Apparatus
1 x Measuring cylinder (250cm³)
1 x Conical flask (250cm³)
1 x Rubber bung
1 x Tube
1 x Container
1 x Electronic scale
1 x Distilled water
1 x Lithium
1 x Goggles
1 x Gloves
1 x Lab coat
The apparatus should be set up like this –
Method
- Weigh a piece of lithium on the scale and record the weight.
- Clean the piece of lithium properly removing any oil from the
surface.
- Use the measuring cylinder to measure 100 cm³ of distilled
water and pour it in the flask.
- Fill the container with water three quarters of the way up.
- Fill the measuring cylinder up with water, ensure there are no
oxygen bubbles in the cylinder, turn it upside down and place it in the container like in the diagram.
- Place the tube underneath the cylinder.
- Add the lithium to the flask and place the bung with the tube
back onto the flask.
- Collect the gas and write down the final volume of hydrogen
collected.
In method 1 I reacted 100 cm³ of water to 0.08g of lithium. And from that reaction 150 cm³ of hydrogen was given out.
I must firstly work out the number of moles in the hydrogen collected, to do this I must use this formula –
The n represents the number of moles. The v is the volume collected and the c is the concentration of the hydrogen. The n is unknown to us, the volume is the amount of hydrogen collected which is 150 cm³ and it is widely accepted that 1 mole of any gas occupies 2400 cm³ at room temperature. With this information we can replace the symbols with actual numbers.
The new formula would be as follows –
n = 0.00625
From this we can deduce the number of moles of lithium which was reacted. To do this we would have to look at the mole ratio of lithium and hydrogen. To get the mole ratio we would need to examine the formula of the reaction. The formula is as follows –
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
As you can see 2 moles of lithium was used to produce 1 mole of hydrogen. So therefore the mole ratio of lithium and hydrogen is 2:1. To deduce the moles of lithium used we must multiply the number of moles of lithium collected.
0.00625 x 2 = 0.0125
Now that the mole of the lithium has been calculated, we can use this to work out the relative atomic mass of lithium. To do this we must use this formula –
m = mr
n
The m in the formula represents the mass of the lithium, which would be 0.08g, the n represents the mole of the lithium which is 0.00625. Once the figures are in place in the formula, the formula should look like this.
0.08 = mr
0.0125
mr = 6.40
the relative atomic mass of lithium according to this experiment is 64
Method 2 : Titration
Apparatus
1 x Stand
1 x Clamp
1 x Burette
1 x Funnel
1 x Pipette filler
1 x Pipette (25cm³)
1 x Beaker (250cm³)
1 x Lithium hydroxide
1 x Hydrochloric acid (0.1 mol dm³)
1 x Phenolphthalien
1 x Goggles
1 x Gloves
1 x Lab coat
The apparatus should be set up as showed in the diagram above.
Method
- Firstly the apparatus must be set up according to the diagram
- Place the pipette into the pipette filler and use it to measure
25cm³ of the Lithium hydroxide and pour it all into the beaker. And place the beaker under the burette.
- Now place the funnel on top of the burette and fill the burette up
with hydrochloric acid till the initial value reads 0. Ensure that the tap on the burette is not open whilst pouring the acid. Take off the funnel once done.
4. Now the indicator needs to be placed into the Lithium hydroxide, in this case it is Phenolphthalien. The indicator will tell us when the solution is neutralised. The initial colour of the base with the Phenolphthalien should be a pinkish colour.
5. Now open the tap on the burette so that the acid flows into the base. While the tap is running try to simultaneously shake the beaker.
6. Let the acid flow, but keep an eye on the base in the beaker, once the colour of the contents of the beaker changes from pink colour to colourless, immediately close the tap.
7. Record the amount of hydrochloric acid that was required to neutralise the Lithium hydroxide. Clean all apparatus with distilled water and perform the test another three times. This will insure a more accurate result, as the average of the four tests will be used.
Fair Test
The key to this experiment being a fair test is measurement. If I can ensure that all my measurements are accurate then the results will be fair and accurate, the experiment could give false results. I must conduct all the experiments under the same conditions. Whether that means using the same apparatuses in each experiment of the keeping the temperature constant.
Risk Assessment
There are always risks involved when handling any kind of chemicals. Both the lithium hydroxide and the hydrochloric acid are irritants. To ensure that there is maximum safety, all personnel involved in the experiment should wear proper eye protection, like goggles and as well as gloves. Not only should they take care in handling the chemicals is there are any spillage’s then the person responsible should clean up the chemicals, so that it is not left for others to find.
