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Introduction

Materials

MSDS Safety Sheets

Composition of a Compound Procedure

Iron Determination Procedure

Observations

Discussion Applications IvanF's Daily Journal Drew's Daily Journal Alana's Daily Journal Bibliography Links

As we mentioned in our introduction, one of the most important parts of chemistry is getting

the recipe exactly accurate for chemical reactions. But just as it is in the kitchen, the recipe cannot always come out perfect. In the discussion, we will discuss the strengths and weaknesses of our results.

In general, Composition of a Compound is an experiment about the Kinetic Molecular

Theory. As we heated the contents of the Erlenmeyer Flask, the Potassium Chlorate absorbed the heat or kinetic energy from the Bunsen burner. Each Potassium Chlorate molecule began to move faster and faster from the increase in kinetic energy, and the attractive bonds between molecules began to weaken. (5, pg. 27) Eventually, enough kinetic energy is added that the Potassium Chlorate molecules actually separated far enough from each other to change state. The Potassium Chlorate began to boil and bubble after it's melting point was reached, and expanded into a liquid form.(4, pg. 207)

While the space between molecules continued to expand, the space between the atoms of

each molecule began to extend as well. After the melting point of the Potassium Chlorate had been reached, each atom within the molecules began to break apart due to high levels of kinetic energy. Oxygen seemingly has the lowest boiling point, and therefore was the only element that was able to completely break apart from the rest of the compound.

The oxygen and chorine atoms were bonded with each other in a covalent bond. In this

type of bond, electrons are shared between the two atoms.(4, pg. 80) When heat was added, the oxygen easily broke away from the chlorine atoms because the oxygen did not need to retrieve any of it's electrons. In other words, the chlorine had not taken any of the oxygen's electrons, thus the oxygen could leave without having to bother to attract back it's valence electrons.

This breaking apart (sublimation of oxygen) from the compound caused us to see bubbling

within the melting Potassium Chlorate. The Potassium Chlorate had not reached it's boiling point, and thus was not what caused the bubbling. However, the oxygen had reached it's boiling point, and it's attempts to escape from within the Potassium Chlorate formed the bubbles we saw.(4, pg. 215)
Unfortunately, we could not find the melting point temperature of Potassium Chlorate or Potassium Chloride. However, it is can suggested that we had not reached the melting point of Potassium Chloride. After all of the oxygen had boiled and left the compound, there was nothing left in the Potassium Chlorate that had a melting point reachable by a bunsen burner. Thus, the product of the decomposition process was a solid once again as it had been at the start of the experiment.

The process is known as decomposition because we actually decomposed a pure substance

into 2 different parts. We separated the oxygen from the Potassium Chloride in the Potassium Chlorate. This is an effective way to mass the amount of an element within a compound because after all of that element has evaporated or sublimated (in this experiment, the element was oxygen), one can determine the mass of the element liberated by comparing the mass of the product with the mass of the reactant.
In Composition of a Compound, in both successful trials, the resulting combined mass of the Erlenmeyer flask and the Potassium Chloride was 39.5 g. Also, in previous trials that we did not count valid due to interrupted results, the combined mass of the flask and Potassium Chloride was approximately 39.5 g as well. Therefore, in all successful trials and all completed, unsuccessful trials, the mass of the Potassium Chloride was 1.1 g and the mass of the liberated oxygen was 0.9 g.

