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

Calculating the relative atomic mass of lithium.

Extracts from this document...


Chemistry Assessed Practical Write-up Calculating the relative atomic mass of lithium 1) Mass of Lithium (g) Volume of Hydrogen collected (cm3) 1 1.2 129 2 1.1 120 3 1.3 137 Amount of H2 collected = 129 ml Mass of Lithium with oil = 1.2g 2Li + 2H20 --> 2LiOH + H2 Vol H2 = 129 24,000 = 0.005375 moles --> 1:2 --> 0.01075 RAM = Mass Moles RFM Lithium = 0.12 0.01075 RFM Lithium = 11.16g The RFM of Li that I worked out is = 11.16g RAM Li = 6.9g Therefore the mass of oil is: 11.16 - 6.9 = 4.26g 2) Volume Lithium Hydroxide = 25.00 cm3 exactly (using a volumetric pipette) Volume HCL (0.1 molar) to neutralize Lithium Hydroxide = See graph Start (cm3) Finish (cm3) Value (cm3) 1 0.0 27.1 27.1 2 1.2 28.2 27.0 3 0.7 27.8 27.1 LiOH + HCl --> LiCl + H2O Volume HCL = 27.1 Concentration of ...read more.


Using the volume of hydrogen I was able to calculate the moles of hydrogen. I used this work out the moles of Lithium from which I could work out the mass and RAM of Lithium. The second experiment was the titration of the lithium hydroxide that was a by-product of the first experiment. By measuring the amount of hydrochloric acid (of a known molarity) I was able to work out the RAM of Lithium. The average of the two RAM's that I found is: 10.835 + 11.16 = 10.99 2 Evaluation Both the results for the RAM that I recorded are around 11g, so any errors that have occurred have obviously affected all the results. Though the second result could be affected by more errors because it has the errors of the first results, and the risk of more errors in the second experiments. ...read more.


The error in the amount of HCl is about 0.1cm3 because the graduations on the pipette are 0.1cm3 apart. Taking into account all these errors the results that I obtained aren't very accurate, as the results show as they are about 4g off the proper RAM of Lithium. These results could be improved by using a 100cm3 volumetric pipette to measure the distilled water. This would make the error almost negligible. You could devise a system where the Lithium could be added when the bung was already in place, therefore there would be no gas escape. You could eliminate the error in HCl by using burette with smaller graduations. The overall percentage error is: 10.99 * 100 = 159 % 6.9 This experiment has basically gone smoothly, there were no major hiccups along the way, and although my results aren't entirely accurate, at least they are consistent. I conclude that the reason for this is that the apparatus that I used was not appropriately accurate for the experiment. 1 1 ...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. Determine the relative atomic mass of lithium.

    1 x Beaker (250cm�) 1 x Lithium hydroxide 1 x Hydrochloric acid (0.1 mol dm�) 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.

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

    Instances in which percentage errors come into account are highlighted and calculated below: Mass of Lithium used was measured by a balance. The value that was recorded alongside the absolute value was 0.099 � 0.001g which means using the formula mentioned above, we reach a percentage error as below to

  1. Determination of the Relative Atomic Mass of Lithium

    I think that my titration was much more accurate than the results I got from when I collected the gas. This is because there were more places where I am sure I made mistakes when I was collecting the gas, than when I was titrating the Lithium Hydroxide and also my results from my titration are all very close.

  2. Determination of the relative atomic mass of Lithium

    In concern of method 1 I found that there was a difference in the results of the two different volumes. The most accurate reading of the volume of hydrogen gas was 76 cm as it gave a relative atomic mass of 6.98, where as 78 cm gave a relative atomic mass of 6.46.

  1. Determination of the relative atomic mass of lithium.

    So, the number of moles of lithium hydroxide is 1.867 x 10-3. I now need to calculate the number of moles of lithium hydroxide present in 100cm3 of the solution from Method 1.

  2. Determination of the Relative Atomic mass of Lithium

    10 10.95 Concentration (mol dm-3) 0.1 Moles 0.001095 0.001095 The Moles of HCl is decided by the equation: Moles = concentration / volume (in dm3) so therefore in this case the equation would be: 0.1 x (10.95 / 1000) = 0.001095 moles and since the HCl and LiOH are in

  1. Investigation to determine the relative atomic mass of lithium

    time to obtain an average > I will use the same equipment each time > Values should also be read at eye level to ensure that the correct value is being seen, viewing from other angles can cause the incorrect value to be seen.

  2. Determination of the relative atomic mass of lithium.

    So I will not be using it in my average titre. Average titre = 18.67cm3 (to 2 decimal places is sufficient accuracy) The chemical equation for this reaction is: LiOH(aq) + HCl(aq) � LiCl(aq) + H2O(l) The mole ratio of lithium hydroxide to hydrochloric acid is: 1:1 To find the

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