Chem 1101 PDF

Title Chem 1101
Author Ronit Gulati
Course Chemistry for Engineering Students
Institution Carleton University
Pages 5
File Size 126.1 KB
File Type PDF
Total Downloads 75
Total Views 148

Summary

Gravimetric analysis of a Chloride Salt ...


Description

Gravimetric analysis of a Chloride Salt CHEM 1101

Partner : Mostafa Deiab Group : CHEM 1101 Friday AM, Group B

Date performed : 8 February 2019 Date submitted : 1 March 2019

Name : Ronit Gulati Purpose

Using gravimetric analysis and by using certain techniques to determine the amount on chloride ion in an undefined salt experimentally. Theory Since many substances are highly soluble in water and or other solvents can be precipitated gaugeable from a solution by doing a precipitation reaction and to completion , we can use the basis of analytical procedures. We assume the reaction undergoes completion due to the fact that silver chloride is highly insoluble. AgCl(s) → Ag+(aq) + Cl-(aq) Since the products and the reactants had a one to one ratio , the molar amount of AgCl was used to calculate the amount of chloride ion in the solution Ksp = [Ag+(aq)] x [Cl-(aq)] = 1.6 x 10-10 * The amount of precipitate that was washed was 0.12 g and 0.1017 g respectively Ksp ( Solubility product constant ) is the equilibrium between a solid and its respective ions seen in a solution. THe value of this specific constant can identify the degree of which the compound can dissociate in water ( H 2O ). Coming into conclusion the ksp of AgCl is low so we can see it as insoluble. To capture out unwanted ions in out precipitate we can use NO3 , to stop the co - precipitation of other ions in the solution. A moderate excess of silver ion is crucial towards the end - point in order to minimize the solubility of the silver chloride. In that case a large excess must be sidestepped due to the fact that this would lead to a high percentage of co - precipitation of other ions. We also heat the solution to increase the rate of the reaction of the precipitation. With the presence of light dry silver chloride decomposes into silver and chlorine respectively AgCl(s) → Ag+(s) + ½Cl-(g) 3Cl2 + 3H2O) + 5Ag+(s) + ClO-3 + 6H+(aq) Photo decomposition in the reaction took place and because of this in the solution analytical results are higher than expected. This is due to the fact that AgCl(s) decomposing and leaving behind Ag(s) and Cl(g). The Cl(g) afterwards continues to go on to create more precipitate with the excess AgNO 3(aq) in the solution. This then results in more precipitation being formed , which then will eventually skew analytical results. Procedure To begin with , we obtained an unknown salt with out unknown recorded sample number being 355. Next assuming the analytical balance is as precise as possible 0.12 g and 0.1017 g of salt was measured out and was put into 2 different 250 ml beakers , making sure we are able to differentiate the 2 beakers. 100 ml of distilled water was added next and 1 ml ( 6 M ) nitric acid was added to both beakers containing the salt. The solutions were then stirred. Next , approximately 23.6 ml and 20.6 ml of ( 0.1 M ) AgNO 3 were then slowly added to both of the beakers and simultaneously stirred. The beakers containing the solutions were heated. To check for completeness of the preposition , a few small drops of silver nitrate was added to both solutions , and no precipitate formed , showing that its complete. Finally the vacuum filtration apparatus was set up , and two sintered glass crucibles were weighed on an analytical balance. Using the sintered glass crucible , the clear solutions were filtered out of the beakers one

by one. Once the solution was completely filtered out , and the precipitate only remained , 10 ml of ( 0.01 M ) nitric acid was added to precipitate. Next , 5 ml of ( 0.01 M ) nitric acid was added to the precipitate. A wash bottle was used to get all the remaining precipitate out of the breaker. To detect any remaining silver present , washings were collected into a test tube and dilute HCl was added. Since no silver was detected , three 5 ml portions of acetone were use to filter the precipitate further. Then the first beaker was placed into the oven at a temperature of 137 degrees celsius for 30 minutes and the same procedure was done for the second beaker ( partner’s beaker ). Both trials / beakers are put into and taken out at the exact same time and both were measured on an analytical scale. Observations - The unknown salt #355 was white and powdery , crystalline and fine - When AgNO3 was added to the solution , a precipitate was seen to form almost immediately and the solution appears milky. - While the 250 ml beaker was being heated and stirred , the solution seemed to clear up up over time - TO check if all the precipitate was formed small amounts of AgNO3(aq) was added , after some time it was clear and concise that there was no more precipitation being formed in the solution - The precipitate seemed to fall to the bottom of the 250 ml beaker , which then slightly looked purple Data Trial

Trial # 1

Sample mass ( g )

Precipitate mass (g)

Oven Temperatures ( degrees celsius )