However, prior to the experiment, we hypothesized that the amount of oxygen to be

released during the experiment would be 0.783 g. This is because O3 has a formula weight of 48.00 g/mol. Potassium Chlorate has a formula weight of 122.6 g/mol. Therefore, the theoretical percentage of oxygen in Potassium Chlorate is 39.15% (48.00 g/mol divided by 122.6 g/mol = 0.3915). 39.15% of 2.0 grams (of Potassium Chlorate) equals 0.783 g or 7.83 x 10-1 g of oxygen (2.0 g x 0.3915 = 0.783 g). Unfortunately, this theoretical result does not concur with the results tabulated during the experiment. The difference between the theoretical and actual amount of oxygen in Potassium Chlorate is 0.1 g. Therefore, the theoretical amount of oxygen in 2 g of Potassium Chlorate is 0.01631 mol or 1.631 x 10-2 mol (n mol = 0.783 g divided by 48.00g/mol), while the actual amount of oxygen molecules in 2 g of Potassium Chlorate is 0.19 mol or 1.9 x 10-1 mol (n mol = 0.9 g divided by 48.00 g/mol).

How is it possible to have such a large discrepancy between the results of the two

calculations? It is possible that because the Potassium Chlorate was exposed to the air when we opened the jar (we had opened it several times since we completed 6 trials of the experiment), moisture from the atmosphere may have dissolved into the Potassium Chlorate, thus making the reactant impure. In other words, when we weighed 2 g of Potassium Chlorate on the electronic balance, it is plausible that we were actually weighing 1.9 g of Potassium Chloride with 0.1 g of Dihydrogen Oxide and other atmospheric elements.

Although the mass of the product of the combustion process, Potassium Chloride,

remained constant throughout all of our trials, the measured volume of oxygen liberated (calculated in mL) did not. In our 4th trial for instance, no water in the oxygen-catching test tubes was displaced by the oxygen being liberated from the Potassium Chlorate. The combined mass of the flask and the Potassium Chloride was approximately 39.5 g by the time that we ended that trial, but not a single air bubble of oxygen had formed within the water basin. Therefore, no oxygen had even reached the water basin. We later checked the tubing for any leaks and found none. We checked the rubber stopper for a leak and found none. Besides, we later proved that there could not be a leak because in the following trial, the experiment worked perfectly with the same stopper and apparatus configuration as before. Our only explanation is that perhaps we did not seal the flask tight enough with the rubber stopper (we had removed the rubber stopper after the unexplainable trial to relieve the pressure inside the flask). Or it is possible, however unlikely, that the oxygen liberated had chemically bonded with the Dihydrogen Oxide in the tubing and never made it to the water basin. The high electronegativity of hydrogen in the Dihydrogen Oxide may have attracted oxygen molecules into bonding with it.(4, pg. 264)

The Dihydrogen Oxide (H2O) in the tubing was most likely the cause of our problems in

trial 5. In this trial, the oxygen liberated from the Potassium Chlorate seemed to be of an unlimited amount. By the time that we ended the fifth trial of the experiment, we had already gathered over 800 mL of oxygen, and still the oxygen bubbles continued to surface in the water basin! After weighing the flask and the Potassium Chloride at 39.5 g, we repeated the combustion procedure and only the usual 50 mL of air (the atmospheric air that occupied the flask and the tubing) displaced the Dihydrogen Oxide in the test tubes. No noticeable amount of oxygen evaporated from the remaining Potassium Chlorate in the flask. What could have caused the oxygen volume levels to exceed our estimates by seemingly an infinite amount?

One explanation was hinted at while we let the apparatus cool down after the 5th trial.

We noticed that water was being pulled upwards through the tube and all the way to the rubber stopper. However, as soon as the water hit the stopper, the Dihydrogen Oxide flowed back down to the water basin, only to repeat this water-suction phenomenon a minute later. This occurred most likely because the combustion of the air within the flask caused a low pressure system around the Potassium Chloride. The lack of gas particles created a point of low molecular concentration within the flask. The water in the basin was vacuumed into the flask through diffusion (from a point of high concentration to a point of low concentration). Therefore, we hypothesize that during the experiment, due to the low pressure system created by combustion, the Dihydrogen Oxide within the tubing was vacuumed so close enough to the rubber stopper, and to the Bunsen burner, that it evaporated. The water molecules were instantly converted into a gas by the increase in kinetic energy, and this constant evaporation of Dihydrogen Oxide resulted in an unlimited supply of air displacing the water in the test tubes.

