Synthesis, Recrystallization, and Equation Determination of an Unknown Coordination Compound

7700 words (31 pages) Essay

8th Feb 2020 Chemistry Reference this

Tags:

Disclaimer: This work has been submitted by a university student. This is not an example of the work produced by our Essay Writing Service. You can view samples of our professional work here.

Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UKEssays.com.

Part A: Synthesis and Recrystallization

Part B: Determining the Equation and Calculating Percent Yield

 

Abstract

 The overall goal of this laboratory experiment was to determine an equation for a coordination compound, with the most accurate percent yield. This equation would resemble KaFeb(ox)c

dH2O. The compounds provided in order to complete this activity were K2C2O4

H2O, FeCl3

6H2O, KMnO4, and NaOH. To find the unknown equation, a synthesis of crystals was made by combining 8.31 g K2C2O4

H2O, and 4.11 g FeCl3

6H2O, with several other technical steps consisting of isolation, filter systems, hot water, and an ice bath. From these crystals, an equation could be found with titrations of NaOH and KMnO4. The mass percent of the reaction resulted to be 55.55% for oxalate, 19.25% for potassium, the iron was 15.36%, and the water was 9.84%. The equation calculated resulted to be K3Fe(ox)3

3H2O, and the percent yield was 40.28%. This finding suggested the crystallization could have been much more accurate in the determination of the equation, due to the low percent yield.

 

Introduction

 A coordination compound is a neutral compound made when a complex ion combines ligands and occasionally one or more counter ions. This compound results from the combination of a Lewis acid and a Lewis base (Tro G-5). The bonds in this reaction are coordination covalent bonds. This bond is in result of a ligand donating lone pairs of electrons to an empty orbital of a metal in a complex ion (Tro G-5). Whereas a counter ion is an ion that bonds with ionic species in order to maintain neutrality (Tro G-5). The number of ligands bound to the metal ion is called a coordination number. The compounds can also form more than one coordination bond with deems them as either bidentate or polydentate. Coordination compounds prove crucial to a variety of different living organisms due to the role they play in daily life. Coordination compounds in society include vitamin B12, hemoglobin, chlorophyll, dyes and pigments, and catalysts. Vitamin B12 is a water-soluble coordination compound that is essential to a number of organisms, including humans. Vitamin B12 is responsible for aiding in the development of red blood cells. Vitamin B12 also is responsible for cellular metabolism in enzyme forms. Additionally, vitamin B12 works with folic acid in the development of DNA. Vitamin B12 also is responsible for the creation of fatty acids in the myelin sheath that surrounds nerve cells (Encyclopedia Britannica). Another important role of coordination compounds is hemoglobin and chlorophyll. Chlorophyll is the molecule that absorbs sunlight to transform CO2 and water into simple glucose sugars. This process is responsible for sustaining the life of all plants (May). Another one of the best coordination compounds is hemoglobin. Hemoglobin is a red protein responsible for transporting oxygen in the bloodstream. Hemoglobin carries oxygen to the bodies organs and tissues, and then sends it back through as carbon dioxide to your lungs (Mayo Clinic). Coordination compounds can also be found in the fight against pest control. These compounds are used as pesticides in common agriculture (Singh). Another role of coordination compounds is in chelating agents. These coordination compounds are either bidentate (having two attachment sites) or polydentate (having more than two attachment sites). A chelating agent used in medicines to capture heavy metals that poison the body, such as lead. This specific agent is called EDTA. EDTA collects free lead ions and removes them from the blood tissue and exerts it from the body (Craven). Overall, coordination compounds are very helpful and are found more commonly than one would expect in society, the community, and in industry. Coordination compounds can only fulfill their specific duties in society with the correction ratio and equation. In the experiment assigned, we were suppose to create and analyse a coordination compound of iron involving a bidentate oxalate with the basic formula of KaFeb(ox)c

dH2O, to determine its overall equation. In order to find this, we used an ice bath, titrations, and ion exchange. Ice baths are commonly used in chemistry in order to take heat from the reaction in order to slow it down or reduce solubility. Additionally, ion exchanges are used in chemistry to remove ions from a solution and separate ions from each other. These separations are used commonly in scientific practice to effect purifications and to aide in the analysis of unknown mixtures (Walton). Additionally, titrations in chemistry is when a known concentration is used to determine an unknown concentration. In this experiment, we prepared this compound by simply mixing potassium oxalate and iron (III) chloride solutions. This compound is not water-soluble so crystals synthesized. We determined the equation through tedious work of titrations, and ion exchanges.

