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To determine the melting point, first we had to calibrate the thermometer. We measured the temperature of ice water which is supposed to be zero(0) degrees Celsius. Out thermometer read zero degrees which it was supposed to read. Then we measured the temperature of boiling water which is supposed to be 100 degrees Celsius. Our thermometer measured 100 degrees which was correct. We had to make a boiling point correction by using the formula: BP correction(760mm-atm pressure)*.037c/mm. The number we got was +.074 degrees Celsius which means that water would boil at 100.074 degrees Celsius. This was a negligible difference from 100 degrees so we used the normal boiling point of water. Then we graphed the two measured points against the temperatures we were supposed to get and connected the two points. This graph can be used to correct the melting point of the unknown. The first unknown we received was lactose(number 26). First we set up a hot water bath and waited to see if the substance would melt below 100 degrees and it didn’t. We had to setup an oil bath to get it to melt. We observed the substance to melt at 224 degrees. The actual melting point range was 222 to 228 degrees. We didn’t find the range for three reasons. First of all we didn’t notice that the substance started to melt at 222 because the temperature was rising too fast. If we removed the heat the temperature would stop rising but once we put the burner back under the oil the temperature would shoot up. Secondly when the temperature reached 224 degrees it looked like the substance melted. The third reason was that maybe the temperature of the oil was rising faster than the thermometer can register it. The real temperature could have been a few degrees higher than what the thermometer read. The percent error we made was from .8% from the low number in the range to 2.7% from the high number. The second unknown we received was sodium thiosulfate(number 42). When we set up the water bath, the substance melted at 60 degrees. The actual melting point range was 40 to 45 degrees Celsius. A small fraction of the error (3 to 5 degrees) can be accounted for by faulty equipment, and experimental data. The other 15 to 20 degrees can be accounted for by our carelessness. The percent error for the second unknown was from 33.3% from the low number of the range to 25% from the high number in the range. Lactose(#26) Sodium Thiosulfate(#42) Observed Melting Point/Range (Celsius) 224 60 Corrected Melting Point/Range(Celsius) 224 60 Actual Melting Point/Range(Celsius) 222-228 40-45 Percent Error .8% to2.7% 25% to 33.3% |
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We added some barium chloride to the crucible and weighed it. By subtracting the weight of the crucible from the weight of the crucible and barium chloride, we got the weight of barium chloride we started with. Then we heated the BaCl2 to evaporate all the water. We weighed the BaCl2 and heated it again. We continued this until the weight didn’t change. By subtracting the weight of the anhydrous BaCl2 form the crystalline BaCl2 we got the weight of the water. We lost .69 grams of water. To get the percent of water we divided the mass of water lost by the weight of crystal Barium Chloride and multiplied by 100. We got 13.8 percent. The actual percent of water is 14.8 percent. Our results had a 6.8 percent error. The error we received can be accounted for by inaccurate equipment. Another reason for the error could have been the fact that we didn’t heat the BaCl2 long enough and not all the water evaporated. The last two parts of the lab required us to find the GMW of the crystallized BaCl2 and write its formula. Knowing the GMW of anhydrous BaCl2, we were able to calculate the GMW of the crystals. The proportion of the grams of the crystal to the anhydrous BaCl2 is the same as the proportion of the GMW of the crystal to the anhydrous Barium Chloride. The proportion we used was 4.28/3.56= n/206 where ‘n’ is the GMW of the crystal. The ‘n’ we calculated was 246 grams where as the actual GMW was 243 grams. Our percent error was 1.22 percent. To figure out the formula of the crystallized barium chloride, we used the equation: GMW of water =grams of water lost/X where GMW of water and the grams of water were known and ‘x’ was the number of moles of water in the crystal. This gave us the moles of water. The formula GMW of BaCl2 =grams of anhydrate/X, where the GMW and the grams of BaCl2 known and ‘x’ was the number of moles of the barium chloride in the crystal. The ratio we received was .02 BaCl2 to .038 H2O. This simplified to a 1 BaCl2 to 1.9H2O and we rounded to 2H2O. We then counted the GMW of our formula and when it was the same as the GMW of the real formula, we knew we had the correct formula. |
| go back The purpose
of this lab was to learn real-life applications of stoichiometry. Stoichiometry
can be used to find the empirical formula of a substance. It can also be
used to calculate the percentage of a substance in a mixture.