It is also possible that the low pressure system caused the Dihydrogen Oxide in the tubing to

evaporate at extremely low temperatures. At 1.2 kPa, water can boil at just 284 K. This is known as vacuum distillation and the low pressure system created in our experiment may have allowed the warmth of the tubing to evaporate the Dihydrogen Oxide.(4, pg. 207)

The results of the 6th trial were the most similar to our estimates and hypothesis out of all

conducted trials. Prior to the experiment, we predicted that 2 g of Potassium Chlorate contains 0.365 L or 365 mL of oxygen @ [STP] ( V [L] = 0.01631 mol x 22.4 L). This estimate was achieved after realizing that because there is 0.01631 mol of Potassium Chlorate, there is also 0.01631 mol of oxygen (as long as our chemical equation is properly balanced). We realized that during the decomposition process, one oxygen molecule evaporates from each Potassium Chlorate molecule. Therefore, for every Potassium Chlorate molecule, one oxygen molecule is liberated.(4, pg. 180)

In our 6th trial, 49 mL of Dihydrogen Oxide was displaced by the atmospheric air in the flask

and tube. That is 1 or 2 mL less than the volumes of air recorded on previous successful and unsuccessful trials. Our final volume of displaced Dihydrogen Oxide was 389 mL. By subtracting the 49 mL of air from the end result, we conclude that 340 mL of oxygen was liberated from the Potassium Chlorate, which equals 0.01414 mol or 1.414 x 10-2 mol at 1 atm pressure (101.33 kPa) and at room temperature, 20 degrees Centigrade[n mol = (101.33 kPa x 0.340 L) divided by (8.314 {kPa x L} divided by {mol x K} x 293 K)]. We did our calculations at room temperature because the experiment was done within the confines of a 293 K or 294 K chemistry room.

Once again, there is a large discrepancy between the two values of mols. There is a

difference of 2.170 x 10-2 mol between the theoretical and actual amount of oxygen according to the volume measurements. One likely reason for this was hinted at when we measured only 49 mL of air during the 6th trial of Composition of a Compound, instead of the usual 50 or 51 mL. It is possible that a leak in the apparatus allowed a mL or two of the atmospheric air to escape, and therefore makes it feasible to suggest that some of the liberated oxygen was able to escape the apparatus as well. It is also conceivable that the errors in our results are largely due to moisture in the Potassium Chlorate. Lastly, it may also be possible that some of the oxygen liberated from the Potassium Chlorate was cooled down by the Dihydrogen Oxide in the basin or tubing. This cooling may have allowed the oxygen to chemically bond with the water or elements within the used, non-distilled water to produce Dihydrogen Oxide or other products. Thus, the oxygen would lose it's gaseous form and not displace any water in the test tubes.

However, the most likely reason for the discrepancy is that we did our calculations and

estimates at [STP]. At the time that we did our estimates, we could not predict the pressure and temperature of the atmosphere on the day of the experiment. [STP] includes the pressure of 1 atm and a temperature of 273.15 K. If the temperature in the room of the experiment was about 293 K rather than 273 K, it would result in faster moving particles that, according to the KMT theory, would loosen the attractive bonds between molecules and cause the volume of the gas to expand further. However, this theory explains how we may be able to obtain more oxygen volume than anticipated, but does not explain why we received less volume than expected. After redoing our estimates for the volume of oxygen, this time with room temperature in mind, we calculate that 392.1 mL of oxygen would be liberated theoretically [ V = (0.01631 mol x 293 K x universal gas constant) divided by 101.33 kPa]. This is 52.1 mL higher than the actual volume of oxygen liberated.

The amount of oxygen molecules liberated, according to the weighing of mass after

combustion, was closer to our original estimations than the captured volume of liberated oxygen, and therefore the weighing of mass would be a more accurate way of determining an element's weight in a compound. This is because the weighing of mass does not depend on annoying variables such as pressure, temperature, and leaks in the apparatus.

