- No. moles of LiOH = 0.004235 mol
Calculate the number of moles of Li:
Using the equation:
2Li (s) + 2H2O (l) → H2 (g) + 2LiOH (aq)
2 : 1
As reacting ratios of LiOH and Li are the same:
- No. moles of Li = 0.004235 mol
Calculate the number of moles of Li present in 25cm3 solution:
As the mass of Li and No of moles used are know, it is possible to calculate the relative atomic mass of Li.
- Mass of Li = 0.12g
- No Moles of Li = 0.00416 mol
- Relative atomic mass = mass/no moles
- 0.12/0.00416 = 28.85
However since only 0.03g of Lithium was used, the result for 25 cm3 will represent 4x. to much Li.
- 28.85/4 = 7.21
- Relative atomic mass of Li = 7.21
Potential Hazards.
Lithium is highly flammable and therefore should be kept well away from any flame source and will react readily and violently with water. This means it must be kept in oil for as long as possible, until ready to be used, minimising exposure to the air. Gloes must be used when handling.
Hydrochloric acid is corrosive and an irritant, even though only weak concentrations are being used eye protection and lab coats should be used throughout.
Phenolphthalein indicator is an irritant and is thought to be mildly carcinogenic gloves should be used when handling.
When using glassware caution should be used. On filling burettes/pipettes it is imperative that they be filled below eye level.
Investigation Evaluation.
In conclusion the experiments were fairly accurate with the second experiment being the most accurate. This can be shown by calculating the percentage accuracy:
Percentage accuracy = actual relative atomic mass / calculated relative atomic mass x 100
Using this formula and the actual mass of 6.94 it can be shown:
Exp 1 = 94.2% accurate
Exp 1 = 96.6% accurate
Errors in procedure:
The main sources of qualitative error in the first procedure were:
- Inaccuracies in measuring out the Li, due to residual oil present and the speed at which the transference of the Li had to be conducted
- Reaction o Li with air before transference to flask
- Loss of hydrogen from flask before bung properly installed
- Loss of hydrogen from gas cylinder, if incorrect placement of tube beneath cylinder
The main sources of qualitative error in the second procedure were:
- Misjudgement of endpoint, leading to inconsistencies in titre readings
- All cumulative inaccuracies of the previous experiment in preparing the LiOH.
- Incorrectly transferring the correct amount of LiOH.
Errors in measurements:
The main sources of quantitative error were:
- Limitations in the equipment used
- Misinterpretation/inaccurate readings taken
Absolute error is the maximum amount by which a result may hbe out but would still give the same answer (that will round to give same answer), i.e.
A balance reads to 2 decimal places which means reading of 13.22 could be a value greater that or equal to 13.215 or less than 13.225. So the absolute error for the balance is ± 0.005.
Relative errors are more useful as the represent the error in relation ti the original value. However it is even easier to understand the result as a ‘percent relative error’. This is calculated by dividing the absolute error by the original value and then multiplying by 100.
Using this it is possible to evaluate the percent relative error for the equipment used.
We can calculate the cumlative percentage relative error of all the equipment.
Thus the actual result with quantitative errors taken into account can be calculated.
- 4.802% of calculated mass of Li = 7.21/100 x 4.802% = 0.346
- actual result is 7.21±0.346
- this is between 6.86 and 7.56
To try and compare this with the qualitative error of H2 loss the amount of gas that would have to be lost to produce this change can be calculated by working through the method backwards.
- Method 1 calculated relative atomic mass = 7.38
- 7.38/100 x 4.802 = 0.355
- actual result is between 7.57 and 6.85
Now calculate the amount of H2 loss that would account for this difference in masses.
- 0.12/7.57 = 0.01585 moles of Li
-
0. 01585 /2 = 0.00802 moles of H2
Since 24 dm3 is one mole of hydrogen gas at room temp
- 0.00802 x 24 = 0.19258 dm3 = 193cm3
- 195-193 = 2cm3
2cm3 is a very small amount of gas to be lost and as even such a small amount of gas loss constituted for the same level of inaccuracy as all of the equipment one would expect that more tha 2cm3 would be lost and therefore the mains source of inaccuracy is not the inaccuracy of the glassware but that the gas lost in the first procedure is more significant. Especially since it is possible to compensate for the quantitative errors but not for the qualitative.
Suggested Procedure Improvements.
As the loss of hydrogen from the flask is of most significance it is imperative that were the procedure repeated some alternative method was devised so as to minimise the loss of gas. The best way to do this would be to use a specialist ‘airlock’ flask, which has a sealed compartment with two doors which can be opened alternately to insert the Li without losing gas. However in a normal classroom environment it is more practical to attach the Li to a piece of thread and then insert it into the flask, suspended so it is not in the solution. Then insert the bung and cut the thread so that when the Li enters the solution the flask is sealed.
Other methods for improvements in the procedure are listed here:
The main source of quantitative error came from the balance and the measuring cylinder. The glassware used was not of the very highest quality. Better results could have been achieved by using better quality equipment with smaller absolute error. While it would be impractical due to costs to have a more accurate balance there is a simple solution to give better results of gas reading that a gas cylinder. Instead we can use a gas syringe which has a much smaller absolute error.
Due to their only being about 100cm3 of LiOH produced after the first procedure. It is only really possible to conduct 3 titrations (maximum of 4). This is not sufficient to ensure a good average is achieved, Any anomalies would seriously compromise the integrity of the results. If the experiment were repeated I would use more initial water and Li to produce enough LiOH for at least 5 titrations (ideally to produce 3 within .1cm3 of each other). Also if repeated I would attempt to run through he experiment 3 times, then It would be possible to calculate and average result for the relative atomic mass. This should provide a more accurate representation of the actual value.
