250cm3
Water Measuring
Conical Flask Cylinder
100cm3 Tray Stand
Distilled
Water
Lithium
(0.125g)
Treatment Of Results – Method one
The results collected were as shown below;
Assuming that 1 mole of gas at room temperature and pressure occupies 24,000cm3.
Mass Of Lithium Volume Of H2 Collected
0.125g 195cm3
-
Calculate the number of moles of H2
195cm3
24,000cm3 = 0.008125 moles
Using the balanced equation for the reaction it was possible to deduce the number of moles of Lithium:
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
As can be seen from the calculation above, there was 8.1x10-3 moles of H2 produced. This therefore means that as there is a 2:1 ratio of Li to H2, simply multiplying the number of moles of H2 will show how many moles of Lithium were used
0.008125 x 2 = 0.01625
To deduce the relative atomic mass (Ar) of Li from this, the following equation must be used:
No. Of Moles = Mass this must be rearranged to; R.A.M. = Mass
R.A.M. No. Of Moles
It is then possible to input values; R.A.M. = 0.125 = 7.6923 or 7.69 (2.d.p)
0.01625
Upon completion of the analysis of method one results it was calculated that the R.A.M of Li was 7.69 a.m.u (although we know this to be incorrect)
Method two – Procedure
This procedure was a basic titration of acid against alkali (HCl against the LiOH from method one).
- 25.0cm3 of the LiOH solution was pipetted into a clean 250cm3 conical flask and 5 drops exactly of phenolphthalein indicator added.
- This was titrated with 0.100 mol dm3 HCl(aq).
- The volume of HCl required to neutralise the LiOH was recorded (when the pink colour of the phenolphthalein disappeared and the solution was colourless) and the titration repeated twice more for reliability and accuracy.
Treatment of results - Method two
The first titration was an approximate one where 0.50cm3 of HCl was added at a time to establish approximately what volume of HCl was required for neutralisation. Titrations two and three were carried out with greater precision so that a figure accurate to two decimal places could be obtained.
To determine the R.A.M of Li from these results first the number of moles of HCl used and the balanced equation for the reaction must be examined so that it is possible to deduce the number of moles of LiOH involved in the reaction.
LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l)
No. Of Moles = Vol(dm3) x Conc.
No. Of Moles Of HCl = avg. titre x 0.100 M
1000
= 40.50 x 0.100 M = 0.004055 moles of HCl
1000
As there is a 1:1 ration of HCl to LiOH in this equation then in the 25.0cm3 portion of LiOH used there were 0.004055 moles of LiOH, to obtain the number of moles in the original 100cm3 this value must be multiplied by four.
0.004055 x 4 = 0.01622 moles in 100cm3
From this the R.A.M could be calculated by the equation as in method one,
No. Of Moles = Mass Which must be rearranged to: R.A.M = Mass
R.A.M No. Of Moles
Therefore, R.A.M = 0.125 = 7.7065
0.01622
This value as can be seen is only 0.01 a.m.u away from the value obtained by method one.
Evaluation
Method one –
There were several built in errors within this procedure. The first was the measuring of the mass of the Li, this was carried out on a pan balance which was only able to measure to two decimal places (I was able to obtain a measurement to three decimal places due to the pan flicking constantly from 0.12 to 0.13g) and was also subject to measurement faults due to local pressure changes in the air, however it is possible to counteract this by using an encased top pan balance to prevent air movement affecting the balance. Also, because the Li is stored in oil it is not practically possible to remove all of it which in turn will distort the mass reading. On top of this the Li oxidises in the air very rapidly so this will affect the mass. This is difficult to avoid and only possible in an inert atmosphere which is not possible to create with the apparatus available. In relation to this R.A.M is always determined at room temperature (200C) and pressure (1 atmosphere) where by the procedure implemented this was not possible as temperature fluctuated.
The next was that of placing the Li in the conical flask, the risk exists that some of the H2 gas would escape before the bung could be placed in the neck. This would influence the input and consequently the output of the equation. The most efficient method to exclude this is to use a bung combined with a ‘ladle’. The bung would be put in place and then the Li placed into the water thus allowing no gas to escape.
Thirdly the problem of accurate reading of the measuring cylinder was encountered although it was possible to avoid this by filling the cylinder with air so the meniscus is at the uppermost measurement (rather than to the very top) so that the reading is as accurate as possible. It is further possible to remove this fault by collecting the gas in a gas syringe which can ensure no leaks and has much more refined degrees of measurement so readings are more accurate.
Method two –
Due to method two involving only titrating acid against the LiOH there are very few possibilities for error although human error must be taken into account when reading the level of the burette and visually checking for the point at which the solution turned from pink to colourless. This fault is almost impossible to eradicate.
It was required that only 5 drops of indicator were added, this must be done carefully and accurately as it may affect the result if not.
Determination Of The Relative Atomic Mass Of Lithium – Assessed Practical 1
January 2002