The first of the two parts of the experiment was to find the empirical formula of potassium chlorate (KClO3). To do this we had to get rid of the oxygen. When KClO3 is heated it decomposes to form KCl and O2. For the potassium chlorate to decompose during the 40-minute lab we have, we had to add a catalyst (manganese dioxide). We weighed the test tube and catalyst (9.89g.). Then we added the KClO3 and weighed the test tube again (10.6g). After heating the potassium chlorate, the residue weighed 10.2g. This meant that we lost .4 grams of oxygen. Using the balanced equation 2KClO3 ? 2KCl+ 3O2. We were able to determine that we lost .0125 moles of oxygen (.4/32). From the weights we figured that we had .31 grams of KCl. .31 grams of KCl is .0042(4.2x10-3) moles. Since the formula for potassium chloride is KCl there are equal number of moles of potassium and chlorine atoms in the substance. There were .0042 moles of each molecule to start with. Then number of moles of oxygen in KClO3 is three time as much or .0125 moles. Knowing all this we were able to write the formula K.0042Cl.0042O.00125. After simplifying this we received an empirical formula of KClO2.98 which is very close to KClO3. For the second part of the experiment we started the same way. Out unknown was unknown number 4. We weighed the test tube and catalyst (9.08g.). Then we added the unknown mixture of KClO3 and KCl and weighed the test tube again (9.86g). After heating the potassium chlorate, the residue weighed 9.68g. This meant that we lost .18 grams of oxygen. Using the balanced equation 2KClO3 ? 2KCl+ 3O2. We were able to determine that we lost .01125 moles of oxygen (.18/16). From the same equation we were able to determine that we started with .00375 moles of KClO3 (2/x=6/.00125). Multiplying this by the GMW of KClO3 we received the number of grams of KClO3 we started with (.4596g). Dividing this by the mass of the unknown, and multiplying by 100 we got 58 percent of KClO3 (.4596/.78x100). We were supposed to receive 33 percent not 58 percent KClO3. This error could be accounted for by many things. Faulty equipment and incorrect weighing procedures could account for a lot of the mistake. The catalyst and unknown not being completely dry and water being evaporated could account for some more of the error. Side reactions are another problem. The test tube or the scupula could have contained traces of a compound, which reacted with the unknown, the catalyst or both to change some of it to something else. Water being present in the test tube would account for more. Not heating the unknown long or strong enough can account for more of the error. Heating the mixture too strongly in the beginning could have forced some of the unknown to “jump out” of the test tube. The last reason for the error could be that the teacher made a mistake in making the mixture and instead of it being the 33 percent it should have been, it was 35 percent or so. QUESTIONS 1) If we had the wrong formula for potassium chlorate in part one, it would be hard to predict how it would affect the results in part 2. First we would begin with the incorrect balanced equation. For example, if we start with KClO2 as the correct formula instead of KClO3, the balanced equation would be KClO2 ?KCl+O2. That means instead of having the mole ratio of KClO2 to KCl to O2 being 2:2:3 we would have a 1:1:1 mole ratio. Also the molecular weights would be incorrect. The correct GMW of KClO3 is 122.553g/mol while the weight of KClO2 is 106.5g/mol. This would cause an even further deviation from the correct percentage. 2) If we didn’t get rid of all the oxygen by not heating the potassium chlorate long enough or strong enough, we would get a different empirical formula. It would have a smaller mole ratio of O2 to K to Cl. Instead of the correct 3:1:1 ratio we could have gotten something like 2:1:1 or 1:1:1. 3) To check if manganese dioxide is really a catalyst, we would need to know how much of it we put in the test tube in the beginning of the experiment. Then we would take the residue and dissolve it in water. The KCl is soluble in water, it would dissolve and the manganese dioxide would form a precipitate. We would collect the precipitate by pouring the mixture through filter paper. The KCl solution would go through and manganese dioxide would be left on the filter pap 4) er. We would heat it to get rid of the water and weight it. If the weight in the beginning of the experiment is the same as the weight of manganese dioxide in the end of the experiment, we would know that it was a catalyst and didn’t participate in any reactions. |
| go backThe purpose of this
lab was so that we can learn to determine the molar volume of a gas at
STP. All the gasses at STP have a molar volume of 22.4 liters. That is
one mole of any gas at STP takes up 22.4 liters
