Identification and Characterization of Com-pounds X and Y

by:Matthew Lee

Abstract

We were instructed to run a variety of tests upon compounds X and Y to determine their chemi-cal make up. The procedures run are listed as the following: to create compound Y through the boiling of compound X with hydrochloric acid, a flame test to determine the presence of chlo-rine, a melting point test, the solubility properties and the pH of both compounds, the creation of an empirical formula, the titration of both compounds to determine the number of grams per mol of molecules that are attached to each acidic hydrogen, the freezing point of each compound and the application to the molecular weight formula discovered by Raoult, and the application of the mass spectra and infrared spectra analysis of compounds X and Y. Through all of these tests, we had discovered that compound X is Maleic acid and that compound Y is Fumaric acid.

Introduction

The purpose of this experiment was to identify compounds X and Y through a series of tests which include creating compound Y by boiling compound X with hydrochloric acid, conducting a simple flame test to determine the presence of chlorine, performing a melting point test on both compounds, finding the solubility and pH properties of X and Y, using these tests to discover the empirical formula of X and Y, titrating both compounds to determine their gram equivalent weights, determining the molar weight through the freezing point properties of both compounds, and finally analyzing the mass spectra and infrared spectra for further structural elucidation. Through these tests, we have discovered that both compounds X and Y burn with an orange/yel-low flame determining the absence of chlorine; that compound X melts at around 135-142 de-grees Celsius and that compound Y melts at roughly 300 degrees Celsius; that both compounds are very acidic with X having a pH of 1 and Y having a pH of 2 (likely a presence of an acidic functional group); that compound X is soluble in water, meaning that it is polar covalent while Y was not soluble, meaning that it is non polar covalent; that both compounds have the same em-pirical formula of CHO (there was a chance that both would differ in the number of each ele-ment); that both compounds have roughly the same average number of grams per mole H^+ (64.26 grams/mol X and 63.73 grams/mol Y) and the same number of moles H^+ per mole of compound (1.836 mol H^+ per mol X and 1.852 mol H^+ per mol Y); that the molar weight of compound X and Y is 118 grams/mol, meaning that both likely have the same molecular for-mula; and that certain functional groups such as hydroxide and carbonyl are present, and that be-cause both are acidic, that carboxylic acid is present. From all of these results, we were able to determine that compound Y is Fumaric Acid (pubchem) and that compound X is Maleic Acid (chemspider).

Methods / Materials

Within part I of the experiment, we were instructed to create compound Y through boiling com-pound X with hydrochloric acid. This would produce a white powder. In order to produce this re-sult, we first had to dissolve compound X completely. After this was finished dissolving, we added Hydrochloric acid to the solution to create the mixture. In order for a reaction to occur, we had to bring the mixture to a boil, which would produce a solid. After filtering this new com-pound out, we were to next dry the material within a drying oven. This new compound was com-pound Y.

Part II of the experiment was to determine what color flame the compounds had produced. We had accomplished this test through the use of copper wire and a flame. We first dipped the wire within the different compounds (X and Y, respectively and separately). From here, we held the wire up to the flame. Both compounds, surprisingly at the time, had created an orange / yellow flame. This result can be interpreted as the absence of chlorine within both compounds, for chlo-rine would have burned with a blue / green flame. This is because the orange color is typical of a hydrocarbon fueled fire (85, lab manual).Part III was done to determine the melting point of compound X. We were instructed not to do the same test for compound Y for the apparatus would not be able to sustain 300 degrees Celsius (the melting point of compound Y) without damage. We placed a small amount of the compound X crystals within a tube and observed it for any signs of melting. This had resulted in a range of roughly 135-142 degrees Celsius for both trials.

Part IV was done to determine the solubility of the compounds within the water and to determine the pH of the compounds. This test had determined that compound X is soluble within water, meaning that X is in fact polar covalent. We were also able to deduce that compound Y was not very soluble in water, meaning that it is non polar covalent. We had then dipped litmus paper within the solutions and had discovered that compound X has a pH of 1 while compound Y has a pH of 2.

Part V was a test that had already been completed for us. We had used the given results to isolate the weight of carbon dioxide and of water. Once these weights were isolated, they were used to determine the empirical formula of both X and Y which was CHO.

Part VI involved a series of titration tests to determine the gram equivalent weights of each acidic hydrogen within our compounds X and Y. We had accomplished this through titrating NaOH into a solution of KHP with phenolphthalein indicator to standardize the exact molarity of the NaOH solution. Once this was accomplished, we could now titrate our compounds X and Y with the NaOH with the knowledge that both compounds are acidic. This is to determine the gram equivalent weight which is how much mass is attached to each acidic portion of the com-pound itself.

