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Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added to the mixture which will emphasize the changes. This chemical also returns iodine to iodine ions. 2S2O32-aq + I2aq -----------> S4062-aq + 2I-aq As the reactants of this reactions are colourless it is difficult to tell when the colour change will happen, and when it does it is quite sudden. This happens because the thiosulphate is being used up in the reactions, and when it is all used up the starch-iodine complex is formed resulting in the colour change. Equipment and apparatus needed: "¢ Test tubes "¢ Boiling tubes "¢ Thermometer 0-110°C "¢ Stopwatch "¢ Burettes "¢ 15cm3 Potassium iodide solution,KI, 1.00 mol dm-3 "¢ 10cm3 Potassium peroxodisulphate solution,K2S2O8, 0.0400 mol dm-3 "¢ 10cm3 Sodium thiosulphate solution, Na2S2O3, 0.0100 mol dm-3 "¢ 5cm3 fresh Starch solution "¢ Test tube rack Method "¢ Read and understand carefully the table below illustrating the different mixtures of concentrations for individual experiments. mixture Volume of KLaq/ cm3 Volume of water / cm3 Volume of Na2S2O3aq / cm3 Volume of starch solution/cm3 Volume of K2S2O8aq / cm3 1 5 0 2 1 2 2 4 1 2 1 2 3 3 2 2 1 2 4 2 3 2 1 2 5 1 4 2 1 2 "¢ When making the mixtures, make sure that all chemicals are added to a boiling tube, except the potassium peroxodisulphate which must be added separately as this is what will trigger the reaction. "¢ Make up mixture 1 and place a thermometer in the boiling tube. Add the potassium peroxodisulphate at begin the timing with the clock "¢ Stop timing when the solution turns blue-black "¢ Constantly stir the solution. "¢ Take a temperature reading "¢ Repeat procedure for each mixture of different concentrations. Results A Table mixture Concentration of I-aq / mol dm-3 Clock time / seconds Rate / mol dm-3 s-1 Temperature / °C 1 0.5 28.8 3.45 X10-7 23 2 0.4 40.9 2.44 X10-7 23 3 0.3 58.0 1.72 X10-7 26 4 0.2 136.7 7.32 X10-8 26 5 0.1 455.5 2.20 X10-8 24 Questions B Rate = concentration of I2 Time = [ I2 ] = mol dm-3 s-1 S [ I2 ] = 1.0 X10-5 Mixture: 1. = 1.0 X10-5 = 3.45 X10-7 28.8 2. = 1.0 X10-5 = 2.44 X10-7 40.9 3. = 1.0 X10-5 = 1.72 X10-7 58.0 4. = 1.0 X10-5 = 7.32 X10-8 136.7 5. = 1.0 X10-5 = 2.20 X10-8 455.5 C The Iodine ions are in excess in the reaction mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced. D Moles = mol dm-3 x dm3 = 0.04 x 2/1000 = 8x10-5 mols of S2O8 2- E i 2cm3 of Na2S2O3 0.0100 mol dm-3 Moles of Na2S2O3 = 0.01 x 2/1000 = 2x10-5 moles of S2O3 ii Moles of = 1.0 X10-5 I2 iii Percentage % of reaction = 1.0 X10-5 x100 8.0 X10-5 = 12.5 %
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Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added...
mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced.

