• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Relative atomic Mass of Lithium

Extracts from this document...

Introduction

Determination of the relative atomic mass of Lithium Evaluation I was given a task in which I was had to investigate and determine the relative atomic mass of Lithium. I was given the opportunity to use two different methods of which I could use to get my final result. The practical takes a vast amount of concentration and there are a number of measurement errors that can occur while carrying out the practical. Problems that affected the measurement such as; the equipment being used, not being 100% accurate when measuring out the substances being used, which usually gave an outcome of the readings given being slightly higher or slightly lower than the amount needed. This is evident in the use of the pipette, which expands and contracts due to the temperature and surrounding, so to get what is seen as an accurate reading would be nearly impossible. This degree of uncertainty is referred to as the tolerance. The tolerance could be a problem when the measurements are carried forward to the calculations, for example; when calculating the number of moles of Lithium Hydroxide in the second method. ...read more.

Middle

Uncertainty (%) = Multiplied by 100 Measuring Cylinder Uncertainty (%) = Multiplied by 100 Burette Uncertainty (%) = Multiplied by 100 Balance Uncertainty (%) = Multiplied by 100 From my results it is clear that the important area of measurement source is the balance, this is due to the fact that it has the highest uncertainty. From my calculations I also saw that even though the tolerance of the measuring cylinder is greater to that of the pipette and the percentage uncertainty is lower which is due to the fact that the volume that has been measured in the measuring cylinder is larger than the pipette, meaning the measurement error produced by the pipette is more significant. There are two ways in which you can minimise the measurement errors; one being to measure a larger amount, but by doing this I may need to alter my method, which can be done by either using alternative equipment such as a larger burette or increased concentration of Lithium Hydroxide solution. Another way can be to attempt to use a piece of equipment with a lower tolerance value, but this is not always possible. ...read more.

Conclusion

To prevent this from happening I could have used an inert solvent to wash all the oil off the Lithium providing me with all the correct results. After carrying out and completing my experiment, and going through two methods in which to get the result of my task, I can see why it is essential to have more than one method, as one method will not show the mistakes and effectiveness another method will. I feel that the most effective method was method 2. This is because it was most effective in finding the relative atomic mass of lithium. I feel this is because the most inaccurate piece of equipment, which is the balance, was not included in this method which would have gave a better chance of getting accurate results. Another reason is that within this method the Burette was used which had the most accurate functions to get the most reliable results, which put this method at a steady advantage as the least accurate apparatus was excluded and the most accurate was included. ?? ?? ?? ?? Tino Chingwaru ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. How much Iron (II) in 100 grams of Spinach Oleracea?

    solute in the solution had begun to sink to the bottom causing the solution to be at different concentrations in the curvet. This would mean that when the light from the colorimeter was shun onto the curvet it would have been absorbed at different levels at different depths.

  2. Determine the relative atomic mass of lithium.

    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

  1. to determine the relative atomic mass of lithium. We will be doing this via ...

    I think that with improvement in procedure and method this could have been closer to the true value of 6.94. It all depends on the accuracy, limitations in a procedure and the uncertainties in an experiment, taking all the evidence into account, such as the results and readings along with

  2. Determination of the Relative Atomic mass of Lithium

    However it is not possible to remove every trace of oil from the lithium, and to a similar extent not all the ether can be removed without waiting for an extended period of time, and doing this would increase the size of the oxide layer on the lithium further increasing the weight of the lithium.

  1. Investigation to determine the relative atomic mass of lithium

    I added 5 drops of phenolphthalein indicator, which made the solution, turn pink. I then titrated the solution with 0.1moldm3 HCl. I then started to run the HCl and when it came near the end point (where it turned colourless)

  2. Determination of the relative atomic mass of lithium.

    that the number of moles of lithium is equal to the number of moles of lithium hydroxide. So the number of moles of lithium is also 7.468 x 10-3. Using the same equation as before: Relative atomic mass = mass (g)

  1. Determination of the relative atomic mass of Lithium

    there was more lithium, hence there was a lot more hydrogen released. This therefore means that I would have a greater amount of hydrogen released when using 0.11 g of lithium, than if I was using 0.10 g of lithium.

  2. Determination of the relative atomic mass of lithium.

    So to help me do this I will use the triangle for this formula. 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.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work