• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determination of the relative atomic mass of lithium

Extracts from this document...

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. Marked by a teacher

    Enthalpy of Neutralisation.

    3 star(s)

    1000 ?H = 1646.4 J 0.025 = 65856mol -1 To convert to Joules = 65856 J 1000 = -65.86 Kj mol -1 Note: I used 0.025 moles because 0.05 moles was not the value I got for the neutralisation point and I did not use 50cc to neutralise it.

  2. Determination of the relative atomic mass of lithium.

    I will use a burette because it is very accurate as it is measured to the nearest 1.0 cm3 compared to a measuring cylinder, which is measured to the nearest 2cm3. 3. I will then weigh a piece of lithium making sure that there is as little petrolatum in which it is stored as possible on it.

  1. The Determination of an Equilibrium Constant.

    react with I mole of OH- ?There are 1.5 * 10-3 * 250 = 0.375 mole H+ in the 250 cm3 mixture 3 Number of moles of HCl in the mixture = 1 /1000 * 25 = 0.025 mole Number of moles of CH3COOH at equilibrium = 0.375 - 0.025

  2. Determination of the relative atomic mass of lithium.

    Analysis (method 2):) * (1) Calculate the number of moles of HCL used in the titration. Conclusion so far: Unlike the first experiment I did get to perform the implementation more than once (method 2). I performed the experiment three times which means I have three sets of results.

  1. Determine the relative atomic mass of lithium.

    Method 1. Firstly the apparatus must be set up according to the diagram 2. Place the pipette into the pipette filler and use it to measure 25cm� of the Lithium hydroxide and pour it all into the beaker. And place the beaker under the burette.

  2. To determine the relative atomic mass of Lithium

    Burette Final 36.59 36.55 36.51 Readings/cm3 Initial 0.00 0.00 0.00 Volume used (titre)/cm3 36.59 36.55 36.51 Mean titre/cm3 36.55 The mean titre was obtained by added the volume used to titre and divide it by the number of time the titration was done.

  1. to determine the relative atomic mass of lithium. We will be doing this via ...

    We used 10.0 cm3 of the solution from Method 1. Since there are 10 portions of 10.0 cm3 in 100cm3, to find the number of moles of LiOH present in 100cm3, we need to multiply the number of moles of LiOH in 10.0 cm3 by 10: Number of moles of

  2. To find the relative atomic mass of a sample of Lithium.

    By using a Burette and an Acid Base indicator, an exact amount of reactant 1 can be added to a known amount of reactant 2 until a colour change of the indicator indicates the solution has become neutral. Weight of Li=0.10grms Volume of H2 gas produced=203cm3 Analysis Part 2 Titer

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work