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Determination of the relative atomic mass of lithium. - page 1
Keywords: Experiment evaluationrelative atomic mass lithium accuracy results
By Jenny on 02/07/2009
Level: A Level (Year 13)
Page Number: 1 of 2 pages: 1 2Evaluation
My experiment was the determination of the relative atomic mass of lithium. There were several areas in this experiment in which there were errors which will affect the accuracy of my results.
Errors in procedure
To start with I only took one reading of the hydrogen gas given off when the lithium reacted with the water, I could improve my accuracy here by taking 3 or more readings and using the average. Then, after I dropped the lithium in the water there was a slight delay while I put the bung on allowing some hydrogen to escape, which also affects the accuracy of this reading. To prevent this I could loosely wrap the lithium in wire and push one end of the wire into the bottom of the bung, so that once the bung had been inserted, the flask could be shaken to release the lithium.
The initial volume of gas in the tube could also be a source of error as it was difficult to make sure there was no air in there.
The weight of the lithium might not be accurate as there may have been some oil on it. Additionally you could tell the outside of the lithium had been oxidised as it was a dull grey colour not shiny, which would also have affected the weight. To reduce the error here I could use lithium which had been freshly cut so that there would be no oil or lithium oxide on it.
Diagram of new apparatus
Errors in measurement
During the experiment I had to measure several substances in various methods each of which had different errors associated with them.
250 cm³ measuring cylinder – measurement error = +/- 2 cm³
% error = (2 / 206) x 100 = 0.971 %
Burette - measurement error = +/- 0.1 cm³
% error = (0.1 / 42.75) x 100 = 0.234 %
Pipette – N/A
Balance - measurement error = +/- 0.02 g
% error = (0.02 / 0.11) x 100 = 18.2 %
100 cm³ measuring cylinder – measurement error = +/- 1 cm³
% error = (1 / 100) x 100 = 1 %
Total Error = 0.971 + 0.234 + 18.2 + 1 = 20.405 %
Over 20 % is a large error which will have significantly affected the accuracy of my experiment. By far, the main source of this is the balance measurement error. To improve my experiment I would need to reduce this. One way of doing that
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