As part of my AS chemistry unit, I was given an assessed practical to do. The aim of this practical is to determine the relative atomic mass of lithium. I will do this by the combination of two different methods.Method 1: – By measuring the volume of hydrogen producedMethod 2: – By titrating the lithium hydroxide producedMethod 1I set up my experiment as previously shown with the correct size equipment in the correct places.
But before I began the actual practical part of the assessment, I had measure the amount of lithium I had. Lithium is reactive with air, so I had to measure it in a specific way. Firstly I sliced an appropriate part of lithium with a knife and put it in a Petri dish.
I then weighed the Petri dish occupied by the lithium which was 3.40g and to calculate the amount of lithium I had, I weighed the Petri dish on its own which turned out to be 3.31. Finally I subtracted the weight of the Petri dish and lithium by the weight of the Petri dish which resulted in its own.3.40g – 3.31g = 0.
09g(Petri + lithium) (Weight of Petri dish) (Weight of lithium)After I had the amount of lithium recorded, my next objective was to find out the amount of gas being given off as a consequence of the reaction of lithium and the distilled water. I recorded the amount of water in the measuring cylinder so I would know where the start point of my reaction was. As soon as I added the lithium to the distilled water the reaction would begin. So I removed the stopper and added the lithium before reuniting the two back together, I did this as quickly as I could. After about 1 minute the reaction had stopped and all the gas had been collected and I read it off as being 95cm3, but to record the actual volume of gas collected, I subtracted the two.250 – 95 = 155(Volume collected) (Volume at end) (Actual volume of hydrogen produced)Now that I have both the weight of the lithium used and the amount of gas produced I can start on the technical calculations.
1 mole of gas occupies 24 000 cm3 at room temperature and pressure2Li (s) + 2H20 (l) + 2LiOH (aq) + H2 (g)* Calculate the number of moles of hydrogen.The No of moles of hydrogen is the volume divided by 24 000 cm3So No of moles of hydrogen = volume 15524 000 24 000 = 0.0064 ï¿½ 2 = 0.
0128* Deduce the number of moles of lithium.Number of moles = massRAM0.0128 = 0.09RAM = 7.03* Using your values above and the original mass of lithium, calculate the relative atomic mass of lithium.Method 2After recording the amount of hydrogen gas and lithium and using both values to find the number of moles of both hydrogen and lithium and finding the RAM for lithium, I will now do the titration method.During the titration, I kept 25.0 cm3 of the solution from method 1 in a conical flask and added 5 drops of phenolphthalein indicator, as I did this the solution turned a fluorescent pink colour.
I then had to titrate the solution with 0.0100 mol dm3 of aqueous hydrochloric acid to see how much acid was needed to turn the solution from a bright pink colour, back to a colourless solution.The concluded results were: -Amount of HCL at start = 28Amount of HCL at end = 42.
15So the amount of HCl used is the amount at the start minus the amount at the end.42.5 – 28 = 14 .3 cm3(Start) (End) (Result)LiOH (aq) + HCl (aq) LiCl (aq) + H2O (l)I will now attempt to work out the technical parts to method 2.* Calculate the number of moles of HCl use in the titration.Number of moles of HCl = 0.1 ï¿½ 14.
3 = 0.001431000The number of moles of hydrochloric acid (HCl) used in the titration is 0.00143* Deduce the number of moles of LiOH used in the titration.The number of moles = 0.00143 ï¿½ 10 = 0.090.00143 = 6.29The number of moles of LiOH used in the titration is 6.
29* Calculate the number of moles of LiOH present in 100 cm3 of the solution from method 1.* Use this result and the original mass of lithium to calculate the relative atomic mass of lithium.EvaluationThe experiment I was given was not really accurate because there was random errors which affected the experiment and therefore affected the results. Also the normal mass of lithium is 7 but the result that I concluded was slightly bigger. The main systematic errors were as follows: -1) Large measuring cylinder had an error of +/- 2 cm32) Small measuring cylinder had an error of +/- 1 cm33) Burette had an error of +/- 0.
05 cm34) Measuring balance has an error of +/- 0.005g5) Pipette has an error of +/- 0.005 cm3I will no work out the percentage error of each systematic error which took place.
1) 2 ï¿½ 155 ï¿½ 100 = 1.29% error on the large measuring cylinder2) 1 ï¿½ 100 ï¿½ 100 = 1% error on the small measuring cylinder3) 0.05 ï¿½ 14.3 ï¿½ 100 = 0.34% error on the burette4) 0.005 ï¿½ 0.
09 ï¿½ 100 = 5.5% error on the measuring pipette5) 0.005 ï¿½ 10 ï¿½ 100 = 0.
5% error on the measuring pipette1.29 + 1 + 0.34 + 5.5 + 0.5 = 8.63%The calculated overall error is 8.63% and the actual main error is the measuring balance which has an error of 5.
5%. So this is a factor which has affected the aim of the investigation. The actual random atomic mass of lithium is 7 so method one was closest as the random atomic mass that I concluded was 7.03 so it is quite accurate. I think a reason for this would be because as I was carrying out the investigation in method one, the piece of lithium was reacting in a closed space until all the hydrogen was given off so I could accurately measure how much hydrogen was given off. But in method two, I was titrating so I had to look at the liquid and construe how much acid was needed.
This is not very accurate because you have to estimate and different people have different eye sights which affects it. Another factor is the temperature.If we kept the temperature of the room constant at about 20-25 o C, the results would be more accurate.
To prevent the errors we could minimise reactions between lithium and air, by putting lithium in an organic solvent such as cyclohexane which won’t react with the lithium. Another factor is that in the beginning part of the investigation we could put the bung on quicker to prevent any hydrogen gas escaping. Better measuring cylinders, measuring balances and more accurate pipettes would help too. Also in method two instead of deducing the end point of the reaction by human eyesight, we could accurately measure the end point by some sort of colour receptor. Anther detrimental factor is the batch of lithium used. If the lithium has aged then some or most of its reactivity may be lost.