Experimental

 The purpose of this lab was to determine the equation of a coordination compound with the base equation KaFeb(ox)c

dH2O. First, using a dial-a-gram we weighed out 8.31 g of K2C2O4

H2O onto a wax paper and transferred it into a 125-mL Erlenmeyer flask. Next we added 25-mL of deionized water into the same Erlenmeyer flask and shook the flask gently until it was dissolved. We then weighed out 4.11 g of FeCl3

6H2O, KMnO4 and placed it in a separate 150-mL beaker. From there, we added 10-mL of deionized water to the 150-mL beaker and stirred it until it was completely dissolved. Next we added the oxalate solution to the iron (III) chloride solution in the 150-mL beaker and stirred thoroughly. We then used 3-mL of deionized water to wash the aqueous K2C2O4

H2O remaining in the original 125-mL Erlenmeyer flask and transfer it into the 150-mL beaker. Then we set the 150-mL beaker into one of the drawers and covered it with a square cardboard piece. We checked the solution every 15 minutes to observe the changes, over a total interval of 45 minutes. While the solution was isolated, we obtained a glass baking dish and when it was close to the times end we added ice and water to make an ice bath. Next, we placed the solution in the ice bath for 30 minutes, with the cardboard still on. After the ice bath, we filtered the solid crystals with filter paper and a Buchner funnel. We sprayed the 150-mL beaker with deionized water into the filter paper just just the crystals remained on the paper. Then we heated up 15-mL of water in a 150-mL beaker on the hot plates to 80-90

.

Next we recrystallized the products to remove impurities. We did that by placing the solid crystals back into the 150-mL beaker and added the hot water to the solution in small volumes to minimize the overall volume. We then stirred vigorously to dissolve the crystals while the beaker was on the hot plate at very low temperature to keep warm. Once that was finished, we covered the beaker with a watch glass and stored it in an isolated location. Next, using the analytical balance we weighed out the mass of our green crystal sample which ended up being 3.1564 g. Next, using an analytical balance we weight out two samples of our overall crystal product. One sample was .1249 g, and the other was .1252 g. We then placed the samples in two different Erlenmeyer flasks. To both of the separate flasks, we added approximately 60-mL of deionized water, 1 mL of 85% concentrated H3PO4, and 6-mL of 6 M H2SO4. We then obtained about 100 mL of the KMnO4 solution in a beaker and covered it with an upside down beaker. We then heated one of the solutions in the Erlenmeyer flask on a hot plate (the .1249 g solution). Using a thermometer, the solution was heated until it reached 80

.

During the heating process, one person held the thermometer in the solution, making sure not to touch the bottom on the flask so it would measure the temperature of the solution, not the glass. Also during that time, a burette was arranged above the flask in the clamps so that the titration would go directly into the solution. Additionally, we rinsed the burette with three small 2-3-mL portions of the KMnO4 solution, twisting the burette so the solution covered the whole inside. Then we filled the burette with 50.03-mL of KMnO4 and released any air bubbles in the tip of the burette. When a temperature of 80

