Investigation to determine the relative atomic mass of lithium

                                                         

Method

I got  25cm3 of lithium hydroxide solution and put it into a clean 250cm3 conical flask. I filled the burette with HCl making sure that the tap is closed. I then rinsed out the burette by opening the tap and allowing the HCl to run, this will release any air bubbles and will also clean the tip of the burette. I added 5 drops of phenolphthalein indicator, which made the solution, turn pink.  I then titrated the solution with 0.1moldm3 HCl. I then started to run the HCl and when it came near the end point (where it turned colourless) I started to release the HCl down drop by drop for more accurate results.  Once this is all done I repeated the results twice more (for accuracy) and remembering to swirl it after each drop

I will repeat this experiment until I am confident that my results prove to be correct. I

Method 2 Results

For maximum titration accuracy a rough titration was done first. This involves finding the approximate neutralisation point of the solution by letting the HCl fall freely, when the solution (containing indicator) becomes clear the approximate volume of HCl needed is recorded. Next time, this allows the HCl to be let out within 15ml of the predicted amount, and then the titration is carried out with accuracy.

Fair test and accuracy

• I will try to repeat my experiment within my given time to obtain an average
• I will use the same equipment each time
• Values should also be read at eye level to ensure that the correct value is being seen, viewing from other angles can cause the incorrect value to be seen.
• When filling the burette always remove the funnel because a drop of reagent could fall into the burette altering the results because a burette can measure accurately enough for 1 drop to make a difference.

ANALYSIS

Calculating the relative atomic mass of lithium from method 1

• No. of moles of hydrogen = volume  

                                                        molar volume

  = 176

    24000

=0.0073

So here we can find the no of moles of hydrogen molecule in 176cm3 of hydrogen gas which was collected for the experiment. We divided it by the molar volume 24000cm3 (or 24dm3)

• No of moles of Li that reacted

      2Li + 2H20                                      2LiOH + H2

This stiochemistry mole ratio is 2 to 1 so for every mole of hydrogen we have two moles of lithium that reacted.

So 0.0073 x 2

= 0.014

• R.A.M of Lithium

The equation we are all familiar with is:   no.moles = mass

                                                                                      r.a.m

So if we rearrange this formula we get;

R.A.M =   Mass                                                                               

              No. moles

= 0.10

   0.014

= 7.1g

As we know that the mass of the piece of lithium weighed was 0.10g and we then divided this by its number of moles to get its relative atomic mass.

So our calculated relative atomic mass for method 1 was 7.1, we know that this is not too far off as the actual relative atomic mass of lithium is 6.9g

Calculating the relative atomic mass of lithium from method 2

• Average Titre = 34.4 +34.1

                                   =68.5

                                       2

                                   =34.25cm3

As we had two results, I added these and then divided by two to give the mean result. I did not include my trial titre value as it was an anomaly so I had ignored this during the calculation preocedure.

No of moles of HCl used

We then needed to work out the no of moles of HCl, this is not difficult knowing that we have the volume and concentration values.

• Moles= volume x concentration

                                 1000

= 34.25 x 0.1

     1000

=0.003425

• No of moles of LiOH

LiOH  +  HCL                                       LiCl  +  H20

The stiochemistry mole ratio is 1:1 so for every one mole of HCl we have one mole of LiOH. So the number of moles of LiOH are

=0.003425 moles

• No of moles of LiOH in 100cm of the solution

We had used 25cm so 100

                                     25

                                   = 4

Since we are trying to find the no. of moles used in 100cm3 we will multiply our value by 4

= 0.003425 x 4

= 0.0137 moles

= 0.0137

• So the R.A.M of Lithium

= mass

no.moles

=0.10

 0.0137

= 7.29927003

=7.3(1.d.p) 

We can see that both results were higher than the actual relative atomic mass of lithium (which was 6.9) although we were not too far off as you can see. for method 1 we obtained 7.1 and for method 2 we obtained 7.3cm. Also in method 2 I had my two titre values were very similar and 0.3 off, although one titre value (which was the trial run) was an anomoly so I did not place it in my average titre.

