Determination of the Relative Atomic mass of Lithium
Determination of the Relative Atomic mass of Lithium
Implementing
Method 1: Gas Syringe
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
From a Mass of 0.0463g of Lithium, I measured that 55.0cm3 of Hydrogen gas was given off when the Lithium reacted quite vigorously with the Distilled water.
Method 2: Titration
LiOH(aq) + HCl(aq) LiCl(aq) + H20(l)
When titrating the Lithium Hydroxide solution against Hydrochloric acid I have obtained the following results:
Initial Volume (cm3)
Final Volume (cm3)
Difference in Volume (cm3)
Rough
0.05
20.04
09.99
Titration 1
0.00
20.95
0.95
Titration 2
09.95
20.90
0.95
Titration 3
0.00
20.90
0.90
From these results I have obtained 3 results within 0.5 cm3 of each other, but seeing as I have 10.95 two times, I can leave my result of 10.90 out of my calculation to obtain the average difference, and so 10.95cm3 will be used when I calculate the Relative atomic mass of Lithium.
Throughout both experiments I wore safety goggles and a lab coat to ensure my safety around the Hydrochloric acid and the Lithium Hydroxide as both can be dangerous to your skin and eyes if it comes into contact with them. If I did come into contact with these substances I would have washed it out using distilled water or using the emergency eye wash station. Hydrogen gas is also given off so I ensured there were no naked flames in the lab as hydrogen is flammable.
Analysis
Method 1:
I will now use my results from the previous section to calculate the Relative Atomic Mass of the Lithium sample.
2Li
+
2H2O
2LiOH
+
H2
Ratio
2
2
2
Volume (cm3)
55
Mass (grams)
0.04630
Moles (moles)
0.004580
0.002290
Mr
0.11
Since 55cm3 of Hydrogen was given off in the experiment, I can use the equation
Moles = Volume / 24dm3 as 1 Mole of any gas occupies 24dm3 of space. So for this experiment, the equation is 55cm3 / 24dm3 = 0.00229 moles of Hydrogen.
Therefore since the Lithium and Hydrogen are in a 2 : 1 ratio, we simply multiply the moles of Hydrogen by 2. So 2 x 0.00229 mol = 0.00458 moles of Lithium.
From this we can then work out the Relative Atomic Mass by using the equation :
Moles = Mass / Mr. So therefore the equation for Mr would be Mass / Moles, so in this case the equation would be 0.0463g / 0.00458mol = 10.109 or 10.11.
So therefore, for the first experiment method I have found the RAM of Lithium to be 10.11
Method 2:
I have taken the Volume of HCl to be 10.95cm3 as I found from the experiment that I had 2 titrations where 10.95cm3 was the answer, ...
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From this we can then work out the Relative Atomic Mass by using the equation :
Moles = Mass / Mr. So therefore the equation for Mr would be Mass / Moles, so in this case the equation would be 0.0463g / 0.00458mol = 10.109 or 10.11.
So therefore, for the first experiment method I have found the RAM of Lithium to be 10.11
Method 2:
I have taken the Volume of HCl to be 10.95cm3 as I found from the experiment that I had 2 titrations where 10.95cm3 was the answer, so therefore the average is 10.95 cm3
LiOH
+
HCl
LiCl
+
H2O
Ratio
Volume (cm3)
0
0.95
Concentration (mol dm-3)
0.1
Moles
0.001095
0.001095
The Moles of HCl is decided by the equation: Moles = concentration / volume (in dm3) so therefore in this case the equation would be: 0.1 x (10.95 / 1000) = 0.001095 moles and since the HCl and LiOH are in a ratio of 1 : 1 the moles of LiOH = 0.001095 moles as well.
This also means that in the 50cm3 solution of LiOH, the number of moles =
0.001095 x 5 = 0.005475 moles
In the 1st Method, we see that 2Li and 2LiOH are in a ratio of 1 : 1 and so we can use this to gain our second RAM for Lithium. To do this we simply use the equation Mr = Mass / Moles and so the equation would be: 0.0463 / 0.005475 = 8.46 or 8.50 and so from our second method I have found that the RAM of lithium is 8.46 ~ 8.50.
I have also gained the results of all my classmates and so I can use their data compared to mine, and I can also use this data to show how far away I / the class were from the actual RAM of Lithium.
Method 1 (Gas Syringe)
Method 2 (Titration)
Chris
0.79
05.02
Mike
1.06
09.59
Kavit
09.40
09.02
Dipen
3.60
0.50
Arjan
2.56
07.64
Raif
1.05
08.64
My Result
0.11
08.50
Matt
4.03
09.44
From these results I will calculate the class average excluding my own results:
Method 1: 82.49 / 7 = 11.78 or 11.80
Method 2: 59.85 / 7 = 8.55 or 8.60
I can now compare my results to the class's results, by using the equation:
% difference = Difference / My result x 100.
