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# to determine the relative atomic mass of lithium. We will be doing this via two methods:

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Introduction

The aim of this experiment is to determine the relative atomic mass of lithium. We will be doing this via two methods: * The first method will be to collect the gas evolved when lithium is reacted with distilled water and consequently calculate the relative atomic mass. * The second method will be to calculate the relative atomic mass through titration with Hydrochloric Acid (HCl). Method 1 Results: Mass of lithium (g) 0.099 Volume of hydrogen produced (cm3) 178 The volume of hydrogen collected from the reaction of 0.099 grams of lithium with 100.0cm3 of distilled water was 178 cm3. Treatment of results: We assume that 1 mole of gas occupies 24000 cm3 at room temperature and pressure. The chemical equation for this reaction is: 2Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g) We must firstly calculate the number of moles of Hydrogen produced using the following equation: Moles of gas = Volume (in cm3) 24 000 Therefore, Moles of hydrogen = 178 = 7.4167 x 10-3 moles 24 000 According to the mole ratio in our chemical equation, there are twice as many moles of lithium than hydrogen present. ...read more.

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Conclusion

The value that was recorded alongside the absolute value was 10.0 � 0.04 cm3 which means using the formula mentioned above, we reach a percentage error as below to 3 s.f.: Percentage error = 0.04 x 100 = 0.4% 10 As can be seen, the greatest percentage error is clearly with the measurement of the volume of HCl used (3.38%). To lower this percentage error and consequently improve the accuracy of our results, a larger volume of HCl would need to be used to neutralise the LiOH so therefore a larger amount of LiOH would be need to used. Not only would this decrease the percentage error in the volume of HCl used but will also lower the percentage error in the volume of LiOH used seeing as a larger measurement of LiOH would need to be used. The mass of the lithium contributes the second largest percentage error (1.01%), this can also be lowered by using a larger amount of substance, in this case lithium, which would result in a lower percentage error and a more accurate result. ?? ?? ?? ?? AS - Assessed Practical (Skills I, A and E) - Determination of the Relative Atomic Mass of Lithium 23rd March 06 ...read more.

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