1 : 2
That is twice of mole of Lithium is in the reaction.
0.0079167 x 2 = 0.015833 moles of Li.
n = m / M
Number of moles = Mass / Molar mass
Molar mass = Mass / Number moles
Where mass of Li = 0.11g
Number of moles Li = 0.015833
- / 0.015833 = 6.947
The relative atomic mass of Lithium is 6.95 when measuring the volume of Hydrogen gas produced.
When I reacted Li metal with water, Lithium Hydroxide was the remaining substance in the conical flask since the H2 gas has been given off. I used this Lithium Hydroxide to do the next way for determining the relative atomic mass of Lithium by titrating the Lithium Hydroxide with aqueous Hydrochloric acid of 0.100 moldm3. I set up the apparatus I’ll be using, as I have illustrated in the diagram below.
After set up the apparatus, I took a clean conical flask of 250cm3; I then took the aqueous LiOH that I made and I shake it for about ten times for complete mix, I then took a 25cm3 pipette and rinse it with some LiOH and then pipette 25cm3 of LiOH from the same bottle into the clean conical flask. Then I added about 5 drops of phenolphthalein indicator. Later on I took some HCl acid from the bottle into a beaker, then took a burette and rinse it with some of the HCl acid in the beaker and then fill it up with the same HCl acid. After doing this I then started doing the titration, but before I stated I took the readings of the burette and recorded it on a table. When I added the phenolphthalein indicator, it changes the LiOH solution into pink. Therefore after the titration the colour will change from pink to colourless.
When the colour changes from pink to colourless I took the burette’s reading. I repeated the titration of only three times because the LiOH solution I made was only 100cm3,
which is only for three titration. In order to be more accurate I took the reading of the burette to the nearest 0.05cm3.
The balanced chemical equation shows the reaction between aqueous Lithium Hydroxide and aqueous Hydrochloric acid.
LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l) .
The table below shows the result of my titration.
Pipette Solution Lithium Hydroxide ? moldm3 100cm3
Burette Solution Hydrochloric acid 0.100 moldm3
Indicator Phenolphthalein
1 2 3
Burette Final 36.59 36.55 36.51
Readings/cm3 Initial 0.00 0.00 0.00
Volume used (titre)/cm3 36.59 36.55 36.51
Mean titre/cm3 36.55
The mean titre was obtained by added the volume used to titre and divide it by the number of time the titration was done. That is
Mean titre = sum of volume used (titre)
3
= 36.59 + 36.55 + 36.51
3
= 109.65
3
= 36.55cm3
Concentration = Number of moles
Volume
Concentration of HCl = 0.100moldm3
Volume = Average titre
Number of moles = ?
C = n/v
n = c x v
n = 0.100 x 36.55 x 10-3
= 3.655 x 10-3 moles of HCl
Looking at the stoichiometry ratio in the reaction, the ratio of HCl : LiOH is 1:1, therefore the number of moles of LiOH will be the same as the number moles of HCl in the reaction. That is 3.655 x 10-3 moles of LiOH.
Knowing the number of moles of Lithium Hydroxide, I can work out the concentration of the Lithium Hydroxide in the reaction, which will be the same as the concentration when it was produced, by the reaction of water and Lithium metal. That is
C = n/v
Volume of the Lithium Hydroxide is 25 x 10-3dm3, which is the one I used in the titration
C = 3.655 x 10-3
25 x 10-3
= 3.655 = 1.462 x 10-1 moldm3 of LiOH.
25
Going back to the first way the volume of LiOH that was produced was 100cm3 and the concentration will the same as the one during the titration, which is 1.462 x 10-1 moldm3. With this I can work out the number of moles of LiOH in the first reaction I did and then use the stiochiometry ratio of LiOH to Lithium and then use the original mass of Lithium again. The calculation below shows the steps.
Concentration of LiOH = 1.462 x 10-1dm3
Volume of LiOH = 100cm3
C = n / v
n = C x v
= 1.462 x10-1 x 100 x 10-3
= 0.01462 moles of LiOH
From the balanced equation in the first reaction.
2 Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
The stiochiometry ratio of LiOH to Li is 2:2, that is 1:1
Therefore number of moles of Li metal is 0.01462moles.
