Unknown Compound

3. The IR spectrum of the starting material shows a medium/strong C-O bond at around 1500cm-1, also the starting material shows a strong C-H bond at around 3000cm-1 and another medium C-H bond at 2865cm-1 indicating an aldehyde group whereas the product does not. The IR spectrum of the product shows a two weak broad O-H peaks at around

The primary goal of this laboratory is to correctly identify an unknown substance. To achieve this task, one may use various tests that reveal both chemical and physical properties of a substance. By comparing the results of a known substance and the unknown substance, one may eliminate alternative possibilities and more accurately predict the undisclosed compound. Furthermore, by performing these tests, data can be collected and verified regarding chemical and physical properties of the unknown. Understanding the chemical properties of a known substance aids one’s understanding of the unknown based on comparative analysis of the results of the tests.

Unknown white compound (823U) was discovered in the lab. In order to dispose of it correctly, the substance and its physical and chemical properties had to be identified. The unknown white compound was one of a list of 15 compounds. 5g of the unknown compound were given in order to correctly identify and discover its physical and chemical properties. In order to do so, a solubility test, a flame test, and ion tests were conducted. From the results of these initial tests and the given list of compounds, the unknown white compound was thought to be composed of sodium and a halide (I-, Br-, or Cl-). Of the list, NaCl was the appropriate compound, however NaC2H3O3 was also tested out of skepticism. To verify the identity of the substance, the solubility and flame tests were performed again along with a pH test. The pH tests of NaCl and NaC2H3O2 did not match that of the unknown white compound. The list of compounds had been entirely ruled out. The identity of the unknown white compound was revealed to be calcium chloride. To synthesize at least a gram (calculated to produce 1.2g) of CaCl2, the following reaction was completed.

The GC data for the product produced graph with a signal level out-of-range in peak. This gave a retention time 2.952 minutes. This would not indicate any of the possible ester products. However, after appropriate dilution, a retention time of 1.753 minutes was obtained. This retention time indicates that the ester product was ethyl benzoate.

The guiding question of this ADI lab was, “What are the identities of the unknown compounds?” The goal of this lab was to understand the relationships between moles and molar mass to find the identity of unknown compounds. The mole can be used to measure small amounts of a substance or is used to convert from unit to unit using dimensional analysis. One mole is equivalent to the molar mass in grams of that substance. If you start with the moles of an unknown substance, multiply it by a given compound’s molar mass, and then divide it by however many moles are in the compound of your choice, you will get the mass of the compound. With that answer you can then compare with mass of the compound in the bag to determine its identity. We first started

Primary and secondary alcohols react with the Jones reagent because they can be oxidized to carboxylic acids and ketones respectively. When they oxidize the red color turns green. Tertiary alcohols do not react with the Jones reagent because they cannot be oxidized. Ethanol and 2-propanol both reacted by turning green. 2-methyl-2-propanol did not react because it is tertiary. Our product turned brown, but it can be assumed that there are some impurities in it. We can conclude that some 1-butanol is left in our product. 1-butanol is primary, so it makes sense that is would react a little. The last test to be performed was the iodoform test. Alcohols that have an alpha methyl group and hydrogen react with I2/KI and NaOH by producing a yellow precipitate. Ethanol, 1-propanol, 2-propanol, and 2-methyl-hexanol reacted with the iodoform test by producing a yellow precipitate. Our product did not react which is expected because although it has an alpha methyl group it does not have a hydrogen. Lastly, the boiling point of 2-methyl-2-hexanol is 143 degrees Celsius. The boiling point of our product is 117 degrees celsius. These discrepancies may have been caused by our product not being completely

The objective of this lab was to identify different substances based off their physical properties, which are characteristics of a substances that enable us to distinguish it from other substances.If each of the substances that are used are different then they will be distinguishable by their physical properties. To determine the differences we used; solubility, which is the property of a solid, liquid, or gaseous chemical substances (solute) to dissolve in a solid, liquid, or gaseous solvent to form a solution. We also used density, the degree of compactness of a substance, and the melting and boiling point of the substances which gave the degree temperature at which each substances would melt or boil.