D Moles = mol dm-3 x dm3

= 0.04 x 2/1000

= 8x10-5 mols of S2O8 2-

E i 2cm3 of Na2S2O3 0.0100 mol dm-3

Moles of Na2S2O3 = 0.01 x 2/1000

= 2x10-5 moles of S2O3

ii Moles of = 1.0 X10-5

I2

iii Percentage % of reaction = 1.0 X10-5 x100

8.0 X10-5

= 12.5 %

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In this experiment I will... In this experiment I will be using three methods to determine the relative atomic mass of lithium. The three methods are, first by adding water to lithium, then measuring the amount of hydrogen given off and finally using stoichiometry appropriately to find out the relative atomic mass of lithium. The second method would be to titrate; therefore neutralise 25 cm of lithium hydroxide from that produced in method 1 with HCl, using an indictor to know when the reaction has being completed and finally using stoichiometry appropriately to find out the relative atomic mass of lithium. Health & Safety issues to be observed during every part of the experiment * All chemicals should be labelled with their names and correct safety hazard label for accidents to be prevented. * I will always wear my coats and gloves during the experiments as I will be using HCL, which could burn the skin. * I will always wear my goggles during the experiments as I will be using corrosive compounds like HCL and lithium hydroxide which is a very corrosive alkali. * I will be very careful when handling or pouring the chemicals. * Cover the experimenting table with a blanket in other for fatal accidents to be prevented as I would be using fragile materials I the whole experiments. Fair test for all experiments * I should follow the procedure I have planned for each method correctly. * For all my results I have decided to use one decimal place because the apparatus I was using for the experiment gave me results to one decimal place. * I could repeat each experiment twice or thrice if easy, in order to get a more reliable set of results. Experiment 1 Fair test for experiment 1 * The lithium should be wiped as dry as possible with tissue. * The lithium should be cut as quickly as possible, in other for less possible chances of getting impurities by dust or it reacting with oxygen. * The graduated cylinder should be placed vertically always during the experiment. Health & Safety issues to be observed while carry out experiment 1 * I should weigh accurately as possible 0.10 g of lithium using the weighing balance. * The lithium should be added into the conical flask as quickly as possible and the bung should be replaced to exclude possibilities of gases being lost. * I should make sure the delivery tube is straight throughout the experiment and that it is working well by testing it before beginning the experiment. * I should wipe with tissue the forceps which would be used to transfer the Lithium to avoid contamination. * I should make sure no bubbles are in the graduated cylinder after transferring the cylinder into the trough. I would do this by covering the top of the cylinder using a wet paper and very quickly inverting the cylinder into the trough. Observations * There was an air bubble in the graduated cylinder before the experiment which was hard to remove, so I looked at the amount of volume it occupied in the cylinder and I will deducted it from final result 0.2. Analysis of Results From my AS level, module 1 course I know that 1 mole of a gas occupies 24000 cm3 at room temperature and pressure. Also I know that 1 mole of an element equals to its relative atomic mass. I could use this information to work out the relative atomic mass of lithium by linking them together as done below. 2Lis + 2H2Ol => 2LiOHaq + H2 g 2/2 2/2 2/2 ½ 1 Li => 1/2 H2 I can now calculate the relative atomic mass of lithium as I know that 1 mole of a gas occupies 240003 cm or 24 dm3 at room temperature; so I shall be linking the amount of final volume of gas collected to be able to find the number of moles. And finally take the number of moles and I know the mass of lithium used, I could use the formula below to find the relative atomic mass of lithium. Number of moles = mass / Ar and re-arranging this to Ar = mass/ number of moles. 140 cm3"“ 0.2 = 139.8 1 mole of gas => 24 dm3 in r.t.p x => 139.8 cm3/1000 = 0.1398 I could see that the two figures on the right hand side of the equation is in the same units, so I divided the bottom figure 139.8 cm3, by 1000, to get it in dm3. I would now cross multiply them; as done below x => 0.1398/24 => 0.005825 From looking at the equation, I know that 1 mole of lithium reacted with 0.5 moles of hydrogen. So the stoichometry is 2 to 1, and to get the number of moles of lithium, I would just need to multiply that of hydrogen by 2. 