"Finding The Relative Atomic Mass Of Lithium."

1350 words - 5 pages

Determination of the relative atomic mass of lithiumIntroduction:For this investigation I will determine the Relative atomic mass (Ar) by using two different methods. In the first method I will dissolve a piece of lithium of a known mass in water, I will then collect the hydrogen gas produced, which can be used to calculate the relative atomic mass of Lithium. In the second method I will titrate the resulting solution of lithium hydroxide with a known concentration of hydrochloric acid, this can also be used to calculate the relative atomic mass of lithium.Readings:*100 cm3 of water*37.47 g of lithium plus the watch glass*37.41 g of Watch glass*122 cm3 of hydrogen gas collectedCalculations from method 1Moles of hydrogen produced:Assuming that 1 mole of gas occupies 24000 cm3 at pressure and room temperature!I will now calculate the number of moles of the hydrogen gas produced. I will do that by dividing the volume of gas produced by 24000cm3 :Moles (n) = Volume (cm")24000 cm"So:n = 122 cm" = 0.005083 mol of H224000 cm"Equation:2 Li(s) + 2 H2O (L)2LiOH(Ag) + H2 (g)By looking at the Stoichiometry ratio in the reaction I can see that the ratio of Li: H2 is 2:1, therefore the number of moles of Li will be twice as the number of moles of H2 in the reaction.So:0.005083 x 2= 0.010166 mol of LiMass of Lithium =Mass of the Lithium plus the watch glass - the mass of the watch glassSo:37.47g - 37.41g = 0.06g·0.06g of lithium was usedRelative atomic mass of lithium:I will now calculate the relative atomic mass of the piece of lithium that I used by rearranging a formula.Molar mass= relative atomic mass (Ar)If moles= Mass in g then Ar = mass in gMolar mass (Ar) molesI already know that the mass of lithium that I used in 0.06 grams, and that I used 0.010166 moles of lithium:So: Ar (also known as RAM) = 0.06 = 5.90.010166Therefore the relative atomic mass of lithium used in this investigation is 5.9.Calculations from method 2:Titration123Final burette readings21.621.521.5Initial burette readings000Volume of acid Used cm321.621.521.5Average volume of acid used:21.6+21.5+21.5=21.5 cm33Number of moles of HCl used in titration:HClVoume: 21.5cm3Concentration: 0.1 moldm"215cm3 = 0.0215 dm"Moles = volume x concentration so:n = 0.0215 x 0.1 = 0.00215Therefore the number of moles of HCl used is 0.00215.Equation:LiOH (aq) + HCl (aq) LiCl (aq) + H2O (l)By looking at the stoichiometry ratio in this reaction, the ratio of HCl: LiOH is 1:1, therefore the number of moles of LiOH will be the same as the number of moles of HCl in the reaction. That is O.O0215 moles of LiOHNumber of moles of LiOH present in 100 Cm3 of the solution from method 1If 0.00215 moles of LiOH for 25cm3 then for 100cm3 I will need to multiply the numbers of moles of LiOH times four.So:0.00215 x 4 = 0.0086Equation:2 Li (g) + 2H2O(L)2LiOH(aq) + H2 (g)By looking at the stoichiometry ratio in the reaction, the ratio of LiOH: Li is 1:1, therefore the number of moles of Li will be the same as the number...

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