the next lab, Tue Mar 31 08:30:00 am 2020 (EDT).

Now, let's finish the calculation and the determination of the formula of the iron compound:

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

1. Determination of molecular formula of KxFe(C2O4)y · zH2O, i.e. determination of x, y and z:

% composition any species means grams of that species present per 100 grams of the complex.

Therefore,

% Potassium (K) in compound : 11.50 % means 11.50 g of K in 100 g of KxFe(C2O4)y · zH2O

Moles of K in 100 g complex

% Iron: 14.70 % means 14.70 g of Fe in 100 g of KxFe(C2O4)y · zH2O

Moles of Fe in 100 g complex

% Oxlate : 40.56 % means 40.56 g of C2O4  in 100 g of KxFe(C2O4)y · zH2O

Moles of C2O4 in 100 g complex

% H2O: [100 - (11.50+14.70+40.56)]% = 33.24% mean 33.24 g of H2O in 100 g of KxFe(C2O4)y · zH2O

Moles of H2O in 100 g complex

Generally, the number of any species in the complex = (moles of K)/ (lowest moles among all the species in the complex)

Here, it is already given that the compled KxFe(C2O4)y · zH2O has 1 Fe per molecule of the complex.

Therefore, we have tp set 1 = moles of Fe (calculated above) = 0.263 mol and then compared to this convert the moles of all other species to numbers, i.e. x, y and z.

x = (moles of K)/ (moles of Fe) = (0.294 mol)/ (0.263 mol) = 1.12 1 [Since number of a atom or group in the complex cannot be a fraction.]

Similarly, y = (moles of oxalate)/ (moles of Fe) = (0.461 mol)/ (0.263) = 1.75 2

z = (moles of water)/ (moles of Fe) = (1.846 mol)/ (0.263) = 7.02   7

Therefore, the simplest formula of the Iron Oxalate Complex Salt: KFe(C2O4)2 · 7H2O

All of the above data are tabulated below:

Species	% composition	g per 100 g of sample	Number of moles per 100 g sample (mol)	Number per molecule of complex (x, y, z)
Potassium (K)	11.50%	11.50 g	0.294 mol	x = 1
Iron (Fe)	14.70%	14.70 g	0.263 mol	1 (given)
Oxalate (C2O4)	40.56%	40.56 g	0.461 mol	y = 2
Water (H2O)	33.24%	33.24 g	1.846 mol	z = 7

2. Determination of percent yield:

Calculate the moles of FeCl3 used in preparation:

Mass of FeCl3 : 1.60 g and molar mass of FeCl3 = 162.2 g/mol

Therefore, the moles of FeCl3 used in preparation = (Mass of FeCl3 in g)/ (molar mass of FeCl3 )

= (1.60 g)/ (162.2 g/mol)

= 9.86* 10-3 mol

Calculate the theoretical moles of KFe(C2O4)2 · 7H2O:

From the balanced reaction above, it is evident that 1 mole of FeCl3 is used up to produce 1 mole of the complex.

Therefore, moles of FeCl3 used = moles of KFe(C2O4)2 · 7H2O complex produced = 9.86* 10-3 mol

The theoretical moles of KFe(C2O4)2 · 7H2O = 9.86* 10-3 mol

Calculate the actual moles of KFe(C2O4)2 · 7H2O synthesized:

Mass of KFe(C2O4)2 · 7H2O prepared, i.e. experimental yield = 6.000 g

Molar mass of KFe(C2O4)2 · 7H2O = [(39.10 g/mol)+(55.845 g/mol)+(2* 88.02 g/mol)+(7* 18.01 g/mol)]

= 397.055 g/mol

Therefore, the actual moles of KFe(C2O4)2 · 7H2O synthesized

= (mass of KFe(C2O4)2 · 7H2O prepared)/ (Molar mass of KFe(C2O4)2 · 7H2O)

= (6.000 g )/ (397.055 g/mol)

= 0.0151 mol (answer)

Calculate the percent yield:

Mass of KFe(C2O4)2 · 7H2O prepared, i.e. experimental yield = 6.000 g

The theoretical moles of KFe(C2O4)2 · 7H2O = 9.86* 10-3 mol

Therefore, the theoretical yield (in g) of KFe(C2O4)2 · 7H2O

= (theoretical moles)* (molar mass of KFe(C2O4)2 · 7H2O)

= (9.86* 10-3 mol)* (397.055 g/mol)

= 3.915 g

Therefore, percent yield = [(theoretical yield)/ (experimental yield)] * 100 %

= [(3.915 g)/ (6.000 g)] * 100 %

= 65.25 % (answer)