Drying / Cooling time ( minutes )

0.12 土 0.0001

0.0543 土 0.0001

137 土 0.02

15 min

0.1017 土 0.0001

0.1833 土 0.0001

137 土 0.03

30 min

Trial # 2 ( partners )

Calculations Volume of AgNO3 ( mass of the sample ) x ( assumed percent Cl ) ➗ ( M.W Cl ) ➗ ( molarity of AgNO3 ) * (1000 ml ) + 5 ml Trial 1 ( 0.12 g ) * ( 0.55 ) ➗ ( 35.5 g/mol ) ➗ ( 0.1 mol / L ) * ( 1000 ml / L ) + 5 = 23.5915 ml 土 0.2 ml = 23.49 ml 土 0.2 ml Trial 2 ( 0.1017 g ) * ( 0.55 ) ➗ ( 35.5 g/mol ) ➗ ( 0.1 mol / L ) * ( 1000 ml / L ) + 5 = 20.7536 ml 土 0.2 ml = 20.75 ml 土 0.2

% of Chlorine in Sample Moles AgCl = ( mass precipitate ) ➗ (M.W AgCl ) = Moles Cl Mass Cl = ( moles Cl ) * ( M.W Cl ) % Cl = ( mass Cl ) ➗ ( mass sample ) * 100 ∴ % Cl = ( mass precipitate ) ➗ ( M.W AgCl ) * ( M.W Cl ) ➗ ( mass sample ) * 100 Trial 1 ( 0.0543 g ) ➗ ( 143.32 g/mol) * ( 35.5 g/mol ) ➗ ( 0.12 g ) * 100 = 11.21 % Cl Trial 2 ( 0.1833 g ) ➗ ( 143.32 g/mol) * ( 35.5 g/mol ) ➗ ( 0.1017 g ) * 100 = 44.64 % Cl Uncertainty % uncertainty precipitate = ( unc mas crucible ) + ( unc mass crucible & precipitate ) ➗ mass precipitate % uncertainty sample = ( unc mass sample ) ➗ mass sample % uncertainty = ( % uncertainty precipitate + % uncertainty sample ) * 100 Trial 1 % uncertainty precipitate = ( 0.0001 + 0.0001 ) ➗ 0.0543 % uncertainty sample = ( 0.0002 ) ➗ 0.12 % uncertainty = [(0.0001 +0.0001 ➗ 0.0543 + ( 0.0001 ) ➗ 0.12 ]*100 % uncertainty = 0.452 % Uncertainty = 0.00452% x 11.21% = 0.051% Therefore % Cl in sample = ( 11.21 土 0.051 ) % Trial 2 % uncertainty precipitate = ( 0.0001 + 0.0001 ) ➗ 0.1833 % uncertainty sample = ( 0.0002 ) ➗ 0.1017 % uncertainty = [(0.0001 + 0.0001 ➗ 0.1833 + ( 0.0001 ) ➗ 0.1017 ]*100 % uncertainty = 0.207 % Uncertainty = 0.00207% x 44.64% = 0.0925

Average Uncertainty 0.452 + 0. 207 ➗ 2 = 0.329% uncertainty Average [%Trial 1 + % Trial 2] ➗ 2 = [ 11.21 + 44.64 ] / 2 = (27.93 土 0.1 ) Accuracy [ average - actual ] / average * 100 = 27.93 - 56.52 / 56.52 * 100 = - 50.58 Relative Error Relative error = ( average deviation / average value ) x 100 Relative error = 1.1779 % Precision Higher %Cl - Lower % Cl / average x 100 = 119.69 ppt Discussion The true value of the chloride percentage in the salt was known to be 56.52 % which which was fairly close to my partners data but was insufficient with mine. The value we obtained could have been caused by lack of drying , which would have left moisture in the precipitate and added mass to the crucible. Another cause could have been stray molecules which may have increased the mass of the precipitate in the crucible. Lower values could have been caused by a multitude of things with the most probable one being that not all of the precipitate was successfully cleaned from the beaker Another reason could be that some of the precipitate was accidentally flushed away when cleaning the precipitate. Conclusion Sample 355 was 56.52 % chlorine , and the results obtained were not that accurate. The average percentage from the experiment 27.93% which was not that close to curry at hand. The accurate was way off of a reasonable answer , with a relative spread of 119.69 parts per thousand. Bibliography Kolthoff , E.B. Samdell, Textbook of Quantitative Inorganic Analysis, 3rd. Ed., MacMillan, New York, 1952, chapters V - VIII, XVII D.A.Skoog, D.M. West, Fundamentals of Analytical Chemistry, 2nd. Ed., Holt, Rinehart and Winston, New York, 1969, Chapters 5-8....


Similar Free PDFs