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Determination of the relative atomic mass of Lithium

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Determination of the relative atomic mass of Lithium In this investigation I am going to determine the relative atomic mass of lithium by two different methods. Method 1: I will measure the volume of hydrogen produced when a known mass of lithium reacts with water. Method 2: I will titrate the solution of lithium hydroxide produced. After I have done the two methods and established my results. I will compare the two methods and conclude on which method was more appropriate and reliable. Apparatus * Safety goggles * Protective gloves * 2 � 250 cm3 conical flasks * Delivery tube * 250 cm3 measuring cylinder * Clamp stand * 2 bosses * 2 clamps * Trough * 100cm3 measuring cylinders, burettes with clamps and stands or 25.00 cm3 pipettes and fillers * Pre-weighed lithium pieces of between 0.08g and 0.13g and stored in oil * 100cm3 distilled water * Forceps * Filter paper * 25.00cm3 pipette * Pipette filler * 50.00cm3 burette * Small funnel for filling burette * White towel * 50cm3 0.10moldm3 hydrochloric acid * phenolphthalein indicator * Balance accurate to 2 decimal places Method 1: Procedure 1. I am going to set up the apparatus as shown below. The 250 cm3 conical flask must contain exactly 100 cm3 of distilled water. 2. ...read more.


30.9 31.9 30.8 Initial reading (cm3) 0.00 0.00 0.00 Volume used (titre) (cm3) 30.9 31.9 30.8 I worked out the mean average like this: Mean = 30.9 + 31.9 + 30.8 � 3 = 31.2 cm3 Summary On average, 25.0 cm3 of LiOH (aq) required 31.2 cm3 of 0.100 dm3 HCI. Treatment of results The equation for the neutralisation reaction in my titration is; LiOH (aq) + HCL (aq) LiCI (aq) + H2O (I) * Calculate the number of moles of HCI used in the titration. Number of moles = morality � volume (cm3) 1000 = 0.1 � 31.2 = 0.00312 mol 1000 * Deduce the number of moles of LiOH used in the titration. With reference to the calculation above � 0.00312 mol of LiOH reacted with HCI. According to the above equation the ratio between LiOH and HCI is 1:1 Therefore, the number of moles of LiOH that reacted with HCI = 0.00312 mol * Calculate the number of moles of LiOH present in 100 cm3 of the solution from method 1. The 25 cm3 solution of LiOH was 0.00312 mol Therefore in 100 cm3 � 0.00312 � 4 If 25 cm3 � 0.00312 mol then 100 cm3 � 0.01248 mol * Use this result and the original mass of lithium to calculate the relative atomic mass of lithium. ...read more.


* If bigger pieces of lithium was used it would minimise how much hydrogen was released before the bung was replaced. * By using the sink it limited the amount of inaccuracies. The cylinder was filled in the sink to minimise air bubbles then returned to the trough. After this the results were consistent. There were no air bubbles. The small trough increased inaccuracies. * The use of bigger pieces of lithium will decrease the errors in measurement. * By using a gas syringe it will overcome the limitation of the air bubbles. The result would be more consistent and reliable than the results of the measuring cylinder. There will be no need for the trough, which was the main cause of errors. * A balance with a higher certainty with readings up to 3 decimal places would make the results more reliable and accurate. Main Sources of Error * Mass is a major source of error, as it happens in both experiments. Errors are caused by oil and oxidation of the lithium. Comparing the two methods, I can state that the titration is more accurate because the readings repeated and an average titre is used of the consistent readings. 1 Ransford Asare Chemistry ...read more.

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