was reached by the heating solution, we placed a magnetic stir bar in the flask and turned the hot plate to low. We titrated without stopping until the solution turned clear. Once the solution was clear, we titrated in 1-mL increments until the solution turned a light pink color and stayed that way for a minimum of 30 seconds. We then recorded the final amount titrated to be a volume of 19.14-mL. We repeated this entire process for the second solution of .1252 g. The burette this time was instead filled to 40.45-mL. The final burette volume resulted to be 9.50-mL. We then calculated the number of moles of oxalate in the sample. That was done by using the amount titrated of KMnO4 in the two samples and converting that to two different MnO4 mole values. Next we converted moles of MnO4 to moles of oxalate from the balanced equation. We then turned those moles into grams by using the molar mass of oxalate. Next we divided the mass amount calculated by the whole sample to get the percent of oxalate in the compound. Next we mounted the ion exchange column on the ring stand. Using a 10-mL graduated cylinder we rinsed the column by pouring 4-mL of deionized water into it and collected the liquid that came from the bottom of the column in s a small 50-mL beaker. Using a piece of pH paper, we tested the pH that came from the column to make positive that it was acidic. Then we let the level of the rinse liquid to fall to the level of the resin and repeated this process three times. When the level of the water during the last rinse dropped to the level of the resin, we tested a drop of it onto pH paper to make sure the pH was now neutral. This was repeated three times in order to reach a neutral pH. We weight out 0.1623 g of the green crystals into a 50-mL beaker using the analytical balance and wax paper. Next, using a 10-mL graduated cylinder we added 4-mL of deionized water. We then gently swirled the solution until it was fully dissolved. We placed a clean, dry 150-mL beaker under the ion exchange column and transferring the weighed out green crystals in. We collected the excess in the 150-mL beaker below the column. When the level of the liquid dropped to the top of the resin, we added another 4-mL of deionized water to the 50-mL beaker that contained the crystals. We transferred the new solution into the exchange column and collected the excess in the same 150-mL beaker. This procedure was then repeated exactly with two more 4-mL deionized water samples. The vernier system, a burette, and a pH probe were then arranged. Before use, we calibrated the ph probe using buffers with a pH of 7 and a pH of 4. We then placed the 150-mL beaker on a stir plate with a stir bar inside, placing the pH probe inside the beaker, making sure it did not touch the stir bar. We obtained approximately 25-mL of NaOH and placed it in the burette. We titrated in 1-mL increments. We titrated 11-mL of NaOH and two equivalence points were reached at 4-mL and 7-mL. We continued the titration until the solution reached a pH of 11.12. We then repeated the titration with the second sample of the green crystals. The titration took 12-mL to reach a pH of 11.26, and the equivalence points were at 5-mL and 8-mL.

Results and Discussion

Overall, this experiment perceived to be a success. The broad goal of this experiment was to create a coordination compound through titrations, determine the equation, and have the highest percent yield possible. In the end, we completed all of the aspects of the goal. We correctly followed each step and determined an overall equation for the coordination compound. The equation ended up to be K3Fe(ox)3

3H2O. We also determined the percent yield to be 40.28%.

In the beginning of this experiment, we combined 8.31 g of K2C2O4

H2O and 4.11 g of FeCl3

6H2O. After isolation, we noticed the green crystals started to form in nearly 10 minutes, because the reaction in insoluble in water. After the crystals were purified and began to form fully, we were quite pleased with our resulting crystals. There appeared to be a good amount of crystals. The only problem with this portion of the experiment, was when we were recrystallizing our crystals. When we recrystallized, instead of pouring the small amounts of hot water into the beaker with the crystals as needed for them to dissolve, we poured all of the 15-mL of water into the crystals. They clearly dissolved quickly, but I am mostly positive that the reason behind our low percent yield was due to this error. Because there was too much water, not as many crystals formed.

