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SYNTHESIS AND ANALYSIS OF K3Fe(C2O4)3*3H20

Damini Ram CHEM 3B 10/10/2020

The synthesis of potassium ferric oxalate and the verification of the purity of the substance we produced were included in this series of experiments. This experiment was divided into three sections in the laboratory. We measured the percentage of oxalate ions in the potassium Ferric Oxalate Trihydrate sample using titration during this experiment. This would include applying the solution of Potassium Permanganate to an oxalate solution before the reaction is complete between both species. By understanding the stoichiometric reaction.

K Fe(C O ) •3H O 3

2

4 3

2

The compound potassium ferric oxalate will be performed, produced by: Fe(NH4)2(SO4)*6H2O(aq) + H2C2O4(aq) ----> FeC2O4*2H2O(s)+2(NH4)HSO4(aq)+4H2O

As ferrous oxalate is a yellow insoluble solid when all other products are soluble. Ferrous oxalate will be oxidized by hydrogen peroxide in presence of oxalate ion to produce potassium ferric oxalate which is produced: 2FeC2O4*2H2O(s)+H2O2(aq)+H2C2O4(aq)+3K2C2O4(aq)----> 2K3Fe(C2O4)3*3H2O(s)

When the two products will be analyzed in two different ways. Oxalate ion in the product will be determined by titration of the solution with the potassium permanganate. The reaction will occur with the reaction from permanganate ion produced by: 5C2O4-2 + 2MnO4- + 16H+ ---> 10C2(g) + 2Mn+2 +8H2O Procedure: Part 1: Synthesis of an Iron Compound On an analytical balance, accurately measure 5 g of ferrous ammonium sulfate,Fe(NH4)2(SO4)2•6H2O. Record the exact mass in your notes. Obtain 15 ml of deionized water in a clean squeeze bottle and transfer the deionized water to a clean 250-mL flask. Add the 5 g of ferrous ammonium sulfate to the 150-ml beaker and stir to dissolve. Add 5 drops of 3 M H2SO4to the solution of ferrous ammonium sulfate. Stir to mix. Clamp the 250-ml flask to a ring stand and support the flask with an iron ring and wire gauze. Heat the mixture to boiling. Be sure to continuously stir the contents to prevent bumping. Turn off the Bunsen burner. Carefully remove the 150-ml beaker using crucible tongs or the utility clamp. When the solution has cooled down to room temperature, add 10 m; of ethyl alcohol to the solution and mix it thoroughly. Store the product in a drawer for the following week's reaction for crystals. Next meeting, collect data information on your product. Wash crystals with 5mL ethyl alcohol and 5mL deionized water. Dry out crystals with filter paper and weigh your product. Calculate your percent yield for the Fe(NH4)2(SO4)2*6H2O as the limiting agent.

Procedure: Part 2: Analysis of an Iron Compound Preparation of stock solution, Collect the solid my simple filtration. Wash the crystals on the filter paper with a solution composed of 5 ml of ethyl alcohol and 5 ml of deionized water. Transfer the crystals to a dry piece of filter paper and blot them dry. Then obtain the weight of the product. Place this solid in a vial bottle with your name, and the name of the compound. Store this in your drawer.Calculate the percent yield assuming Fe(NH4)2(SO4)2•6H2O is the limiting reagent. This begins the preparation of the stock solution of the sample.Weigh precisely 1 g of your product and place into a 100 ml volumetric flask.Add 50 ml of deionized water to the flask and mix the solution until the solid dissolves. Add 20 ml of 3M H2SO4 to the solution. Carefully add deionized water to the flask until the liquid reaches the calibration mark on the neck of the flask. Place a stopper on the top of the flask, and mix the solution well. Transfer the solution to a clean and dry 250 ml flask. Label this solution ‘A’.Rinse out the 100 ml volumetric flask with deionized water and discard the rinse water. Transfer 5 ml of the solution ‘A’ back into the volumetric flask using a transfer pipette. Carefully fill the flask to the 100 ml mark with deionized water. Mix the solution as before. Label this flask as ‘B’.Use ‘B’ for the spectrophotometric analysis of the iron ion. Place 5 ml of solution ‘B’ into a 100 ml volumetric flask with a transfer pipette. Add 10 ml of 1M of Sodium Acetate solution and 2 ml of 10% hydroxylamine hydrochloride solution.Add 3 ml of 0.25% 1,10 phenanthroline solution and mix the contents well. Let the solution sit for 5 minutes then dilute to the mark with deionized water. Mix well. Read the absorbance of the test solution and of the standard solutions at 508 nm on the spectrophotometer. Turn to page 33 of the manual to find instructions. Procedure: Part 3 Calculate how much of 0.1 M Potassium Permanganate you will need to make 250 ml of a 0.02 M solution. According to your calculations, use the appropriate amount of deionized water to dilute the solution in a 400 ml beaker. Mix this solution very well. Calculate the mass of oxalic acid that is required to make 250 ml of a 0.50M oxalic acid solution. Weigh this amount of oxalic acid into a clean and dry beaker. Transfer the solid into a 250 ml volumetric flask. To this flask, add 100 ml of deionized water, and swirl to help dissolve the solid. Once all the solid is dissolved, add deionized water until it reaches the calibration mark on the neck of the flask. Mix well. Accurately place 20,0 ml of the standard oxalic acid solution in a clean 250 ml flask with a transfer pipette. Add 10 ml of deionized water and 15 ml of 3M sulfuric acid. Heat the solution to 80˚C. Remove the flask from the heat and titrate with the KMnO4 solution until the solution shows a faint pink color of excess KMnO4 for 30 seconds. Repeat this titration 2 more times with fresh samples of oxalic acid solution to determine the percent oxalate in your compound ‘A’, substitute 20.0 ml of ‘A’ for the 20.0 ml of standard oxalic acid that you

