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Reactivity of metals hypothesis
Introduction to Laboratory Techniques
General laboratory techniques 1
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In this experiment my lab partner and I collected an unknown sample (sample A) and performed a series of tests and analyses to determine the chemical composition of the unknown. One of our learning goals was to practice performing the basics of analyzing an unknown; this entails measuring boiling point, melting point, and noting its physical properties (color, smell, etc.) In addition to this, we also practiced reading basic spectra, looking at and interpreting test results, synthesizing derivatives of our unknown, and testing the melting point of said derivatives to help us further identify our unknown. When we obtained our sample, we noted it’s physical characteristics. It was a clear liquid, and had a particular odor; I thought it smelled …show more content…
like crushed up vegetables (spinach?), while my lab partner thought it smelled of cardboard. We performed our first test by measuring the boiling point, which came out to approximately 147°C. i.
Our first test to perform was the potassium permanganate / Baeyer test, to test for unsaturation. We mixed in ethanol, our unknown, and some of the KMnO4 solution. The initial pink color of our mixture turned to a golden brown color which indicated a positive test result. If our test gave a negative result, the solution would have remained a pink color (small amounts of brown precipitate might have formed). Below is a diagram of the positive test result: ii. Our second test to perform was a chromic acid test, to test for the presence of alcohols. We mixed in acetone, our unknown, and some of the chromic acid reagent. After mixing, our initial orange color changed to a blue-green hue pretty quickly; this indicates a positive test result. Because we had a positive test result, this indicates the presence of primary and secondary alcohols in our unknown. If our test yielded a negative result, the solution would have remained an orange color indicating the presence of tertiary alcohols or no alcohols. Below is a diagram of the positive test …show more content…
result: iii. Our third test to perform was an Iodoform test, to test for the presence of methyl ketones. We mixed in methanol, NaOH, our unknown, and the iodine/KI solution. After mixing, we had a pretty solid yellow color which indicated a positive test result. If our test yielded a negative result, we would have had a brown solution and no yellow precipitate. Below is a diagram of the positive test result: iv. Our fourth and final test to perform was a Schiff’s Fuchsine Aldehyde test, to test for the presence of aldehydes. We mixed in the Schiff’s reagent and a drop of our unknown. After mixing the solution and letting it sit for 10 or so minutes, we had a faintly pink-purple solution which indicated a positive test result. If our test yielded a negative result, we would have had a colorless solution. Below is the structure of the Schiff’s Fuchsine reagent: When observing the IR spectrum of our Unknown A, we observed three distinct peaks.
One was the typical alkane peak at ~2900/3000 cm-1. Another peak was slightly to the right of our first peak at ~2850 cm-1 which indicates an aldehyde (the C-H bond). Another peak was observed at around ~1710 cm-1 which indicates yet again the presence of an aldehyde (the C=O bond). After observing that all of our test results were positive, we knew that in our compound unsaturation was present, there were primary or secondary alcohols, methyl ketones, and an aldehyde. Keeping this in the back of our minds, we grabbed the derivatives and tested the melting points of each. When determining the melting point of the 2,4-dinitrophenylhydrazone derivative, we observed it to be exactly at 106°C. With this in mind, we also tested the melting point of our semicarbazone derivative, which turned out to be around ~101°C. After obtaining these numbers, we took a look at the Table J.1 and found the numbers that matched up best with our IR results, various tests, and melting point results. We determined that our Unknown sample A was actually Octanal; the structure of which is drawn
below: After completing this experiment, my lab partner and I both felt that we learned a lot about the entire process of analyzing. We felt that our analysis skills were very much solidified; skills ranging from observing IR spectra, to performing various tests on the unknown. After observing what we did, I think it is safe to say that we are confident in our inference. This lab provided a good first step into the practical applications of organic chemistry, and was exciting overall.
Solid A was identified to be sodium chloride, solid B was identified to be sucrose, and Solid C was identified to be corn starch. Within the Information Chart – Mystery White Solid Lab there are results that distinguishes itself from the other 4 experimental results within each test. Such as: the high conductivity and high melting point of sodium chloride, and the iodine reaction of corn starch. Solid A is an ionic compound due to its high melting point and high electrical conductivity (7), within the Information Chart – Mystery White Solid Lab there is only one ionic compound which is sodium chloride, with the test results of Solid A, it can be concluded that is a sodium chloride. Solid B was identified as sucrose due to its low electrical
The purpose of the Unknown White Compound Lab was to identify the unknown compound by performing several experiments. Conducting a solubility test, flame test, pH paper test, ion test, pH probe test, conductivity probe test, and synthesizing the compound will accurately identified the unknown compound. In order to narrow down the possible compounds, the solubility test was used to determine that the compound was soluble in water. Next, the flame test was used to compare the unknown compound to other known compounds such as potassium chloride, sodium chloride, and calcium carbonate. The flame test concluded that the cation in the unknown compound was potassium. Following, pH paper was used to determine the compound to be neutral and slightly
The unknown bacterium that was handed out by the professor labeled “E19” was an irregular and raised shaped bacteria with a smooth texture and it had a white creamy color. The slant growth pattern was filiform and there was a turbid growth in the broth. After all the tests were complete and the results were compared the unknown bacterium was defined as Shigella sonnei. The results that narrowed it down the most were the gram stain, the lactose fermentation test, the citrate utilization test and the indole test. The results for each of the tests performed are listed in Table 1.1 below.
