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The determination of the relative atomic mass (Ar) of lithium.

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The determination of the relative atomic mass (Ar) of lithium Aim For this experiment we will determine the relative atomic mass of lithium using two methods: - * Collecting Hydrogen gas * Performing a titration Apparatus * Conical flask * Clamp stand * Plastic tub * 250cm3 measuring cylinder * Bung containing apparatus Diagram Method to collect Hydrogen * Set up apparatus as shown above * Clean all equipment with water and/or the solution that will be passed through it. * Take lithium out of the oil, with forceps. Wipe off all the oil * Weigh about 0.10 grams of lithium or write down actual weight correct to 2.d.p Actual Weight = 0.10g * Add the lithium to the conical flask containing 100cm3 of distilled water * Quickly replace the stopper and Shake the solution vigorously, ensures an even concentration/diffusion of lithium. ...read more.


Make sure the tap is closed when filling up the burette. * Set up the equipment as shown above * Fill the burette to the 0cm3 mark, with the bottom of the meniscus touching the top of the line. Allow the tap to run, releasing any air bubbles and will also clean the tip of the burette. * Once it has been filled with the acid (HCl). Measure out 10cm3 of the LiOH using a 10cm3 pipette and put it in to the conical flask * Add five drops of phenolphthalein indicator, once placed in the alkali it will turn a pink/magenta colour. * Begin by adding the acid 2 or 3cm3 at a time. The first titration should be a rough titration. ...read more.


Moles = 0.1/0.014 RAM = 7.142857143 Method 1 Accuracy = +-7.06/6.95 x 100 = 101.58% = 1.58% error Method 2 Accuracy = +-7.143/6.95 x 100 = 102.7769784 = 2.78% error Percentage error - Method 1 Measuring cylinder to measure 100cm3 water = 1/100 x 100%= 1% Measuring cylinder to measure 170cm3 H2 gas =1/170 x 100% = 0.59% =+-1.59% Method 2 Burette to measure 3 titres of 25 + 3 readings of about 14cm3 = 0.15/25 x 100% x 3 = 0.6 x 3 = 1.8% = 0.15/14 x 100% x 3 = 1.07 x 3 = 3.21% Pipette to measure 3 x 10cm3 = 0.05/10x100% = 0.5 x 3 = 1.5% =+-6.51% Evaluation * The overall results of my experiments are quite both accurate. Allowing for the inaccuracy of the equipment I was using, informs me that there was no human error in my results. The main sources of error were the equipment. ...read more.

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