There were also many limitations the procedure. One very important limitation was the scale of the measuring apparatus. The scale of the apparatus was not very accurate. Each point of the apparatus went up by a number; I believe that the number should have been less as it went up thus making readings more accurate.
5 3
4 2.5
3 2
2 1.5
1 1
Another error which comes to mind is the bung and the delivery tube. When I say this I mean am I actually sure that the bung was totally air tight and that the delivery tube was properly secured by the bung? If the answer is no then this would mean that some of the hydrogen in the tube would have escaped. Obviously this would alter the result which means it is an error, it is an unfair test and that my results are wrong.
Each and every source of error affects the results by different amounts. For example one of the most damaging errors in my opinion has to the bung along with the delivery tube. If the gaps are not air tight, then the hydrogen will escape by a small amount. Even though it would have only been a small amount of hydrogen escaping, it would affect the results the most. This is because the actual amount of gas I am trying to measure is escaping. This escaped gas could have gone on to increase the amount of hydrogen I record. If some is escaping then the result won’t be true.
Flask
Bung (Birds eye view)
Delivery tube
Hydrogen escaping from the edges
Another factor of error will be from the inverted measuring cylinder (the one used to measure the hydrogen). When we start the implementing we are supposed to fill the cylinder with water and somehow get it in the water tub WITHOUT any air bubbles getting in. the method we used was actually very good. After filling the cylinder with water we wrapped cling film all around the end of the cylinder. Now even if you shook the cylinder slightly it would not loose any water. However I am still not sure that I had a FULL cylinder of water. My theory is that when you placed it in the water indeed it was full, but when you placed the other end of the delivery tube under the cylinder, air from the tube would escape and a few small bubbles may well travel into the cylinder. This would mean that the cylinder is not full of water and in fact have a little gas in it. Thus this would add to the end result making an error. This factor will affect the result by only a minute amount. This is because if the bubbles did enter the cylinder and they were big enough to notice, then I would have started again. However if the bubbles are small it is unlikely that would have even noticed them. So with the amount of air entering only being minuet, the results will not be affected by a huge amount.
Ways to improve to method is to simply use better equipment. The equipment used in school is not extremely good. If the apparatus was of a higher standard then the level of error would decrease. Another improvement could done is too stop the bung being off the flask for that period of time when we had to add the lithium.
Since there was insufficient time to do the experiment more than once, there cannot really be a large number of errors. This is due to that fact that we cannot make an average of any results.
Analysis (method 2)☺
- (1) Calculate the number of moles of HCL used in the titration.
Conclusion so far:
Unlike the first experiment I did get to perform the implementation more than once (method 2). I performed the experiment three times which means I have three sets of results. Usually this means that you get maybe one or two anomalous results. However by looking at my table (later) you can see that there are not any anomalous results because the results are very close together (42.6, 42.7, 42.7 cm3). Even though I had the opportunity to perform the experiment more than once, I feel a fourth time would have made my result much more accurate, however I am satisfied that there was enough time to perform the implementation more than once. So now I can make an average and I can conclude that the result is accurate.
The balanced equation I will be using for this experiment is:
LiOH(aq) + HCL(aq) LiO(aq) + H2O(l)
So to calculate the number of moles of HCL used in titration I will use this equation gained from a source:
n = Volume X Molarity
1000
n = this is the number of moles of HCL (hydrochloric acid), what I am trying to find.
Volume = this volume I am going to put here is the average (or mean titre cm3) of my three sets of results.
Molarity = this was the molarity of the HCL which was given to us.
Divide by 1000 = this is how the equation is done.
n = 42.67 X 0.1 = 0.00427 M
1000
The average (mean titre) has to be calculated because I have done the experiment three times. First of all it will give me a much more accurate reading plus I cannot use all three results at the same time. To figure out the mean average I need to use method:
42.6 + 42.7 + 42.7 = 42.67
3
The three figures are from my results, the volume of HCL used, and dividing it by three is done because I have three results.
- (2) Deduce the number of moles of LiOH (lithium hydroxide) used in titration.
