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Chemistry chemical reactions lab report
Chemistry chemical reactions lab report
Laboratory reports 4 chemical aspects
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Following my research using the molecular formula, I dove deeper into my research and looked closely at my calculated data. My molecular formula is C_4 H_9 Cl. I used ChemSpider to match my formula to millions of compounds. The compounds that match my molecular formula are 1-Chlorobutane, Tert-Butyl Chloride, 1-Chloro-2-Methylpropane, 2-Chlorobutane, (2S)-2-Chlorobutane, and (2R)-2-Chlorobutane. (ChemSpider) C36843 has a density of 0.8601 g/mL. (Table 1) All of the calculated values of C36843 are shown in the above, Table 1. The density of 1-Chlorobutane is 0.886 g/Ml, the density of Tert-Butyl Chloride is 0.846 g/mL, the density of 1-Chloro-2-Methylpropane is 0.883 g/mL, the density of 2-Chlorobutane is 0.87 g/mL, the density of (2S)-2-Chlorobutane …show more content…
C36843 has an average molecular weight of 96.835 g/mol. (Table 1) 2-Chlorobutane has an average mass of 92.039 g/mol. Meanwhile; 1-Chloro-2-Methylpropane, (2S)-2-Chlorobutane, and (2R)-2-Chlorobutane all have one of 92.567 g/mol. (ChemSpider) Even though the mass calculations are off, mass is not a defining factor because it is the average of all the compounds found in the environment and I still decided that my compound is 2-Chlorobutane. My mass calculation was off 4.796 g. Some factors that could lead to my error in calculating the weight was not boiling the compound to the right evaporation state, having trouble with the balance, or contaminating the sample before measurement. No experiment is perfect, which is why most experiments require more than 2 trials. My 2 trials helped get my experimental molar mass close to the actual molar mass, however more trials would benefit the experiment and make it even closer to the actual value. Performing the experiment in a closed system would also heavily prevent contamination and balance issues should only occur at a very miniscule amount. The biggest analyzing factor that affected my calculations was not having enough trials because simple human error is common in this experiment, as I stated …show more content…
The molar mass of my unknown is 96.835 g/mol; the density is 0.8601 g/Ml; the boiling point is 68.5℃ and the molecular and empirical formulas are C_4 H_9 Cl. (Table 1) The molar mass of 2-Chlorobutane is 92.039 g/mol; the density is 0.87 g/mL; the boiling point is 69.2℃; and the empirical and molecular formulas are C_4 H_9 Cl. (ChemSpider) The properties all match up and they helped me identify my unknown. During the experiment and the later research that followed, I learned how to identify what a compound using its intrinsic properties and how isomers function in the real world. I found out that 2-Chlorobutane has an estimated amount of 6 isomers and that shape sometimes does define how a compound works. The isomers all have the same formulas, however the differently arranged bonds change the whole compound. I also learned how to measure the atmospheric pressure using a barometer in the classroom and this experiment really improved my understanding of Chemistry concepts involving mass calculations. One of the most important concepts I have learned from this research paper and my data analysis, is that good Chemistry experiments require more than 2 trials and precise procedures that can be
Some possible errors raised during the synthesis and spectrometric analysis of TPCP include the insufficient mixing of the hexane and TPCP, in which will result in the low absorbance of the compound. Additionally, the low yield is contributed from the loss of product during filtration.
Extraction is a separation method that is often used in the laboratory to separate one or more components from a mixture. Sucrose was separated at the beginning because it is the most immiscible and it’s strongly insoluble. Next Acetylsalicylic Acid was separated which left Acetanilide alone. Variety steps could have led to errors occurring. For example the step of separation, when dichloromethane layer was supposed to be drained out, it could be possible some aqueous layer was drained with it. Which could make the end result not as accurate. Also errors could have occurred if possibly some dichloromethane was not drained out. Both way could interfere with end result of figuring the amount of each component in the mixture. The solids percentage were 22.1% more than the original. That suggests that solids weren’t separated completely which clarifies the reason the melting points that were recorded were a slightly lower than the actual component’s melting point. The melting point for Acetylsalicylic Acid is 136 C but that range that was recorded during the experiment was around 105 C to 118 C. The melting points were slightly lower than the literature value. Sucrose was the purest among all component due to its higher melting point which follows the chemical rule that the higher the melting point the more pure the component
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
In this experiment there were eight different equations used and they were, molecular equation, total ionic equation, net ionic equation, calculating the number of moles, calculating the theoretical yield and limiting reagent, calculating the mass of〖PbCrO〗_4, calculating actual yield, calculating percent yield (Lab Guide pg.83-85).
