• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9

The relative atomic mass of lithium

Extracts from this document...


COURSEWORK Background information I'm going to produce a piece of coursework, which determine the relative atomic mass of lithium. Two methods are going to be used to approximate the relative atomic mass of lithium. 1) In the first method the RAM (Relative atomic mass) of lithium is determined by the Volume of hydrogen gas produced. 2) In the second method the RAM of lithium is determined by titrating the lithium hydroxide produced. To some extend, I think that the first method is the most practical and less complicated, because the less calculations has to be done. Hence it is more easer to measure the volume of hydrogen gas. Test-1 My mission was to find out the atomic mass of lithium. I took a sample and weighted it, removed parts of the sample so that it would weigh approximately 0.100g. The oil was removed from the sample by filter paper wiping it with,so no more oil can be seen. The scale was not a very sensitive one and we could weigh with a margin of error of 0.01g. Then I placed the sample in a beaker that contained 100 cm3 of distilled water. ...read more.


5) Then the average titre is recorded. Results I gathered a large enough range of results to make accurate observations from. The experiment was repeated until consistent results were obtained; Run 1 Run 2 Run 3 Initial volume (cm3) 1.3 2.7 1.4 Final Volume (cm3) 39.7 38.4 40 volume Titre used (cm3) 38.4 35.7 38.6 Mean titre (cm3) 56.35 Burette reagent HCl mol dm -1 Pipette solution LiOH mol dm-1 indicator Phenolphthalein mol dm -1 Analysis Calculation: 1st test- by measuring the volume of hydrogen produced A mass of 0.07 of lithium was used, which gave 116Cm3 of H2 * Calculate the number of moles of hydrogen 116/24000 = 0.0048 mol of H2 2Li (s) + 2H2O(l) ----->2LiOH(l) + H2(g) so, 2* 0.0048 = 0.0097mol of Li using the formula to find the RAM of Li --------> Mole = mass/Ar or Mr so, 0.07/0.0097 = 7.2g of Li The answer is very close to the standard RAM of Li Error calculations, 7.2*100/7=102.86 - which means that there is 2.86% error in the first method 2nd test - by titrating the lithium hydroxide with HCl. From the results table the mean titre is found by adding the runs altogether that divide the by 2 Mol =vol/1000*concentration = 38.4+35.7+38.6=56.35 /2 = 28.175cm3 = 28.175/1000*0.100=0.002818mol = ...read more.


if the methods had more details in it then this will inpruve the accuracy of the experiment. The method didn't give enough information of how to calculate the results, so i strugeled a bit, but in the end i worked out with the help of the book. To minimise the errors the method should have explained in detailed the performers that the experimenter has to carry out. To make the experiment the most accurate possible the equipment should be the more advanced ever, and variables should be pure with out any mixture. For further improvement a freshly cut pure lithium should be used, this could be done by putting the lithium in the vacuum before transferring it into the conical flask. A freshly cut pure lithium means having accurate results (out comes). Also after introducing the lithium into the conical flask the stopper should be replaced very quickly so a low minimum gas escapes. Then to measure this gas a more accurate apparatus should be used like(Eudiometer ).The mass balance despite taking quick readings, was not the most accurate possible, so to improve the accuracy of the experiment a more advanced electronic mass balance should be used. I think more time is also needed. If i was doing the experiment again i would like a bite more time to perform the experiment as this would improve the accuracy of the experiment ...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 Classifying Materials 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 Classifying Materials essays

  1. The rates of reaction between CaCO3 and HCL

    24.94 24.69 00.32 165 24.84 24.25 24.92 24.67 00.34 180 24.84 24.25 24.92 24.67 00.34 195 24.83 24.24 24.91 24.66 00.35 210 24.82 24.23 24.90 24.65 00.36 225 24.81 24.22 24.90 24.64 00.37 240 24.80 24.22 24.89 24.64 00.37 Analysis: Trends and patterns: The lines shown on the graph are

  2. Affect of concentration on reaction

    quickly at first, then it gradually slowed down and if the experiment was carried out longer, the reaction would have stopped. This is because when calcium carbonate was first added to acid, it had the most surface area, so it reacted fastest at start.

  1. The role of mass customization and postponement in global logistics

    (http://highered.mcgraw-hill.com/sites/0072394668/student_view0/chapter2/glossary.html) * Shorthand for high variability in marketing. Uses the power of the database to vary the marketing message - or the actual product - to fit the characteristics of an individual customer or prospect. (http://www.unitedwire.net/buzzwords/buzzdf01.htm) * A highly streamlined and flexible approach to production that enables quick and efficient production of customized products and/or services at low cost and high volume.

  2. Relationship between mass of MgO and its formula

    to get a ratio of 1.11:1, which is very close to the ideal ratio of this experiment. This shows that I did do the experiment well in some areas. The results I received from the class are very varied but in some places do show some good relationships.

  1. Bonding Practical

    Here is a list of the different tests and I will explain how I will perform each one. Is it a solid I believe this is the first test that should be done because if a substance is a liquid then it must be a small covalent compound for example substance C.

  2. Determining the Relative Atomic Mass of Lithium.

    This means that the timing must be perfect in order to have accurate results. The overall average Ar was close to the actual Ar of Lithium, the overall average Ar was 7.1. There are other factors that could have contributed to the inaccuracy in the results, these are: - *

  1. Determination of the relative atomic mass of lithium

    * My first method I will take several measurements of the volumes of the hydrogen gas produced when I react a 0.1g of lithium with distilled water. * My Second method will be to carry titration on the lithium hydroxide I will obtain from the first method.

  2. Determining the relative atomic mass of lithium.

    However, before I calculate the relative atomic mass of lithium, I would define the terms that would be used the formulae. The mole (n) of any substance is the amount which contains the same number of particles of the substance, as there are atoms in a 12g of Carbon-12 atom.

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