The overall mass of our crystals after recrystallization was 3.1564 g. Additionally, to go off from what was completed in the part one, from these crystals we were able to find the overall equation for the coordination compound and the percent yield. To determine the amount by percent of oxalate in the compound, samples of the crystals (.1249 g and .1252 g) was placed into a combination of 60-mL of deionized water, 1 mL of 85% concentrated H3PO4, and 6-mL of 6 M H2SO4. This sample was then titrated with KMnO4 in order to find the percent by mass of oxalate in the compound. We determined the average amount of oxalate in the compound to be 55.55% by mass. The reason we were able to find the mass percent by titrating it with KMnO4 is because when the crystals reacted with KMnO4, it was the oxalate that reacted directly with the titration. From there, we determined the amount of potassium percent by mass and iron percent by mass in the compound. We did that by first using an ion exchange to get a correct neutral pH of the compound being titrated (the crystal samples). We then titrated it with NaOH until the pH was roughly 11. Our overall pH in this process resulted to be 11.2. From there the amount by mass of the compounds could be calculated by determining equivalence points. The percent by mass of Fe3+ resulted to be 15.36%. The percent by mass of K+ resulted to be 19.25%. These amounts could be found because the amount of moles of K+ was directly proportional to the amount of H+ given off by the NaOH reacting with the crystals. The percent by mass of water was found by adding up the percents already calculated of potassium, oxalate, and iron, then subtracting them by 100%. This percent by mass ended up being 9.84% H2O. From these percents, we then transformed those to grams (which was the same number just changed to mass). We then transformed the amount in grams of each substance to moles. The amounts of moles of each substance was then individually divided by the smallest amount of moles out of the substances, which ended up being iron. That is because iron was the limiting reactant. That final calculation allowed us to determine the overall equation which was  K3Fe(ox)3

3H2O. This resulting equation makes sense because it is a common coordination compound equation used in laboratory activities. Potassium ferrioxalate trihydrate is commonly synthesized because it is a less complex compound to synthesize. This compound is also a very good coordination compound. The oxalate (C2O42-) has points of attachment at two spots on the iron, so it is bidentate. Three oxalate ions place themselves around the iron in an octahedral geometry. The iron clearly has a +3 charge in order to balance out the -3 charge of the ion. The iron chelate crystalized out of the solution as the potassium (Ohlone).

Possible percent errors that could have occured in this experiment were incorrect addition of hot water to the crystals, obtaining the KMnO4 to quickly, not having the pH probe perfectly calibrated, human error when measuring, and more accurate data could have been determined by doing a second titration with the NaOH determining K+ and Fe3+. In the beginning of the first part of the experiment, when creating our crystals, we added in all of the hot water instead of small increments in order for it to dissolve and create the purest crystals. This could have lowered the amount of crystals created, making the mass we obtained smaller than it should have been. In the second part of the experiment, the KMnO4 was suppose to be obtained directly before it was used in titrating because the surrounding oxygen could have messed with it. We obtained our KMnO4 about 15 minutes before actually using it, but once we noticed this error we placed an upside down beaker over the top opening to cut off oxygen flow. This could have affected our first titration in determining the oxalate. Additionally, the pH probe could have been off, indicating incorrect equivalence points when determining K+ and Fe3+. Also, incorrect human eye measurement could have effect the amount of substance was being titrated. The volume amounts when titrating could have been off by a little, making our overall moles slightly different as well. Finally, we were initially supposed to run two titrations with both the KMnO4 and NaOH in order to get an average. We did not have time to complete a second NaOH titration, and because of that we only had one data trial. Our trial could have been off or incorrect, and because we do not have a second trial it is the only one that our data is based off it. Overall, there were several issues that could and did happen when completing this experiment. In the end though, we came up with an equation that made sense of K3Fe(ox)3

3H2O , and a decent percent yield of 40.28%.

Table 1:

 Table 1 describes the amount of mass we used of K2C2O4

H2O and FeCl3

6H2O when creating our crystals. The amounts used were 8.31 g and 4.11 g.

Reactant

Mass

K2C2O4

H2O

8.31 g

FeCl3

6H2O

4.11 g

 

Table 2:

 Table 2 represents the volume of substance used when titrating. During the titration to find the oxalate, we titrated 30.89 mL of KMnO4 in the first titration, and 30.95%. To find K+ moles, we used 4.12 mL of NaOH. To find Fe3+ moles, we used 7.443 mL of NaOH.

