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Determination of the relative atomic mass of Lithium.

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Shimona Madalli 12A 15/02/03 Determination of the relative atomic mass of Lithium (Skills I, A and E) In this investigation, the aim is to determine the relative atomic mass of lithium, Li by two different methods: 1) By measuring the volume of hydrogen gas, H2 produced - METHOD 1 2) By titrating the lithium hydroxide, LiOH produced - METHOD 2 Skill I Hazards of method 1 and method 2 - All apparatus must be handled with care especially the glassware (e.g. conical flask) as can be broken easily. - Wear labcoat and safety goggles to avoid skin and eye contact with these chemicals. - The hydrochloric acid was dilute and as only 0.1M concentration was used the chemical acts as an irritant only. If spilled wash thoroughly with water and rinse eyes. - Hydrogen gas, H2 is a colourless and non - toxic, it is also flammable but as there were no naked flames this will not cause any risks. - Lithium is a very reactive metal. It is exothermic as it generates heat when involved in reactions. It is an irritant and so contact with this chemical should be avoided. - Lithium hydroxide, LiOH, is an alkali as well as an irritant. It also causes burns but as the solution is dilute this does not pose any problems. ...read more.


+ 2H2O (l) 2LiOH (aq) + H2 (g) Step 1 : Moles of hydrogen gas = volume produced 24000cm� 188cm� 24000cm� = 7.83 x 10-3 mol Step 2 : Mole ratio H2 : Li 1 : 2 7.83 x 10-3 mol : 0.016mol Step 3 : RAM of lithium = mass 0.09 = = 5.63 (prediction 6.92) moles 0.016mol Calculations : Method 2 - RAM of lithium Average titre = 12.6cm� + 12.4cm� + 12.5cm� = 12.5cm� (prediction 12.9cm3) 3 LiOH (aq) + HCL (aq) LiCl (aq) + H2O (l) Number of moles of LiOH present in 100cm� : Step 1: Moles of HCL = volume (cm�) x concentration (M) 1000 = 12.5cm� x 0.1M 1000 = 1.25 x 10-3 mol Step 2: Mole ratio HCL : LiOH 1 : 1 1.25 x 10-3 mol : 1.25 x 10-3 mol Step 3 : In 100cm� = 10 x (1.25 x 10-3 mol) = 0.0125mol I will use this result to calculate the RAM Step 4 : RAM of lithium, Li = mass Moles = 0.09g = 7.20 (prediction 6.92) 0.0125mol Skill E Overall the accuracy of my experiment was fairly consistent, for example all readings were to 3 significant figures. As you can see from the percentage errors (see below) method 2 was much more accurate. ...read more.


- this will surely increase the accuracy of the results. Limitation - It was very unclear when measuring initial and final amount of water in the measuring cylinder and so this would produce inaccuracies in the final volume of hydrogen gas produced. Improvement - Can use another piece of apparatus to improve the accuracy such as a gas syringe. Limitation - As the lithium metal was stored in oil, it was very difficult to remove all the oil and so this could of affected the actual mass of the metal. Improvement - Only solution to this is to measure the lithium metal many times. Or in an inert atmosphere, mix with hexane to remove oil and following this blow helium or nitrogen to remove the hexane. Comparison of methods 1 and 2 Comparing the two methods, method 2 which was the titration experiment was much more accurate. By using a graduated burette, the accuracy of my obtained results improved greatly which can be seen in my results table. Both methods are linked, if there were inaccuracies in method 1 then this would affect the accuracy of method 2. Relative error Relative error = Difference in figures Standard figure (in this case the standard figure is the standard RAM of lithium, which is '7') METHOD 1 : 7 - 5.63 X 100% = 19.6% 7 METHOD 2 : 7.20 - 7 X 100% = 2.86% 7 1 ...read more.

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