Identification of Unknown Compounds

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Identification and Synthesis of Unknown White Compound Santiago Charry, Christopher Stanley, Abigail Harrison, Micah Perwien Abstract: The purpose of the lab was to identify compound 488Q, verify the identification, and synthesize one gram of 488Q. In order to do this, a series of tests including physical observation, flame test, solubility test, reaction test, pH test, and conductivity tests were conducted. The synthesis involved a neutralization reaction between two reactants and the dehydration of the product formed. The results of testing proved 488Q to be NaC 2 H 3 O 2 , which was synthesized through the reaction between NaOH and HC 2 H 3 O 2 . Introduction An unknown white compound was analyzed to identify its properties and what exactly the compound was. After identifying the compound, one gram of the compound was synthesized. Such a technique is vital for chemists to develop; unknown compounds can be encountered in situations such as investigating crime scenes, powders that pose a biological threat, analyzing environmental

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Learn how to identify unknown white compounds.

Transcript of Identification of Unknown Compounds

Identification and Synthesis of Unknown White CompoundSantiago Charry, Christopher Stanley, Abigail Harrison, Micah Perwien

Abstract: The purpose of the lab was to identify compound 488Q, verify the identification, and synthesize one gram of 488Q. In order to do this, a series of tests including physical observation, flame test, solubility test, reaction test, pH test, and conductivity tests were conducted. The synthesis involved a neutralization reaction between two reactants and the dehydration of the product formed. The results of testing proved 488Q to be NaC2H3O2, which was synthesized through the reaction between NaOH and HC2H3O2. IntroductionAn unknown white compound was analyzed to identify its properties and what exactly the compound was. After identifying the compound, one gram of the compound was synthesized. Such a technique is vital for chemists to develop; unknown compounds can be encountered in situations such as investigating crime scenes, powders that pose a biological threat, analyzing environmental soils, and more. Identification of unknown compounds by scientists has previously been done in various ways, such as through high-performance liquid chromatography{Formatting Citation}1, in which components in a mixture are separated in order to identify each individual component and its relative amount to the others. By revealing all of the components and their amounts, the compound itself can be identified. In addition to this technique, scientists often use time-of-flight mass spectrometry to identify unknown compounds2, which reveals an ions mass-to-charge ratio by analyzing how long it takes an accelerated ion to reach a detector. In this report, readers can expect to learn various tests used for identification of unknown compounds, and how to synthesize an exact amount of a compound from other compounds. ExperimentalIdentificationPhysical Properties: The unknown white compound provided at the beginning of the lab was titled 488Q. It was to be compared with fifteen other compounds for identification. These compounds were NH4Cl, (NH4)2SO4, CaCO3, CaCl2, Ca(NO3)2, MgCl2, MgSO4, K2CO3, KCl, KNO3, K2SO4, NaC2H3O2, Na2CO3, NaCl, and Na2SO4. Compound 488Q was initially compared to the provided compounds based on its physical attributes. Flame Test: The flame test was used to identify the cation of compound 488Q, since each cation (regardless of anion) of the salts burned a distinct color when held to the flame. A scupula held a small arbitrary amount of a salt to a flame created by a Bunsen burner, causing the flame to burn a different color. Between tests the scupula was cleaned with deionized water and air dried to ensure that solely one compound was being tested at once. Solubility Test: Next a solubility test was held. This was done using a 50-mL buret and a 100-mL beaker. One gram of compound 488Q was placed in the beaker. The buret was filled with deionized water, and water was slowly let into compound 488Q .2 milliliters at a time. Before adding additional .2 milliliters of water, the beaker was swirled for 30-60 seconds to ensure that the compound had a chance to dissolve as much as it could with the current amount of water. Verification of 488Qs Identity as NaC2H3O2 Reaction and pH Test: Verification of 488Qs identity as NaC2H3O2 was first done through the use of a reaction test. 488Q and NaC2H3O2 were each reacted with an excess amount of HCl in separate tests to compare the results. This reaction would yield NaCl and CH3COOH, a principle ingredient of vinegar that gives it its distinct odor. When both reaction tests were conducted, the product of the reaction with 488Q and the product of the reaction with NaC2H3O2 both shared a distinct smell of vinegar, providing qualitative support. A pH test using a pH meter on the products of the reactions was conducted as well to provide further quantitative support. The pH probe was first calibrated using known pH values of HCl and NaOH3. Conductivity Test: The final test conducted to verify 488Qs identity as NaC2H3O2 was the conductivity test. 488Q was dissolved in a 50mL beaker with water. The same amount of NaC2H3O2 was dissolved in the same amount of water in a different 50mL beaker, ensuring identical densities. The conductivity of each of the dissolved compounds were then found through the use of a conductivity probe.Synthesis of 488QThe synthesis of compound 488Q, or NaC2H3O2, was done through the neutralization reaction between HC2H3O2 and NaOH, which yielded H2O and NaC2H3O2. 12.2 mL of 1.0M NaOH and 12.2 mL of 1.0M HC2H3O2 were measured into 50-mL graduated cylinders. These were then combined into a 200-mL beaker, and reacted to form H2O and NaC2H3O2. The aqueous solution was then boiled on a hot plate allowing the H2O to evaporate and leave the NaC2H3O2 behind. The remaining salt was then dehydrated in an oven to leave only anhydrous NaC2H3O2 behind. This was verified through the flame test, in which a small amount of the product on a scupula was held to the flame of a Bunsen burner. To further verify that our created product was NaC2H3O2, the same reaction test with HCl was conducted as previously described. The product of the reaction had its pH tested as before to ensure duplicity. Waste management throughout the lab involved pouring chemical compounds into respectively titled chemical waste cartons, and rinsing glassware and equipment with deionized water and allowing it to air dry. Results and DiscussionIdentification of Unknown Compound 488Q The data table below displays the results of each identification test with each compound. When a test for a given compound had different results than that of compound 488Q, it was eliminated from further testing, with the negative test shown in italics. Tests continued until all but one compound had been eliminated (shown in bold), proving 488Qs identity as that compound. SubstancePhysical AttributesFlame TestSolubility Test

