Synthesis of stontium Iodate monohydrate Lab PDF

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Objectives To synthesize strontium iodate monohydrate with the intention of obtaining as much of the product as possible.

Data Volume of Sr (NO3)2 solution = 40mL Molarity of Sr (NO3)2 solution used = 0.05 M Volume of KIO3 solution used = 50 mL Molarity of KIO3 used = 0.1 M

Calculations Mass of product, watch glass and filter paper = 43.6993g Mass of watch glass and filter paper = 42.532 Mass of product g = 43.6993 – 43.2203 = 0.475g Mole of Sr (NO3)2 used = Molarity*Volume = 0.05Moles/L *0.04L = 0.002 moles Moles of KIO3 used = 0.1Moles/L*0.05 = 0.005 moles Limiting reagent in this case is Sr (NO3)2, since KIO3 is available in excess. All of Sr (NO3)2 Will take part in the reaction Sr (NO3)2 = 437.43 g*0.002 = 0.875g Molar mass of strontium iodate monohydrate = 455.44g Ratio of Sr (NO3)2 to Sr (NO3)2.H20 = 437.43/455.44 = 0.9605 Actual mass of Sr (NO3)2 formed = 0.475*0.9605 = 0.4562g Percentage yield = (0.4562g/0.475)*100 = 96.05% Molar mass of

Analysis and Conclusion My results were fairly close to what they should have been. A mean percent yield of 96.05% is probably good considering all the factors that can cause the percent yield to be less than 100%. The Sr(IO3)2 could get too warm and wash away in the water, the precipitate could not completely form when mixing the reactants, and some precipitate could become stuck in the beaker and not wash out. Those are factors that are not all easily controlled, so overall my percent yield of 96.05.% seems plausible when looking at those factors that could affect the results.

If I were to repeat this experiment, I would probably take more time in letting the reactants mix and form the precipitate. That way I could be sure almost all the precipitate actually formed. I would also keep the wash bottle in colder conditions to make sure none of the precipitate washed away.

Answers to questions Pre-Lab Assignment 2(1) Supernatant liquid: the usually clear liquid overlying material deposited by settling, precipitation, or centrifugation. 2(2) Limiting reagent: the reactant in a chemical reaction that limits the amount of product that can be formed. 2(3) Theoretical yield: amount that can be produced as found from correct computations. 2(4) Decantation: the process of pouring the supernatant liquid into the funnel assembly without disturbing the precipitate at the bottom of the beaker. 3(1) Ca(NO3)2(aq) + 2 NH4F(aq) ------> CaF2(s) + 2 NH4NO3(aq) 3(2) (45.00 mL) x (4.8724 x 10^-1 mol/L Ca(NO3)2) = 21.9258 mmol Ca(NO3)2 3(3) (60.00 mL) x (9.9981 x 10^-1 mol/L NH4F) = 59.9886 mmol NH4F 3(4) 21.9258 millimoles of Ca(NO3)2 would react completely with 21.9258 x (2/1) = 43.8516 millimoles of NH4F, but there is more NH4F present than that, so NH4F is in excess and Ca(NO3)2 is the limiting reactant. (21.9258 mmol Ca(NO3)2) x (1 mol CaF2 / 1 mol Ca(NO3)2) x (78.0748 g CaF2/mol) = 1711.85 mg = 1.712 g CaF2 in theory (1.524g CaF2 / 1.712gg CaF2) * 100 = 89.01 % yield 4(1) The cold water prevents the precipiate from redissolving 4(2) The filter paper is moistened so that it will stick to the funnel which doesn't let the air pass through the passage, which increases the speed of filtration. 5 I will pour the supernatant liquid into the funnel assembly without disturbing the precipitate at the bottom of the beaker Post Lab Question

1. Spills Incorrect measuring Calculation errors like improper rounding of numbers 2(1) Ba(NO3)2 + 2 NaIO3 → Ba(IO3)2 + 2 NaNO3 (30.00 mL) x (5.912 x 10^-1 mol/L Ba(NO3)2) = 17.736 mmol Ba(NO3)2 (50.00 mL) x (9.004 x 10^-1 mol/L NaIO3) = 45.02 mmol NaIO3 2(2) 17.736 millimoles of Ba(NO3)2 would react completely with 17.736 x (2/1) = 35.472 millimoles of NaIO3, but there is more NaIO3 present than that, so NaIO3 is in excess and Ba(NO3)2 is the limiting reactant. (17.736 mmol Ba(NO3)2) x (1 mol Ba(IO3)2 x 1 mol Ba(NO3)2) x (505.1476 g Ba(IO3)2*H2O/mol) = 8959.29 mg = 8.959 g Ba(IO3)2*H2O in theory Theoretical yield = (6.895/8.959)*100 = 76.9617% 2(3) (30.00 mL) x (4.912 x 10^-1 mol/L Ba(NO3)2) = 14.736 mmol Ba(NO3)2 (50.00 mL) x (9.004 x 10^-1 mol/L NaIO3) = 45.02 mmol NaIO3 14.736 millimoles of Ba(NO3)2 would react completely with 14.736 x (2/1) = 29.472 millimoles of NaIO3, but there is more NaIO3 present than that, so NaIO3 is in excess and Ba(NO3)2 is the limiting reactant. (14.736 mmol Ba(NO3)2) x (1 mol Ba(IO3)2 x 1 mol Ba(NO3)2) x (505.1476 g Ba(IO3)2*H2O/mol) = 7443.85 mg = 7.444 g Theoretical yield = (6.895/7.444)*100 = 92.6249 % 2(4) Error = ((92.6249-76.9617)/ 92.6249)*100 = 16.9104%

3(1) x- 0.10*(20/100) = 6.895g x = 6.91g 3(2) x-0.28*(125/100)=final yield = 6.56g, where x = 6.91g Percentage error = ((6.91 – 6.56)/6.91)*100 = 5.065 %...