• 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 Results To determine the relative atomic mass of lithium I cut a piece of lithium, weighing 0.82g and reacted it with 100cm3 of distilled water. This reaction gives off hydrogen gas, which I collected in a 250cm3 measuring cylinder. I collected 129cm3 of hydrogen gas. To calculate the number of moles of hydrogen that I collected, assuming that 1 mole of gas occupies 24000cm3 at room temperature and pressure, I must divide the amount of gas I collected by 24000. 129 � 24000 = 0.005375 moles I can use this information to calculate the number of moles of lithium that reacted. I will multiply the number of moles of hydrogen I collected by 2 because of the 2:1 ratio of the lithium used to hydrogen given off. 0.005375 � 2 = 0.01075 moles Now, by dividing the amount of lithium I used in the reaction, which was 0.082g, by the number of moles of Lithium that reacted, 0.01075 moles, I can work out an estimate for the relative atomic mass of lithium. 0.082 � 0.01075 = 7.628 This is close to the actual relative atomic mass of Lithium but not as accurate as I had planned to get. After doing this I then titrated 25.0cm3 of the lithium hydroxide made by the reaction, with a 0.100 moldm-3 solution of hydrochloric acid. ...read more.


When Lithium reacts with water it makes Lithium Hydroxide, which is an alkali. Although it was not a very strong alkali, because of the low concentration, I was careful not to get it on my hands. To neutralise the Lithium Hydroxide in the titration, I used 0.100 moldm-3 Hydrochloric acid. This is also not very strong, but can cause irritation to the skin and eyes, so I was again, careful not to get it on my hands. In order to stop the chemicals going in my eyes or on my clothes, I wore an apron and goggles through the whole experiment. The last chemical I used was the Phenolphthalein indicator, which I used to see when the chemicals had neutralised. This is not a high hazard, but can stain my clothes so I wore the apron and made sure not to get it on my hands. Evaluation Overall I am quite pleased with my experiments, but I would have liked them to be more accurate. Before I started the experiments I rinsed out all my apparatus with distilled water to get rid of any substances that could have been in them. These are the factors which I think affected the accuracy of my experiments. Firstly, when I was doing method 1, there were a few areas where I think I lost some accuracy. ...read more.


I am sure that for both methods I measured out the chemicals with a high degree of accuracy, making sure that the bottom of the meniscus was on the line at eye level when measuring, It was only when collecting results from method 1 that I made mistakes. I think the most successful part of my experiment was my titration where I managed to get very accurate results. To minimise error, I should have taken much more care in method 1, when I was collecting the gas. This is where I got the inaccuracy of my results. I could have put the top on the conical flask faster to reduce the amount of gas that escaped. Also, to increase the reliability of my results I could have asked someone to check my measurements to make sure I was reading them correctly. A way of minimising errors caused by escaping gas could be to use a gas syringe instead of collecting the gas in a measuring cylinder, because there would be less error from bubbles in the measuring cylinder and the results would be much more accurate. If I were to repeat the experiment I would take much more care when carrying out the experiments and collecting the results, but overall the experiment went as well as I expected. ?? ?? ?? ?? Rob Ayres ...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. Marked by a teacher

    Enthalpy of Neutralisation.

    3 star(s)

    0 22 5 24 10 25 15 27 20 28 25 28 30 27 35 27 40 26 45 25 50 25 CALCULATION. ?H = m x c x ?T = (26.5 + 25) x 4.2 x (30 - 21)

  2. Determination of the relative atomic mass of lithium.

    the lithium before using it, although care must be taken to make sure the surface of the lithium does not oxidise. Another factor that may have caused the anomaly is error in measurement. This could be a misjudgement through human error for example; an error in misjudging the watermark in a burette or measuring cylinder could cause slight inaccuracy.

  1. Determination of the relative atomic mass of lithium.

    When we start the implementing we are supposed to fill the cylinder with water and somehow get it in the water tub WITHOUT any air bubbles getting in. the method we used was actually very good. After filling the cylinder with water we wrapped cling film all around the end of the cylinder.

  2. The Determination of an Equilibrium Constant.

    = 0.35 mole Number of moles of CH3COOH produced = 0.35 mole According to the equation CH3COOH + C2H5OH CH3COOC2H5 + H2O The amount of C2H5OH produced equals the amount of CH3COOH produced ?Number of moles of C2H5OH at equilibrium = 1.288 + 0.35 = 1.638 mole The amount of CH3COOC2H5 (H2O)

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

    present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5. 0.005216666 mol dm-3 X 5 = 0.026083333 mol dm-3 Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II)

  2. Determine the relative atomic mass of lithium.

    Add the lithium to the flask and place the bung with the tube back onto the flask. 8. Collect the gas and write down the final volume of hydrogen collected. In method 1 I reacted 100 cm� of water to 0.08g of lithium.

  1. Determination Of The Atomic mass of Lithium.

    o Put your thumb over one end of the tube and blow in the other to make sure there are no holes in the tube where hydrogen could leak out. o Remove as much oil as possible before adding the lithium to the water.

  2. Determination of the relative atomic mass of lithium.

    The equation for this is concentration of lithium (0.10g) multiplied by volume of hydrochloric acid divided by 1000. Number of moles of HCl = concentration x volume/1000 0.10 x 39.27/1000 = 0.0393 The answer to this calculation is 0.00393 The ratio of this experiment is going to be: Ratio HCl

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