The Synthesis of an Iron Oxalato Complex Salt (“Green Crystals”)
In this experiment you will synthesize a compound that will contain the elements potassium, iron, carbon, hydrogen, and oxygen. Carbon and oxygen will be present in the compound as oxalate (C2O4-2) whereas hydrogen and oxygen will be present as H2O. The final product may be given the formula KxFe(C2O4)y . z H2O, where z H2O is called the water of hydration. This is the first experiment of a series in which you will synthesize the compound and then determine its simplest (empirical) formula (i.e., x, y, z) using a variety of analytical techniques.
One important factor in any chemical synthesis is the actual quantity of desired product obtained compared to the theoretical amount predicted on the basis of the stoichiometry of the reaction. The ratio of the mass of product obtained to the theoretical quantity, expressed as a percentage is referred to as the “percent yield” or more simply the “yield”.
There are many reasons why the actual yields are not 100 percent. Possibly the reaction reaches equilibrium instead of going to completion. Maybe the reactants are involved in other reaction that the one that produces the desired product. Probably some product is lost in crystallizing the separating crystals from supernatant liquid, etc.
In this experiment you will react an aqueous solution containing 3.20 grams FeCl3 (162 g/mol) with an aqueous solution containing excess K2C2O4 to produce the product KxFe(C2O4)y . z H2O.
xK+(aq) + Fe3+(aq) + yC2O4-2(aq) + zH2O KxFe(C2O4)y . zH2O(s)
A new technique in this experiment is recrystallization. This allows the purification of solid crystals by dissolving the crystals in a minimum quantity of hot solvent and then cooling. The purified crystals can then be harvested by filtration. Larger, purer crystals are obtained if the hot solution is allowed to cool slowly without being moved or disturbed. A second “crop” of crystals could be obtained from the filtrate by evaporating a fraction of the solvent by heating followed by cooling the remaining solution. The second crop of recrystallized product is generally less pure that the first.
Objective: To prepare several grams of pure crystals of KxFe(C2O4)y . zH2O.
Apparatus: Two 50 mL beakers, centigram balance, crucible tongs, 250 mL beaker, watch glass, vacuum filtration apparatus with Buchner funnel, small brown bottle or bottle with aluminum foil.
Chemicals: FeCl3 solution (0.400 g FeCl/mL), K2C2O4 . H2O, ice, acetone (flammable!)
1. Obtain (in a clean 50 mL beaker) 8.00 mL of stock solution containing 0.400 g
2. Weigh 12.0 to 12.5 g K2C2O4 . H2O into a clean dry 50 mL beaker using a centigram
balance. Add 20 mL of distilled water to dissolve the K2C2O4 . H2O. Heat on the hot
plate stirring constantly until the K2C2O4 . H2O is completely dissolved. S.T.O.P.
3. Using the crucible tongs to handle the hot beaker, pour the hot solution into the beaker containing the iron (III) chloride solution. Stir. S.T.O.P.
4. Cool the solution of 30-45 minutes by placing the beaker in a 400 mL beaker containing ice and water. Crystals should form during this time. Take care that the beaker of product does not sink in the water the ice melts.
5. After giving the crystals ample time to form, carefully pour off and discard the solvent without removing any crystals, a process called decantation.
6. Now is the time to recrystallize. Add 20 mL of distilled to the crystals, heat gently with stirring to dissolve the crystals completely. If some dark residue remains undissolved, carefully decant the clear solution into another beaker and discard the residue.
7. Cover the beaker with a watch glass and set it in the AP lab cabinet until the next laboratory period to allow crystals to form. If the crystals are allowed to form slowly without being disturbed, large crystals will be obtained. If the solution is moved or stirred while crystals are forming, smaller crystals will result.
8. Next day. Set up a vacuum filtration and filter the crystals using a Buchner funnel and clean filter flask. If your crop of crystals appears to be quite small, save the filtrate so that you can obtain a second crop of crystals.
9. Wash the crystals twice with ice water. Use less than 5 mL of ice water for each wash and work quickly to avoid dissolving the product in the wash water. Finally wash the crystals twice with 5 mL portions of acetone.
10. Obtain the mass of a piece of filter paper and a clean, dry beaker. Spread the crystals on the filter paper in the bottom the beaker and place in the AP lab cabinet to air dry.
11. When the crystals are dry and no longer have the odor of acetone, weight them on a centigram balance. Place them in a bottle covered with aluminum foil (the crystals will decompose upon exposure to light!). A minimum of 3.5 grams of crystals is needed for upcoming experiments.
1. Assuming that all of the Fe originally in FeCl3 ends up in the product,
KxFe(C2O4)y . zH2O, how many moles of product should be obtained?
2. What additional information would you need to calculate a percent yield?
3. What is decantation?
4. Why does the use of ice-water reduce the amount of crystals that will be dissolved during the rinsing process?
5. What are the hazards associated with working with FeCl3, K2C2O4 and acetone?
6. Suggest a method to determine the amount of water contained in your crystals as hydrate.
7. Suggest a method to determine the amount of oxalate contained in the crystals you created.
I don't think I want to do your pre-lab work for you but I can get you started by answering #1.
mols FeCl3 = grams FeCl3/molar mass FeCl3 = mols KxFe(C2O4)y.zH2O
thank you that was the only one that i couldn't get. that and number 7. im a little confused on how somebody could do that?posted by Colin
There are two standard lab procedures for determining oxalate. One is a gravimetric procedure in which the C2O4^2- is put into solution and precipitated as CaC2O4, filtered, dried, and weigh the CaC2O4, then calculate percent. The other is a volumetric procedure in which the oxalate is placed into solution and the solution is titrated with KMnO4 in an acid solution. The KMnO4 is reduced to Mn^2+ and the oxalate is oxidized to CO2. Both procedures work very well.
Thanks a lot Dr.Bob222posted by Anoymous