Living organisms undergo chemical reactions with the help of unique proteins known as enzymes. Enzymes significantly assist in these processes by accelerating the rate of reaction in order to maintain life in the organism. Without enzymes, an organism would not be able to survive as long, because its chemical reactions would be too slow to prolong life. The properties and functions of enzymes during chemical reactions can help analyze the activity of the specific enzyme catalase, which can be found in bovine liver and yeast. Our hypothesis regarding enzyme activity is that the aspects of biology and environmental factors contribute to the different enzyme activities between bovine liver and yeast.
Mass Pb(NO2)3/ MW Pb(NO3)2 to find the mass of SO4, which equaled 0.1394g, then plugged into the percentage was exactly 55.75% SO4. The begging of the experiment requested that the percentage of SO4 in their fertilizer sample be found and that is exactly what was accomplished by this experiment, and the outcome, demonstrating the ability of my group to preform the requested task presented by the
Our procedure though was not without its mistakes. These mistakes are vital because they affect the data we conclude. Theoretically, according to the balanced chemical equation, for every mole of hydrated cobaltous chloride that is being heated, the decomposition ensures that the compound decomposes into one mole of cobalt(II) chloride and six moles of gaseous water vapor. Thus, in theory we should lose the mass equal to six moles of water vapor in each trial. Unfortunately, this is not the case because we don’t have perfect lab conditions and factors such as the time heated, utilization of the same crucible, and the inconsistency of magnitude of the flame from the Bunsen burner all contribute to differences in mass percent change for each
The synthesised ethyl cinnamate weighed 0.56 g, which was equivalent to 140% of the theoretical yield of 0.414 g.
The purpose of this lab was to calculate the percent composition by mass of oxygen in potassium chlorate.
Number of moles of cyclopentadiene = 0.2g cp* (1 mol cp)/(66.10g cp) = 0.00303mol cp
Regarding the densities of Coke and Diet Coke, I believed that the density of coke would be greater than the density of Diet Coke. Because the content of Coke contains more sugar than Diet Coke, it would contain more mass and since density is mass dependent, Coke would be denser than Diet Coke. From the results of the experiment, there was a slight difference between the densities of Coke and Diet Coke. The measurements obtained from the pipette and the graduated cylinder demonstrated that Coke is denser than Diet Coke while Diet Coke was shown to be denser than Coke using the burette. With the pipette, the average density of Coke is 1.02 and the average density of Diet Coke is 0.99. With the graduated cylinder, the average density is 0.976968 and the average density of Diet Coke is 0.95. With the burette, the average density of Coke is 0.99 and the average density of Diet Coke is 1.0. Among the three instruments, the most precise was the graduated cylinder and the most accurate was the volumetric pipette. Since density is defined as mass/volume, changing the volume of Coke or Diet Coke would have changed.
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...
...bromebutane. Unfortunately, our group was only able to obtain the chromatograph for 2-bromobutane and the rest of the three chromatographs were provided by our T.A. Some possible reasons why the chromatographs for 2-butanol, 1-butanol, and 1-bromobutane were unable to be displayed properly is due to the malfunction of the syringes. If the syringe is not air-tight, the gaseous products can escape before being inserted into the injection port. In addition, the collection tube may have had a minor gas escape from the rubber septum, resulting in less concentrated gaseous products being inserted into the injection port. A possible solution is sealing the collection tube with parafilm. All in all, the provided data chromatographs and the rendered chromatograph by the 2-bromobutane in the lab session did match the expected results for the distribution of gaseous products.
Using stochiometry, the moles of HCl were also.0006575 at the equivalence. point. Then there is the. Finally, our concentration of HCl was calculated out to be. 06575.
Determining the Relative Atomic Mass of Lithium An experiment has been carried out to determine the relative atomic mass of Lithium by using two different types of methods The first method that was carried out was to determine the volume of Hydrogen produced. In this experiment a fixed amount of Lithium was used, in my case it was 0.11g. At the end of this experiment, the volume of Hydrogen gas I collected was 185cm³. Then using the solution of lithium hydroxide made from experiment one, I used it in the titrating experiment, to find out the total volume of Hydrochloric acid used to titrate the lithium hydroxide. RESULTS TABLE Experiment Initial Volume ( cm³) Final Volume ( cm³) Total volume Of HCl used ( cm³) Rough 0.2 30.3 30.1 1 6.3 35.8 29.5 2 2.7 32.0 29.3 Average 29.6 CONCLUSION Method 1 [IMAGE]2Li (s) + 2H20(l) LiOH(aq) + H2(g) Number of moles of Hydrogen. Volume of hydrogen gas was 185 cm³. Weight of Lithium was 0.11g. N = __V__ _185_ = 0.0077 MOLES 24000 24000 Number of moles of Lithium.