Analysing
From the results table I can see that I got concordance results that is the results that I got were quite close to each other. With the help of the results that I got in experiment 1 and experiment 2, I will now calculate the relative atomic mass of lithium.
However, before I calculate the relative atomic mass of lithium, I would define the terms that would be used the formulae.
The mole (n) of any substance is the amount which contains the same number of particles of the substance, as there are atoms in a 12g of Carbon-12 atom.
The concentration (c) is a measure of how much solute there is dissolved in a solvent. It is measured in moles per unit volume. The unit of volume used could be dm³, cm³ or ml. It is important to know that 1 ml of a substance equals 1 cm³.
The relative atomic mass (Ar) is the mass of an atom measured on a scale where a single atom of Carbon-12 equals 12g.
2Li (s) + 2H2O (l) → 2LiOH (aq) + H2 (g)
To calculate the number of moles of hydrogen we should know that 1 mole of any gas occupies 24 dm³ or 24000 cm³ at room temperature and atmospheric pressure. Hence,
Number of moles 24 = volume of gas in dm³
n 24 = V
n = V 24
134 cm³ = 0.134 dm³
n = 0.134 dm³ 24
n = 5.583 10³ mols
To deduce the number of moles of lithium we can see from the equation above that the mole ratio of hydrogen to lithium is 1:2. Hence,
Number of moles of Lithium = number of moles of hydrogen 2
n = 5.583 10³ mols 2
n = 0.011 mols
Using the values above and the mass of lithium I used in the experiment I can now calculate the relative atomic mass of lithium.
Relative atomic mass = mass of lithium in grams number of moles
Ar = m n
Ar = 0.09g 0.011 mols
Ar = 8.18 (3 sig. fig)
Therefore, the relative atomic mass of lithium for experiment 1 is 8.18 (3 sig. fig). Now I will calculate the relative atomic mass of lithium again by using the results I got in experiment 2.
LiOH (aq) + HCl (aq) → LiCl (aq) + H2O (l)
To calculate the number of moles of HCl that is hydrochloric acid, I would use the following formulae:
Number of moles = concentration in mol dm³ Volume in dm³
n = c V
Concentration of HCl is 100 cm³ = 0.100 dm³
Volume is 27.0 cm³ = 0.027 dm³
n = 0.100 mol dm³ 0.027 dm³
n = 0.0027 mols
To deduce the number of moles of LiOH, I know that the mole ratio of HCl to LiOH from the equation is 1:1. Hence,
Number of moles of LiOH = Number of moles of HCl
n = 0.0027 mols
The number of moles of LiOH in 25 cm³ of a solution is 0.0027 mols. Hence, the number of moles of LiOH in 100 cm³ of a solution can be calculated as follows:
Number of moles of LiOH in 25 cm³ = 0.0027 mols
Number of moles of LiOH in 1 cm³ = 0.0027 25
n = 0.000108 mols
Number of moles of LiOH in 100 cm³ = 0.000108 100
n = 0.0108 mols
Using the results above from method 2, I can now calculate the relative atomic mass of lithium as follows:
Relative atomic mass = mass in grams number of moles
Ar = m n
Ar = 0.09g 0.0108 mols
Ar = 8.3 (2 sig. fig)
Hence, the relative atomic mass of lithium that I calculated from this experiment is 8.3 (2 sig. fig)
Evaluation
Now that I have done the experiment and calculated the relative atomic mass of Lithium, I would evaluate the overall accuracy of the experiment. By looking at the final results that is the relative atomic mass of lithium, I can see that I could have improved my result in several different ways as the final result that I got was 8.3 which is more than the original relative atomic mass of lithium.
The most accurate piece of equipment that I used was the balance that weighed the piece of lithium in grams. It weighed to the nearest 0.01g and so minimised the chance of inconsistency of the final results. However, there were different sources of error that affected my overall accuracy in the experiment. The most significant error could have been produced in the experiment in two different ways. Firstly, small amount of hydrogen gas could have escaped when I replaced the bung in experiment 1. This would therefore result in the volume of gas less than expected due to gas loss. This could be improved by sealing the bung so that no gas could escape. Secondly, the measuring cylinder that I used in experiment 1 to measure the amount of hydrogen gas collected, had bigger divisions and measured 2 cm³ between divisions. This could result in more chances of an error. I could have improved it by using a measuring cylinder with smaller divisions.
There might have been other errors as well. The piece of lithium used in the experiment might lack purity as it was not completely silver and it might have reacted in the air before it was used to react with distilled water. This would also affect the overall accuracy of the results. When I did the titration experiment, I did not wash out the burette with hydrochloric acid first so there might have been some impurities in the burette from the previous time it was used. This could have been improved by rinsing the burette first with hydrochloric acid. Before I did the final titration experiment, I could have a done a rough titre to ensure that the titration results were closer to each other than they are. It is also possible that some of the drops of hydrochloric acid were left on the filter funnel when I poured it down the burette. This could also affect the overall accuracy of the final result.
To minimise inbuilt uncertainties from the experiment, I could have used large quantities of chemicals as large quantities reduce the risk of percentage error. I could also have used equipment that had smaller divisions e.g. the measuring cylinder to decrease the percentage error. These uncertainties, if improved, could enhance the final result as there would be less risk of an error in the practical done and the final result might then be closer to the original relative atomic mass of lithium and the result produced would hence be more valid.
A substitute method of calculating the relative atomic mass of an element accurately is by using the mass spectrometer. It is an instrument that measures the mass of atoms and molecules. The vaporised sample of an element being tested is introduced into the instrument and is then ionised by heating or is bombarded with electrons. The charged particles are then accelerated towards the electromagnetic field. The individual ions differ slightly in mass and charge. The lighter the ions or the greater the charge the on the ion, the greater will be the deflection. The ions are detected by means of a photographic plate. The ion detector then prints out a chart which is called a mass spectrum. Although, this is an improved method of calculating the relative atomic mass of elements, it is usually used in industrial chemical laboratories and not in educational institutions.