Titration

Amount in mL Titrated

Trial One: KMnO4

30.89 mL

Trial Two: KMnO4

30.95 mL

Trial One: Using NaOH to find K+

4.12 mL

Trial Two: Using NaOH to find Fe3+

7.443 mL

 

Trial 3:

 Table 3 represents the percent by mass of each compound in the equation K3Fe(ox)3

3H2O. These percents are found by the amount of mass compare to the sample mass being experimented on.

Substance

Percent by Mass

C2O42-

55.55%

K+

19.25%

Fe3+

15.36%

H2O

9.84%

 

Table 4:

 Table four represents the overall percent yield of the crystals. This is found by using the amount of mass created of the crystals, divided by the theoretical amount that could have been produced if completely perfectly, then multiplied by 100.

Percent Yield

40.28%

Titrations: Determining the Percent Mass of  C2O42-

Titrations in this experiment were used to determine the moles of a specific compound. For the first titration, the goal was to find the oxalate concentration by titrating the KaFeb(ox)c

dH2O crystals with KMnO4. Once the substance turned pink, we observed the amount of KMnO4 used to reach that point. From that volume in mL, we were able to find the moles of MnO4. From the moles of MnO4we were able to calculate the moles of C2O42- from the balanced equation 16H+(aq) + 5C2O42- (aq) + 2MnO4 (aq)

10CO2(g) + 2Mn2+ (aq) + 8H2O (l). We then took the moles of C2O42- and converted it to grams of oxalate using the molar mass of oxalate. From there, we divided the amount of grams of oxalate we calculated by the amount in grams of the crystals to get the percentage of C2O42- in the compound. We completed this cycle with both samples of the crystal compound. Then after the mass by percent was found for both titrations, we added them up and divided by two to find the average.

1st Titration: initial burette volume: 50.03 mL      final burette volume: 19.14 mL

-          Amount of KMnO4 used: 30.89 mL

-          30.89 mL KMnO4 (

1 L1000 mL

)(

0.01021 mol1 L

) = .0003154 mol MnO4

.0003154 mol MnO4 (

5 mol C2O4 22 mol MnO4 

) = .0007885 mol C2O42-

.0007885 mol C2O42- (

88.018g1 mol

) = .06939 g C2O42-

.06939 g C2O4 2.1249 g crystals

x 100 = 55.56% of compound

2nd Titration: Initial burette volume: 40.45 mL           final burette volume: 9.50 mL

-          Amount of KMnO4 used: 30.95 mL

-          30.95 mL KMnO4 (

1 L1000 mL

)(

.01021 mol1 L

) = .0003159 mol MnO4

.0003159 mol MnO4 (

5 mol C2O4 22 mol MnO4 

) = .0007899 mol C2O42-

.0007899 mol C2O42- (

88.018g1 mol

) = .06953 g C2O42-

.06953 g C2O4 2.1252 g crystals

x 100 = 55.53% of compound

-          Average: 55.53 + 55.56 = 111.098 / 2 = 55.55%

Determining the Percent Mass of K+

 We determined the K+ concentration by titrating it with NaOH. Once the equivalence point was reached, the amount of NaOH titrated was equal to the amount of H+ in the solution. The amount of H+ moles in the solution is equal to the amount of K+ moles in the solution.We began by titrating the solution to find the equivalence points. Once the equivalence points were found, we took the first derivative of the amount in mL at the equivalence point, and the first derivative of the amount after the equivalence point. We then added those two derivative amounts together. From there, we divided the amount of the first derivative at the equivalence point by the total of the two together. We then added the number in mL to that calculated value to get an overall mL of NaOH approximation for the equivalence point. From that equivalence point in mL, we used stoichiometry to determine the amount of moles of NaOH. From the moles of NaOH, we changed to moles of H+. From there, we knew that the moles of H+ was equal to moles of K+. From the moles of K+ we changed that into grams using the molar mass of potassium. From there we divided the amount of K+ calculated by the amount of green crystals in our sample to get the percent of K+ in the sample.

