Determination of the Relative Atomic Mass of Lithium
...(g) In order to calculate the Ar of lithium the number of moles of hydrogen produced had to be calculated. This was done by using the following formula: Moles H2= amount of H2 produced 24000 (It was assumed that one mole of gas occupies 24000cm³ at room temperature and pressure). This figure was then used to calculate the number of moles of lithium that reacted; it was assumed that two atoms of lithium are roughly equal to one atom of hydrogen. Hence, the number of moles of H2 produced was simply multiplied by two. The Ar of lithium could then be calculated. The formula for calculating Ar is as follows: Ar = mass of substance used moles All of the calculations above were conducted for all five trials and a mean Ar was taken. Figure One presents these results. Method Two A measure of 25cm³ of the LiOH produced in Method One was then taken from the conical flask. This was done by using a 25cm³ pipette with a pipette filler, which sucked the solution in. The solution in the pipette was then transferred into a 250cm³ beaker. A burette was then filled with 50cm³ of 0.100 mol dmˉ³ hydrochloric acid (HCl) and attached to a clamp on a stand. Five drops of phenolphthalein indicator was then added to the LiOH in the beaker, which turned it a pale pink colour. This was then placed under the burette and the tap opened, which allowed the HCl to flow into the beaker. The solution was then titrated with the HCl until it went clear. The solution was agitated whilst the HCl was flowing into the beaker. The amount of HCl added (titre) was then recorded, to the nearest 1cm³, to be used in our calculations (refer to Figure Two). This process was repeated five times, in conjunction with Method One, in order to give a number of readings. The equipment was cleaned and used chemicals disposed of in between. New chemicals were used on each occasion. This was so that fir results could be obtained. Once the five readings had been obtained the Ar could be calculated. The equation for the neutralisation reaction in the titration is as follows: LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l) The average titre was then calculated using all five readings. This figure was then used in the calculation of the number of moles of HCl used in the titration. The formula used was: Moles H2 = molarity × mean titre 1000 (The mean titre had to be divided by 1000 because it was measured in cm³, rather than dm³). As there is a 1:1 relationship between LiOH and HCl, this figure also informed us of the number of moles of LiOH used in the titration. The number of moles in 100cm³ of the LiOH solution then had to be calculated. To obtain this figure the number of moles in 25cm³ of LiOH was multiplied by four. The Ar of lithium could then be calculated using the following formula: Ar = molarity moles of LiOH in 100cm³ of solution Results Figure One – Results for Method One Trial Amount of Li used (g) Amount of H2 produced (cm³) Moles of H2 collected Moles of Li that reacted Ar of Li 1 0.040 64 0.00266 0.00532 7.52 2 0.031 53 0.00221 0.00442 7.01 3 0.039 72 0.00300 0.00600 8.17 4 0.040 50 0.00208 0.00416 9.61 5 0.034 59 0.00246 0.00492 6.91 Mean Ar of Li = 7.844 Median Ar of Li = 7.520 Range = 2.7 Figure Two – Results for Method Two Trial Amount of Li used in 100cm³ of distilled water (same as Method One) Amount of 0.100 mol dmˉ³ HCl used in titration (cm³) 1 0.040 22 2 0.031 28 3 0.049 26 4 0.040 31 5 0.034 20 In the average titre, 25cm³ of LiOH required 25.4cm³ of 0.100 mol dmˉ³ HCl. Number of moles of HCl used in titration: Moles = 0.100 × 25.4 1000 = 0.00254 As there is a 1:1 relationship between LiOH(aq) and HCl (aq), the number of moles of LiOH used in the titration is 0.00254. Number of moles of LiOH present in 100cm³: 0.00254 × 4 = 0.01016 Ar of Li = molarity/moles of LiOH (in 100cm³) = 0.1/0.01016 = 9.84 Evaluation The results obtained do not appear to be accurate. The percentage accuracy of the results obtained for the two methods can be calculated by dividing the Ar obtained (in the case of Method One, the mean Ar for the five trials) by the known Ar for lithium (6.9) and then multiplying this figure by 100: Method One: (7.844/6.9) × 100 = 113.68% Method Two: (9.840/6.9) × 100 = 142.61% These figures show that Method One was more reliable than Method Two because the Ar obtained was closer to the known Ar of 6.9. However, these figures also show that the results were not accurate as they were not within a five percent margin either side of 6.9. The median Ar value obtained in Method One, 7.520, was closer to 6.9, being 108.98% of this figure, but is still significantly d...