Prior to the experiment, we were given the question of determining which of the following

mines would yield the greatest amount of iron per ton of ore mined: a mine of FeO, a mine of Fe2O3, and a mine of Fe3O4. The mine of Fe2O3 would yield the most iron per ton of ore mined because oxygen makes up 30% of this compound's (named Ferric Oxide) formula weight, compared to 22.3% for FeO (named Ferrous Oxide) and 27.7% for Fe3O4. Also, the ratio of Oxygen to Iron is the greatest in Fe2O3; the ratio is 3:2, compared to 4:3 for Fe3O4 and 1:1 for FeO, and therefore, Ferric Oxide will yield more iron per ton of ore mined than any of the other iron compounds.

Iron Determination was a simple experiment to complete, but a difficult experiment to

control all of the variables. In other words, it was easy to cook the recipes needed in this experiment, but our greatest problem was the uncertainty of whether or not we had prepared our ingredients acceptably or not.

Iron Determination is an experiment based upon titration: a process that requires a solution

with a known molarity of compound and mixing it with a solution with an unknown molarity of a compound.(4, pg. 287) This experiment involved the slow dropping of a Potassium Permanganate solution into a 10 mL sample of an Iron (II) Sulphate solution. During this process, a Permanganate ion (with the ability to lose as many as 7 valence electrons) would enter the Iron solution and be surrounded by 5 Ferrous ions (with each Ferrous ion having 2 electrons in their valence shell).

Surprisingly, the elements do not bond together, perhaps because the bond would be merely

a covalent bond (difference of only 0.3 in electronegativities). Iron has an electronegativity of 1.8, which is 0.3 units higher than the Permanganate. In theory, a difference of 0.3 would mean that the electrons would be shared among the molecules. Instead, the Ferrous ions accept valence electrons from the Permanganate. Each of the 5 Ferrous ions that surround the Permanganate ion receive one valence electron. Each of the Ferrous ions leave with a charge of +3 electrons, while the Permanganate ion is left with just 2 valence electrons. At this point of the titration, the purple colour of the Permanganate does not stay because it loses it's original characteristics when it bonds and exchanges electrons with the Iron molecules.(4, pg. 604)

The equivalence point occurs when the Permanganate ions can no longer find any Ferrous

ions to bond with. It is at this point that all Ferrous ions have accepted an extra electron from the Permanganate, and cannot hold any more electrons in it's valence shell. Because the Permanganate has no Iron left to interact with, the Permanganate simply sits in the Iron solution with nothing to do. The purple colour of the Permanganate stays in the Iron solution for the first time because it does not bond with any Iron molecules, thus informing us that we have reached the equivalence point of the titration.(4, pg. 604)

Titration is an excellent way to determine the molarity of another solution because we have

a method of comparison. For instance, if 1 mole of Permanganate was needed to titrate 10 L of Iron solution, we would know that there was 5 moles of Iron in the solution (because for every 1 Permanganate ion, there are 5 Ferrous ions that surround it).

In the first trial of Iron Determination, the amount of "empty" volume (volume that was not

occupied by Potassium Permanganate) in the buret at the start of the experiment was measured at 11.2 mL. After titration with the Ferrous Sulphate solution, we measured 19.5 mL. That calculates to 8.3 mL of Potassium Permanganate that was used for titration, which is equal to 4.2 x 10-5 mol [n mol = (0.005 mol divided by L) x 0.0083 L]. This calculation was done by assuming that the Molarity of the Potassium Permanganate was exactly 0.005 M. It requires 5 Ferrous Sulphate molecules to be titrated by a single Permanganate ion, thus 2.1 x 10-4 mol of Iron (II) Sulphate reacted. Therefore, the mass of the titrated Iron in the 10 mL of solution equals 32 mg of Ferrous Sulphate [W (g) = 0.0002.1 mol x (151.9 g divided by mol)].

