From knowing that 1 mole of gas occupies 24,000cm³ at room temperature and pressure I can deduce the amount of moles that I collected from the experiment
Moles = 170 ÷ 24,000 = 0.00708
With these figures I can deduce the number of moles from lithium that reacted by using the ratio of lithium to hydrogen in the reaction
2Li (s) +2H2O (l) 2LiOH (aq) +H2 (g)
The ratio of moles of Lithium to Hydrogen is 2:1, to get the moles you have to multiply the number of moles hydrogen produced by 2
Moles of hydrogen X 2 = 0.142mol-1
These results allow me to determine the relative atomic mass of lithium using the formula for working out relative atomic mass:
Relative atomic mass = Mass ÷ Moles.
This allowed me to find that the relative atomic mass is 7.06mol-1
Method 2
For my second experiment I will be doing a titration this will allow me to find out how much lithium hydroxide will need to be used to neutralise HCL this experiment will be repeated 3 times so that results are accurate.
For this experiment I will be using
LiOH from the first experiment
HCl (100 mol-1)
Phenolphthalein
250cm3 conical flask
Pipette 100cm3
Burette 100cm3
Funnel
- Set up the equipment as shown above
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Using a pipette, drain 25cm3 of solution and place it into the 250cm3 conical flask.
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Fill the burette with 100cm3 of HCl
- Add 2 drops of Phenolphthalein to the LiOH
- Add HCl from the burette one drop at a time until the LiOH becomes clear
- Take a measurement of how much HCL was used
On average 25cm3 of LiOH needed 33.63cm3 of 0.100moldm HCl to neutralise
Moles = Concentration X Volume
Moles = 0.100 X (33.63cm3 ÷1000) =0.03363
Since this reaction has 1:1 ratio the number of moles of LiOH was also 0.03363
Using this information and the original mass of lithium reacted I am able to calculate the relative atomic mass of lithium
Relative atomic mass = Mass ÷ Moles.
Relative atomic mass= 0.01 ÷ 0.01266 = 0.789 mol
Evaluation
The overall accuracy of my experiment was OK but there were a few things that could have been changed to make the results more reliable.
The first experiment may have been less accurate than my titration because it was only done once whereas the titration was repeated three times.
While I was placing lithium inside the conical flask, gas would have escaped in the time it took to take the lid off and place the lithium inside. To make sure that this doesn’t happen again we could place a device within the flask such as a ‘shelf’ within the flask for the lithium to sit on and when the lithium needs to be added slant the flask slightly so the lithium falls in this would therefore mean that the flask is airtight.
If I had been able to use a bigger sample of lithium then the surface area would have been smaller and there would have been less oil on the sample making the mass more accurate and also the overall results. Also I could try and clean the lithium more thoroughly and store it in an area where it can avoid oxidation.
As I cannot be sure if any residue had been left on the equipment from previous experiments, in the future I will make sure that all equipment has been thoroughly sterilised before I use the equipment.
Improving the scale of the experiment will in the future allow me to have more accurate results, for example if I had used equipment that was more accurate than 1 decimal point it would allow smaller errors in the results, but using the schools equipment it was not possible.
