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Determination of the relative atomic mass of Lithium.

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Determination of the relative atomic mass of Lithium Implementing For all my results I have decided to use three significant figures. This is because the apparatus I was told to use for the experiment gave me results to three significant figures. There was, therefor, no point in calculate to any further accuracy as this would just define a figure that has already been calculated to a specific depth of accuracy. Results Method 1 Mass of Lithium used 1.09 g Volume of gas produced 175 cm3 Method 2 1 2 3 Final Burette Reading 42.1 42.3 42.3 Initial Reading 0 0 0 Volume added (cm3) 42.1 42.3 42.3 Average 42.2 cm3 Analysing Method 1 Calculate the number of moles of hydrogen. 1) 175 = 0.00729 Number of moles = Amount of gas produced 2400 2400 This was calculated to find out the number of moles in the Hydrogen gas that had produced. Deduce from this the number of moles of Lithium. 2) The number of moles of Lithium = 0.00729 x 2= 0.0146 If you look at the chemical equation- 2Li(s) ...read more.


100 cm3 is 4 times as much as 25cm3, therefor if I multiply 0.00422 by 4 then this would give me the number of moles on 100 cm3 Use the result and the original mass of lithium to calculate the relative atomic mass of Lithium. 4) 0.104 = 6.15 Relative atomic mass = Mass used in g 0.0169 Number of moles I used the formula above to calculate the relative atomic mass. My average atomic mass for Lithium = 6.64 The actual relative atomic mass for Lithium = 6.90 Difference = 0.26 % difference = 3.77 Risk assessment The first safety hazard I encountered in my experiment was the Li that I placed into the waster in the first method. I was told it reacted violently with water to form lithium hydroxide. The Li was contained at almost all times throughout the experiment but as there was a chance of me being in proximity of the reaction then I wore goggles. Also I was shown that I was not to use a cube that has sides of more than 3mm. ...read more.


Also the only effective way to remove the oil from the Li is to cut the outside layer off the Li and this will ensure there is no oil left on the Li when it is weighed. Also this will stop the oil from being in the Li hydroxide and creating any inaccuracies in the second method. Hydrogen is another source for an inaccuracy. This is because in can be effected and lost. It can be lost in two ways, either when the bung is replaced onto the conical flask when the Li starts to react or when the hydrogen goes through the rubber tubing it can be lost. There are no ways to improve the experimental procedure for this and so they are inaccuracies that I cannot change. The third inaccuracy is the fact the gas expanded when it came from the Li and water as it was hot so the final gas reading would be greater than it should be and so it effected my results. I would of used a percentage error type of evaluation but as I don't have specific measurements that could of shown me how inaccurate the data I collected was I cannot use percentage errors for some data's and not for others. ...read more.

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