A Qualitative Analysis Experiment

Friday, July 15, 2011

Experiment 17

Here is a list of compulsory experiment report in Form 6 :





Experiment 17


Topic :         Technique - synthesis.



Purpose :       To determine the percentage of aluminium in a sample X by means of the

preparation of a complex compound of aluminium with

8 – hydroxyquinoline.



Materials :    KA 1  is a solution of 8 – hydroxyquinoline in ethanoic acid.

                        KA 2  is 2.0 mol dm־³ aqueous ammonium ethanoate.

                        KA 3  is sample X.

                        KA 4  is 2.0 mol dm־³ hydrochloric acid.

                       

Apparatus :  Electric balance ± 0.01g (common use).

                        Oven (common use).

                        Magnetic stirrer.                                 

One 250cm³ beaker.

One 50cm³ measuring cylinder.

One Bunsen burner.                              

One tripod stands and wire gauze.

One retort stand and clamp.

One thermometer 0 ºC to 100 ºC.

One crucible and lid.

One filter tuner and filter paper.

One wash bottle filled with distilled water.







Introduction : Aluminium in a certain sample can be determined quantitatively by

                           means of the process of formation of complex compounds, for example,

                           an alum complex is produced by the reaction of an alum with

                           8 – hydroxyquinoline.





 Procedure :   (a) Weight out accurately between 0.25g and 0.30g of  KA 3 . The

                              readings are recorded in the table below.



Mass of container +  KA 3 / g
104.69
Mass of empty container / g
104.43
Mass of  KA 3 / g
0.26



                              All of the  KA 3  are placed into a 250cm³ beaker, and 150cm³ of

                              distilled water are added followed by one drop of  KA 4 . The mixture

                              is warmed to a temperature between 60 ºC and 70 ºC. Then 25cm³ of

                              solution  KA 1  are added, slowly followed by 50cm³ of solution

                              KA 2 . A precipitate will be formed. The beaker is removed. This

                              solution is stirred for half an hour by using a magnetic stirrer.



                              An empty crucible is weighed and the readings are recorded in the

                              table below. The solution above is filtered and the precipitate is

                              washed with cold distilled water. The precipitate is transferred into an

                              empty crucible and the precipitate is dried in the oven at about 120 ºC.

                              The crucible is cooled and weighed together with its contents.



                       

Mass of crucible + precipitate / g
30.43
Mass of empty crucible / g
30.28
Mass of precipitate / g
0.15









Question :    



(c)  Explain.

            (i) Why the mixture of solutions  KA 1  and  KA 2  needs to be stirred for half an

                  hour using the magnetic stirrer?

                        To make the solution  KA 1  and  KA 2  to be uniform to get better

                        reaction and transmit heat uniformly.



            (ii) Why the complex precipitate is washed with cold water?

To separate out the fluorine particle with the pure precipitate.





(d)  How would you ensure whether the complex precipitate obtained is completely dried?           First, randomly take the mass of precipitate in different time after 1 hour and then

            until get the constant mass to ensure the complex precipitate obtained is

            completely dried.



(e)  Calculate the mass of 1 mole of complex compound with the formula Al(C9H6NO)3.

            Mass = 27 + 3[9(12) + 6(1) + 14 + 16]

                     = 27 + 432

                     = 459 g





(f)  Calculate the percentage of aluminium in the complex compound.

                                                         27_

Percentage of aliminium =  459   x  100 %

                                         =  5.882 %





(g) Calculate the percentage of aluminium in sample X.

            459 g of Al(C9H6NO)3 contains 27 g of Al.

            0.15 g of sample X contains Y g of Al.



                                27 x  0.15

            Y =            459

            Y =  0.00882 g



                                                                                                    0.00882

            Percentage of aluminium in sample X =     0.26      x  100 %

                                                                          =  3.392 %









                       

Conclusion :   The percentage of aluminium in sample X is 3.392 %.



Comment :     (i) Temperature must be constant between 60 ºC and 70 ºC when

                             continue the reaction to prevent error occurred.

                        (ii) We should carry out technique – synthesis in right way, do not need

                              do not put too many salts into the distilled water.


2 comments:

muz said...

very2 useful..tq very much

Unknown said...
This comment has been removed by the author.