Determination of the relative atomic mass of lithium.

Authors Avatar

Ali Omar                Page

Determination of the relative atomic mass of lithium

Introduction

As part of my AS chemistry unit, I was given an assessed practical to do.  The aim of this practical is to determine the relative atomic mass of lithium.  I will do this by the combination of two different methods.

Method 1: - By measuring the volume of hydrogen produced

Method 2: - By titrating the lithium hydroxide produced

Method 1

        I set up my experiment as previously shown with the correct size equipment in the correct places.  But before I began the actual practical part of the assessment, I had measure the amount of lithium I had.  Lithium is reactive with air, so I had to measure it in a specific way.  Firstly I sliced an appropriate part of lithium with a knife and put it in a Petri dish.  I then weighed the Petri dish occupied by the lithium which was 3.40g and to calculate the amount of lithium I had, I weighed the Petri dish on its own which turned out to be 3.31.  Finally I subtracted the weight of the Petri dish and lithium by the weight of the Petri dish which resulted in its own.

3.40g                    –    3.31g                               = 0.09g

(Petri + lithium)         (Weight of Petri dish)        (Weight of lithium)

        After I had the amount of lithium recorded, my next objective was to find out the amount of gas being given off as a consequence of the reaction of lithium and the distilled water.  I recorded the amount of water in the measuring cylinder so I would know where the start point of my reaction was.  As soon as I added the lithium to the distilled water the reaction would begin.  So I removed the stopper and added the lithium before reuniting the two back together, I did this as quickly as I could.  After about 1 minute the reaction had stopped and all the gas had been collected and I read it off as being 95cm3, but to record the actual volume of gas collected, I subtracted the two.

Join now!

250                          –    95                      = 155

(Volume collected)      (Volume at end)    (Actual volume of hydrogen produced)

        Now that I have both the weight of the lithium used and the amount of gas produced I can start on the technical calculations.

1 mole of gas occupies 24 000 cm3 at room temperature and pressure

2Li (s) + 2H20 (l) + 2LiOH (aq) + H2 (g)

  • Calculate the number of moles of hydrogen.

...

This is a preview of the whole essay