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The aim of our experiment was to find out the molar mass of the lithium sample and compare the molar mass to the actual molar mass of lithium from the periodic table.

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Introduction

Chemistry Write up # 4: Molar mass of lithium Table showing masses and measurements of chemicals in the experiment Description Value Unit Uncertainty Percentage Uncertainty Molar mass literature value 6.941 Mass of lithium # 1 0.035 g ?????? 2.86% Mass of Lithium #2 0.040 g ?????? 2.50% Air pressure 100.6 Kpa ????? 0.00994% Gas constant 8.31 Temperature of water #1 and #2 22.0 Celsius ??0.5?mL 2.27% Gas displaced # 1 40.0 Ml ??0.5?mL 1.25% Gas displaced # 2 50.0 Ml ??0.5?mL 1.00% Concentration of HCl 0.1 Mol dm-3 Observations: * There was a grey layer on the outside of the lithium which was the Lithium that had already reacted with oxygen to form lithium oxide. * Some gas escaped in the time it took to put the lid on after adding the lithium to the water. * The Lithium fizzed slightly when put in the 100ml of H2O. * While titrating the indicator change the lithium hydroxide blue and when titrated with the HCl the solution turned clear. * There was no smell throughout the experiment. Titration raw data Table showing the volume of (0.1) hydrochloric acid needed to neutralise the Lithium Hydroxide solution ml) ...read more.

Middle

% deviation = Experimental value - Literature value X 100 Literature value 1 % deviation = 9.8 - 6.941 X 100 6.941 1 % deviation = 41% Titration C = and N= N (HCl) = C * V N (HCl) = 0.1 * 0.0079 N (HCl) = 0.00079 1.26% + 0.00% =1.26% LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(aq) LiOH and HCl have a 1:1 ratio So N (LiOH) = 0.00079 N= Mr = m/N Mr = 0.035 / 0.00079 = 44.3 / 4 0.25ml used Mr = 11.1 1.26% 2.86% + 1.26% =4.12% � 0.5 % deviation = Experimental value - Literature value X 100 Literature value 1 % deviation = 11.1 - 6.941 X 100 6.941 1 % deviation = 60% C = and N= N (HCl) = C * V N (HCl) = 0.1 * 0.0090 N (HCl) = 0.00090 1.11% + 0.00% =1.11% LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(aq) LiOH and HCl have a 1:1 ratio So N (LiOH) = 0.00090 N= Mr = m/N Mr = 0.040 / 0.0090 = 44.4 / 4 0.25ml used Mr = 11.1 1.11% 2.50% + 1.11% =3.61% � 0.4 % deviation = Experimental value - Literature value X 100 Literature value 1 % deviation = 11.1 - 6.941 ...read more.

Conclusion

Even 10ml of gas escaped and not displaced would dramatically change our results. More displacement means more moles which in turn means a lower molar mass. The lower molar mass would be closer to the accurate value which shows that escaping gas was a major problem. Another problem with the experiment is that some of the Lithium had already oxidised. This Lithium oxide will not react with the water meaning that the actual mass of Lithium that reacts is smaller than the mass of Lithium that was weighed. A small mass would mean that the molar mass would be less. Because of the equation Mr = m / N . A smaller molar mass would mean that it would also be closer to the actual value of 6.941. Improvements: To solve the escaping gas problem an idea would be to suspend the Lithium on a piece of paper over the conical flask. You would then shove the bung into the paper with the lithium and into the conical flask. This would mean there would be no time for any gas to escape meaning the results would be more accurate. Another improvement is we could have used absolute pure non oxidised lithium because the lithium oxide or oxidized Lithium does not react with the water. ?? ?? ?? ?? Thomas Bishop IB13VF ...read more.

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