To be able to determine the molar volume of a gas, we had to find out the volume of a certain weight of a gas at a known temperature and pressure. To do this we had to set up an apparatus with which we can measure the weight and volume of a gas. Since it’s nearly impossible to measure the volume of a gas directly, we had to use displacement of water to measure the volume of a gas. We setup the apparatus as follows: 1) The oxygen generator consisting of potassium chlorate, and a catalyst (manganese dioxide) in a test tube connected by a rubber tube to a Florence flask with water in it. 2) The flask was connected by rubber tubing to a collection beaker. There were three major steps in this experiment. The first one was setting up the oxygen generator. First we heated about 1 gram of manganese dioxide for a few seconds to drive out al the moisture. We then added about 1 gram of potassium chlorate (KClO3) and mixed it with the catalyst. The last part of this step was weighing the generator. Ours weighed 26.39 grams. The second part was to make the pressure in the Florence flask equal to the atmospheric pressure. To do this we had to get the water to flow through the tube and into the collecting beaker. Then we clamped of the entrance to the system. The other entrance, being submerged in water, mad the system airtight. We moved the collecting beaker up and down until the level of the water in the beaker and the flask was equal. We poured the water out of the beaker and dried it. The clamped end was placed into the oxygen generator and unclamped it. The final step of the experiment was the collection of data. We heated the test tube to have the potassium chlorate decompose into potassium chloride (solid) and oxygen (gas). The oxygen would go into the flask, displace the water into the collection beaker. When the beaker was half full we removed the heat and waited for the water flow to stop. When the apparatus returned to room temperature we equalized the pressure in the beaker and the flask and clamped it off. Measuring the volume of the water we received 328 milliliters. The atmospheric pressure was 796 torr and the temperature of the system was 24C(297K). Weighing the generator we received a value of 25.81 grams. With all the data gathered we had enough information to do all the calculations. According to our calculations, we figured the molar volume of oxygen to be16.94 liters. Compared to the actual 22.4 liters, our result had a 24 percent error. Many things can account for this error. First of all the pressure inside the flask was probably not exactly the same as the atmospheric pressure. This may have caused some of the oxygen released, to be used to equalize the pressure. This oxygen was not accounted collected. Secondly our apparatus might not have been airtight and some oxygen was able to escape and not be collected. Water vapor being in the test tube and some water in the potassium chlorate that was present at the start of the experiment would have evaporated and for dome problems in the weighing procedure. We received slightly more oxygen lost by mass than was actually lost because of this. Also not all the water displaced by the oxygen was collected. Some water was probably left in the rubber tube. The equipment not being top quality would account for some more of the error. An incorrect temperature reading of even one degree would have caused some more of a deviation from the actual number. Since temperature and temperature related constants (partial pressure of water vapor) are both used in the calculations, this may have caused relatively large declination from the correct result. |
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The numbers of Nuclear Reactor plants have grown sufficiently. Electricity is being generated in a number of ways, it can be generated by using Thermal Power. Thermal Power is produced using two basic systems, the Steam Supply System and the Electricity Generating System. The Steam Supply System produces steam from boiling water by the burning of coals and the Electricity Generating System produces electricity by steam turning turbines. The Nuclear power plants of this century depend on a particular type of Nuclear Reaction called fission. Fission is the splitting of a heavy nucleus like the uranium atom to form two lighter atoms, as well as less massive particles as the Neutrons. In the Nuclear Reactors this splitting is induced by the interaction of a neutron with a fissionable nucleus. The energy released from the fission reactions provide heat, part of which is ultimately converted into electricity. The produced heat is then removed from the nuclear fuel by water that is pumped through the past rods that contained fuel. The basic feature of the nuclear reactor is the release of a large amount of energy from each fission event that occurs in the nuclear reactors core. On the average, a fission event releases about 200 million electron volts of energy. A typical chemical reaction on the other hand releases about one electron volt. The complete fission of one pound of uranium would release about the same amount of energy as the combination of 6000 barrels of oil or 1000 tons of high quality oil. The reactors cooling fluid serves dual purpose. Its most urgent function is to remove the heat from the core, produced when the energy is released from the Nuclear reactions and transformed by the collisions into the random nuclear motion. Its function is to transfer this heat into an outside core, typically for the production of electricity. The produced fluid may be used directly to drive a turbine generator. Alternately, it may be used to heat a secondary fluid that drives the turbine. In most all the commercial systems that fluid is vaporized water. The products formed during