Discussion of Results and Scientific Explanations of Claims
My group had been given a specific task to investigate, according to the lab manual1. We were employed by the Environmental Protection Agency as chemists. Our task is to identify an unknown compound that was found in a nearby landfill. In this laboratory, the unknown compound was given to us by our TA. The main goal of this experiment was to correctly identify the unknown compound. We also wanted to discover as many chemical and physical properties as possible. Physical properties are those that can be determined without performing any changes to the identity of the substance. Chemical properties are properties that can only be determined after a chemical reaction is performed in order to change the identity of the substance according to "Physical"2 .Discovering multiple physical and chemical properties of this compound will ultimately help determine how the compound behaves. Also determining some of the physical and chemical properties will aid us in identifying the unknown compound. The final goal was to come up with two syntheses of the unknown compound. In using these two syntheses, the goal is to compare it to the compound. We wanted to compare it based on cost effectiveness, safety, and potential yield of the compound. The physical state of the unknown compound was that it was a white solid that had a very weak odor. The conductivity of the compound was tested and the reading on the voltmeter was .053 J. This means that the unknown compound does conduct electricity. The compound was tested and had a pH of 5. This means that the compound is acidic according to the pH scale. Being acidic means that the substance has more hydrogen ions than hydroxide ions according to "pH"3 .The chemical and physical properties are stated in Table 1. The compound was soluble in water and hydrochloric acid, but was not soluble in sodium hydroxide, toluene, and acetone as indicated by Table 2. For a compound to be soluble it means that the compound must be able to dissolve in a certain liquid. This was proven by using the solubility tests. Since the unknown compound was soluble in water, it means that it is a polar or ionic compound, and since it is soluble in hydrochloric acid it means that it is probably an organic base. It was soluble in water because water is a polar molecule and since it’s polar, the molecules interact using hydrogen bonds and dipole-dipole intermolecular forces. This then leads to the fact that the partial negative and positive parts of the water molecule can break the bonds of ionic and polar molecules which causes them to dissolve within the water. This compound can be ruled out of being a nonpolar compound since it was not soluble in toluene. It was not soluble in toluene because the bonds within the compound could not be broken therefore it could not dissolve. Also, since it was soluble in water but not in acetone, it could not be a polar compound. This concludes that the compound is an ionic compound according to "Solubility"4. An ionic compound is a compound that is formed between cations and anions. Cations are positively charged atoms and anions are negatively charged atoms. The quantitative solubility was then calculated and is indicated by Table 3. In order to calculate the quantitative solubility the amounts of solvent and solute had to be determined. After performing the test it was determined that the quantitative solubility was .5g/mL which means that the solubility is 500 g/L in 1 M H2O. Using the anion test, the compound was tested for the presence of multiple ions as indicated in Table 4. It was tested for a chloride ion, sulfate ion, carbonate ion, and an acetate ion. While testing for the presence of a chloride ion, a white precipitate was formed which verified that there was a chloride ion present. The precipitate was formed because the chloride was insoluble. In order for there to be an acetate ion present there had to be a fruity smell development while performing the test. During the test there was a fruity smell which clarified that there was an acetate ion present. The fruity smell was of ethyl acetate, which proved that there was in fact a presence of acetate. The test for a sulfate ion and a carbonate ion came up negative. The analysis of cations involved the flame test and the test for ammonium. There was a red result after the flame test was performed which proves that calcium is present in the compound. This happens because when the electron is heated it becomes excited and in its excited state it becomes unstable. It then tries to fall back down to the ground state, which causes a release in energy. The wavelength of the energy determines the color of the flame. The test for ammonium was negative, which proved that there was no ammonium ion present as indicated in Table 3. The test was negative because there was no smell of ammonia when the mixture was wafted. Also when we performed a pH test on the solution, the pH was not greater than seven which meant there was no ammonium present. If ammonium was present then it would have dissolved into the water that was on the Litmus paper and gave a pH greater than seven.
After all the above tests were completed, the compound could now be identified. The identity of the compound was calcium chloride (CaCl2). After determining the compound, the MSDS and LD50 could be distinguished. The MSDS stated that safety googles, gloves, and protective clothing should be worn at all times while handling this compound. It could cause irritation with eyes and possibly burns. Also it may cause respiratory and digestive tract irritations. If your eyes come in contact with this compound, they should be immediately flushed for fifteen minutes. If it is ingested you should drink two to four cups of water or milk. If it comes in contact with skin you should immediately flush your skin for at least fifteen minutes. Also, if it comes in contain with skin you should remove all contaminated clothing. The LD50 tells you the amount of the specific that will kill one half of a tested subject. The LD50 was 1940 mg/kg for mice, 1384 mg/kg for rabbits, and 1 gm/kg for rats. ("Material Safety Data Sheet")5
After the compound was identified, a test for comparisons was performed. A comparison of reactions was formed when both the known and unknown compounds were placed in several other compounds. They were placed in an acid, base, potassium nitrate, silver nitrate, and sodium carbonate. The reactions with the following compounds were the same for both the known and unknown compounds as indicated in Table 6. The known and unknown compounds both formed a precipitate in the acid, base, silver nitrate, and sodium carbonate. Neither the known nor unknown reacted with the potassium nitrate. These tests for comparisons helped verify the identity of the unknown compound. The reactions mentioned above are indicated in Table 7. All of the reactants undergo a double displacement reaction. A double displacement reaction occurs when half of one reactant is exchanged with half of the other reactant in order to form the product according to "Single"6 .Calcium chloride has acidic properties. Calcium chloride can form an insoluble salt when it reacts with silver nitrate and sodium carbonate. We then came up with two syntheses with are listed in Table 8. The first reaction listed in the table is better than the second reaction because it doesn’t produce CO2 which is a greenhouse gas and is bad for the environment. Also after the compound was identified we were able to calculate the percent yield which is indicated in Table 9. Percent yield is the amount of the product that is expected after a reaction that forms precipitate. In order to form precipitate we mixed our compound with sulfuric acid. After forming the precipitate we then used a centrifuge. A centrifuge is a machine that separates different densities from each other. We did this three times with our unknown compound and two times with the known compound. After using the centrifuge we had to scrape out the solid at the bottom of the test tube and weigh it to find its mass. The mass of the three unknowns and the two knowns were very similar. They were not exactly the same because of some possible errors. The percent yields of each of the compounds were different from each other but the average was about 19.06%.
Many errors could have occurred during this experiment. One for example is that the compound was tested and had a pH of 5 which means that it is an acid, but according to the solubility test it was supposed to be an organic base. When in fact CaCl2 is an inorganic salt so it has neutral properties. Some error but have occurred when either recording the pH or while performing the solubility test of the compound in hydrochloric acid. Another error could have occurred while calculating the percent yield. An error could have occurred because we had to scrape out the solid from the bottom of the test tube and some of the solid got stuck on the sides of the test tube, which caused the mass of the solid to not be correctly calculated. Other errors could have occurred due to measurements. Some of the measurements could have been off by just a small amount and that could have caused an error within the results.
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