To find the relative atomic mass of lithium:
Relative atomic mass = 0.05g
7.525 x 10-3
= 6.6445
Conclusion: The relative atomic mass of is 6.64
Method 2
I titrated 25.0cm3 of the solution from method 1 with 0.1M hydrochloric acid. The results of this titration are:
18.5cm3
18.7cm3
20.1cm3
18.8cm3
I treated the 20.1cm3 result as an anomaly, as it was very different from the other results. So I will not be using it in my average titre.
Average titre = 18.67cm3 ( to 2 decimal places is sufficient accuracy)
The chemical equation for this reaction is:
LiOH(aq) + HCl(aq) à LiCl(aq) + H2O(l)
The mole ratio of lithium hydroxide to hydrochloric acid is:
1:1
To find the relative atomic mass of lithium, I must first calculate the number of moles of hydrochloric acid used in the titration. This can be done by using the following equation:
Concentration (M) = number of
volume (dm3)
The volume of hydrochloric acid (dm3) = 18.67cm3
1000
= 0.01867dm3
The concentration of hydrochloric acid is 0.1M
The above equation can be rearranged to make the number of moles the subject:
Number of moles = concentration (M) x volume (dm3)
This gives:
Number of moles of HCl = 0.1M x 0.01867dm3
= 1.867 x 10-3 moles
I can now deduce the number of moles of lithium hydroxide using the mole ratio. The mole ratio of lithium hydroxide to hydrochloric acid is 1:1. This means that the number of moles of lithium is the same as the number of moles of hydrochloric acid.
So, the number of moles of lithium hydroxide is 1.867 x 10-3.
I now need to calculate the number of moles of lithium hydroxide present in 100cm3 of the solution from Method 1. I used 25cm3 of the solution from Method 1. Since 25cm3 fits into 100cm3 four times, to find the number of moles of lithium hydroxide present in 100cm3, I will need to multiply the number of moles of lithium hydroxide in 25cm3 by four:
Number of moles of lithium hydroxide in 100cm3 = 1.867 x 10-3 moles x 4
= 7.468 x 10-3 moles
Using this result I can now calculate the relative atomic mass of lithium. For this I must go back to the chemical equation from Method 1:
2Li(s) + H2O(l) à 2LiOH(aq) + H2(g)
The mole ratio of lithium to lithium hydroxide is:
2:2
This means that the number of moles of lithium is equal to the number of moles of lithium hydroxide. So the number of moles of lithium is also 7.468 x 10-3.
Using the same equation as before:
Relative atomic mass = mass (g)
Number of moles
The mass of lithium used in 100cm3 is 0.05g.
So, the relative atomic mass of lithium = 0.05g
7.468 x 10-3
= 6.6952
Conclusion: The relative atomic mass of lithium is 6.70
EVALUATION
I feel that overall the results of my experiment were fairly accurate. I can test the accuracy by calculating the percentage of accuracy for each experiment. This is done by dividing the calculated result of the relative atomic mass by the actual atomic mass (6.9) of lithium, then multiplying this by 100:
Method 1: 6.64 x 100 = 96.2 % accurate
6.9
Method 2: 6.70 x 100 = 97.1 % accurate
6.9
This shows that my results have a high percentage of accuracy.
There was however an anomalous result in the titration in Method 2. It was very different from the other results so, for accuracy I ignored the anomaly when calculating the average titre. This anomaly may have been caused by a number of factors. The lithium is kept in oil while in storage to prevent it from oxidising. When using the lithium the oil must be cleaned off to ensure that it reacts to its full potential. If the oil is not cleaned of the lithium will not produce as much hydrogen, also if there is oil left on the lithium when weighing it may affect the weight, making the experiment less accurate. It would improve the accuracy and reliability of my experiment to completely clean of the lithium before using it, although care must be taken to make sure the surface of the lithium does not oxidise.
Another factor that may have caused the anomaly is error in measurement. This could be a misjudgement through human error for example; an error in misjudging the water mark in a burette or measuring cylinder could cause slight inaccuracy. The measuring devices that I used are all fairly accurate, the scales have an accuracy of ±0.005g and the measuring cylinder, burette and pipette have accuracies of ±0.05cm3, used accurately this equipment have sufficient accuracies for the purposes of my experiment. To ensure accuracy and reliability, measuring equipment should have the highest accuracy that is available and immense care must be taken when measuring. Another example of human error that may affect results, is the decision of exactly when the end point of titration is and the reaction time between realising the end point and stopping the acid. This error can be minimised by taking care and slowing the acid towards the end point.
During Method 1, while calculating the number of moles of hydrogen it was assumed that the experiment was being carried out under standard room temperature and pressure, although this was not checked. To improve the accuracy of my results the room temperature should be checked. Because, if the temperature is not standard, one mole of the hydrogen would not take up 24dm3, which would make the calculations inaccurate.
According to the percentage accuracies, Method 1 is more accurate than Method 2. There are a number of factors that could be responsible for the lower percentage accuracy of Method 2; error in measurement of the lithium or distilled water, misjudgement of water mark in the measuring tube, oxidation of the lithium. Also, I repeated Method 2 three times, allowing me to disreguard anomalies and take an average. But, because I had to use the solution from Method 1 in Method 2, I did not repeat it. Any error in measuring the lithium or distilled water or fault in the lithium during Method 1 is likely to affect the results of Method 2. This means that ideally the solution from Method 1 would have an accuracy as near to 100% as possible. This could be done by repeating Method 1, at least three times, each time keeping the solution. Calculate the atomic mass from each volume of gas produced, the solution with the highest accuracy can then be used for Method 2. The calculation of the atomic mass of lithium from Method 2 would then be expected to produce the highest accuracy and most reliable result, according to the calculated percentage accuracies.
