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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 To calculate the relative atomic mass of lithium I am going to use two methods. Method 1: I am going to measure the volume of hydrogen produced when a known mass of lithium reacts with H2O Method 2: I will now titrate the solution of lithium hydroxide produced. Method 1 results Mass of lithium (g) Volume of hydrogen produced (cm3) 1: 0.09 173 2: 0.09 176 Treatment of results Assuming that one mole of gas occupies 24000cm3 at room temperature and pressure. 2Li (s) + 2H2O (l) � 2LiOH (aq) H2 (g) I am going to calculate the number of moles of hydrogen, H2 that I had collected: No. ...read more.

Middle

titration is: LiOH + HCl � LiCl +H2O 1 mole of LiOH + 1 mole of HCl give us 1 mole of LiCl + 1 mole of H2O Now I'm going to calculate the no. of moles of HCl used in the titration: No. of moles = ( m(0.100) x v (40) ) � 1000 LiOH = 25 cm3 HCl = 0.100dm-3 = 4.0 � 1000 = 0.004 moles Now I will deduce the number of moles of LiOH used in the titration This is the same number of moles of HCl used in the titration = 0.004 To calculate the number of moles of LiOH, I will have to calculate how many moles present in 100cm3 of the solution from method 1: 100cm3 � 0.004 x 4 = 0.016 moles Now I am going to use the original mass of lithium and this result so calculate the relative atomic mass (RAM) ...read more.

Conclusion

I read wrong because I have checked my calculations so it must have been in either the titration of the amount of hydrogen I collected. Whilst doing the experiment I encountered other problems as well: : The oil that the lithium had soaked up could have affected the mass of the lithium/ 2: The gas inside the measuring cylinder could have been released and therefore giving me the wrong amount and the wrong calculations. 3: during the titration stage I could have been more accurate by having some help to take down the readings while I stop the titration instead of me doing both. 4: The pipette is not 100% accurate in its measurements so that also gave me an in accurate reading. ?? ?? ?? ?? Lee Palser centre no: 10534 Chemistry candidate no: 9110 ...read more.

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