-          Titration One:

-          First equivalence point: 4 mL = 0.243     5 mL = -0.347

0.243 + 0.347 = 0.59

0.243/0.59 = 0.412 + 4 = 4.412 mL

-          4.412 mL NaOH (

1 L1000 mL

)(

.1109 mol1 L

) = .000489 mol NaOH

.000489 mol NaOH (

1 mol H+3 mol NaOH

) = .000163 mol H+

.000163 mol H+ = .000163 mol K+

.000163 mol K+ (

39.0983 g1 mol

) = .06377 g K+

.006377 g K+/ 0.1623 g  x 100 = 19.25 %

Determining Percent Mass of Fe3+

 We determined the K concentration by titrating it with NaOH. Once the equivalence point was reached, the amount of NaOH titrated was converted to the amount of Fe3+ moles in the solution. We determined the Fe3+ concentration by titrating it with NaOH. We began by titrating the solution to find the equivalence points. Once the equivalence points were found, we took the first derivative of the amount in mL at the equivalence point, and the first derivative of the amount after the equivalence point. We then added those two derivative amounts together. From there, we divided the amount of the first derivative at the equivalence point by the total of the two together. We then added the number in mL to that calculated value to get an overall mL of NaOH approximation for the equivalence point. From that equivalence point in mL, we used stoichiometry to determine the amount of moles of NaOH. From the moles of NaOH, we changed to moles of Fe3+. From the moles of Fe3+ we changed that into grams using the molar mass of Fe3+. From there we divided the amount of Fe3+ calculated by the amount of green crystals in our sample to get the percent of Fe3+ in the sample.

-          Titration One:

-          Second equivalence point: 7 mL = -0.191   8 mL = 0.240

0.240 + 0.191 = 0.431

0.191/0.431 = 0.443 + 7 = 7.443 mL

-          7.443 mL NaOH (

1 L1000 mL

)(

.1109 mol1 L

) = .000825 mol NaOH

.000825 mol NaOH (

1 mol Fe3 mol NaOH

) = .000275 mol Fe3+

.000275 mol Fe3+ (

55.845 g1 mol

) = .01536 g Fe3+

.01536 g Fe3+/0.1623 g x 100 = 15.36%

 

Determining Percent Mass of Water

 The number of water molecules was calculated by taking the percent by mass of everything that was found in the titrations ( Fe3+, K+, and C2O42-) and adding them together, and subtracting it from 100%.

-          100% – (15.36% + 19.25% + 55.55%) = 9.84% H2O

Determining Overall Formula KaFeb(ox)c

dH2O

The overall goal of this experiment was to find the equation of the coordination compound KaFeb(ox)c

dH2O. In order to do this, we converted each of the percents of the compounds in the equation to grams (the numbers stayed the same, just switched to mass). We then transformed the amount of grams into moles of each substance using each one’s specific molar mass. Next we divided the amount of moles by the smallest amount of moles which was Fe3+. From that we ended up getting the overall amount of each compound in the coordination compound.

-          c = 55.55 g C2O42- (

1 mol88.018g

) = 0.6311 mol C2O42- / 0.2750 Fe3+ = 3

-          a = 19.25 g K+ (

1 mol39.098g

) = 0.4924 mol K+ / 0.2750 Fe3+= 3

-          b = 15.36 g Fe3+ (

1 mol55.85g

) = 0.2750 mol Fe3+/ 0.2750 mol Fe3+ = 1

-          d = 9.84 g H2O (

1 mol18g

) = 0.5467 mol H2O / 0.2750 Fe3+= 3

-          KaFeb(ox)c

dH2O

Formula:K3Fe(ox)3

3H2O

Percent Yield

 Another goal of the experiment was the create the highest possible percent yield. The percent yield is how close one was to the actual. For our percent yield, we took the total mass of our crystal product we created and divided it by the actual mass of the crystal product.

