Determination of the relative atomic mass of lithium.

Moles = mass

            Ar

Ar =       0.11

      0.01716666666

Ar = 6.40776699g

Ar = 6.4g

Ionic Equation

Li+(aq) + OH-(aq) + H+(aq) + Cl-(aq)                     Li+(aq) + Cl-(aq) + H2O

By cancelling out the spectator ions (Li+(aq) and Cl-(aq)), we get the ionic equation for the neutralisation reaction.

H+(aq) + OH-(aq)             H2O

This ionic equation applies to all neutralisation reactions.

METHOD 2

Moles of HCL = volume x concentration

                   1000

                  = 41.05 x 0.1

                        1000

                  = 4.105 x 10-3

Ratio HCl : LiOH

         1  :    1

∴Moles of LiOH = 4.535 x 10-3 (in 25cm3)

Moles of LiOH (in 100cm3) = 4 x 4.105 x 10-3

                                   = 0.01642cm3 

Since one mole of Lithium produces one mole of LiOH the moles of Lithium = 0.01814cm3

0.01814 moles of Lithium weigh 0.11g

So: 1 mole of Lithium weighs = 0.11 x 1

                                          0.01642

                                      = 6.69g

∴Ar of Lithium = 6.69g

From my two experiments the average value for the Ar of Lithium is 6.69+6.4 = 6.55g

     2

Evaluation

My experiments are quite accurate. This is because the agreed Ar for Li is 6.9

% difference

(6.9 – 6.55) x 100 = 5 %

        6.9

My experiment was not 100% accurate due to the following errors:

Points to note:

• The measuring cylinder is not very accurate because the error here can be ±2cm3. A gas syringe is more accurate to measure the volume of hydrogen gas; here the error could be ±1cm3.
• The gas that had been collected is not at atmospheric pressure (1 atm). However 1 mole of gas can only occupy 24000cm3 at r.t.p.

Errors in Method 1 (collecting gas)

• A 25cm3 measuring cylinder was used to measure the volume of gas. However a measuring cylinder is only accurate to

% error = 2 x 100   = 0.97 %

              206

• A 100cm3 measuring cylinder was used to measure the volume of distilled water. Here the error is ±1cm3.

% error = 1 x 100   = 1.0 %

              100

• An electric balance was used to weigh the mass of Lithium. Here the error is ±0.01g.

% error = 0.01 x 100   = 9.0 % (biggest error)

                0.11

• Some H2 gas my have escaped while the bung was being replaced, after the Lithium was added.

• The pressure of the H2(g) in the measuring cylinder is not at atmospheric pressure. The gas collected is lower that A.P.   Note: the figure of 24000cm3 if at r.t.p.

• The temp of the gas may not be 298K (room temperature).

• The Lithium used had a layer of oxide on its surface.

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

• % overall error for method 1 = 0.97 + 1.0 + 9 = 10.97 %

Errors of Method 2 (Titration)

• A 25cm3 pipette was use to transfer the LiOH solution. Here the error is ±0.1cm3.

% error = 0.1 x 100 = 0.4 %

                25

• The accuracy of a burette is ±0.05cm3. My average titre was 41.05cm3.

% error = 0.05 x 100 = 0.1 %

                41.05

Overall % error for method 2 = 0.1 + 0.4 = 0.5 %

Overall % error = 0.5 + 10.97 = 11.47%

Improvements for Method 1

• To prevent any H2 gas escaping the apparatus below could be used:

        

To start the reaction, the conical flask would need to be tilted so that the water goes inside the test tube.

• A gas syringe is more accurate in measuring the amount of gas. The accuracy is ±1cm3.

• Also a gas syringe should be used to eliminate any possibility of H2 gas dissolving in the water.

The apparatus below could be used for above improvements:

• To make sure the pressure of the gas is at 1 atm, lower the measuring cylinder until water level, inside and outside are the same.

     

Improvements for Method 2

• If I had more time I could’ve done another titration so that the average would be more accurate.