If a person does come in contact with hydrochloric acid on the skin then plenty of water should be applied to the effected part. If the chemical come in contact with the eye, immediately rinse the eye with water, and seek medical attention straight away. If the chemical is swallowed the drink plenty of water and seek medical attention straight away.
If inhaled lithium hydroxide can cause severe irritation and corrosive tissue damage, similar to other strong bases. Like sodium hydroxide and potassium hydroxide. If inhalation of lithium hydroxide occurs, remove source of contamination or move victim to fresh air and seek medical attention.
Lithium hydroxide can be very irritating to the skin. Solid lithium hydroxide or concentrated solutions may cause severe tissue damage. Avoid direct contact with this chemical, wear protected gloves.
Lithium hydroxide can cause severe irritation and corrosive damage to eyes. If there is contact with eyes, gently blot or brush away dust quickly. Immediately flush contaminated eye with luke worm and run with water, seek medical attention. To prevent this from happening wear appropriate eye protection.
Results of experiment
From the results in the table I can say that it takes on average 28.73cm³ of hydrochloric (0.1moldm³) to neutralise 25cm³ of lithium hydroxide.
From that I can use this formula to work out the relative atomic mass of lithium.
Firstly I have to calculate the number of moles of hydrochloric acid used in the titration process. To do this I must use this formula –
The n represents the moles used, the c represents the concentration of the acid and the v stands for the volume of the acid used in the titration process. Once all the values are in place the formula would look like this –
n = 0.10 x 29.1
1000
n = 0.00291
From that we can deduce the number of moles of lithium hydroxide used in the titration. To do this I have to work out the mole ratio of the lithium hydroxide and the hydrochloric acid. To get the mole ratio we would need to examine the formula of the reaction. The formula is as follows –
LiOH(aq) + HCl(aq) LiCl(aq) + H2O(l)
As you can see the mole ratio is 1:1 as I mole of lithium hydroxide is required to react with 1 mole of hydrochloric acid. So the number of moles of lithium hydroxide present is 0.00291
After that I would need to work out the number of moles of lithium hydroxide present in the 100 cm³ of the solution from method 1.
I know that in 25 cm³ of lithium hydroxide there is 0.00291 mole so in 100 cm³ there would be –
0.00291 x 4 = 0.01164
Now that we know the mole of the lithium hydroxide we can use this and the original mass of the lithium to work out the relative atomic mass of lithium. To do this I must use this formula –
The m represents the mass of the lithium and the n represents the mole of the lithium hydroxide.
The formula could be rearranged with the actual figures like this-
Mr = 0.08 .
0.01164
This would mean that –
Mr = 6.87
After the calculations I have come to the conclusion that the experiment shows that the relative atomic mass of lithium is 6.96.
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:
6.40 x 100 = 92.75 % accuracy
6.90
Method 2:
6.87 x 100 = 99.5 % accuracy
6.90
This shows that my results have a high percentage of accuracy.
Even though the percentage of accuracy is extremely high in both of the experiments, I however feel that method one was the inferior method of the two. As you can see the difference in accuracy is a big margin.
Even though the percentage of accuracy of method 2 is 99.5%, it is not 100%, which would, indicate that there were some anomalies in the method. As the margin of error is only 0.5% it is difficult to pinpoint where exactly the mistake occurred. 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 properly 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. The accuracy of the experiment would be improve if all the oil was cleaned off of the lithium
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 watermark in a burette or measuring cylinder could cause slight inaccuracy. 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 deciding when the end point of titration is and the reaction time between realising when the end point and stopping the flow of acid from the burette. If the reaction time is slow, there could be more acid used than that that is needed. This error can be eradicated by slowing down the pace at which the titration is performed.
Method 1 was very inaccurate. I think one of the mistakes that were made during the experiment was to assume that the temperature was standard. Measurements of the room temperature wee not taken during the experiment so we cannot be sure if the room temperature was standard. If the room temperature is not standard, one mole of the hydrogen would not take up 24000cm³, this would mean that all the calculations would be incorrect.
From observing the percentage accuracy’s, I can see that Method 2 is more accurate and effective than Method 1 . There are a number of factors that could be responsible for method 1 being less accurate. These could be errors, there could be errors in measurement of the lithium or distilled water, oxidation of the lithium occurring when lithium is out in open air.
Also, I repeated Method 2 four times, which would give me better results as I could use an average of the different tests. This was not possible in Method 1 as I had to preserve the lithium hydroxide solution for method 2, which meant that method 1 could only be performed once.
All in all I think I have carried out the experiments to the best of my abilities.