I calculated the percentage accuracy to analyse just how accurate I was;

This is done by dividing the calculated result of the relative atomic mass by the actual atomic mass (6.9) of lithium, then multiplying this by 100:

Method 1 Accuracy = 7.1/6.9 x 100 = 102.8985507% = 2.89% error

Method 2 Accuracy = 7.3/6.9 x 100 = 105.7971014 = 5.79% error

Evaluation

When comparing my resulst to the actual relative atomic mass of lithium I have shown that my reuslts are actually higher than that of the actual relative atomic mass of lithium.

I believe that my results were quite accurate in experiment 1 I was 0.3 off my answer whereas in experiment 2 I was 0.4 off the real answer.

The pipette, volumetric flask and burette have an accuracy of approximately 0.05cm3; this is far more accurate when compared to that of the beaker, which has an accuracy of 0.05cm3. So any error through equipment is minimal, yet it still has to be taken into consideration.

My expeirimentwas fairly accurate although not 100%. Their may have been many reasons for this such as:

• The lithium was at first kept in oil to prevent it from forming its oxide layer. Although once it is taken out of the oil it will slowly react with air to form lithium oxide so we had to work quite quickly although it had an effect of increasing the weight of the lithium.
• When using the lithium we have to take off all the oil. If it isn’t it will not react to its full potential so will not produce as much hydrogen.

• Their may have been an error due to human error of misjudgment of the water mark in the beurettte or the measuring cylinder.

• Also the decision of the end-point of the titration and the reaction time between realising the point and stopping the acid.

• Also it was assumed that the experiment was being carried out under standard temperature and preassure. Although we did not check this. This would have effected our resulst during method 1 becasause if the temperature was not standard then one mole of hydrogen would not take up 24000cm3 which would have added to further innacuracy.

• Hydrogen is almost insoluble in water. However if some hydrogen dissolves then this can affect my results.

I think that the significant errors where;

• End-point of the titration- in titration it is impossible for one to judge the solution during the end-point of the indicator to be the exact same colour all three times with just the naked eye.

The above is always quite difficult to controll as it is due to human error.

• Lithium was exposed to the air for a short period of time causing the surface atoms of the lithiumto react with the surrounding air.
• The oil was also part of the measurement of the weight however I dried most of it off but this again would have caused some of the surface to come in contact with the air.

I feel that their were quite a bit of improvement which could have been made.

• We could have used a gas syringe instead of a measuring cylinder to collect the hydrogen as a pocket of air can be left in a measuring cylinder when filling it with water , this would alter the readng
• More HCl than necessary may have been added, this was quiet difficult to overcome  as phenolethein will still go clear if excess HCL is added. It was suggested that a more sensitive indicator be used but the other indicators available (methyl orange and bromothhynol blue) were not suitable for the experiment as they are for strong aid and weak base.
• We can see that method 1 is more accurate than method2 ,any error in method 1 is likely to effect the error in method 2. Since I had used the Lithium Hydroxide from lithium 1 all the innacuracys from experiment 1 would contribute to the overall innacuracy in experiment 2. To overcome this we could have used accurate manufactured lithium hydroxide solution.
• In titration it is impossible for one to judge the solution during the end point of the indicator to be the exact same colour all three times with just the naked eye. This is another limitation in the procedure. The experiment can be improved by using a computer and a real time video camera. If the computer could hold the colour of the initial titration then one could do the second titration while looking at the computer screen, so both titrations would be stopped at exactly the same colour. This would totally reduce the human error when it comes to the colour change of the indicator, making the results more fair, accurate and reliable
• Unreliable results are caused by random errors, repettition of the experiment can cause these errors to spread evenly around the correct value therefore giving a true impression of the answer. So if I could do the experiment again I would definitely take more results allowing me to gain a more precise and reliable set of results.

I feel that the experiment could have been extended if we had also looked at other elements to see if their relative atomic mass could also have been worked out using the same two methods which I had used.

Bibliography

•  find out general information on lithium and its properties
• Chemsitry 1
•