Although high-performance liquid chromatography (HPLC) analysis is provided by most companies to show that the material is pure, yet, doesn’t really reveal much as the product supplied was purely an incorrect compound.

The use of a derivative test was completed as another set of evidence in the process of determining the unknown, by providing another melting point because it is more accurate than obtaining a boiling point. A derivative test is a chemical reaction that forms a specific set of compound that is solid. Unknown 31A’s derivative test results were inconclusive.

Inorganic and organic compounds are both necessary for the human body. The definition of an inorganic compound is a compound that does not contain hydrocarbon groups. An inorganic compound is composed of several other elements, such as salts, metals, and other elemental compounds, but can contain hydrogen or carbon. If they have both, carbon and hydrogen molecules, they are organic. Organic compounds are defined as compounds containing hydrocarbon groups. Organic compounds will include things like the nucleic acids, found in DNA, lipids and fatty acids found in the cells of living organisms, proteins and enzymes that are necessary for cellular processes to take place, and more. There are numerous effects to the body if there are disturbances

The functional group testing in table 4 indicates there was a positive test with the 2,4- DNPH and chromic acid test concluding that the unknown could be an aldehyde. This supports the previous conclusion that the unknown could be Hexanal. Compared with the identification of the unknown, which was cyclohexanone. The results of the identified unknown doesn’t match the information found in the functional group testing. What was supposed to happen was that the chromic acid should’ve tested negative but the results came out positive. A possible error could’ve been missed labeling of the test tubes.

For this experiment, Alcohol D and Acid 2 reacted in the presence of concentrated sulfuric acid, resulting in a colorless solution with brown layer on top. After washes with sodium bicarbonate and brine, the pale-yellow liquid product was dried and then distilled. Distillation resulted in two colorless fractions, the second of which had a boiling point of 69-70 ˚C. This boiling point is unrealistic for any compound obtained in this experiment, so it was not used in identifying the product. After distillation, both fractions were spectroscopically analyzed. The IR and NMR spectra obtained for both fraction were identical, meaning both fractions contained exactly the same substance. Both fractions also smelled the same, like piña colada, therefore confirming this conclusion. This outcome also meant that the amount of product synthesized was 5.7393 g.

The purpose of the lab was to separate a mixture containing methanol, acetone, methyl ethyl ketone, ethyl acetate, propyl acetate and 2-pentanone using gas chromatography. The effect of temperature and pressure variance of the Mini GC on the chromatogram was analyzed and the best temperature-pressure profile was determined based on the results. Temperature programming was determined to give better separation than using a constant temperature. The gas chromatogram was analyzed and peaks were identified based on their behavior within the column. The first two peaks were identified as either acetone or methanol. The next two peaks were identified as 2-pentanone and Methyl Ethyl Ketone and the last two peaks were identified as Ethyl Acetate and Propyl Acetate.

The concentration of ethanol in vodka was determined by back titration. A known amount of excess potassium chromate was added to the vodka sample to convert ethanol into acetic acid by oxidation. The amount of leftover chromate was determined by titration with iron (II) solution, prepared from dissolution of ferrous ammonium sulfate. Using the amount of chromate added to the vodka solution, the amount of chromate reacted was determined and used to find the amount of ethanol in the sample. A potentiometer in the mV mode was used to create first and second derivative curves for determining end points. The concentration of ethanol determined was 39.91%, with standard deviation of 0.13%. The percent error was 0.22%, with the true concentration being 40.00%.

If I was to do an experiment like this again I could use a different variable instead of alcohols. Instead I could investigate a chain of alkanes or alkenes, “in order to test alkenes I would have to look into the association double carbon bonds have with one