The percent yield of products that was calculated for this reaction was about 81.2%, fairly less pure than the previous product but still decently pure. A carbon NMR and H NMR were produced and used to identify the inequivalent carbons and hydrogens of the product. There were 9 constitutionally inequivalent carbons and potentially 4,5, or 6 constitutionally inequivalent hydrogens. On the H NMR there are 5 peaks, but at a closer inspection of the product, it seems there is only 4 constitutionally inequivalent hydrogens because of the symmetry held by the product and of this H’s. However, expansion of the peaks around the aromatic region on the NMR show 3 peaks, which was suppose to be only 2 peaks. In between the peaks is a peak from the solvent, xylene, that was used, which may account to for this discrepancy in the NMR. Furthermore, the product may have not been fully dissolved or was contaminated, leading to distortion (a splitting) of the peaks. The 2 peaks further down the spectrum were distinguished from two H’s, HF and HE, based off of shielding affects. The HF was closer to the O, so it experienced more of an up field shift than HE. On the C NMR, there are 9 constitutionally inequivalent carbons. A CNMR Peak Position for Typical Functional Group table was consulted to assign the carbons to their corresponding peaks. The carbonyl carbon, C1, is the farthest up field, while the carbons on the benzene ring are in the 120-140 ppm region. The sp3 hybridized carbon, C2 and C3, are the lowest on the spectrum. This reaction verifies the statement, ”Measurements have shown that while naphthalene and benzene both are considered especially stable due to their aromaticity, benzene is significantly more stable than naphthalene.” As seen in the reaction, the benzene ring is left untouched and only the naphthalene is involved in the reaction with maleic
2-ethyl-1,3-hexanediol. The molecular weight of this compound is 146.2g/mol. It is converted into 2-ethyl-1-hydroxyhexan-3-one. This compounds molecular weight is 144.2g/mol. This gives a theoretical yield of .63 grams. My actual yield was .42 grams. Therefore, my percent yield was 67%. This was one of my highest yields yet. I felt that this was a good yield because part of this experiment is an equilibrium reaction. Hypochlorite must be used in excess to push the reaction to the right. Also, there were better ways to do this experiment where higher yields could have been produced. For example PCC could have been used. However, because of its toxic properties, its use is restricted. The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols. This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present. Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Also, my 2,4-DNP test was positive. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol. This could not have been my product because the Tollen’s test. My test was negative indicating no such aldehyde. Also, the textbook states that aldehydes show 2 characteristic absorption’s in the range of 2720-2820 wavenumbers. No such absorption’s were present in my sample. Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketone
Introduction: Someone in a restaurant has suddenly fallen ill and a mystery powder has been discovered with the victim. As the chief investigator, your duty is to identify the mystery substance through a lab. In this lab, it will consist of five known compounds and one unknown compound. Your job is to distinguish which one out of the five substances is the mystery powder. To figure out the mystery matter you will have to compare their physical and chemical properties and match them with the appropriate compound. Furthermore, you will compare their reactions with water, universal indicator, vinegar and iodine solution to discover what the mystery powder is.
The purpose of conducting experiment was to determine the identity of white compound. Based on the 5 gram of unknown white compound several experiment conducted including solubility test, pH test, flame test, and ion test. Several materials including chemicals used throughout experiment and will be described through paragraphs.
The IR spectrum that was obtained of the white crystals showed several functional groups present in the molecule. The spectrum shows weak sharp peak at 2865 to 2964 cm-1, which is often associated with C-H, sp3 hybridised, stretching in the molecule, peaks in this region often represent a methyl group or CH2 groups. There are also peaks at 1369 cm-1, which is associated with CH3 stretching. There is also C=O stretching at 1767 cm-1, which is a strong peak due to the large dipole created via the large difference in electronegativity of the carbon and the oxygen atom. An anhydride C-O resonates between 1000 and 1300 cm-1 it is a at least two bands. The peak is present in the 13C NMR at 1269 and 1299 cm-1 it is of medium intensity.