2 Li => 1 H2, so is 0.005825*2 = 0.01165 Ar of lithium = 0.10g / 0.01165 = 8.6 Result Analysis The result of the first method is actually 8.6, which is 1.7 marks above 6.9, which is the relative atomic mass of lithium in the periodic table. Second Experiment Fair test for experiment 2 * I should run the tap on slowly during the experiment in order for me to know the accurate end point. * I should use enough drops of the indicator so as to obtain enough colour of the solution to be able to be certain when the reaction is complete. * I should repeat the titration twice or trice to obtain consistent results and get an average titre. * I should make sure the inside of the conical flask is clean by wiping it with tissue in order to avoid impurities. * I should make sure 5 full drops of phenophatelum are added to the solution; as I am following the given procedure. Health & Safety issues to be observed while carrying out experiment 2 * The lithium hydroxide which would be transferred into the conical flask is a corrosive alkali, thus it could irritate the skin. * I should take care lithium hydroxide does not splash out into my eyes by wearing googles as it could cause severe irritation. Observations * After I added 5 drops of phenophatelum to the solution, it turned purple. Then during the course of the reaction the colour of the hydrous lithium chloride started changing to plane white. Analysis of Results Experiment Volume of HCl used / in cm3 1 28.0 2 32.0 3 30.5 Average reading 30.2 I am looking for the relative atomic mass of lithium by using this method and lithium is an element that consists as part of the base in this reaction. As I have learnt from AS level chemistry, module 1, mole calculations, I could calculate the number of moles of the base because I already know the concentration of the acid, volume of the acid and by using stoichiometry of the equations. I would then link the number of moles of acid to that of the base using this equation, Number of moles = concentration * volume. HCl s.t.d soln = 0.10 mol Volume used => 30.2 cm3 / 1000 I could see that the volume used is not in dm3, so I will divide it by 1000, to get it in dm3. I would now use the equation below to find the number of moles. Number of moles = concentration * volume = 0.10 * 0.0302 = 0.00302 This volume was found from 25cm3 of Lithium hydroxide, so to get that of 100 cm3, I will multiply it by 4. 0.00302 *4 = 0.01208 Now that I know the number of moles of Lithium to be 0.01208, I could use the following equation that I have learnt from my Essential AS Level chemistry book to find the relative atomic mass of lithium. Relative atomic mass Ar = mass of element in grams / number of moles, as Mass of Li = 0.10g, Number of moles = 0.01208. Ar of Li = mass in grams / number of moles Ar = 0.10g / 0.01208mol Ar = 8.2781457 = 8.3 Result Analysis This result is the most correct result compared to that of the first method. It is actually 1.4 marks above 6.9, which is the relative atomic mass of lithium in the periodic table. Evaluation of Results Overall in general I could say the accuracy of the 1st and 2nd experiments was good, however with practice I think I would have done it better. I still think I carried out all the experiments well as I possibly could at the time with the given conditions and apparatus. For instance I was able to carry out three tries for the 2nd experiment; to be able to get an average result and I also finished within the time period allowed for me as this would make my result more reliable. However, looking back from my result analysis section, I could see that the relative atomic mass values I found in both of the two methods used were not exactly 6.9, like that of lithium in the periodic table, but the results were quite close. Using method 1, I got the relative atomic mass of lithium to be 8.6; a difference of 1.7 to 6.9, which is that in the periodic table. Using method 2, I got the relative atomic mass of lithium to be 8.3; a difference of 1.4 to that in the periodic table. So looking at the slight inaccuracy of results, of which greatest is 1.7 higher than 6.9 expected, in method 1. Therefore method 2, which is of titrating 25 cm3 of lithium hydroxide with HCl proved to be the most accurate, then method 1, which is of adding water to lithium gave a close result to that achieved in method 2. This wasn't what I expected though as method 2 as the most possibility of errors, but I could see by the results that I had followed the fair test measures well in order to cut this errors. I thought method I would be the most accurate because I was using standard apparatus to carry out the method, but the method has a lot of huge errors compared to method 2's one. For example when putting the lithium inside the conical flask, I noticed that before I could fit the bung back, I had lost a lot of gas. There are more sources of error that I noticed written below. Sources of error In the experiments, I believe that the main sources of error must be human and equipment error. But there are other errors which are neglible because they are small. For example a source of error is in measuring the amounts of the substances or materials. In method 1, did I truly measure out 25.0 cm3 of the method 1 solution LiOH in the conical flask? Or where some of the solution still left in its container and I may have been looking at the cylinder incorrectly- not from a parallax level. Another one is the measuring apparatus, for instance I noticed the pipette that I used in experiment 2, was quite dirty and I had no choice but to use it. Automatically I assumed that this would