This is done relatively simply. By looking at the equation you can see that there is no large number in front of any of the reactants or products. As in method one I had to times the number of moles of hydrogen by 2 to get the number of moles of hydrogen. However in this case the ratio is 1 : 1. This then means that the number of moles of LiOH will be that same as the HCL.
Number of moles of HCL = 0.00427
Number of moles of LiOH = 0.00427
- (3) Calculate the number of moles of LiOH present in 100 cm3 of the solution from method 1.
Again this is done in a simple manor. If you read the first step of the procedure it says “Pipette 25.0 cm3 of the solution in the conical flask from method 1 into a clean 25.0 cm3 conical flask and add 5 drops of phenolphthalein indicator.” However at this stage I need to find the number of moles present in 100cm3 whereas at stage (2) it was only for 25cm3. This means all I have to do I times the answer from part (2) and that will gives me the answer this calculation.
At 25cm3 the number of moles of LiOH is 0.00427 M.
At 100cm3 the number of moles of LiOH will be (0.00427 X 4) 0.01708 M.
The reason I times it by 4 is because to make 25 into a 100, you need to times by 4.
- (4) Use this result and the original mass of lithium to calculate the relative atomic mass of lithium.
This again will be slightly different to the methods used in the earlier stages. In this part of the calculation I will need some of the information and results from parts (1), (2) and (3). Just like in method one I had decide upon a formula to help me find the relative atomic mass of the lithium. The formula I have chosen is:
n = m
Mr
n = the number of moles.
M = the mass of the material or substance.
Mr = the relative molar mass.
This whole process is very similar to method 1. I studied the formula and again found that it cannot be used because I am trying to find the Mr and not the mass. So I need to re – arrange the formula so that it suites what I need to do. I need to use the triangle to find out how to re – arrange it.
m
n Mr
Now using this triangle I can get the equation that I need to calculate my result (the relative atomic mass). If I want to figure out the Mr, then what I do is look at the other letters. I can see that the mass (m) is over the number of moles (n). This means I have to calculate the answer by using the mass over the number of moles. So when I re – arrange the formula it will look like this:
Mr = m
n
With the figures in it will be:
Mr = 0.13 = 7.61124 M
0.01708
The figures that you can see all came from what they should be. First I am trying to figure out the Mr so that is not there yet. The mass came from my original mass of the lithium from method 1 (as seen on above table). Finally I have got the number of moles from my answer to part (3) because I need to use this result to calculate the original mass of lithium. This is where I have received my figure for the number of moles. Plus this is why I had to do part (3) to get this figure.
Evaluation (method 2)☺
In this method I believe that there could have been many errors performed. One of these errors includes the measuring apparatus. I noticed the pipette that was given to us was very dirty. Automatically I assumed that this would cause an error in my results. I assumed this because the dirt would influence my solution. It may slightly change the concentration or it may well change the amount of solution I originally thought was in there. This would happen because the dirt would take up some space in the pipette thus changing the amount you think is in there. A way to improve this problem is to basically use your common sense. After use you should always clean out all the apparatus and ensure that it is properly cleaned. Plus make sure that it is not left lying around to collect dust and become dirty.
In this experiment, I believe that there could have been another error. The main sources of error, in my opinion, must be human and equipment error. For example a source of error is in measuring the amounts of the substances or materials. Did I truly measure out 25.0 cm3 of the method 1 solution (LiOH) in the conical flask? When I was measuring the amount of the method 1 solution in the measuring cylinder did I truly use the correct method? I may have been looking at the cylinder incorrectly. I should have been looking at it right inline with it, even though this is what I did, it is still very hard to record an accurate reading. This is known as the meniscus level.
A way to improve this problem is to simply ensure that you are very careful in measuring the results or if unsure of what to do then simply ask your teacher or your supervisor. Also make sure the equipment you use is not faulty.
Not only should one be worried about what mistakes not to make or how to improve or fix the problem, but also the total accuracy of the experiment as a whole. There are a number of useful ways in which to make the experiment more accurate. First of all you can make sure that the measuring apparatus you use is at a high accuracy. For example the measurements should be much more closer instead of it going up by large numbers.
