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Determination of the relative atomic mass of Lithium

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Introduction

A Level Chemistry Coursework Determination of the relative atomic mass of Lithium Implementing The following shows my observations in the form of tables. Results from Method One Mass of lithium 0.11 grams Volume of hydrogen produced 0.186 dm� Results from Method Two Reading ? 1 2 3 Burette Initial (cm3) 0 0 0 Reading Final (cm3) 38.8 38.6 38.6 Titre (cm3) 38.8 38.6 38.6 Analysis Method One Results Using the method given I found the volume of hydrogen produced to be as stated below: Mass of lithium 0.11 grams Volume of hydrogen produced 0.186 dm� We were told to assume that 1 mole of gas occupies 24000 cm� at room temperature and pressure. Using these readings the aim was to do the following: * Calculate the number of moles of hydrogen. * Deduce the number of moles of lithium, and * Using these values and the original mass of lithium, calculate the relative atomic mass of lithium. Therefore with the use of the formulas below, I can complete the aim. 1. Number of moles = mass � Ar. (n = m � Ar) 2. Number of moles = volume � 24 (used for gases). (n = V � 24) So I then used these formulas in the equation below to work out the number of moles of hydrogen and lithium. 2Li (s) + 2H2O (l) 2LiOH (aq) + H2 (g) ...read more.

Middle

* The plan had told me to weigh about 0.10 grams of lithium on a scale, it should have given a range, because how can I tell how much about really is? * The lithium had to be added to the distilled water as quickly as possible, without losing any hydrogen gas. It was all about your reaction time, if too slow then a lot more gas escapes than if you were fast. * There was already air bubbles present in the measuring cylinder prior to the reaction, this meant that the end volume of hydrogen gas would be a percentage higher than it would be if there were no bubbles. * The delivery tube may have had some substance inside (left over from a previous experiment), so the volume of gas entered cannot therefore, be to its maximum. * The fact that the methods did not tell me to rinse the apparatus, i.e. the burette with HCl meant that there may have been other substances still in adhesion with the glass. EVALUATION I am confident that my results are reliable because my final relative atomic mass came to be 7.11, which is 0.21 units away from the actual relative atomic mass of 6.9. I think that the overall accuracy of my experiments was fair because the Ar was 7.11, but I think that with improvement this could have been closer to 6.9. ...read more.

Conclusion

Smaller divisions in measuring apparatus means that if errors are made then they will be to a small percentage error. Using equipment like syringes can be easier to use and more accurate depending on the scale divisions. The uncertainties in the evidence in terms of their effect on the measure of confidence, which can be placed on my final conclusion drawn, are: * The percentage errors, below I have drawn a table which states the percentage errors along with the absolute values: Quantity Measured by ...... Value � absolute value Percentage error Volume (cm3) Measuring cylinder (186 � 0.1) cm3 � 0.053 % (2sf) Volume (cm3) Burette (38.6 � 0.1) cm3 � 0.26 % (2sf) Mass (g) Top pan balance (0.11 � 0.001) g � 0.91 % (2sf) So therefore the percentage error can affect the validity of the conclusions drawn and also determine how misreading the scale (random errors (reading errors)) or reading a measurement through the wrong point (systematic errors (parallax error)) can make an error no matter how small or big the error is. For equipment such as the pipette is hard to work out the percentage error, because all I knew was that the mark on the neck meant that the volume of liquid up to that point was 25cm3. As there are no other markings such as scales, I couldn't work out the absolute value. Secondary sources used: * The Usbourne science dictionary, * Chemistry 1 * Encarta By Priyesh Patel 12O ...read more.

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