In our second trial, the buret measured 19.5 mL of unoccupied volume at the start of the

titration and ended at 28.7 mL after the titration. This calculates to 9.2 mL of Potassium Permanganate being used to titrate the 10 mL sample of the Iron Sulfate solution, and converts to 4.6 x 10-5 mol of Permanganate (n mol = 0.005 M x 0.0092 L), or 8 mg of Potassium Permanganate. Therefore, because it requires one Permanganate molecule to titrate five Iron (II) Sulphate molecules, 2.3 x 10-4 mol of Iron (II) Sulphate reacted as well. Thus, we calculate that 34 mg of Iron (II) Sulphate reacted [W (g) = 0.00023 mol x (151.9 g divided by mol)].

In our third trial, the buret began the experiment with 28.7 mL of unoccupied volume

and ended with 38.4 mL after the titration process. This calculates to 9.7 mL of Potassium Permanganate being used to titrate the 10 mL sample of Iron Sulfate solution, and converts to 4.9 x 10-5 mol once again. This calculates 2.5 x 10-4 mol of Ferrous Sulphate, and to 38 mg of Ferrous Sulphate.

However, there is a discrepancy of 6 mg between the results of the first and third trials.

Although it may not seem significant at first, it actually is a difference of 18.9% (percentage = 38 mg divided by 32 mg). As we mentioned earlier, this experiment is very sensitive to the slightest molarity change in the titrate or the analyte. It is very possible that the Potassium Permanganate was not 0.005 M, and therefore ruined our results. This is possible because to prepare the solution, we attempted to measure 0.158 g of Potassium Permanganate, but were forced to use 0.16 g because the scale was not sensitive enough to measure in mg. We dissolved the 0.16 g of Potassium Permanganate into 200 mL of distilled water to give us approximately 0.001 mol of Permanganate and a Molarity of 0.005. However, even the slightest increase in the amount of dissolved Potassium Permanganate may result in less of it being required to titrate with the Iron. This is because the Permanganate ions would be more concentrated and would be more abundant per mL of solution than if the mixture were 0.005 M.

We also had problems with accuracy in the amount of Permanganate used, which may have

caused our large difference in the results of the 1st and 3rd trials. For instance, after we had thought we had completed a trial of titration, we shook the analyte and most of the purple colour of the Permanganate dissolved back into the solution. It is possible that for certain trials, we did not evenly dissolve the Permanganate ions throughout the solution and thus ended up using less Permanganate than required. However, for the final two trials, several drops of Permanganate solution entered the Iron solution even after the purple colour began to stay. It is conceivable that the results from the final two titrations are higher than they should be.

It is also possible that the distilled water was no longer completely distilled because we

conducted the experiment over a week after the bottle was opened. This could result in foreign elements existing in the water that could titrate with the Potassium Permanganate or Iron, or perhaps even add extra Iron to the solution.

We created our 1 Molar of Sulfuric Acid by mixing 11.1 mL (0.2 mol) of 18 M Sulphuric

Acid into 200 mL of distilled water. Problems arose when we chose to dissolve the grinded Iron tablets into the solution when we only had nearly 150 mL of solution. Unfortunately, we did not realize that the proper procedure specifically asks us to add the tablets after we had exactly 200 mL of the solution in the first place. This either may have caused the tablets to dissolve quicker than expected (since the Sulphuric Acid was at a higher molarity than 1 M while in 150 mL of solution), or ruined the molarity of the solution (the solution was to be 1 M before the addition of the tablets; the volume of the tablets caused us to have 200 mL of solution in total when we were to have both 200 mL of solution and the volume of the dissolved tablets). In other words, we were to have 200 mL in total plus the volume of the tablets, but we ended up having 200 mL of solution including the volume of the tablets. We may have caused the molarity of the solution to be higher than 1 M since we did not add as much distilled water as we should have.