cause an error in my results as it could either contaminate my solution or may slightly change the concentration I originally thought was in there because the dirt would take up some space in the pipette thus changing the amount I think is in there. A way to improve this problem is to basically clean out all the apparatus and also make sure that it is not left lying around to collect dust and become dirty. The following sources of error I found to be evident and which would have affected the experiments very much. 1 During the method 1, there where three errors caused by the lithium been stored in oil, and could have affected the other method. First when weighing the lithium, I calculated that the weighing machine would have had a very high percentage error of 10%. 0.01/0.1 * 100 = 10%. * Secondly, as a source of error, weighting of lithium will not be accurate because of the following reasons. 1 All the oil might not have been removed from the lithium by just cleaning it with tissue but there was no way of distinguishing this apart from the human eye. This method of parallax is therefore inaccurate as distinguished above. * Thirdly, when I tried to transfer the lithium into the conical flask with distilled water, some pieces of the lithium got stuck on the conical flask and this caused an error as to my whole experiment as I wasn't able to place the bung as quickly as possible. Hydrogen would have been lost in the time taken for me to add the sample of lithium to the flask with distilled water and replacing the bung. Therefore, the final volume of gas evolved in the measuring cylinder would not have been correct; therefore, the second and third experiment as well would have also given a slightly wrong result, as the same solution was used for them. 2 Lithium was given to us; we students had to remove small amounts of lithium, weigh it on measuring scales, and cut the lithium using a knife and tweezers in order to get 0.10g. By the time we had measured out of the lithium for the first experiment; the lithium would have already reacted with the air, and would have become slightly impure. * During the method 2, it wasn't easy to determine the end point, because when there is a colour change, the colour doesn't change throughout the solution and this would have caused errors in the method, as the aim of method 2 is not to put excess Hcl, but as the colour didn't change through out the solution, I had to keep the tap open slightly until it did. Reliability of Result * In method 1, the procedure I used was good but the equipment I used was not quite reliable as some unevitable errors were made as I have explained before in sources of error. The biggest problem in this method is if the transferring tube has holes, but I test it before experiment started. Apart from these ones there were only some small possibilities of errors and this was proved by the relative atomic mass I found was just 1.7 greater than the expected 6.9 for lithium. * In method 2, the procedure I used wasn't accurate, as the method is quite complex because I had to handle hydrochloric acid and because of this I took very good care during this method by following the fair test. This would have a great possibility of accuracy in this method and this was proved by the relative atomic mass I found was 1.4 greater than the expected 6.9 for lithium. Improvements * An improvement to method 1, is to use an hydrocarbon substance to remove all traces of the oil as this is more effective and can remove impurities from all sides of the lithium sample, unlike a human using tissue. * An improvement of method 1 is to do the experiment again and then use all the improvements I have noted but there was no time left. * An improvement of method 1, is to use a magnifying glass to look at the lithium pieces if the oil on them as all gone and if possible while doing this place the lithium pieces in a vacuum, so that they would be less contaminated. * An improvement of to method 2 is to ensure that all the titrations are fairly tested, thus the same conditions and equipment are used. They are all washed and dried well after each repeat. * An improvement of this method 2 is to ensure a method that all the colour of the solution in the burette would change, as when to stop the tap is not no more obvious. The only way to this is by first opening the burette tap fully then as soon as a small colour change as been seen, the tap should be closed slightly, so that only drops of the acid are coming through and the point when all the colour of the solution changes could be seen more clearly. Conclusion In this experiment I was using two methods to find the relative atomic mass of Lithium and method 2 was the most accurate one has I found out an answer of 8.3, which is 1.4 greater than the Ar or lithium.   

In this experiment I will be using three methods to determine the relative atomic mass of lithium. The three methods are, first by adding water to lithium, then measuring the amount of hydrogen given off and finally using stoichiometry appropriately to find out the relative atomic mass of lithium....

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