used in the standardization procedure. Add deionized water and the sulfuric acid as before Heat the solution to 80˚C and titrate with the potassium permanganate as before. Repeat twice. RESULTS: Synthesis of Potassium Tris(oxalato) chromium(III) Trihydrate ● ● ●

Theoretical yield Mass Percent yield

: 5.9709 g : 4.2231 g : 72.38 %

Synthesis of Copper (II) Oxalate Complexes ● ● ●

Theoretical yield Mass Percent yield

: 2.8386 g : 2.1753 g : 17.76%

Synthesis of Potassium Tris(oxalato) Ferrate(III) Trihydrate ● ● ●

Theoretical yield Mass Percent yield

: 6.2895 g : 0.7217 g : 11.47 %

CALCULATIONS:

A. Synthesis of Potassium Tris(oxalato) chromium(III) Trihydrate 8H+ + 3H2C2O4 + 6K2C2O4.H2O + K2Cr2O7 → 2K3[Cr(C2O4)3].3H2O + 6CO2 + 9H2O + 8K+

The number of moles H2C2O4 = mass MM = 5.0 90.04 = 0.0555 mol

The number of moles K2Cr2O7 = mass MM = 1.7 374.14 = 6.1168 x 10-3 mol (limiting reactant) 1 mol of K2Cr2O7 = 2 mol of K3[Cr(C2O4)3].3H2O

6.1168 x 10-3 mol of K2Cr2O7 = 0.01224 mol of K3[Cr(C2O4)3].3H2O The theoretical yield

= mol x MM = 0.01384 x (487) = 5.5708 g

Mass of K3[Cr(C2O4)3].3H2O obtained is 4.2331g Percentage yield of K3[Cr(C2O4)3].3H2O

=

actual yield x 100% Theoretical yield = 4.3221 x 100% 5.9709 = 72.38%

Solution

[Fe ] (M)

Abs. (a.u.)

2+

STANDARD

2.0 x 10

.217

STANDARD

4.0 x 10

0.410

STANDARD

6.0 x 10

0.620

STANDARD

8.0 x 10

.874

My Test

-5

-5

-5

-5

Solution

.218

B. Synthesis of Copper (II) Oxalate Complexes CuSO4.5H2O + 2K2C2O4.H2O → K2[Cu(C2O4)2(H2O)2] + K2SO4 + 5H2O

The number of moles CuSO4.5H2O

= mass MM = 2.0 249.61 = 7.0125 x 10-3 mol (limiting reactant)

The number of moles K2C2O4.H2O

= mass MM = 6.22 184.2 = 0.0337 mol

1 mol of CuSO4.5H2O = 1 mol of K2[Cu(C2O4)2(H2O)2] 8.0125 x 10-3 mol of CuSO4.5H2O = 8.0125 x 10-3 mol of K2[Cu(C2O4)2(H2O)2] The theoretical yield

