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# Determination of the relative atomic mass of lithium.

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Introduction

Determination of the relative atomic mass of lithium. Aim: To determine the relative atomic mass of lithium. Analysis (method 1):) * (1) Calculate the number of moles of hydrogen. Conclusion so far: Since I have only this experiment once, method 1, there are no other results, only one set of results. Due to this fact there cannot be any anomalous results because there are not any other results to compare it to. The reason why I only did the experiment once was that was a lack of time, thus no average can be made. Plus I can conclude that since I have only done the experiment once, it could be an anomalous result itself. * During this experiment I will assume (from source of information) that 1 mole of gas occupies 24000 cm3 at r.t.p. (room temperature and pressure). The balanced formula I will be using for this experiment will be: 2Li(s) + 2H20(l) 2LiOH(aq) + H2(g) So to figure out the number of moles of hydrogen I am going to use an equation gained from my knowledge. n = Vcm3 24000 n = 224 = 0.00933 Moles 24000 The "224Cm3" or the volume was obtained from my results table (later on ). * (2) Deduce the number of moles for lithium. From looking at the balanced equation from above, we can simply look at it and tell that the ratio is 2 : 1, which means I can say there are two moles of Lithium and one mole of Hydrogen. So with this information I know what the mole for hydrogen is (0.00933M). Since there are double moles of lithium (2 moles) I simply need to times the number of moles for hydrogen by 2 to get the number of moles for lithium. Number of moles for hydrogen = 0.00933 X 2 = the number of moles for Lithium = 0.01866 M Now I have the number of moles for hydrogen and lithium. ...read more.

Middle

To figure out the mean average I need to use method: 42.6 + 42.7 + 42.7 = 42.67 3 The three figures are from my results, the volume of HCL used, and dividing it by three is done because I have three results. * (2) Deduce the number of moles of LiOH (lithium hydroxide) used in titration. This is done relatively simply. By looking at the equation you can see that there is no large number in front of any of the reactants or products. As in method one I had to times the number of moles of hydrogen by 2 to get the number of moles of hydrogen. However in this case the ratio is 1 : 1. This then means that the number of moles of LiOH will be that same as the HCL. Number of moles of HCL = 0.00427 Number of moles of LiOH = 0.00427 * (3) Calculate the number of moles of LiOH present in 100 cm3 of the solution from method 1. Again this is done in a simple manor. If you read the first step of the procedure it says "Pipette 25.0 cm3 of the solution in the conical flask from method 1 into a clean 25.0 cm3 conical flask and add 5 drops of phenolphthalein indicator." However at this stage I need to find the number of moles present in 100cm3 whereas at stage (2) it was only for 25cm3. This means all I have to do I times the answer from part (2) and that will gives me the answer this calculation. At 25cm3 the number of moles of LiOH is 0.00427 M. At 100cm3 the number of moles of LiOH will be (0.00427 X 4) 0.01708 M. The reason I times it by 4 is because to make 25 into a 100, you need to times by 4. * (4) Use this result and the original mass of lithium to calculate the relative atomic mass of lithium. ...read more.

Conclusion

The concentration could have been changed thus altering the method 2 result. Implementation:) (Method 1) In this part of the course I will explain the correct procedure on how to perform the experiment. Starting off, set up your apparatus as shown in the diagram. 250cm3 measuring cylinder Lithium 100cm3 distilled water At this stage be sure that you have filled the conical flask with exactly 100cm3 of distilled water. Plus you must also make sure that the distilled water is pure and is not mixed with other substances. Now at this stage you can get your lithium ready. Either measure out around 0.10g of lithium or just use what is given to you. In my case I used 0.13g. Draw out a table to record your original mass of lithium and the initial and final volume of hydrogen used. THIS IS THE CRUCIAL POINT. Remove the bung and add your lithium. As soon as you can, replace the bung tightly to ensure no loss of gas. Finally you simply record the final volume of hydrogen. Safety is very important. Make sure that you are wearing goggles, your hair is tied back and that your bags are tucked under the table so that no one falls over. (Method 2) First get your pipette and measure out exactly 25.0cm3 of your lithium hydroxide (method 1 solution). Pour this into a clean 250cm3 conical flask and add around five drops of phenolphthalein indicator. Place this under the titration apparatus and titrate with a HCL solution with the concentration of 0.100dm3(aq). Record the results in an appropriate format (above) showing the final burette reading, initial burette reading, the volume used and the mean titre. You need to repeat the process about three times to gain more accurate results. On your table show all your results. Finally record the average (mean) titre. Safety is extremely important as you are handling hydrochloric acid. Goggles must be worn at all times, hair must be tied back and bags must remain under the desk. If anything goes wrong, consult the teacher immediately. ...read more.

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