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# Calculating the relative atomic mass of lithium.

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Introduction

Chemistry Assessed Practical Write-up Calculating the relative atomic mass of lithium 1) Mass of Lithium (g) Volume of Hydrogen collected (cm3) 1 1.2 129 2 1.1 120 3 1.3 137 Amount of H2 collected = 129 ml Mass of Lithium with oil = 1.2g 2Li + 2H20 --> 2LiOH + H2 Vol H2 = 129 24,000 = 0.005375 moles --> 1:2 --> 0.01075 RAM = Mass Moles RFM Lithium = 0.12 0.01075 RFM Lithium = 11.16g The RFM of Li that I worked out is = 11.16g RAM Li = 6.9g Therefore the mass of oil is: 11.16 - 6.9 = 4.26g 2) Volume Lithium Hydroxide = 25.00 cm3 exactly (using a volumetric pipette) Volume HCL (0.1 molar) to neutralize Lithium Hydroxide = See graph Start (cm3) Finish (cm3) Value (cm3) 1 0.0 27.1 27.1 2 1.2 28.2 27.0 3 0.7 27.8 27.1 LiOH + HCl --> LiCl + H2O Volume HCL = 27.1 Concentration of ...read more.

Middle

Using the volume of hydrogen I was able to calculate the moles of hydrogen. I used this work out the moles of Lithium from which I could work out the mass and RAM of Lithium. The second experiment was the titration of the lithium hydroxide that was a by-product of the first experiment. By measuring the amount of hydrochloric acid (of a known molarity) I was able to work out the RAM of Lithium. The average of the two RAM's that I found is: 10.835 + 11.16 = 10.99 2 Evaluation Both the results for the RAM that I recorded are around 11g, so any errors that have occurred have obviously affected all the results. Though the second result could be affected by more errors because it has the errors of the first results, and the risk of more errors in the second experiments. ...read more.

Conclusion

The error in the amount of HCl is about 0.1cm3 because the graduations on the pipette are 0.1cm3 apart. Taking into account all these errors the results that I obtained aren't very accurate, as the results show as they are about 4g off the proper RAM of Lithium. These results could be improved by using a 100cm3 volumetric pipette to measure the distilled water. This would make the error almost negligible. You could devise a system where the Lithium could be added when the bung was already in place, therefore there would be no gas escape. You could eliminate the error in HCl by using burette with smaller graduations. The overall percentage error is: 10.99 * 100 = 159 % 6.9 This experiment has basically gone smoothly, there were no major hiccups along the way, and although my results aren't entirely accurate, at least they are consistent. I conclude that the reason for this is that the apparatus that I used was not appropriately accurate for the experiment. 1 1 ...read more.

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