nuclear fission have a slightly lower mass, due to the nuclear mass defect. This nuclear mass defect can be used to determine the nuclear binding energy that held the heavier nucleus together and was released when fission occurred. The energy released by fission can be calculated by finding the difference between the mass of the parent atom and neutron, and the masses of the daughter atoms and emitted neutrons, and converting this mass into energy using. Neutrons released when an atom undergoes fission are capable of causing other nuclei to undergo fission, if a moderator slows the neutrons down. A sustained fission reaction caused in this way is called a chain reaction. Natural uranium ore contains about 0.7% uranium-235. To increase the likelihood of sustaining a chain reaction for uranium, the fissionable isotope of uranium must be increased in its relative proportion through enrichment. An Isotope is one of two or more atoms of an element that differ in the number of neutrons found in the nucleus. A nuclear reactor produces a sustained chain reaction a a controlled rate. The heat energy produced by the reaction is used to drive turbines, generating electricity. Control rods, made of materials such as cadmium, which absorb neutrons, are used to control the rate of a chain reaction in a nuclear reactor. A critical mass of fissionable material is the minimum mass that will produce a nuclear explosion. A sustainable nuclear chain reaction requires more material than the critical mass. Most Reactors today use uranium. It is the form of uranium oxide fuel pellets. To produce electricity the refined uranium oxide fuel pellets are placed in the cylindrical rods. The rods are arranged into a fuel bundle, which is then ready to be placed in special pressure tubes inside the reactor. Nuclear reactors can not explode like a nuclear bomb. Even under a worst-case scenario, with a core meltdown, a critical mass of fuel would not be present and the fuel would burn into the ground. Refueling can be done by removing fuel bundles from the pressure tubes and replacing them with new bundles. Heavy water is used as the moderator in a reactor. Heavy water contains deuterium, an isotope of hydrogen having one neutron in the nucleus. Heavy water also transfers heat from the fuel into a heat exchanger, which heats ordinary water to produce steam. The steam is used to turn turbines, which are connected to the electrical generators. Condensers change the steam back into water so it can be cycled back to the steam generator. If excess heat builds up in the reactor’s vessel, the heavy water can be drained out. This causes the chain reaction to stop, because the moderator is no longer present. Supporters of the use of nuclear energy feel that it is a safe and effective way to produce energy. With the demand for energy increase and the problems associated with burning fossil fuels, such as acid precipitation and the greenhouse effect, they regard the use of nuclear energy as being necessary. Nuclear energy avoids some of the problems of generating hydroelectric power. Flooding land to build dams creates environmental and social problems. The use of nuclear energy may avoid the need for long transmission lines. Nuclear plants can be built in relatively close proximity to where the power is needed. Nuclear energy produces very small amounts of waste by volume. The radioactive materials can be concentrated for storage and monitoring in one place. As a result poisonous metals, such as arsenic, lead, and mercury, also toxic gases, carbon dioxide, and fly ash are not released into the atmosphere. Some believe that the use of nuclear energy cite various problems. The opposition to the use of nuclear energy has grown so strong in recent years that some reactors have been shut down. Other reactors scheduled for development have been delayed or were never completed because of the social and political pressure exerted by the antinuclear lobby. The debate continues. |
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The purpose of this lab was to carry out a chemical reaction in which a metallic element, copper, is converted into blue vitriol. By synthesizing blue vitriol we were also able to characterize and isolate the reaction product. Experimental Procedure See the lab handout. No deviations from the procedure indicated in the lab manual were made other then the following. 6.01grams of copper wire were weighed and cut into little pieces instead of being coiled into a flat spiral. Not all of the mixture was filtered. Because of the constantly ripping filter paper only a part of the mixture was actually filtered and the other part left in the flask. At the end of the lab a PH test of cupric sulfate (aq.) was conducted as it was indicated on the board. Results 6.02 grams of copper wire weighed. Cu+15.6 M HNO3?the solution turned green and copper dissolved in the solution. Copper was oxidized. Cu+ HNO3? Cu2++NO2+2OH- Nitric Acid+Copper+Sulfiric Acid? Solution turned green then started boiling during which brown gas was produced. Eventually the solution went from light green to dark blue. Blue vitriol yield: Beaker mass+30.00 grams Beaker +Blue Vitriol=32.45 grams Blue vitriol mass=32.45-30.00=2.45 grams Characterization
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