• 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 The aim of this experiment is to determine the relative atomic mass of lithium. I will be doing this in two ways: the first method will be to collect the gas evolved when lithium is reacts with distilled water and calculate the relative atomic mass this way. The second method will be to calculate the relative atomic mass through titration. ANALYSIS Method 1 The volume of hydrogen collected from the reaction of 0.05 grams of lithium with 100.0cm3 was 90.3cm3. The chemical equation for this is: 2Li(s) + 2H2O(l) � 2LiOH(aq) + H2(g) The mole ratio of lithium to hydrogen is: 2:1 First I must calculate the number of moles of hydrogen, I can do this using the following equation: Number of moles of gas = volume 24 000 (one mole of any gas has a volume of 24 dm3 RTP) Therefore, the number of moles of hydrogen = 90.3cm3 24 000cm3 = 3.7625 x 10-3 moles According to the mole ratio, there are twice as many moles of lithium than hydrogen present. Therefore the number of moles of lithium equals the number of moles of hydrogen multiplied by two: Number of moles of lithium = 3.7625 x 10-3 moles x 2 = 7.525 x 10-3 Now that I have calculated the number of moles of lithium, I can now calculate the its relative atomic mass using the following equation: Number of moles = mass (g) ...read more.

Middle

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 = 7.468 x 10-3 moles Using this result I can now calculate the relative atomic mass of lithium. For this I must go back to the chemical equation from Method 1: 2Li(s) + H2O(l) � 2LiOH(aq) + H2(g) The mole ratio of lithium to lithium hydroxide is: 2:2 This means that the number of moles of lithium is equal to the number of moles of lithium hydroxide. So the number of moles of lithium is also 7.468 x 10-3. Using the same equation as before: Relative atomic mass = mass (g) Number of moles The mass of lithium used in 100cm3 is 0.05g. So, the relative atomic mass of lithium = 0.05g 7.468 x 10-3 = 6.6952 Conclusion: The relative atomic mass of lithium is 6.70 EVALUATION I feel that overall the results of my experiment were fairly accurate. I can test the accuracy by calculating the percentage of accuracy for each experiment. This is done by dividing the calculated result of the relative atomic mass by the actual atomic mass (6.9) ...read more.

Conclusion

To improve the accuracy of my results the room temperature should be checked. Because, if the temperature is not standard, one mole of the hydrogen would not take up 24dm3, which would make the calculations inaccurate. According to the percentage accuracies, Method 1 is more accurate than Method 2. There are a number of factors that could be responsible for the lower percentage accuracy of Method 2; error in measurement of the lithium or distilled water, misjudgement of water mark in the measuring tube, oxidation of the lithium. Also, I repeated Method 2 three times, allowing me to disreguard anomalies and take an average. But, because I had to use the solution from Method 1 in Method 2, I did not repeat it. Any error in measuring the lithium or distilled water or fault in the lithium during Method 1 is likely to affect the results of Method 2. This means that ideally the solution from Method 1 would have an accuracy as near to 100% as possible. This could be done by repeating Method 1, at least three times, each time keeping the solution. Calculate the atomic mass from each volume of gas produced, the solution with the highest accuracy can then be used for Method 2. The calculation of the atomic mass of lithium from Method 2 would then be expected to produce the highest accuracy and most reliable result, according to the calculated percentage accuracies. ...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.

    The magnesium may have been impure and the oxide layer may have taken some of the mass. I also found that when I was trying to weigh my magnesium the balance was wandering and I had to pick a reading and this may not have been the most accurate.

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

    not included in the equation that I will use to work out the average. I had to repeat this experiment as I ran out of Potassium Manganate (VII) (aq) and when I tried to create a new solution of it I got my calculations incorrect and thus created a solution of 0.025 mols dm-3 for Potassium Manganate (VII)

  1. Determination of the relative atomic mass of magnesium by back titration

    empty touching the end on the surface of the solution to leave the correct amount of liquid in the pipette * Constantly swirl the conical flask whilst running in the solution from the burette. * Towards the end, add in drops to avoid adding an excess solution Repeating the experiment is also required.

  2. Determination of the relative atomic mass of lithium.

    This could have made the experiment inaccurate as air might have entered the measuring cylinder whilst the delivery tube was put up the measuring cylinder. This would have made the amount of hydrogen collected at the end of the experiment, inaccurate.

  1. Relative atomic Mass of Lithium

    To minimize this error I could have attempted to secure the stopper much quicker but this wouldn't have a great deal of effect. I could have also stuck a piece of Lithium on the bottom of the stopper so my results are not lowered.

  2. Determination of the Relative Atomic Mass of Calcium

    / 40.1 * 100 = 61.72 % (4sf) This high percentage error could have been caused because of the following reasons. As the calcium was added to the water I had to put the bung onto the flask as soon as possible.

  1. Determination of the relative atomic mass of lithium.

    relative atomic mass of lithium, I must first calculate the number of moles of hydrochloric acid used in the titration. This can be done by using the following equation: Concentration (M) = number of mole Volume (dm3) The volume of hydrochloric acid (dm3)

  2. Determining the relative atomic mass of lithium

    To overcome this problem the lithium should be added to the distilled water straight after it has been weighed. The oil the lithium is stored in is useful to stop this process but it will contaminate the lithium hydroxide as well as causing errors in measurement.

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