Part VII involved finding the freezing point of our compounds and plugging this number into the formula discovered by Raoult that correlates the depression of the freezing point of a solution to the molal concentration of a solution. This is all to determine the molecular weight for the com-pounds. After placing rock salt and deionized water to a beaker with ice, we were able to place our tubes of the compounds into the glacial acetic acid to measure the freezing point. Our result for this portion of the lab had yielded us with the result of 126.16 grams/mol, but the ideal molar weight of our substance X and Y was actually 118 grams/mol.

The goal of part VIII was to determine the structure of compounds X and Y. After observing the mass spectrometry and the infrared spectroscopy of both compounds, we were able to determine that both compounds had similar, if not identical functional groups. These were hydroxyl and carbonyl groups. Because we had known that both substances were also acidic, we were able to determine that these structures bond together to form carboxylic acid. We had also determined through our solubility test and our titration test that compound X is actually polar, while com-pound Y is non polar. Knowing that both compounds also have similar molecular weights and identical empirical formulas, one can combine this information to determine that both com-pounds have identical molecular formulas, but are actually isomers of each other. The next step in determining the actual structures was to determine the molecular formula itself, which is C4H4O4. The next step is to determine how the isomers are shaped. The only non polar structure was the trans- isomer for compound Y, and the only polar structures left to identify compound X were the cis and geminal structures. We had determined that the cis-structure was compound X, for rearrangement of the geminal- structure creates an entirely different set of characteristics, while the cis-structure would be similar to that of Y.

Results / Calculations

Part I required us to weigh the mass of a clean dry 100 mL beaker, which was 53.265 grams. Next, we added exactly 3.00 grams of compound X to the beaker. 5 mL of distilled water was added to dissolve the substance. In order for X to dissolve, we heated the mixture. We next added 7.0 mL of 12 M HCl to our solution. We added this mixture to a 250 mL beaker filled with DI water, which we had heated for 5 minutes. After this, we had placed the solid within a filter paper (.896 grams) and drained the substance into a 250 mL Erlenmeyer flask. We then placed our filter paper with the substance onto a 94.123 gram watch glass and placed this within the dry-ing oven. After drying, our sample for Y had weighed .723 grams.

Compound X Compound Y

3.00 grams .723 grams

Part II required us to cut off a piece of copper wire and dip it first into compound X. After recording the color in which this burned, we were to do the same for compound Y. Compound X burned with an orange / yellow color, while compound Y burned with an orange / yellow color. This meant that both were composed of hydrocarbons and lacked chlorine, which would burn green/blue. This result occurred in a practice test that we had run on a substance called Parlon.

Compound X Compound Y Parlon

Color Orange / Yellow Orange / Yellow Green / Blue

Part III required measuring the melting point of the various compounds. We were to practice this test first with two substances, 4-aminobenzoic acid and phenacetin. After this, we tested the melting point of compound X.

Trial 1 Trial 2 CRC Handbook

4-aminobenzoic acid 189.2 degrees Celsius 192.1 degrees Celsius 187-189 degrees Celsius

phenacetin 132.7 degrees Celsius 134.5 degrees Celsius 133-136 degrees Celsius

compound X 135 degrees Celsius 142 degrees Celsius 135 degrees Celsius

Part IV required us to dissolve 0.2 grams of compounds X and Y separately within 100 mL beakers, each containing 40 mL of deionized water. Here, we were to measure the pH of the compounds and the solubility. After this test, we had mixed the compounds with Na2CO3. This yielded cloudy and clear because carbonate is not soluble in water.

pH Solubility Mixture with Na2CO3

Compound X 1 Yes Cloudy then clear

pH Solubility Mixture with Na2CO3

Compound Y 2 No Cloudy then clear

Part V of the lab assignment required us to find the empirical formula of the compounds. We were given the following information:

A 1.0542 gram sample of X was burned in a combustion apparatus. The results were:Weight of ascribe tube before combustion 75.2382 gWeight of ascarite tube after combustion 76.8377 gWeight of drierite tube before combustion 81.4128 gWeight of drierite tube after combustion 81.7418 g

Given this data, calculate the empirical formula of compound X

I had solved this problem by first determining exactly how much CO2 was burned off in the as-carite tube, and how much water was burned off in the drierite tube. These are my calculations:

1.5962 grams CO2.329 grams water

I used this information to find the empirical formula of the substance.

