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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

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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