Synthesis of a Coordination Compound Background:

Coordination Compounds

• You should look over the material in your textbook, McMurray and Fay: Chemistry pages 873 - 898. You should also review the description of coordinate covalent bonds on page 253.

Typical anions include nitrate, NO₃-, sulfate, SO₄2- and carbonate, CO₃2-.

The most common such cation is ammonium, NH₄+.

Coordination compounds are somewhat different. They begin with a central metal cation. This cation is bound to a number of surrounding molecules or ions to form a more complex structure. These surrounding species are given the name ligands (from Greek word to tie up.) These ligand species are bound to the central atom with coordinate covalent bonds. In a coordinate covalent bond, both electrons are contributed by the same species. For this to occur the ligand species must contain an unbound pair of electrons. In practice, this electron pair is generally on an atom of oxygen, nitrogen or a halogen.

Common species can serve as ligands: H₂O (water), NH₃ (ammonia), NO₃- and NO₂- (nitrate and nitrite ions), Cl^- (chloride ion.) These each form one bond and are called monodentate ligands (literally, one tooth with which to bite.)

We will be using a bidentate ligand, a species that can form coordinate covalent bonds at two ends of the molecule. This species is the oxalate ion, C₂O₄2-. (The figure does not show the charges or all the electrons.)

Our ion complex will have three oxalate ions bound to a single Fe³⁺ ion. It will have a net charge of minus 3. The cation will be the potassium ion, K⁺. The formula for the compound is

• K₃ Fe(C₂O₄)₃ . 3H₂O

Many inorganic compounds (such as this one) are solids that incorporate water as a normal part of the crystal structure. The crystals are dry; the water is not present as a liquid. The amount of water is quite regular and it clearly affects some of the properties of the solid. It should be included in the molecular formula.

Background-- Chemical Synthesis

Synthesis of this iron compound involves several stages

• We will begin with a solution of Fe²⁺ ions, Iron (II).
• We will acidify and add oxalic acid.
• A reaction occurs producing a precipitate of Iron(II) Oxalate.
• We will heat the mixture to convert the finely powdered precipitate into larger crystals. (They are still microscopic, but they will settle out much faster.)
• We will decant, pouring off as much liquid as possible. This removes most of the other ions present and reduces the volume.
• We will add more oxalate ion and hydrogen peroxide. The H₂O₂ will oxidize the Fe from the +2 (ferrous) to the +3 (ferric) oxidation state.
• Converting the Fe to +3 causes it to bind directly with oxalate ion. The result is the complex anion. It contains an atom/ion of iron, Fe, in the center and three oxalate ions are bound to it.
• The new ion dissolves readily and the solid disappears. A strongly colored solution appears. We will cool the solution and add alcohol. Both of these steps reduce the solubility of the product and crystals will begin to form. This is the potassium salt of the ion described above.
• We will let the crystals grow and during the next period we will isolate them by filtering and drying.
• In subsequent lab periods we will analyze the compound for K, for oxalate, for Fe, and for water of crystallization.

Safety Concerns:

We will generally use electrical hot plates for the heating. Heating solutions on a ring stand with a Bunsen Burner presents several safety problems. The proper technique will be demonstrated.

Experimental Procedure

Glassware: Beakers (150, 250 ml), graduated cylinder (50 ml), buret, thermometer, stirring rod, funnel, watch glass; filter paper, storage bottle. A filtering funnel setup will be provided. A Bunsen burner and ring stand or a hot plate will also be required.

Chemicals: solid: Fe(NH₄)₂ SO₄ . 6H₂O, solutions: 6M H₂SO₄, 3% H₂O₂, 1M oxalic acid, 2M potassium oxalate, ethanol (ethyl alcohol).

Notes and record keeping:

• In your notebook, briefly describe each step. (Try not to simply copy the directions below).
• At each stage, record any changes you see (color changes, precipitates, bubbles...)