However, the greatest and most noticeable problem was the lack of the Iron tablets'

solubility in the solution. The outer layers of the iron tablets would not dissolve, which prevented us from determining if the solution and the tablets had become a homogeneous mixture (when the Iron completely dissolves), or whether the solution remained as a heterogeneous mixture (when the Ferrous Sulphate does not completely dissolve). Also, after the second trial, we noticed a small amount of a precipitate (a solid that forms within a solution) in the Sulphuric Acid, which leads us to believe that not all of the Iron managed to dissolve. However, the table of Relative Solubility tells us that Ferrous Sulphate is very soluble in Sulphuric Acid, therefore most of the Iron should have dissolved. Lastly, even though the Sulfuric Acid that we used was as clear coloured as it should be, other jugs of Sulphuric Acid had turned brownish. It is possible that the Sulphuric Acid that we used was no longer pure as the other jugs had become, and perhaps had a lesser molarity than expected.

Not only did we not have the patience to wait out a slow but accurate titration process, but

we forgot to follow the proper procedure again. We were in too much of a rush to leave for after school activities that we neglected to follow the rule that instructed us to continue the titration trials until all values are within 0.3 mL of each other. We only did 3 trials of Iron Determination when we should have done many more. In fact, that was the reason why we chose to create a 200 mL solution of Sulphuric Acid instead of a 100 mL one: in case we needed more than 10 trials. Therefore, we are now unsure of which, if any, of our results are correct or not. It also would have been helpful to create a new Sulphuric Acid solution and a new Permanganate solution to compare results with the earlier trials and see whether we had created the solutions properly or not.

The chemical equation, MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) Mn2+ (aq) + 5Fe3+ (aq) + 4H2O(l),

is balanced, therefore the ratio of mols that reacted between the Permanganate and Ferrous ions is 1:5. On average, the number of mols of Permanganate that reacted is 4.6 x 10-5 mol. Therefore,
2.3 x 10-4 mol of Ferrous Sulphate reacted to the titration and became Ferric Sulfate. That equals approximately 1.4 x 1020 Ferric Sulphate molecules ( number of particles = 6.022 x 1023 things x 2.3 x 10-4 mol) , or 35 mg of Iron (III) Sulfate in the 10 mL of Sulfuric Acid solution. There was 200 mL of solution in total, therefore if the 4 Iron Sulfate tablets had distributed itself evenly throughout the entire solution, there would be a total of 4.6 x 10-3 mol [n mol = 2.3 x 10-4 mol x (200 mL divided by 10 mL)], or 700 mg of Iron (II) Sulfate ( W = 35g x 20).

Therefore, in each 0.6 g Iron Supplemental tablet, we calculate that there is 175 mg of

Iron (II) Sulphate (W = 700 g divided by 4 tablets), which is equal to 1.2 x 10-3 mol [n mol = 0.175 g divided by (151.9 g divided by mol)], and is equal to 7.2 x 1020 molecules of Ferrous Sulphate.
However, our primary objective was to determine the amount of elemental iron in each tablet. If there is 1.2 x 10-3 mol of Ferrous Sulphate, there is 1.2 x 10-3 mol of Ferrous molecules as well. This equals 0.067 g or 67 mg of Iron [ W (g) = 1.2 x 10-3 mol x 55.8 g divided by mol] in each Iron Supplemental tablet. And fortunately for us, on the label of the Iron tablets that we used, it claims that each tablet has about 60 mg of elemental iron.

Composition of a Compound and Iron Determination were experiments that tested whether

we young chemists knew how to apply our molar formulas to the chemical world. The formulas would be useless without empirical methods of measuring mass or volume, while the empirical methods of obtaining volumes would be useless in applying the results to the real world if we could not convert between values. Our experiments were poorly executed and yielded results sometimes far of our estimations, and thus taught us a valuable lesson in humility. However, we did learn another thing from these experiments: when making wine in the kitchen, make sure you accurately add the right amount of each ingredient, or else it'll go sour.