488QCoarseYellow-orange.5g/mL

NH4ClCoarseOrange

(NH4)2SO4CoarseOrange

CaCO3CoarseRed-orange

CaCl2CoarseRed-orange

Ca(NO3)2CoarseRed-orange

MgCl2CoarseWhite

MgSO4CoarseWhite

K2CO3CoarsePink

K2ClCoarsePink

KNO3Fine

K2SO4CoarsePink

NaC2H3O2CoarseYellow-orange.504g/mL

Na2CO3Fine

NaClCoarseYellow-orange.366g/mL

Na2SO4CoarseYellow-orange.281g/mL

Physical Attributes: 488Q had a coarse, crystal-like texture, thus eliminating two of the compounds, KNO3 and Na2CO3, which had a fine powdery texture, leaving 488Q with thirteen possible compositions. This test was limited by its basis on personal judgment of qualitative factors when analyzing physical properties, making elimination of salts based on this test a possible area of error.Flame Test: When compound 488Q was held to the flame, the color of the flame changed to a yellow-orange hue. The only other compounds that did this were compounds containing sodium as the cation, thus eliminating ten more potential compounds. 488Qs only possible identities after this test were NaC2H3O2, NaCl, and Na2SO4. This test was another area of possible error, due to the unconventional techniques used to carry out the test. The typical technique would use a nichrome wire dipped in a concentrated solution of the wire to carry the solution to the flame3. The technique carried out during this lab, however, held a small sample of the salt on a scupula to the flame instead, resulting in possible errors in results. Furthermore, the data may be limited since this test also required personal judgment of qualitative factors when analyzing flame colors. As a result, elimination of ten compounds from consideration as 488Qs identity based on this test may have resulted in skewed results.Solubility Test: The compound had completely dissolved when 2 mL of water had been added, giving compound 488Q a solubility of .5g/mL. In the solubility charts given for the lab, NaC2H3O2 has a solubility of .504 g/mL, NaCl had a solubility of .366g/mL, and Na2SO4 had a solubility of .281g/mL. Since compound 488Q had solubility of .5g/mL, the data proved that 488Q was more than likely NaC2H3O2.Verification of 488Qs Identity as NaC2H3O2The data table below shows the results of the reaction and pH tests, and the conductivity tests. The purpose of these was to verify that 488Q was indeed NaC2H3O2, so the only compounds tested were 488Q and NaC2H3O2. SubstanceObservation of reaction with HClpH of product from reaction with HClConductivity Test

