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... ... middle of paper ... ... dichloroethane the density was 1.24 (1,3) making it most likely the unknown. Another property was the boiling point. The boiling point of the unknown during the experiment was 86.6 ℃; in comparison 1,2- dichloroethane was ±0.03℃ within the range, 83.3 ℃ (1,3). Lastly, to top it all the IR spectrum was closely related to the IR of the unknown. Researching the identity I also found a physical description of 1,2- dichloroethane that matched my description of the unknown, “Colorless liquid with a pleasant, chloroform-like odor” (1). I learned in the experiment that the numbers from my results for the unknown does not have to be exact for it to be identified. As long it is in a certain range the unknown should be known by using research and background knowledge. I also learned that there are many factors that can affect an experiment whether it is systematic or random.
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
Also, both their Ph level was 7 which mean that they are neutral and not acidic and reaction with iodine solution was exactly the same. Therefore, with all the experiments conducted and analyzed, icing sugar is the mystery substance. c) Q: Which properties, physical or chemical, were most useful in identifying the mystery powder? Explain your answer.
A characteristic property can help identify a substance. A characteristic property will never change even when the volume of a substance is varied. A characteristic property also does not change when a substance changes state in matter. A physical property cannot identify a substance. A physical property will change when the volume of a substance is varied. It can also change when the substance changes state in matter. For example, if the volume and mass of a substance changes then the physical appearance will also change. However, the density, which is a characteristic property, will not change at all. The boiling point of a substance is the temperature that a substance changes from a liquid to a gas. The boiling point of a substance is a characteristic property because the boiling point of a substance will never change even when the volume and mass changes. The only thing that will change is the time that it takes to reach that temperature. If the mass and volume of the substance is small, then it will take a small amount of time for the substance to reach the temperature. However if the mass and volume of the substance is larger, then it will take a longer time to reach the temperature. The purpose of this lab was to see if when the volume of a substance changes so does the boiling point.
In the lab the reaction that took place was a synthesis reaction. A synthesis reaction, is a type of chemical reaction in which two or more simple substances combine to form a new product. The reactants may be elements or compounds. In this case it is a gas and a metal that will react and produce a compound. The general form of a synthesis reaction is, A + B → AB. In order for this lab to be done successful you need knowledge on, percent composition, the empirical and molecular formula, the law of conservation of mass, moles and molar mass, qualitative and quantitative. To begin, the percent composition of a compound is the percent of the total mass that each element has in that compound. Every compound would have a certain percent composition. To calculate percent composition of a compound, you would have to determine the total molecular mass of the compound. For example, for H2O the total molar mass would be 18.00g/mol. You would then input the mass of one of the elements and the molar pass into the equation % by weight (mass) of element = (total mass of element present ÷ total mass of compound) x 100 to find out the percent composition. So for Oxygen it would be, % of O = (16.00g ÷ 18.00g/mol) x 100 which would equal 88.9%. Therefore the percent composition of O in this compound is roughly 88.9%. Furthermore, the molecular formula is the number and types of atoms that are existing in a single molecule of a substance. The empirical formula also known as the simplest formula is the ratio of elements present in the compound. The key difference between these two is that the empirical formula shows the simplest positive integer ratio of atoms of each element present in a compound whereas the molecular formula of a compound is a way ...
In the analysis, the hypothesis was that pennies made before 1982 were made out of a different substance than the pennies made after 1982. The hypothesis turned out to be true. The pre 1982 pennies were made out of copper which has a density of 8.6 g/mL and the post 1982 pennies are made out of zinc which has a density of 7.14 g/mL. So it can concur that mass for pennies made before 1982 are higher than pennies made after 1982. Some errors were when calculating. The densities for the pennies and making the graphs. When making the graphs, it can be suspected that each graph is wrong could be wrong due to not adding enough data points or misplaced points. For the data tables , it can also be suspected that the density for each data table could
We obtained an unknown metal with an identification number (6) from the instructor. Make sure not to get fingerprints on the sample otherwise it will interfere with identifying the density of
To find out what the Mystery Mixture was the group of 8th graders had to use physical and chemical properties. First, the group decided to use physical properties. Physical properties are texture, smell, color, appearance, shape, and size. Citric acid & Baking soda was powdery with varying sized chunks, had a soft light white color, and had a very faint tangy smell. The Mystery
Using the previously calculated Kf, the molar masses of unknown substances A, C, and D were able to be calculated. However, given that the original Kf was slightly larger than the theoretical value, the molar
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.
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
Chloroform is a dense, colorless liquid made up of CHCl3 (Carbon, Hydrogen, and chlorine) and is known for being hazardous. Chloroform can be created by heating a mixture of chlorine and either chloromethane or methane. As a solvent, it is commonly used in the lab because it is relatively unreactive. It forms a
The procedure is very short. First, I constructed the data table. I then massed one piece of Aluminum foil and record it in the data table. Next, pour 25.0 mL of 0.400 M Cu²+ solution and tear up the massed piece of foil. Drop the torn pieces in the solution and record the observations. Mass the other pieces of Aluminum foil, tear it up, and drop it in 25.0 mL of the 0.200 M Cu²+ solution. Record observations and clean up according to the teachers
The goal of Project 11 is to identify an unknown substance that we were given. This project was split up into 2 weeks. The first week my group and I identified our substance by going through multiple tests like: a solubility test, conductivity tests, pH tests, anion test, and cation test. For the second week we had to confirm our unknown substance.
These were all naturally occurring substances. No refinement had occurred, and isolation of specific compounds (drugs) had not taken place.
1. The labels have fallen off of three bottles thought to contain hydrochloric acid, or sodium chloride solution, or sodium hydroxide solution. Describe a simple experiment which would allow you to determine which bottle contains which solution.