• 1. Weigh out 10 g (+ 0.1g) of Ferrous ammonium sulfate, Fe(NH₄)₂ SO₄ . 6H₂O
• 2. Place the solid in a 250 ml beaker and add 30 ml of warm distilled water.
  • Stir until completely dissolved.
• 3. Acidify with 6-8 drops of 6M sulfuric acid, H₂SO₄. (Be sure to note the color.)
• 4. Slowly add, while stirring, 50 ml of 1 M oxalic acid, H₂C₂O₄.
  • A yellow solid (precipitate) will form. This is iron(II) oxalate, Fe (C₂O₄)
• 5. Slowly heat the beaker and its contents to boiling; stir and reduce the heating level to keep the material at a gentle boil.
  • After about 5 minutes remove the heat and allow the beaker to cool.
• 6. Now slowly decant (pour off the liquid layer and retain the wet solid.) Try to avoid the loss of solids.
  • It is a good idea to decant into a clean beaker; if you make a mistake, you can return the liquid to the first beaker and try again . Dispose of the liquid and the washings in step 7 in the container labeled Oxalate Wastes.
• 7. Add another 30 ml of warm water, stir, allow to settle and decant again. Repeat (two washings) This removes excess reagent and impurities.
• 8. Add 20 ml of 2M potassium oxalate, K₂C₂O₄ to the beaker with the solids.
• 9. Heat to 40^oC; stir occasionally.
• 10. From a buret slowly add 35 ml. of 3% hydrogen peroxide, H₂O₂ . Continue to stir and keep the temperature at 40 degrees. You will see some bubbles (oxygen), some color change, and a decrease in the amount of solid.
• 11. Heat the mixture to boiling, and add 13 ml of 1 M oxalic acid-- first adding 9 ml at once and then adding the remainder drop by drop. Keep the mixture hot during this addition.
• 12. Examine the liquid. If your solution shows more than a few tiny crystals, check with your instructor. If any solid remains, filter the solution into a 150 ml beaker and discard any solids. If the solution is clear, simple transfer it hot to the smaller beaker.
• 13. CAUTION: Make sure that there are no open flames near your work place now (this includes other students) Add 20 ml of 95% ethyl alcohol. If crystals form, gently heat the solution to dissolve them (use a hot plate in the hood, not a burner) .
  • Now wait. You should see some crystals form as the solution cools. The process is slow and the crystals will be quite small. Cover the beaker with a watch glass until the next period. The product is slightly light sensitive so you should keep it in dark area like your lab locker.
• 14. [Complete this step in the next lab period.]
  • Use a vacuum filtration setup to separate the crystals from the liquid. Wash the crystals with two 10 ml rinses of 1:1 alcohol: water solution followed by one 10 ml rinse with acetone. Air dry the crystals on the filter paper. When dry, weigh the product..

• From the formula for the compound and the iron in the starting material in step 1, compute the yield. What fraction of the iron ended up in the product?
  • a. How many moles of Fe in your 10 g of Ferrous ammonium sulfate, Fe(NH₄)₂ SO₄ . 6H₂O ?
  • b. how many moles of K₃ Fe(C₂O₄)₃ . 3H₂O could this produce if there are no losses ?
  • c. And how much would that weigh, ideally?
  • d. The %yield is 100 times your sample weight divided by the answer in part c.

Chemical Synthesis --

The goal in synthesis is to produce a pure chemical compound from other, readily available chemicals. There are no general rules that cover all chemical synthesis. The synthesis could be a simple one step process, but multistep synthesis is also quite common. We will focus on synthesis where we might prepare 1-10 grams of material; other chemists may be satisfied with a few milligrams of material or they might need to make several kilograms. Chemical engineers develop processes that synthesize tons of material.

Obviously one needs suitable reactants -- species that undergo reaction to produce the desired product. In addition, it is important that the product can be isolated from solvents, excess reagents, and other reaction products. We try very hard to find synthesis conditions which produce good yields, involve few side reactions, and provide relatively easy conditions for isolating and purifying the product.