= mol x MM = 8.0125 x 10-3 x (353.82) = 2.8350 g

Mass of K2[Cu(C2O4)2(H2O)2] obtained is 2.1753g Percentage yield of K2[Cu(C2O4)2(H2O)2]

= actual yield x 100% Theoritial yield = 2.1753 x 100% 2.8386 = 17.76%

C. Synthesis of Potassium Tris(oxalato) Ferrate(III) Trihydrate Fe(NH4)2(SO4)2.6H2O + H2C2O4 → FeC2O4 + H2SO4 + (NH4)2SO4 + 6H2O

Number of mole Fe(NH4)2(SO4)2.6H2O= mass MM = 5.0 392.21 = 0.0128 mol 1 mole Fe(NH4)2(SO4)2.6H2O = 1 mole of Fe(OH)3 Thus, mole of Fe(OH)3 = 0.0128 mol 3K2C2O4 + 2Fe(OH)3 + 3H2C2O4 → 2K3[Fe(C2O4)3].3H2O + 3H2O

Mole of K2C2O4

= mass MM = 3.50 166.22 = 0.0211 mol

Mole of H2C2O4

= mass MM = 2.0 90.04 = 0.0222 mol

From the overall equation 3 mol of K2C2O4 ≡ 2 mole of K3[Fe(C2O4)3]· 3H2O 0.0211 mol of K2C2O4 = 0.0141 mol K3[Fe(C2O4)3] · 3H2O 3 mole of H2C2O4 ≡ 2 mole of K3[Fe(C2O4)3]· 3H2O 0.0222 mol of K2C2O4.H2O = 0.0148 mol of K3[Fe(C2O4)3]·3H2O 2 mol of Fe(OH)3 ≡ 2 mol of K3[Fe(C2O4)3]· 3H2O 0.0128 mol of Fe(OH)3 ≡ 0.0128 mol of K3[Fe(C2O4)3]· 3H2O Thus, the limiting reactant is Fe(OH)3 . The theoretical yield= mol x MM = (0.0128) x (491.21) = 6.2875 g Mass of K3[Fe(C2O4)3].3H2O obtained is 0.7217 g Percentage yield of K3[Fe(C2O4)3].3H2O

= actual yield x 100% Theoretical yield = 0.7217 x 100% 6.2895 = 11.47%

We analyzed how to synthesize coordination compounds in this experiment. A covalent coordination bond is a covalent bond in which all electrons are produced by an atom. The coordination compound is the product of a lewis acid-base reaction in which, by coordinating covalent bonds, neutral molecules or anions bind to central metal. This form of bonding varies from a standard covalent bond in a bond in which one electron is produced by each atom. The coordination compound in part A of this experiment is K3[Cr(C2O4)3].3H2O that is formed when lewis acid (Cr3+) reacted with C2O4 and H2O lewis base. Since chromium (III) is chromium's most stable oxidation state, six monodentates can be formed in which two oxalate ions, C2O42, two water molecules and [Cr(C2O2)2(H2O)2] are formed. The mass of the obtained dark green crystal is 4.3231 g. 4.3231 g is the potential yield obtained. The percentage yield, then, is 72.49%. The coordination compound is K2[Cu(C2O4)2(H2O)2] in Part B, which is formed when this experiment's lewis acid, Cu2+ and lewis bases, which are oxalate ion and water, react together. Copper (II) is not as abundant as chromium. Typically, with 4 short bonds and 2 long bonds, there are either 4 coordinates or 6 twisted coordinates. The arrangement of 2 oxalates relies on the cation used in this experiment, which is K+, giving a total of 6 coordinates with carbonyl O connections. The mass of copper (II) oxalate complexes obtained is 2.1753 g while theoretically the yield is 2.8386 g. Therefore, the percentage yield is 73.28 %. In part C, when three oxalato ligands (C2O42-) were bonding to each central metal atom, K3[Fe(C2O4)3].3H2O formed. These ligands are bidentate, each of which binds to the metal atom at two separate positions. The Lewis acid is Fe, while the Lewis bases are C2O42- and H2O. In this experiment, a 0.7217 g bright green crystal is formed. The potential yield is 6.2895 g, which gives a cumulative percentage yield of 11.55 percent. There is a significant gap in reaching 100% relative to the theoretical yield, which is due to certain errors during this trial. A precautionary measure should then be taken in order to maximize the percentage yield. First of all, before doing this experiment, it is very important to observe the steps and understand them very well. This is since the commodity type would be smaller if only one stage is reversed....