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Determination of the relative atomic mass of lithium.

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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.

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