? mol C = 1.5962 grams CO2 (1 mol CO2 / 44.009 grams CO2)(1 mol C / 1 mol CO2)(12.011 g C / 1 mol C) = .4357 grams of Carbon

? mol H = .329 g H2O(1 mol H2O / 18.015 g H20)(2 mol H / 1 mol H2O)(1.008 g H / 1 mol H)= .036817 g Hydrogen

? mol O = 1.0542 g - .4357 g C - .036817 g H = .581683 g Oxygen

? mol C = .4357 g C / (1 mol C / 12.011 g C) = .036275081 / .036757081 = 1? mol H = .036817 g H / (1 mol H / 1.008 g H) = .036524802 / .036757081 = 1? mol O = .581683 g O / (1 mol O / 15.999 g O) = .03635746 / .036757081 = 1

Element X’s Empirical Formula = CHO

A 1.4745 gram sample of Y was burned in a combustion apparatus. The results were:Weight of ascribe tube before combustion 80.7821 gWeight of ascarite tube after combustion 83.0196 gWeight of drierite tube before combustion 78.2988 gWeight of drierite tube after combustion 78.7560 g

Given this data, calculate the empirical formula of compound Y

I had solved this problem by first determining exactly how much CO2 was burned off in the as-carite tube, and how much water was burned off in the drierite tube. These are my calculations:

2.2375 grams CO2.4572 grams water

I used this information to find the empirical formula of the substance.

? mol C = 2.2375 grams CO2 (1 mol CO2 / 44.009 grams CO2)(1 mol C / 1 mol CO2)(12.011 g C / 1 mol C) = .61066174 grams of Carbon

? mol H = .4572 g H2O(1 mol H2O / 18.015 g H20)(2 mol H / 1 mol H2O)(1.008 g H / 1 mol H)= .051163764 g Hydrogen

? mol O = 1.4745 g - .61066174 g C - .051163764 g H = .812674496 g Oxygen

? mol C = .61066174 g C / (1 mol C / 12.011 g C) = .050841873 / .050757702 = 1? mol H = .051163764 g H / (1 mol H / 1.008 g H) = .050757702 / .050757702 = 1? mol O = .812674496 g O / (1 mol O / 15.999 g O) = .050795331 / .050757702 = 1

Element Y’s Empirical Formula = CHO

Part VI required us to titrate our unknowns. First, we needed to standardize the molarity of our NaOH. The following is our results:

Trial 1 Trial 2 Trial 3

Mass KHP acid .500 g .502 g .506 g

Moles KHP acid .00245 mol .00245 mol .00247 mol

Volume NaOH added

13 mL 13.2 mL 11.8 mL

Molarity NaOH .185 g/L .185 g/L .209 g/L

Average Molarity .193 g/L

Moles KHP acid was calculated by taking the mass per trial of KHP acid and dividing it by the molecular weight. Ex: ? mol KHP=.500g KHP(1 mol / 204.2 g KHP) = .00245 mol

Molarity of NaOH was calculated by taking the Moles of KHP acid and divide this number by the volume of NaOH added. Ex: ? Molarity NaOH = (.0024 mol / 13 mL)(1mL / 10^-3 L) = .185 M

Average Molarity found by added up all of the Molarity of NaOH and dividing it by the total number of figures..185 g/L + .185 g/L + .209 g/L = .193 g/L

The next step in part VI is the titration of compound X.

Trial 1 Trial 2 Trial 3

Mass X .201 g .205 g .206 g

Volume of NaOH added

16.8 mL 15.7 mL 16.9 mL

Moles OH- .0032424 mol .0030301 mol .0032617 mol

Moles H+ .0032424 mol .0030301 mol .0032617 mol

Grams X per mole H+

61.99 g / mol 67.63 g / mol 63.16 g / mol

Average grams X per mole H+

64.26 g / mol

Mole H+ per mole X (know the molecular weight from part VII)

1.836 mol H+

Moles OH- found by multiplying the volume of NaOH added by the molarity of NaOH.Example: ? Moles OH-=16.8 mL NaOH(10^-3L/1mL)(.193 g NaOH/1L)=.0032424 mol

Find the grams X per mol H+ by dividing the mass by the number of moles.Example: ? grams X per mole H+ = .201 g/.0032424 mol = 61.99 g / mol

Find average grams X per mole H+ by adding up all of the trials of the average grams X per mole H+ and dividing this by the total number of trials.61.99 g / mol + 67.63 g / mol + 63.16 g / mol = 64.26 g / molFind the mole H+ per mole X by dividing the molecular weight by the average grams X per mole H+. (118 g/mol) / (64.26g/mol) = 1.836 mol H+

The next step in part VI is the titration of compound Y.