• Some reactions produce clean products with no need for further separations or purification steps. This might involve the decomposition of a single compound to a solid and a gas.
• CaCO₃ (s) ---heat-->CaO (s) + CO₂ (g)
• Another simple synthesis involves two reactants. An excess of the second reactant is used, and that species is either a gas or liquid and is easily removed by evaporation.
• 2 Fe + 3 O₂ -------> 2 Fe₂O₃ (s)
• CaO + H₂O ----> Ca(OH)₂ (s)
• CaC₂ +2 H₂O ---> Ca(OH)₂ +C₂H₂ (acetylene gas)
• 2 Fe + S₂ ---heat--> FeS (s) the extra sulfur disappears on heating
• CuO (s) + C(s) --heat->Cu (l) + CO (g)
  • liquid metal collects at the bottom of the container and separates from remaining the carbon and unreacted ore.
  • Some reactions that occur in solution produce only a single product or the other products are gases or water. By evaporating the solvent we can collect pure crystalline products. .
• NH₃ (g) + HCl (aq) ---> NH₄Cl (aq) The symbol (aq) means an aqueous solution of the species.
• Fe + 2 HCl(aq) ---> H₂ + FeCl₂ (aq)
• 3 Cu + 8 HNO₃(aq) ---> 2 NO(g) + 4 H₂O + 3 Cu(NO₃)₂ (aq)
  • When we write a compound like Cu(NO₃)₂ (aq) we should remember that in solution this species really exists as free ions, Cu²⁺ and (NO₃)^- , in water. When we evaporate the solvent, crystals of the compound begin to form.
  • If we fully evaporate the solvent, all of the ions present will eventually come out of solution. We will often stop the process after we collect 80-90% of the product, sacrificing the yield to get a purer product. Likewise, rinsing the product may dissolve some of it, producing a lower yield of purer product.
• Another synthesis which is relatively simple involves reactions that produce a relatively insoluble product; the other species, including the solvent, can be removed by filtration. The solid, once collected, is rinsed and dried.
• Ba²⁺ + SO₄2- ---> BaSO₄ (s)
• Fe²⁺ + H₂C₂O₄ --Fe(C₂O₄) (s) + 2 H⁺
  • This reaction of Fe²⁺ with oxalic acid is the first step of today's experiment.
  • To be practical, it is desirable that the solid not be a very finely divided powder. Fine particles either pass through a filter or they tend to clog the filter, making filtration very slow or even impossible. There are several "tricks" such as heating the suspended solid, waiting, or the addition of additional ionic species which convert fine powders into coarser precipitates.
    • A precipitate is a solid that forms from a solution, usually appearing as a fine powder that slowly settles to the bottom of the container.
• In many organic chemical reactions the products can be separated by distillation or by solvent extraction. In both cases we rely on a difference in the physical properties of the components of the mixture.
• • A simple reaction is the fermentation of a sugar solution by yeast. (Think of this as a biotechnical method of synthesis--no one said the chemist must do all of the work personally.) The product is a 8-10% solution of ethanol in water, along with some remaining sugar and a few other species. Since the alcohol is much more volatile than the water, distillation produces a liquid containing 75-80% alcohol.
  • Another biosynthesis is the production of antibiotics, like penicillin, by microorganisms. This is the historic route; most penicillin is now produced by direct chemical synthesis. In this case only a trace of the desired product is present and separation is a major problem. The starting point is to filter the growth medium; the penicillin is water soluble so it remains in solution. We then add a small amount of hydrocarbon, like hexane, and shake the mixture and allow the two layers to settle out. In this case, the penicillin will almost all end up in the hexane layer, but most of the other materials remain in the water layer. (Many other separation steps are required to get a pure product.)
• We should note that many commercial products often do not need to be chemically pure products. Wine and beer, for example, are fermentation products from which the alcohol has not been extracted.
• These examples may look pretty simple, so it is worth noting that many good chemical reactions cause problems when we try to isolate the products. We can oxidize a solution of iron(II) sulfate by mixing it with a solution of potassium dichromate and sulfuric acid, but it will be difficult to obtain a sample of iron(III) sulfate this way.
• 6 FeSO₄ + K₂Cr₂O₇ + 7 H₂SO₄ --(aq)-> 2 Cr^{3 +} + 6 Fe⁺³ + 2 K⁺ +13 SO₄2- +7 H₂O
  • This reaction is quite efficient but we end up with a solution which contains three cations and one anion; evaporating the solution will produce a mixture of three salts. The solubility of the three products is about the same, so we can't easily separate the mixture.

