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

Determination of the relative atomic mass of Lithium.

Extracts from this document...


Chemistry Coursework Stephanie Knowles Determination of the Relative Atomic Mass of Lithium Analysing: We followed the plan set to us and, for method one, attempted capturing the hydrogen gas produced by the reaction of lithium and water. We decided to do this four times to gain more reliable results, and to identify any anomalies. On our first attempt we used 0.10g of lithium, this amount produced 90cm� of hydrogen gas, on our second attempt we used 0.9g of lithium, which produced 130cm� of hydrogen. At this point it was clear that there was a mistake as bigger quantities of lithium would have reacted more, producing more gas, or at least the amount should have been around the same as our first attempt. On the third try we used 0.13g of lithium- this reaction caused 230cm� of gas to be collected. Making the conclusion that one of our 0.9g or 0.10g of lithium results was incorrect, we attempted a 0.10g of lithium again this produced 250cm� of gas, again not agreeing with our result for the 0.13g of lithium. ...read more.


We need to know how many grams produces 1 mole, therefore divide 1 mole by 0.020mols: 1mol - 0.020mols = 5g Giving the answer that the relative atomic mass of lithium, is 5. Procedure Two- Calculate the number of moles of HCl used in the titration- We can do so using this balanced equation: LiOH (aq) + HCl (aq) LiCl (aq) + H O (l) 12.6 cm� (average titration amount) The HCl has a concentration of 0.1mol dm�. We must use this formula- 12.6 1000 x 0.1 = 0.00126 Therefore 0.00126mols HCl were used. Deduce the number of moles of LiOH used- From the above equation we can see that the same amount of HCl is used as LiOH so there is also 0.00126mols of LiOH used. Calculate number mols LiOH present in 100cm �of the solution from procedure one- We know that there was 0.00126mols present of LiOH in 25cm�, to calculate how much was present in 100cm�, multiply by 4: 4 x 0.00126mols = 0.005mols Calculate the relative atomic mass of lithium- The relative atomic mass is the same as 1mole of lithium. ...read more.


Other then human error the only explanation for this huge difference was the lithium we used. Lithium must be stored in oil as after a minute or so in the presence of moisture and oxygen the surface becomes tarnished to a grey colour and perhaps the occurrence of lithium oxide as well. This might well have affected the collected lithium hydroxide in the conical flask therefore affecting the titration. Slicing an entirely silvery coloured fresh piece of lithium and using it instead of a piece from the outside of the lithium 'block' could have prevented this. In conclusion, the experiment was not a successful one- the apparatus problems and the lithium used (which would have also explained the diverse amounts of hydrogen collected in procedure one) would explain the very different results of the relative atomic masses from each other and the correct one. With the adjustments suggested mad the results would have been more reliable and closer to the truth. ...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

    C and O have to be present in CaCO2 thus making it purer. The information stated in the background knowledge it is known that from 10Og of CaCO2 44g of CO2 is produced if the CaCO3 is pure. To calculate the percentage purity divide 100 by Y.

  2. Free essay

    Periodic table

    Mutagens - Factors that trigger a mutation in the cells. Changes in the letter of a word can change the word's entire meaning. Gene mutations ==> uncontrolled cell division which can result in cancerous tumours. Not all mutations are harmful i.e. for insects it is part of their survival mechanism.

  1. Affect of concentration on reaction

    1.1555 30 1.1610 1.1780 1.1695 35 1.1730 1.1900 1.1815 40 1.1840 1.1980 1.1910 45 1.1940 1.2040 1.1990 50 1.2040 1.2120 1.2080 1.2 M Time (s) 1st Mass Loss (g) 2nd Mass Loss (g) Average Mass Loss (g) 0 0.0000 0.0000 0.0000 5 0.8180 0.8380 0.8280 10 0.9050 0.9230 0.9140 15

  2. Relationship between mass of MgO and its formula

    5cm Mg O Mass 0.1g 0.06g RAM 24 16 Moles= mass RAM 0.1= 0.004167 24 0.14= 0.00375 16 0.004167 =1.11 0.00375 0.00375=1 0.00375 Ratio 1.11:1 7.5cm Mg O Mass 0.16g 0.04 RAM 24 16 Moles= mass RAM 0.16= 0.006667 24 0.04= 0.0025 16 0.006667=1 0.006667 0.0025=0.375 0.006667 Ratio 1:0.375 10.0cm

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

    developing a concept into a realisable solution that is both economic and effective Supplier management and supply chain agility Dynamically pool resources to meet the changing interests of customers efficiently and effectively; with decreasing vertical integration, this is increasingly accomplished through relationships with collaborating organisations Production flexibility and capacity management

  2. The relative atomic mass of lithium

    The solution turned pink - The next thing to do was to put to put the HCl into a burette, measure the initial volume then relies the HCl into the conical flask slowly until the solution turns colourless, the final volume is then recorded.

  1. Determination of the relative atomic mass of Lithium.

    Results - LiOH (aq) + HCl (aq) LiCl (aq) + H O (l) � Titre Initial volume of acid (ml) Final volume of acid (ml) Volume of acid used (ml) 1 0 42.2 42.2 2 0 42.7 42.7 3 0 42.1 42.1 4 0 31.8 31.8 Average result of titration is 42.3 ml.

  2. Gravimetric Determination of Phosphorus in Plant Food

    Fold the paper as illustrated in Figure 1 and fit into a glass funnel. Be certain that the filter paper is opened in the funnel so that one side has three pieces and one side has one piece of paper against the funnel, not two pieces on each side.

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