Trial 1 Trial 2 Trial 3

Mass Y .2 g .208 g .199 g

Volume of NaOH added

15.2 mL 17.7 mL 16.6 mL

Moles OH- .0029336 mol .0034161 mol .0032038 mol

Trial 1 Trial 2 Trial 3

Moles H+ .0029336 mol .0034161 mol .0032038 mol

Grams Y per mole H+

68.18 g / mol 60.89 g / mol 62.11 g / mol

Average grams Y per mole H+

63.73 g / mol

Mole H+ per mole Y (know the molecular weight from part VII)

1.852 mol H+

Moles OH- found by multiplying the volume of NaOH added by the molarity of NaOH.Example: ? Moles OH-=15.2 mL NaOH(10^-3L/1mL)(.193 g NaOH/1L)=.0029336 mol

Find the grams Y per mol H+ by dividing the mass by the number of moles.Example: ? grams Y per mole H+ = .2 g/.0029336 mol = 68.18 g / mol

Find average grams Y per mole H+ by adding up all of the trials of the average grams Y per mole H+ and dividing this by the total number of trials.68.18 g / mol + 60.89 g / mol + 62.11 g / mol = 63.73 g / mol

Find the mole H+ per mole Y by dividing the molecular weight by the average grams Y per mole H+.(118 g/mol) / (63.73 g/mol) = 1.852 mol H+

Part VII

We were required to find the freezing temperature of our compound. The following is our re-sults:

Mass of Compound X 1 gram

Mass of Glacial Acetic Acid 10.308 grams

Freezing Temperature of Glacial Acetic Acid 13 degrees Celsius

Freezing Temperature of Solution 10 degrees Celsius

Difference in temperature 3 degrees Celsius

Molecular Weight of Compound X 126.16 g/mol

Class Average Molecular Weight 354.20 g/mol

MW solute = ((i Kf)(grams solute)) / ((kg solvent)(delta Tf))Kf = 3.90 degrees Celsius / mol

? MW = ((3.90 degrees Celsius/mol)(1 mol))/((.010308 kg)(3 degrees Celsius))Molecular Weight = 126.16 g/mol of Compound X

Actual Weight (given by professor) = 118 grams / mole

Part VIII

12.011 g/mol C + 1.008 g/mol H + 15.999 g/mol = 29.018 g/mol CHO (29.018 g/mol CHO)/(118 g/mol of X and Y) = .2459 (4) = C4H4O4

Lewis Dot Structures of Compound X (Maleic Acid) and Compound Y (Fumaric Acid)

Discussion / Conclusion

Throughout the course of the experiment, we were taken through a variety of different test to run upon our unknown compound. The purpose of part I was to create compound Y from compound X. The purpose of part II was to conduct a flame test to test for the presence of chlorine within our compound. Part III was necessary to be able to tell the melting point of our compounds. This is essential in characterizing the differences between compound X and Y. Part IV was essential in determining the structural isomers of X and Y due to polarity and pH of each substance. Part V was also essential due to the fact that we were able to isolate a specific empirical formula for both compounds. This was even more important to our search, for we were able to conclude that both substances were based upon the same empirical formula. Part VI was very important in de-termining the grams per mole of molecules that are attached to the acidic hydrogen groups. Part VII was important in being able to determine the molecular weight of our substances. This was one of the most important steps, for it confirmed that the molecular formulas were the same for both compounds. This also confirmed the fact that we were dealing with structural isomers. Part VIII was important in being able to determine exactly what functional groups were present within our compounds. This was important in being able to determine the fact that both com-pounds had carboxylic acid within their structures. This made the structures very easy to draw out once we had determined the molecular formula. The isomers were then easy determined through all of the information that we had found from previous steps. One of the main points that helped me to determine the isomer shape for compounds X and Y was the polarity of each com-pound. Once Y was determined due to it being the only non polar covalent compound, the com-pound X was found between the cis- isomer and the geminal- isomer due to its flexibility in shape. The geminal- isomer is not able to rearrange itself without becoming an entirely new sub-stance.

A point in which our data had varied from the data that should have occurred was within part I. During the stage in which we had boiled our compound X with the HCl as a catalyst, the water

from within the beaker had overflowed and spilled into our tube. I believe that this had caused some of our compound Y to spill out and be lost from the final product. This is why our amount of Y as a product at the end is so much smaller than the initial amount of X used at the beginning of the experiment.

References

1) Chemistry M01A Laboratory Manual. N.p.:Moorpark College Chemistry Department and L.J. Williamson, 2014. Print.

2) CRC Handbook of Chemistry and Physics. N.p.: n.p., n.d. Web. 5 May 2015.

3) ”Fumaric Acid | C4H4O4 - PubChem." Fumaric Acid | C4H4O4 - PubChem. N.p., n.d. Web. 05 May 2015.

4) ”Maleic Acid." Chemspider. Royal Society of Chemistry, n.d. Web. 4 May 2015.

5) Tro, Nivaldo J. Chemistry: A Molecular Approach. Third ed. Boston: Pearson, 2014, Print.