Solubility

We often rely upon solubility to isolate a product, it is appropriate to examine this topic in more detail.

• The material we dissolve is usually called the solute; a solution may contain several solutes. The liquid is referred to as a the solvent.
• The concentration of a solution is usually expressed in molarity--moles of solute per liter of solution. We might also refer to the concentration of grams of solute per liter of solution or as grams of solute per liter of solvent.
• A saturated solution is one in which no more solute will dissolve. This is the maximum concentration (under normal circumstances.) If you add more solid solute, it just gets wet and remains at the bottom of the flask.

In many chemical reactions a compound of low solubility is produced from the reaction of more soluble species. It is then possible for the amount of product (think of it as solute) to exceed the amount that will dissolve in the available solvent. If the reaction is rapid, the product will be produced as a powdered precipitate. The solution becomes cloudy, then opaque, and gradually the solid settles out. If the reaction is slow, the solid may emerge as a few small crystals which gradually become larger crystals.

We can also obtain a solid product by modifying a solution until it becomes saturated. If the solution is nearly saturated, cooling it might produce a crystalline product.

• For example, the solubility of copper (II) nitrate is 138 g/100 ml of water at 0^oC, but it is 666 g/100 ml at 80^o C.
• We could recrystalize copper nitrate by dissolving 500 grams in 100 ml of water, filtering if any solid remains and then slowly cooling the solution in an ice bath. Since only 138 g will remain in solution, we would collect 362 g of solid copper nitrate crystals.
• Our yield or recovery would be 362/500 = 0.72 or 72%.

• Let's assume a 500 grams sample that contains 480 grams of copper nitrate, 5 grams of copper sulfate, 10 grams of sodium nitrate, and five grams of copper(II) oxide.
• We dissolve the 500 grams in 100 ml of hot water. The copper oxide is not soluble, so it is removed when we filter the hot solution.
• When we cool the solution to 0^oC, 138 g of copper nitrate remains in solution, so 480-138= 342 grams crystallizes out. (that's 342/480 or a 71 % recovery.)
• The 5 grams of copper sulfate will remain in solution because the solubility of CuSO₄ in cold water is much more than 5g/100 ml. (14.3 g/100 at 0^oC)
• Likewise all 10 grams of Sodium nitrate will remain in solution; the solubility is 78 g/100ml at 0^oC.
• The result is that we isolate only the copper nitrate, and recrystallizing greatly improves the purity of the product.

1. Example: A compound's solubility at 20^oC is 100 g/100ml in water, but is 25 g/100 ml in a 1:1 alcohol/water mixture. Let us start with 80 g dissolved in 100 ml of water. (It dissolves fully)
2. Now let us add 100 ml of alcohol. The solubility now drops to 25 g/100 ml. Since we have 200 ml of solvent (100 water + 100 alcohol), a saturated solution will contain 50 grams of the compound. The remaining 30 grams will appear as a solid--cystals if the process occurs slowly or as a powder if the process is rapid.
3. Our yield might be poor. We will only collect 30 out of the original 80 grams for a 30/80 or 38 % recovery. The goal of getting a much purer compound is achieved at the cost of lower yield.

One subtle feature about forming crystals. Tiny crystals have a lot of surface area and it costs energy to produce surfaces. If we put tiny crystals and large crystals in a saturated solution, we will discover that the small crystals dissolve and disappear, while the large crystals get bigger. Since it is much easier to rinse and collect crystals than finely divided powder, we will try to use conditions that favor the larger crystals.

In several of our examples we assumed that the solubility of one compound is not changed by the presence of another solute. This is not really a good assumption and we should expect the numbers to change, but the final conclusions are still valid.

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