Method 1: 1.69 / 10.11 x 100 = 16.72 % Smaller
Method 2: 0.10 / 8.50 x 100 = 1.18% Smaller
From these results we can see that there was a greater % difference between my results and the classes in the 1st Method where the gas syringe was used with 16.72%. Whereas, in the 2nd method where the titration was used there was only a difference of 1.18% between my and the entire classes results. We also see a difference of 15.54% between method 1 and method 2, which shows method 1, is less accurate than method 2.
I can now also compare my results to the actual RAM of Lithium:
The actual RAM of Lithium is 6.9 and so I can find the % difference between my results and the actual results by taking the difference and then dividing by the original value and then multiply that by 100.
Method 1: 10.11 - 6.9 = 3.21
3.21 / 6.9 x 100 = 46.5% Larger
Method 2: 8.50 - 6.9 = 1.6
1.6 / 6.9 x 100 = 23.2% Larger
From this result we can see that Method 1 again has the greater margin of error being almost 50% away from the actual value. Method 2 is the more accurate method as in both cases it shows the least % difference in this case being %23.2 which is quite large, but the % difference in Method 1 is 23.3% larger than method 2 , which shows method 1 is not as accurate as method 2.
Evaluation
We can see from my results compared to the entire classes, that there does appear to be a degree of systematic error, this is because my % difference between my classmates was smaller than my results compared to the actual RAM of lithium.
Method 1 Sources of Error:
The first thing we notice when going to measure out our desired mass of Lithium, is that many of the pieces had formed an oxide layer on the outside as they must have been previously exposed to the air. This provides a problem because this oxide layer can actually add to the overall weight of the lithium, and when measuring out our sample, 4-5 pieces of Lithium with an oxide layer around them would add quite a large amount of weight that would definitely affect my results, especially as 0.5g was the recommended weight, and the smaller the weight we measure out, the larger the percentage error present, as we will have a margin of error of ± 0.05g.
Secondly when looking at the gas syringe apparatus, we can see that the maximum volume is 100cm3, and this meant that we could only measure accurately to ±1cm3 which could definitely affect our results.
When weighing out the pieces of Lithium metal, we had to dip the lithium in ether to remove the layer of oil to supposedly prevent air from getting to the lithium, we then had to clean off the ether on a piece of filter paper. However it is not possible to remove every trace of oil from the lithium, and to a similar extent not all the ether can be removed without waiting for an extended period of time, and doing this would increase the size of the oxide layer on the lithium further increasing the weight of the lithium. So the oil and or ether on the lithium could increase the weight of the lithium which again affects the results of the experiment.
There was also the possibility of gas loss when doing the 1st experiment, this is because there was a certain time delay when placing the bung on top of the conical flask, and during this time some hydrogen gas may have escaped, the extent of the gas loss depends on how close the bung was held to the flask, and also if there was any human error and for example the bung was dropped. This loss of gas could have caused further experimental errors, however I do not think the gas loss could have caused as great an inaccuracy compared to the inaccuracies caused when weighing out the lithium.
Method 2 Sources of Error:
When using the burette we were subject to a measuring accuracy of ±0.1cm3 and while much better than the gas syringe, 0.1cm3 can still affect the results of the experiment.
When using the Phenolphthalein indicator, it was difficult to judge the exact end point of the experiment, and so sometimes more acid that usual may have been released from the burette which would have caused my results to have varied more than usual, and therefore cause an overall inaccuracy in my experiment and therefore cause inaccuracies in my RAM calculations.
When doing a titration, it is best to do a number of titrations to ensure the accuracy of your results, however in this experiment we were limited by the volume of Lithium Hydroxide that was produced from the 1st experiment, this volume was 50cm3, but 10cm3 must be used as the rough titration and chances are that there will not be exactly 40cm3 of LiOH left to do another 4 titrations and so we will only be able to do 1 rough titration and 3 accurate titrations.
Possible Improvements:
I think that it would be best to use a slightly greater mass of Lithium and Water. The greater mass of Lithium means that there will be a smaller margin of error when measuring it out and the larger mass of water will mean that we will have a larger volume of LiOH available to do the next experiment with, which means we can repeat the titrations around 5 times to ensure that our results are concurrent.
It would also be more useful to use a larger gas syringe in this experiment perhaps a 200cm3 gas syringe, this not only allows us to use more lithium in the sample, but it also means that instead of a ±1cm3 margin of error, we now have a ±0.5cm3 margin of error, and this would increase the accuracy of my measurements and therefore the entire investigation.
I think that it would be best to keep the Lithium metal in an inert atmosphere to help prevent the formation of the oxide layer; this could again greatly increase the accuracy of my experiment as the measuring of the lithium provided the greatest errors in this experiment, and if these measurement errors are minimised, then the overall accuracy of the experiment can be maximised. Gas loss in the first experiment is a problem, however it is difficult to minimise gas loss, the best way is to minimise human error, simply is to hold the rubber bung as close to the top of the conical flask and to get another person to drop in the lithium metal into the flask and then place the bung in position as quickly as possible.
