Synthesisof Strontium Iodate Monohydrate
Objectives
1.Synthesis of Strontium Iodate Monohydrate.
2.Determining the percentage yield of Sr(IO3)2in the precipitate.
Introduction
StrontiumIodate Monohydrate is synthesized in the lab through a reaction oftwo aqueous solutions, Strontium Nitrate and Potassium IodateMonohydrate, to precipitate strontium Iodate Monohydrate and anaqueous solution.
Thereaction occurs according to the following equation:
Sr(NO3)2+ 2KIO3—-> Sr(IO3)2+2KNO3(Adam 3).
Byuse of stoichiometry, one can theoretically determine how thereaction occurs but in the lab other factors affect the amount ofyield that precipitated.
Dataand Observations
|
EXPERIMENT 1 |
EXPERIMENT 2 |
|
|
Volume of Sr(NO3)2 solution used, mL |
38 |
38 |
|
Molarity of Sr(NO3)2 solution used, M |
0.05m |
0.05M |
|
Volume of KIO3 solution used, mL |
40 |
40 |
|
Molarity of KIO3 solution used, M |
0.10M |
0.10M |
|
Mass of precipitate, watch glass and filter paper, g |
95.419g |
95.473 |
|
Mass of watch glass and filter paper, g |
94.897g |
94.891g |
|
Mass of product, g |
0.522g |
0.582g |
|
Number of moles of Sr(NO3)2 used |
0.0019mol |
0.0019mol |
|
Number of moles of KIO3 used |
0.004mol |
0.004mol |
|
Limiting reagent |
Sr(NO3)2 |
Sr(NO3)2 |
|
Theoretical yield of product, mole |
0.0019 |
0.0019 |
|
Percentage yield |
60.323% |
67.257% |
|
Mean percentage yield |
63.79% |
DataTreatment and Discussion
Takethe total mass of the product, watch glass and filter paper andsubtract the initial mass of filter paper and watch glass to get theactual mass of precipitate (Adam 6).
Experiment1: 95.419g – 94.897g=0.522g
Experiment2: 95.473g – 94.891g= 0.582g
Usedthe equation: Molarity = Moles/Liters
Therefore:Moles = Molarity × Liters
Experiment1: 0.05 × (38/1000) = 0.0019
Experiment2: 0.05 × (38/0.0019) = 0.0019
0.0019molesof Sr(NO3)2are used.
Experiment1: 0.10 × (40/1000) =0.004
Experiment2: 0.10 × (40/1000) =0.004
0.004moles of KIO3are used.
Toget the limiting agent use the moles involved in the reaction everymole of KIO3used, 0.5 moles of Sr(NO3)2are used. This is according to the balanced equation of the reactionwhich is
Sr(NO3)2+ 2KIO3—-> Sr(IO3)2+2KNO3(Adam 3).
Since0.004 moles of KIO3require 0.002 moles of Sr(NO3)2and only 0.0019 moles of Sr(NO3)2are available, Sr(NO3)2is the limiting agent.
Molarmass of Sr(NO3)2is 437.43g multiplied by 0.0019 moles gives 0.831g which are thetheoretical mass of Sr(IO3)2produced (Adam 7).
Usethe mass of the precipitation (0.522g and 0.582g respectively) toderive the mass of Sr(IO3)2produced. Derive the percentage of Sr(IO3)2that constitutes the strontium iodate monohydrate by dividing molarmass of Sr(IO3)2 by that of strontium iodate monohydrate (Adam 7).
437.43g/455.44062g= 0.95951
Toget percentage: 0.95951 × 100 = 95.951%
Multiplythe percentage by the mass of the precipitation to obtain actualamount of Sr(IO3)2formed:Experiment 1: 0.522g × 0.95951 = 0.50086422g
Experiment 2: 0.582g × 0.95951 = 0.558435g
Divide0.50086422g and 0.558435g from experiment 1 and 2 respectively by thetheoretical mass of Sr(IO3)2,0.875g and multiply by 100 to obtain the percentages. 60.323% and67.257% are the solutions arrived at for experiment 1 and 2respectively.
Forthe mean percentage yield, sum up the two percentages (forexperiments 1 and 2) then divide by two to obtain a quotient of63.79%.
Theexperiment can be summed up as successful considering that severalfactors could have contributed to discrepancies in the experiment.Some Sr(IO3)2may have been washed away, or the precipitate may have partiallyformed. Considering the numerous factors that may have contributed toerror, a 63.79% precipitate is an indication that the experiment wasa success.
Conclusion
Synthesisof Strontium Iodate Monohydrate is achievable and with a high levelof yield depending on your accuracy in the experiment. The percentageyield is highly dependent on numerous factors that affect theaccuracy of the entire experiment.
WorkCited
Adam,Cap. Synthesisof Strontium Iodate Monohydrate.Saint Joseph’s University, 2016. Web. 21 March. 2017.