488QDistinct vinegar odor4.546609-6630 mS/cm

NaC2H3O2Distinct vinegar odor4.456371-6381 mS/cm

Reaction and pH test: When both reaction tests were conducted, the product of the reaction with 488Q and the product of the reaction with NaC2H3O2 both shared a distinct smell of vinegar, providing qualitative support, which is once again limited by personal judgment. The product of 488Qs reaction with HCl had a pH of 4.54, similar to the product of NaC2H3O2s reaction with HCl, which had a pH of 4.45, further verifying that 488Q was in fact NaC2H3O2.Conductivity Test: Through the use of a conductivity probe, the conductivity of compound 488Q was found to be between 6609-6630 mS/cm, and the conductivity of NaC2H3O2 was found to be 6371-6381 mS/cm. The similarity of this test further proved with quantifiable evidence that 488Q was in fact NaC2H3O2. Drawing conclusions across such a gap in each compounds conductivity limits the reliability of this test. Synthesis of 1 gram NaC2H3O2In order to synthesize compound NaC2H3O2, NaOH of 1.0M and HC2H3O2 of 1.0M were reacted. To find the amounts of reactant required to produce one gram of NaC2H3O2, a balanced equation of the reaction was created. NaOH(aq) + HC2H3O2(aq) H2O(g) + NaC2H3O2(aq)The following conversion factors were used to find the amount of reactants necessary for the production of one gram of NaC2H3O2. 1g NaC2H3O2 (1mol NaC2H3O2/81.94g NaC2H3O2) = 0.0122 mol NaC2H3O2. 0.0122 mol NaC2H3O2 (1mol NaOH/1mol NaC2H3O2) = 0.0122 mol NaOH0.0122 mol NaOH (1L NaOH/ 1mol NaOH)= 0.0122L NaOH.0.0122L NaOH (1000mL/1L) = 12.2mL NaOH necessary. 0.0122 mol NaC2H3O2 (1mol HC2H3O2/1mol NaC2H3O2) = 0.0122 mol HC2H3O2.0.0122 mol HC2H3O2 (1L HC2H3O2/ 1mol HC2H3O2)= 0.0122L HC2H3O2.0.0122L HC2H3O2 (1000mL/1L) = 12.2mL HC2H3O2 necessary. After the reaction and steps taken to leave only anhydrous product, the amount of product created was measured and found to be .949 grams. The product created was compared with NaC2H3O2 through the flame test and reaction and pH test to ensure the products identity as NaC2H3O2. SubstanceObservation of reaction with HClpH of product from reaction with HClFlame Test

Product CreatedDistinct vinegar odor4.54Yellow-orange

NaC2H3O2Distinct vinegar odor4.25-5.0Yellow-orange

Test results proved that the product created was in fact NaC2H3O2. Percent yield of the synthesis was then calculated. The calculations are shown below. (0.949g NaC2H3O2 actual/1.0g theoretical) 100 = 95 percent yield.

ConclusionsAccording to the data resulting from the tests conducted, compound 488Q was in fact NaC2H3O2; the dehydration of the product formed from the reaction between NaOH and HC2H3O2 was NaC2H3O2.The data showed that NaC2H3O2 burns yellow-orange in the flame test, has a solubility of .5g/mL, yields NaCl and CH3COOH when reacted with HCl acid, and has a conductivity of about 6609-6630 mS/cm. The conclusion drawn from the conductivity test is uncertain due to the wide data gap between our product and the provided NaC2H3O2 and the flame tests conclusion is uncertain as well due to unconventional technique. The ability to identify unknown compounds is a significant skill because unknown substances can be encountered in a variety of scenarios, such as identifying an unknown compound found in the plasma of breast cancer patients4. The ability to create an exact amount of a compound may be important if resources dont include a compound necessary, but the components of the compound are available. Such a level of resourcefulness and flexibility can be vital for scientists to conduct projects.References(1) Bobeldijk, I.; Vissers, J. P. .; Kearney, G.; Major, H.; van Leerdam, J. . J. Chromatogr. A 2001, 929, 6374.(2) Clauwaert, K.; Vande Casteele, S.; Sinnaeve, B.; Deforce, D.; Lambert, W.; Van Peteghem, C.; Van Bocxlaer, J. Rapid Commun. Mass Spectrom. 2003, 17, 14438.(3) Moodle Lab Techniques https://www.mhlearningsolutions.com/dcs2/dcsflow/index/.(4) Yoshioka, M.; Abu-Zeid, M.; Kubo, T.; el-Merzabani, M. Biol. Pharm. Bull. 1994, 17, 16972.