The C-H (sp3) hydrogens from our product displayed at wavelength 2959 cm-1 correlates to the methyl groups located on the ends of isopentyl acetate4. A really prominent, strong peak located at 1742 cm-1 shows that a C=O ester stretch is located in the product, along with at 1244 cm-1 the spectrum shows a strong peak representing the C(=O)-O stretch that is crucial to the structure of isopentyl acetate. Shown in my IR spectrum is a weak O-H (H-bonded) peak at 3464 cm-1 which shows that I have an impurity of isopentyl alcohol in my product. Isopentyl alcohol has similar boiling points and density as my product so the impurity could have easily boiled out with the isopentyl acetate during distillation. The isopentyl alcohol was also present in my 1H-NMR spectrum backing up the impurity peak at 3464
For the Identification of a Volatile Liquid Data Analysis I had the unknown number A53826. There are many ways that I used to determine the identification of the liquid. Some characteristics of the liquid that will help further in my research are that the liquid was colorless with a harsh strong odor. Before I did my research there were many components I did to help with my research. I first started by performing an experiment that helped determine the mass of my sample. The mass of my sample is one of the key components that will help me determine the moles of my sample. By using the ideal gas law (PV=nRT to n= (PV)/RT) using the mass of my sample in grams, the temperature of my sample after it evaporated during the double-boiling lab, the atmospheric pressure in the room using a barometer, and the volume of the flask I found the moles of the sample (approximately 0.00530 moles). Using the moles I found the molecular weight (mass of sample/ moles of sample). The average molecular weight was 94.9 g/mol. I first started my research by calculating the empirical formula by doing a combustion analysis for 2.0000 grams of my sample unknown (CxHyCLz) yielding 1.7787 grams of carbon dioxide and 0.7294 grams of water. By finding grams of carbon from the amount of grams in carbon dioxide, and finding grams of hydrogen from the amount of grams in water I added the two together (grams of carbon and hydrogen) and then subtracting it from 2.0000 grams of the sample to get my grams of chloride. I then converted the grams to moles and divided it by the smallest amount of mole ratio to get my empirical formula, CH_2 Cl. By using the empirical formula mass, 49.477 grams, and the molecular weight of my sample, 94.9 g/mol, I was able to find my mol...
Our first goal in Project 7 was to determine what our three unknown solutions were. We did this through a series of tests. Our first test was a series of anion tests. We performed anion tests to determine whether any of the following anions were present in our solution: chloride, sulfate, nitrate, carbonate, and acetate. Our first solution, labeled as B, had only the chloride test come out positive. The next solution, C, tested positive for acetate, as did our last solution, E. We next performed anion tests. These included flame test, as well as an ammonium test. For the flame test, certain cations turn flames different colors, so we used this knowledge to test to see which cations could be present in our solutions. During this test, the only solution that appeared to turn the flame any color was solution C, which turned the flame bright orange, indicating the sodium ion was present. This led us to the conclusion that solution C was sodium acetate. We next performed an ammonium test, which involved mixing our solutions with sodium hydroxide, and smelling the resulting solution in order to detect an ammonia smell. Solution B was identified as smelling like ammonia, indicating the presence of the ammonium cation. From this, we identified solution B as ammonium chloride. We next checked the pH of all three of the solutions, first by using litmus paper. Solution C was slightly basic, solution E and B were both acidic, with a pH around 4. Since we knew that solution E had acetate, and was acidic, and did not turn the flame any color, we determined it was acetic acid, as none of the ions in acetic acid would turn a flame any color.
After the IR was obtained, analyzed and compared to reference spectrums, and the boiling point was compared to the boiling points in the chart, it was concluded that the unknown product from the reaction was 2-methyl-2-heptanol. The boiling point of 2-methyl-2-heptanol is 162°C, which is between the range of the boiling point obtained
The flame test produced a lighter orange flame indicating there was not a combustion reaction (shown in Table 2). The colour pink indicates that the cation Cobalt (Co2+) is present in a solution. Since the solution A was originally observed as a pink, transparent liquid, it was determined that the Cobalt was present in solution A. The litmus test results indicated that this solution is neutral (Table 2), therefore the anion found within the solution must abide by this. It was determined from the litmus results and Table 4 that the possible anions present in solution A were Cl-, NO3-, and SO42-. In order to determine the anion in the solution, Cl-, NO3-, and SO42- were compared. The anion Cl- produces precipitate with solutions C, E and F, while the NO3- and SO42- ion do not produce precipitate with all three. Therefore, the solution A must be CoCl2.
A false positive, an error that arises when the test results are positive even though the sample being studied is not present, could have occurred if the solution was cloudy prior to the addition of K2CrO4 and NH4OH.10 A cloudy solution meant that there was a possibility of contamination where remnants of precipitation that were examined in Part A still remained. These remnants, which could have held lead or silver would have posed a problem, especially if lead was suspended in the cloudy liquid, since Pb2+, like Ba2+, can react with K2CrO4, to form yellow precipitate. The consequence to this similarity would have resulted in a potential false positive with the cation in this step to be incorrectly identified as Ba2+ when it was actually mistaken for
Octane with chemical formula C8H18/CH3(CH2)6CH3, molecular mass: 114.22, boiling point :126, melting point : -56.8, Subject no.1262 has many dangers and side effects on workers and firefighters. Therefore, we'll be discussing methods and ways to prevent the hazards that can be caused by the substance.