• 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...


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.


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.


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. Determination of the relative atomic mass of lithium.

    Since 1 mole of gas at room temperature and pressure occupies 24 dm3: Moles of H2 = 158 x 10-3 = 6.6 x 10-3 mol 24 Since 2 moles of Lithium produces 1 mole of Hydrogen: 2 (6.6 x 10-3)

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

    Experiment C 5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II)

  1. Determination of the relative atomic mass of lithium.

    of my three sets of results. Molarity = this was the molarity of the HCL which was given to us. Divide by 1000 = this is how the equation is done. n = 42.67 X 0.1 = 0.00427 M 1000 The average (mean titre)

  2. Determine the relative atomic mass of lithium.

    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.

  1. To determine the relative atomic mass of Lithium

    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

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

    Initial Reading On Burette (cm3) Difference (cm3) 1 (Preliminary) 19.5 4.2 15.3 2 34.30 19.50 14.80* 3 15.10 1.10 14.00 4 29.70 15.10 14.60 5 16.00 1.20 14.80* 6 30.65 16.00 14.65 7 44.60 30.65 13.95 8 19.85 5.05 14.80* * Concordant Results On average, 25.0 cm3 of LiOH(aq)

  1. Determination of the Relative Atomic Mass of Lithium

    cm3 of LiOH Amount of acid = Concentration of acid x Volume of acid Volume of acid = 27.0 x 10-3 dm3 Concentration of acid = 0.103 mol dm-3 Amount of acid = (27.0 x 10-3) x 0.103 = 2.8 x 10-3 mol mol of HCl = mol of LiOH

  2. Determination of the relative atomic mass of Lithium

    and the number of moles of lithium (0.0155 mols) into the formula, which left me with 7.10 correct to 3sf. To sum up: Number of moles of hydrogen 0.00775 mols Number of moles of lithium 0.0155 mols Ar of lithium 7.10 gmol-1 The problems, which could have caused my readings

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