The original mass of Lithium = 0.11g
Number of moles of = Mass / Molar mass
Molar mass = Relative atomic mass
Molar mass = Mass / Number of moles
= 0.11 / 0.01462
= 7.5239
= 7.52
The Relative atomic mass of Lithium is 7.52 when titrating the LiOH with HCl acid.
EVALUATION
I think that my practical went well and was suitable for the task given. This can be prove by the fact that I achieved consistency and proves that the experiment can be accurate and trustworthy.
In the titration, the three results are only 0.04cm3 off. Chose of instruments that are relatively accurate minimizes error through instrument limitations. Pipette and Burette have an accuracy of approximately ±0.05cm3, this far more accurate when compared to a Burette which has an accuracy of ±5.00cm3. Therefore errors through equipment are minor, although they have to be taken into considerations.
The main item that can alter accuracy in this experiment is the Li because of the oil it’s stored in. Oil is a viscous liquid that are generally immiscible with water. The Li can jeopardise results if care is not taken due to presence of the oil, Although the oil will not have any effect because it will not react therefore the not will affect the volume of H2 gas that is been produced but the oil will come as a issue when calculating the result because if the oil is not properly dried from Li and the Li is been weighed therefore the oil is been weighed as well, that is mass of the substance will be for both the Li and oil. During the experiment, I might have not notice that there are still more oil in the Li which will affect my result. I observed during the titration that a splash occurred in the conical flask, which some of it gets on the walls of the conical flask. The might have an effect on my results. And also the pipette I was using was holding back same of the solution each time but I tried to push the drop out by wrapping my hands around the pipette and blocking the top of the pipette so that the drop can be pushed out. When I was measuring the volume of H2 produced the step was very easy because I make sure that when I dropped the Li metal into the distilled water I inserted the stopper immediately, therefore I did not lose many H2 gas, although the gas might have been lost during my process of dropping the Li into the water because Li is a very reactive metal.
Most errors are likely to come from measurements. Such errors are due to the change in position of instruments, that is parallax errors, which would include the meniscus. A parallax error would be absolutely present in the burette and pipette; such errors must be taken into considerations. Another form of error is related to humans. It not quite easy of humans to judge the colour change during titration. During titration, it’s totally impossible for one to judge a solution during the end point of an indicator be exactly the same colour of all three titration.
I can improve this experiment by using a computer. The computer can hold the initial colour of the solution, and then another computer will also hold the initial colour of another solution. While looking at the computer’s screen, the both titrations would be stopped at the same colour. This procedure can reduce the human error when it comes to the indicator’s end point; this therefore makes the result more accurate than before. The experiment can also be improved by repeating the titration many more times or by using more Li metal to measure the volume Hydrogen gas produced. Repetitions of titration can obtain me a better average result. If I was to do this experiment again, the main thing I will improve on is the Li. I’ll focus on the Li because of the oil it’s stored in. When I want to weigh out the Li I’ll make sure that I used many absorbing papers to ensure that the oil is totally dried from the Li. This can achieve me with good result both of the volume of H2 and of the titration.
I feel that my results are reliable as they are closer to each other and I have managed to account of the large amount of errors I’ve made and steps to minimize then in future.
THE TRAGEDY OF MACBETH
In Act 1, Scene 2 Macbeth is described as ‘brave’, ‘valiant’ and ‘worthy’. At the end of the play, Malcolm refers to him as a ‘butcher’. Write an essay showing how and why Macbeth’s character and behaviour change during the play.
In the beginning of the play, Macbeth was presented as a hero due to his victory over the Norwegians rebels. Macbeth and his friend Banquo was the King’s army. Macbeth fought bravely and the King Duncan rewarded him. On his way home Macbeth encountered some witches, who gave him prophecies that affected him throughout the play. He was told that he was going to be the Thane of cawdor and he was going to be King. Macbeth never took the witches predictions serious since when the Thane of cawdor lives and there is no way he was going to be King since the King Duncan is still alive. Later on, news was brought to Macbeth that makes in more confessed. The news was from the King, telling Macbeth that he is the new Thane of cawdor, immediately when the servants told Macbeth t
