Introduction
When two aqueous solutions are mixed together, they often react chemically and form products. The chemical reaction may be visible as a change in color from the reactants to the products, the release of gas in the product, or the formation of a precipitate. A precipitate forms when one of the resulting products is aqueous and the other product is insoluble. The insolubility means that the product will be a solid, which usually settles to the bottom of the resulting aqueous solution.
It is possible to determine how much precipitate will form before actually combining the reactants. Using stoichiometry, one can figure out how much precipitate theoretically will be produced. In the lab there are outside factors, which can affect how much precipitate is actually formed. Only under ideal conditions would the actual amount of precipitate formed match the theoretically amount of precipitate formed.
In this lab, strontium iodate monohydrate was synthesized using the following equation: Sr(NO3)2 + 2KIO3 —-> Sr(IO3)2 + 2KNO3. With the knowledge of the starting amount of Sr(NO3)2 and KIO3 used, it was possible to figure out how much strontium iodate monohydrate should be theoretically produced. Then after finding out how much strontium iodate monohydrate actually was produced, the percent yield was found.
Experimental
First, approximately 40.00 ml of 5.00 x 10-2 M Sr(NO3)2 and 50.00 mL of 1.00 x 10-1 M KIO3 were put into separate graduated cylinders. They were then combined in a beaker sitting in ice. This was to prevent the strontium iodate monohydrate from becoming soluble in water, as its solubility in water goes up with its temperature. The mixture was then stirred for about 10 minutes, so that all the precipitate could form. The precipitate and supernate were then poured into a vacuum filter, and the beaker was rinsed with ice cold distilled water to get all the precipitate out. The filter paper used in the vacuum filter was first weighed before filtration, then once the filtered precipitate was dry, the filter paper and precipitate was weighed. This whole procedure was performed twice.
Results
Volume of Sr(NO3)2 solution used: 39.99 mL (1st run), 40.25 ml (2nd run)
Molarity of Sr(NO3)2 solution used: 5.00 x 10-2 M
Volume of KIO3 solution used: 49.98 mL (1st run), 49.50 mL (2nd run)
Molarity of KIO3 solution used: 1.00 x 10-1 M
Mass of product, watch glass, and filter paper: 32.04 g (1st run), 27.32 g (2nd run)
Mass of watch glass and filter paper: 31.33 g (1st run), 26.47 g (2nd run)
Mass of product: 0.71 g (1st run), 0.85 g (2nd run)
Mass of Sr(NO3)2: 0.68 g (1st run), 0.82 g (2nd run)
Number of moles of Sr(NO3)2 used: 0.00200 moles (1st run), 0.00201 moles (2nd run)
Number of moles of KIO3 used: 0.00500 moles (1st run), 0.00495 moles (2nd run)
Limiting reagent: Sr(NO3)2
Theoretical yield of product, moles: 0.00200 moles (1st run), 0.00201 moles (2nd run)
Theoretical yield of product, g: 0.875 g (1st run), 0.880 g (2nd run)
Percent yield: 78.% (1st run), 93.% (2nd run)
Mean percent yield: 86.%
Calculations
In order to find the mass of the precipitate, took the mass of the product, watch glass, and filter paper minus the mass of the watch glass and filter paper. To find the moles of the reactants used, I used the equation Molarity = Moles/Liters and rearranged it to the equation Moles = Molarity x Liters. Before subbing the volume in, I had to convert mL to L by dividing by 1000. To find the limiting reagent, I first had to look at the balanced equation of the chemical reaction, which was Sr(NO3)2 + 2KIO3 —-> Sr(IO3)2 + 2KNO3. Since I knew for every 2 moles of KIO3 used, 1 mole of Sr(NO3)2 was used, I could substitute the actual number of moles of each that were used in the experiment to find the limiting reagent. If KIO3 was the limiting reagent, then 0.00250 moles of Sr(NO3)2 would be needed for the 0.00500 moles of KIO3, but there were only 0.00200 moles of Sr(NO3)2 used, so that made it the limiting reagent.
In order to find the percent yield, I first took the number of moles of Sr(NO3)2 used because from the balanced equation, I knew that there would be an equal number of moles of Sr(IO3)2 produced. I then found the molar mass of Sr(IO3)2, which is 437.43 g, then multiplied by 0.00200 moles, which is the number of moles of Sr(IO3)2 theoretically produced, to find the number of grams of Sr(IO3)2, theoretically produced (0.875 g). I then took mass of the precipitate strontium iodate monohydrate produced (0.71 g) and needed to find the mass of Sr(IO3)2 produced. So I found the percent of Sr(IO3)2 that makes up strontium iodate monohydrate by taking the molar mass of Sr(IO3)2 (437.43 g) and divided by the molar mass of strontium iodate monohydrate (455.44 g) to get 96.046%. I then multiplied the mass of the product by this to find the mass of Sr(IO3)2 actually formed (0.71 g x 0.96046 g= 0.68 g). I then took 0.68 g and divided by the mass of Sr(IO3)2 theoretically produced (0.875 g) and multiplied by 100 to find get the percent yield of 78.%. I repeated this for the values found in the second run. For the mean percent yield, I took the percent yields found for each run, added, then up, and divided by 2.
Discussion/Conclusions
My results were fairly close to what they should have been. A mean percent yield of 86.% 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 86.% 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.
