• 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...


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.


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.


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

    0.05 X 100 = 0.5 % 10 Thermometer = 0.05 X 100 = 0.071428571 % 70 Procedural Errors The spinach may not have been cut into equally sized pieces and the surface area to volume ratio was not equal which could have resulted in some pieces of the leave not being able to release all the Iron (II)

  1. The Determination of an Equilibrium Constant.

    78.75 / 60 = 1.3125 mole * Mass of C2H5OH = 0.79 * 75 = 59.25 g Molecular mass of C2H5OH = 24 + 5 + 16 + 1 = 46 g/mol Number of moles of C2H5OH = 59.25 / 46 = 1.288 mole * Mass of CH3COOC2H5 = 0.92

  2. copper practical

    The copper oxide heated in a test tube with carbon, which produces impure copper. The word equation to describe this is as follows: 2CuO + 2Cu 2Cu + CO2 Purifying the copper To further increase the level of purity of the copper, electrolysis is used.

  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. 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

    I also had to wear plastic gloves when handling it, and use tweezers to pick it up, to stop it reacting with the moisture on my skin.

  2. Determining the atomic mass of lithium from method one. After setting up my apparatus, ...

    0.12 = 0.7 = 0.84% Evaluation * The experiment went fairly well because my results were not very far from each other. This small difference could have been brought about by a number of factors. This include: * When placing the Li in the conical flask, there is a risk

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