• 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...

Introduction

Determination of the relative atomic mass of lithium Introduction As part of my AS chemistry unit, I was given an assessed practical to do. The aim of this practical is to determine the relative atomic mass of lithium. I will do this by the combination of two different methods. Method 1: - By measuring the volume of hydrogen produced Method 2: - By titrating the lithium hydroxide produced Method 1 I set up my experiment as previously shown with the correct size equipment in the correct places. But before I began the actual practical part of the assessment, I had measure the amount of lithium I had. Lithium is reactive with air, so I had to measure it in a specific way. Firstly I sliced an appropriate part of lithium with a knife and put it in a Petri dish. I then weighed the Petri dish occupied by the lithium which was 3.40g and to calculate the amount of lithium I had, I weighed the Petri dish on its own which turned out to be 3.31. Finally I subtracted the weight of the Petri dish and lithium by the weight of the Petri dish which resulted in its own. ...read more.

Middle

During the titration, I kept 25.0 cm3 of the solution from method 1 in a conical flask and added 5 drops of phenolphthalein indicator, as I did this the solution turned a fluorescent pink colour. I then had to titrate the solution with 0.0100 mol dm3 of aqueous hydrochloric acid to see how much acid was needed to turn the solution from a bright pink colour, back to a colourless solution. The concluded results were: - Amount of HCL at start = 28 Amount of HCL at end = 42.15 So the amount of HCl used is the amount at the start minus the amount at the end. 42.5 - 28 = 14 .3 cm3 (Start) (End) (Result) LiOH (aq) + HCl (aq) LiCl (aq) + H2O (l) I will now attempt to work out the technical parts to method 2. * Calculate the number of moles of HCl use in the titration. Number of moles of HCl = 0.1 � 14.3 = 0.00143 1000 The number of moles of hydrochloric acid (HCl) used in the titration is 0.00143 * Deduce the number of moles of LiOH used in the titration. The number of moles = 0.00143 � 10 = 0.09 0.00143 = 6.29 The number of moles of LiOH used in the ...read more.

Conclusion

But in method two, I was titrating so I had to look at the liquid and construe how much acid was needed. This is not very accurate because you have to estimate and different people have different eye sights which affects it. Another factor is the temperature. If we kept the temperature of the room constant at about 20-25 o C, the results would be more accurate. To prevent the errors we could minimise reactions between lithium and air, by putting lithium in an organic solvent such as cyclohexane which won't react with the lithium. Another factor is that in the beginning part of the investigation we could put the bung on quicker to prevent any hydrogen gas escaping. Better measuring cylinders, measuring balances and more accurate pipettes would help too. Also in method two instead of deducing the end point of the reaction by human eyesight, we could accurately measure the end point by some sort of colour receptor. Anther detrimental factor is the batch of lithium used. If the lithium has aged then some or most of its reactivity may be lost. Ali Omar Page 1 Determination of the relative atomic mass of lithium ...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 magnesium by back titration

    * Rinse apparatus with distilled water, and then with the solution being used * Remove the funnel when titrating * Ensuring the value of the pipette is read at eye level * Movement during transferring should be kept at a minimum * Allowing it to drain under gravity * When

  2. The Determination of an Equilibrium Constant.

    * 75 = 69 g Molecular mass of CH3COOC2H5 = 12 + 3 + 12 + 32 + 24 + 5 = 88 g/mol Number of moles of CH3COOC2H5 = 69 / 88 = 0.784 mole * Mass of H2O = 1 * 25 = 25 g Molecular mass of

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

    The whole solid may not have dissolved fully in the acid/distilled water needed to create the correct concentration of solutions in both the Iron (II) Ammonium Sulphate (aq) and the Potassium Manganate (VII) (aq). If this had occurred the solutions produced would have been less concentrated resulting in any equations done using their concentration being inaccurate.

  2. Determination of the relative atomic mass of lithium.

    I used 25cm3 of the solution from Method 1. Since 25cm3 fits into 100cm3 four times, to find the number of moles of lithium hydroxide present in 100cm3, I will need to multiply the number of moles of lithium hydroxide in 25cm3 by four: Number of moles of lithium hydroxide in 100cm3 = 1.867 x 10-3 moles x 4

  1. Determine the relative atomic mass of lithium.

    Fill the container with water three quarters of the way up. 5. Fill the measuring cylinder up with water, ensure there are no oxygen bubbles in the cylinder, turn it upside down and place it in the container like in the diagram. 6. Place the tube underneath the cylinder.

  2. Determination of the relative atomic mass of lithium.

    The mass I have used is from the original mass of my lithium. This is because I need to figures from the previous stages. The number of moles (0.01866) was derived from the answer to part (2) of my analysis.

  1. Determination Of The Atomic mass of Lithium.

    * Now I will explain ways to ensure the most accurate results in method 1. o Secure the bung in the top of the conical flask as fast as you possibly can, making sure that it is totally secure in the flask.

  2. copper practical

    The dry ore is mixed with sand. The word equation of this reaction is as follows: CuFeS2+ 5O2+2SiO2 2Cu2S,FeS + 2FeSiO3+ 4SO2 2) The matte and lag are tapped off separately. Sulphur dioxide is used to create sulphuric acid. 3) Air and sand is reacted with matte.

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