5 3
4 2.5
3 2
2 1.5
1 1
Another way to make the experiment produce more of an accurate result is to make sure that you have got all the correct amounts of all your solutions. Like you hydrochloric acid or you lithium hydroxide. Try to ensure that they are both exactly what the procedure asks for. If you make the mistake of titrating 27.0 cm3 of lithium hydroxide with the hydrochloric acid, then you will record the incorrect result.
Safety is another issue which must be looked at. The hydrochloric acid can be very dangerous if it goes in the eyes so goggles should always be worn. Or if you have long hair make sure it is tied up because it might go in the chemicals plus bags should be under the table and if something goes wrong you should consult miss immediately.
From looking at my results table you can see that my rough experiment had an anomalous result. This could have been for a number of a number of reasons. Firstly it could have been because I had rushed it slightly due to insufficient time. Or maybe it could have been that I had the wrong amount of lithium hydroxide for my solution because I read it wrong. Plus it could have been because that was the first try at the titration and by the second experiment I had the hang of it. Or it could have been due to simple equipment error such as the equipment not being washed thoroughly and previous substances from past experiments still remaining in the equipment. For one or more of these reasons I had s slight error in my results.
Since I had the opportunity to perform the experiment more than once, it means that I can calculate out an average. Due to this fact, it means that the experiment is MUCH more of a fair one because I don’t have to just rely on one set of results. I can combine the three results to give me an average which means a fairer test than usual.
Personally I find method 2 to be a much more accurate method of experimentation. The reason for this is that in method 1, there are many errors the main one being that too much gas escapes when the lid is removed, plus the fact that slight amounts escape through the bung. Another reason why this is not a good method is that there is simply too much room for error. For example the inverted cylinder is not very reliable because no matter how hard you may try, you will always have a little extra air inside thus giving an incorrect result. Only in a few cases may you not have any gas at all. The reason why I believe that method 2 is more accurate is because there is less room for error. The titration process (if done correctly) can be done very accurately. Plus it is not too difficult to inaccurately measure the amount of lithium hydroxide or hydrochloric acid. However, even though I have stated this fact I still received a better result from method 1 (6.96678). This could have been for a number of reasons. These may include that fact that I must have performed a major error in the implementation of method 2. Or indeed the opposite may have happened and that my LiOH solution was faulty. The concentration could have been changed thus altering the method 2 result.
Implementation☺
(Method 1)
In this part of the course I will explain the correct procedure on how to perform the experiment.
Starting off, set up your apparatus as shown in the diagram.
250cm3 measuring
cylinder
Lithium
100cm3 distilled water
At this stage be sure that you have filled the conical flask with exactly 100cm3 of distilled water. Plus you must also make sure that the distilled water is pure and is not mixed with other substances. Now at this stage you can get your lithium ready. Either measure out around 0.10g of lithium or just use what is given to you. In my case I used 0.13g. Draw out a table to record your original mass of lithium and the initial and final volume of hydrogen used. THIS IS THE CRUCIAL POINT. Remove the bung and add your lithium. As soon as you can, replace the bung tightly to ensure no loss of gas. Finally you simply record the final volume of hydrogen. Safety is very important. Make sure that you are wearing goggles, your hair is tied back and that your bags are tucked under the table so that no one falls over.
(Method 2)
First get your pipette and measure out exactly 25.0cm3 of your lithium hydroxide (method 1 solution). Pour this into a clean 250cm3 conical flask and add around five drops of phenolphthalein indicator. Place this under the titration apparatus and titrate with a HCL solution with the concentration of 0.100dm3(aq). Record the results in an appropriate format (above) showing the final burette reading, initial burette reading, the volume used and the mean titre. You need to repeat the process about three times to gain more accurate results. On your table show all your results. Finally record the average (mean) titre. Safety is extremely important as you are handling hydrochloric acid. Goggles must be worn at all times, hair must be tied back and bags must remain under the desk. If anything goes wrong, consult the teacher immediately.
