Question

From Part A: Mass of KxFe(C2O4)y · zH2O prepared : 4.359 g

Mass of FeCl3 : 1.60 g

From Part B: % Potassium in compound : 22.70 %

% Iron (from ion exchange & titration vs. NaOH) : 10.60 %

From Part C: % Oxlate : 55.69 %

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

Calculate the % water of hydration :

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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)

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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)

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Calculate the following for C2O42-:

g in 100 g sample

mol in 100 g sample

mol/mol Fe (3 sig figs)

mol/mol Fe (whole number)

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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)

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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

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

Calculate the moles of FeCl3 used in preparation:

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Calculate the theoretical moles of KxFe(C2O4)y · zH2O:

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Calculate the actual moles of KxFe(C2O4)y · zH2O synthesized:

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Calculate the percent yield:

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Answer 1

Percent water of hydration = 100 - (% Iron + % potassium + % oxalate) = 100 - (10.60 %+ 22.70 % + 55.69 %) = 100 - 88.99 = 11.01 %

Fe³⁺ (Iron) : 10.60 %

100 gms contain 10.60 gms of Fe⁺³

Molar mass of Fe = 55.485 g/mol

No. of moles of iron in 100 gms = mass/molar mass = 10.60 g/55.485 g/mol = 0.19 moles

No. of moles of iron/NO. of moles of iron = 1.00 (3 significant figures)

No. of iron atoms in the sample = 1 (whole number)

K⁺ (potassium) : 22.70 %

100 gms contain 22.7 gms of potassium

Molar mass of potassium = 39.1 g/mol

No. of moles of potassium in 100 gms = mass/molar mass = 22.7 gms/39.1 g/mol = 0.581

No. of moles of potassium/no. of moles of iron = 0.581/0.19 = 3.05 (3 significant figures)

No. of potassium atoms in the sample = 3 (whole number)

C₂O4₂^- : 55.69 %

100 gms contain 55.69 gms of oxalate

Molar mass of oxalate = 88.02 g/mol

No. of moles of oxalate = mass/molar mass = 55.69 g/88.02 g/mol = 0.633

Moles of oxalate/moles of iron = 0.633/0.19 = 3.33 (3 significant figures)

No. of moles of oxalate atoms in the sample = 3 (whole number)

Water of hydration : 11.01 %

100 gms contain 11.01 gms of oxalate

Molar mass of oxalate = 18.02 g/mol

No. of moles of oxalate = mass/molar mass = 11.01 g/18.02 g/mol = 0.611

Moles of oxalate/moles of iron = 0.611/0.19 = 3.21 (3 significant figures)

No. of moles of oxalate atoms in the sample = 3 (whole number)

Empirical formula of K_xFe(C2O4)_y · zH2O : K₃Fe(C2O4)₃ · 3H2O

Mass of FeCl₃ = 1.6 g

Molar mass of ferric chloride = 162.2040 g/mol

Molar mass of K₃Fe(C2O4)₃ · 3H2O = 491.2427 g/mol

There is one iron atom in FeCl₃ and in K₃Fe(C2O4)₃ · 3H2O .

So molar ratio of FeCl₃ and K₃Fe(C2O4)₃ · 3H2O = 1:1

Moles of ferric chloride = theoretical moles of K₃Fe(C2O4)₃ · 3H2O = Mass/molar mass of iron chloride = 1.6 g/162.204 g/mol = 0.00986 moles

Theoretical mass of K₃Fe(C2O4)₃ · 3H2O = Theoretical moles of K₃Fe(C2O4)₃ · 3H2O x molar mass of K₃Fe(C2O4)₃ · 3H2O = 0.00986 moles x 491.2427 g/mol = 4.846 g

Experimental mass of K₃Fe(C2O4)₃ · 3H2O = 4.359 g

Percent yield = (experimental yield/theoretical yield) x 100 = (4.359 g/4.846 g) x 100 = 89.9 %

or it can also be calculated using theoretical moles and actual moles

Actual mole of K₃Fe(C2O4)₃ · 3H2O = mass/molar mass of K₃Fe(C2O4)₃ · 3H2O = 4.359 g/491.2427 g/mol = 0.00887 moles

Percent yield = (actual moles/theoretical moles) x 100 = (0.00887/0.00986) x 100 = 89.9 %