  1. Determining the limiting reactant

(i) K2(C2O4)

H2O : 8.31g

1 mol184.2309g

= 0.045106 mol

 (ii) FeCl3

6H2O : 4.11g

1 mol270.2957g

= 0.015206 mol

       b. Finding theoretical mass using limiting reactant

 (i) 0.045106 mol x

1 mol FeCl3 6H2O3 mol FeCl3 6H2O

x

491.2427g1 mol

=7.386g

       c. Finding percent yield

 l

3.156 g Theoretical mass7.836g Actual mass

l x 100 = 40.28%

Conclusion

 The creation of crystals in this lab consisted of 8.31 g of K2C2O4

H2O and 4.11 g of FeCl3

6H2O. After synthesizing, recrystallizing, and letting the crystals set, we were then able to take several steps in order to determine the equation of the coordination compound. We began by determining the amount by percent of oxalate in the solution by titration the crystalline with KMnO4. Once the substance turned pink, that was equal to the moles of oxalate in the system. The percent by mass of oxalate in the compound was 55.55%. We then determined the K+ percent by mass by titrating the crystal compound with NaOH. When the equivalence point was reached, the NaOH amount titrated was equal to the H+ in the solution. The amount of H+ in the solution was equal to K+. The percent by mass of K+ resulted to be 19.25%. Another titration was then used to determine the amount of Fe in the solution. The percent by mass of the Fe resulted to be 15.36%. With all of these percents, we were able to calculate the percent of water in the compound by subtracting all the calculated percentages by 100%. The percent by mass of water ended up being 9.84%. These percents were turned to mass and converted to overall moles of the compound. The final equation ended up to be K3Fe(ox)3

3H2O. From there we calculated the percent yield of our crystal product from the limiting reactant moles. The amount of product we received, we divided by the actual mass of the product that could be produced. The percent yield resulted to be 40.28 %.

 Overall, the coordination compound synthesis experiment appeared to be a success. The equation resulted to be K3Fe(ox)3

3H2O, which is a coordination compound with a bidentate ligand. The metal ion was iron, the attaching ligand was the oxalate, and the counterion was the potassium. The oxalate is deprotonated, so that means it is -2 charge for the oxalate, the iron has a +3 charge, so overall that is a charge of -3, because there is two oxalates. To make the coordination compound neutral, three counterions must be added, which was the three potassiums. 

References

  • Britannica, T. E. (2018, November 22). Vitamin B12. Retrieved from https://www.britannica.com/science/vitamin-B12
  • Coordination Compounds. (n.d.). Retrieved from http://www2.ohlone.edu/people/jklent/labs/101B_labs/Coordination Compounds.pdf
  • Craven, R., Dr. (n.d.). Applications of Coordination Compounds. Retrieved from https://study.com/academy/lesson/applications-of-coordination-compounds.html
  • Hemoglobin test. (2017, December 22). Retrieved from https://www.mayoclinic.org/tests-procedures/hemoglobin-test/about/pac-20385075
  • Importance of coordination compounds. (n.d.). Retrieved from http://eguruchela.com/chemistry/learning/Importance_of_coordination_compounds.php
  • May, P. (n.d.). Chlorophyll. Retrieved from http://www.chm.bris.ac.uk/motm/chlorophyll/chlorophyll.htm
  • Singh, R. V., & Srivastava, S. (2013, June). Role of Coordination Compounds as Good Pesticides. Retrieved from https://www.researchgate.net/publication/288045560_Role_of_coordination_compounds_as_good_pesticides
  • Walton, H. F. (2016, October 17). Ion-exchange reaction. Retrieved from https://www.britannica.com/science/ion-exchange-reaction
  • Canvas, Classmates, Mariya Jo Ray, Previous Labs, Chemistry 233 Lectures, Anna Sandstrom, Steven Shelofsky

Cite This Work

To export a reference to this article please select a referencing stye below:

Reference Copied to Clipboard.
Reference Copied to Clipboard.
Reference Copied to Clipboard.
Reference Copied to Clipboard.
Reference Copied to Clipboard.
Reference Copied to Clipboard.
Reference Copied to Clipboard.

Related Services

View all

DMCA / Removal Request

If you are the original writer of this essay and no longer wish to have your work published on the UKDiss.com website then please:

Related Lectures

Study for free with our range of university lectures!