Stoichiometry of a Precipitation Reaction Essay
Stoichiometry of a Precipitation Reaction
Abstract: In this experiment Stoichiometry of a Precipitation Reaction, the main objectives were to use stoichiometry to calculate the theoretical yield of CaCO3 that precipitates from the reaction between Calcium chloride, Dihydrate and Sodium carbonate, and then compare that value to the actual yield. In order to do this accurately, we must be able to measure the products and reactants with accuracy. The last objective of this experiment is to calculate the percent yield of the product produced. When 1.0 g of Calcium chloride was mixed with 0.80 g of sodium carbonate, it produced 0.70 g calcium carbonate, which was very close to the theoretical yield of 0.68 g.
Experiment and Observation: The first step in this experiment was to use the scale to measure out 1.0 gram of the Calcium chloride, Dihydrate and mix that in the 100 mL beaker with 25 mL of distilled water until dissolved to obtain a Calcium chloride solution. Stoichiometry was then used to calculate how much Sodium carbonate was needed. Since one mole of calcium chloride reacts with one mole of sodium carbonate to produce one mole of calcium carbonate, the stoichiometric calculation was carried out as follows:
0.80 g sodium carbonate was then weighed out and mixed with 25 mL of distilled water in a paper cup to obtain a sodium carbonate solution. Then the theoretical yield of calcium carbonate was calculated as follows:
Using 1 g of CaCl2·2H2O and .72 or .8 g (slight excess) Na2CO3 Should give a CaCO3 theoretical yield as follows:
1 g CaCl2·2H2O x 1 mole CaCl2·2H2O x 1 mol CaC03 x 100 g CaC03 = 0.68 g CaCO3 _______________ __________ __________ 147 g CaCl2·2H2O 1 mol CaCl2·2H2O 1 mol CaC03
To double-check, we can calculate CaCO3 theoretical yield by using Na2CO3
0.72 g Na2C03 x 1 mol Na2C03 x 1 mol CaC03 x 100 g CaC03 = 0.68 g CaCO3 __________ _________ __________ 106 g Na2C03 1 mol Na2C03 1 mol CaC03
The two solutions were then mixed together in the 100 mL beaker, and precipitate formed immediately. The whole beaker was filled with a milky white liquid. The small paper cup was then placed in a coffee mug, and the filter was folded in quarters and weighed to be 1.0 g. One of the folded sides was pulled out to make a small filter that would fit in the small paper cup, and the precipitate solution was poured into the filter to drain out the water. After all of the water was drained out of the filter and only the precipitate remained, the filter including the product, was laid on paper towels to dry completely. Once dried, the filter was weighed once again to yield a result of 1.7 g. Since the filter itself weighed 1.0 g, that was subtracted from that value to get 0.70 g of calcium carbonate precipitate that formed from the reaction. The percent yield was then calculated as follows: % yield= actual yield/theoretical yield *100
0.70/0.68* 100= 102%
Calculations and Error: One of the biggest sources of error in this experiment could be that when the beaker with precipitate was emptied into the filter apparatus, some of the precipitate stuck to the sides of the beaker. Even though 5 mL more water was poured into the beaker to get the excess off of the sides, it was near to impossible to actually get all of the excess off of the sides. Also, when pouring the mixture of the solutions in the filter, since I originally was holding the filter up to do this, when I dropped it down to actually lay in the paper cup and coffee mug, it slanted over, and some of the mixture fell out which in both cases means that not all of the precipitate was collected.
Discussion and Conclusion: In this experiment, when the theoretical yield for the combining of 1.0g of calcium chloride with 0.80g sodium carbonate was calculated stochiometrically, it came up with a theoretical yield of 0.68 g calcium carbonate. When the experiment was performed, the actual yield of calcium carbonate was 0.70 g which gave a percent yield of 102% which is basically spot on. Everything used in the experiment was measured to match the exact number the experiment asked for, although since the scale does not show double digits to the hundreds place after the decimal, it was not exactly possible to measure out 0.72g of sodium carbonate, so it was necessary to round up to 0.80 g and just have some excess. If this experiment were repeated in the future, some ways to avoid errors such as the ones listed in the errors section would be to one, have an apparatus that kept the funnel with the filter completely held up that way it would not rest on an angle.
When it is at an angle, and everything gets poured in, it is going to flow over the sides a little bit causing some of the product to be lost. Another way to avoid an error would be to have some kind of sprayer thing that you can use to get all of the precipitate off of the sides of the beaker. If the beaker could be held upside-down and the precipitate sprayed off of the sides, it would be easier to make sure that all of the precipitate went into the filter. The results that the experiment yielded were very close to the expected results which surprised me because I have never been that close to complete accuracy before! At the same time, I did lose some of the precipitate that would have made my final mass more then what I came up with.
A. From your balanced equation what is the theoretical yield of your product? The theoretical yield of the product was 0.68 g of calcium carbonate B. According to your data table, what is the actual yield of the product? The actual yield was 0.70g calcium carbonate
C. What is the percent yield?
The percent yield was 102%
D. A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error.
I had a perfect percent yield, and a perfect degree of accuracy, but some of my precipitate was lost before it was weighed, so my actual yield would have probably been higher. E. How could these errors be reduced in the future?
As discussed in the discussion and conclusion section, using a better apparatus that held the funnel upward would help reduce precipitate loss from spillage, and having a sprayer to get the precipitate off of the sides of the beaker would also help. F. Lets say we decided to run this experiment again. This time we used 1.0 gram of CaCl2·2H2O and 1.0 gram of Na2CO3. a)How many grams of CaCO3 would we produce? Please show/explain how you found your answer. We would produce 0.94 g Calcium carbonate. This was calculated by (1 g* 100 g/mol )/106 g/mol= 0.9434 g CaCO3 b)Of the two reactants, one was the limiting reagent and the other was the excess reagent. Please calculate the grams of the excess reagent still remaining in solution. The LR is the Calcium chloride so there would be an excess of 0.06 g of sodium carbonate. G. Before the advent of Advil and Tylenol, did people simply have to “grin and bear it” when it came to pain? One of the most common ancient medicines for pain, fever, and inflammation came as a byproduct of the willow tree. While the first uses date back to 400 BCE, American historians cite the use of willow bark tea by the Lewis and Clark exploration party in the early 1800’s. Salicylic acid derived from the willow tree’s bark was the key chemical involved with the relief of pain and the reaction to make aspirin is a fairly simple one performed in numerous chemistry classroom nation wide. Aspirin can be made by reacting acetic anhydride (C4H6O3) with salicylic acid (C7H6O3) to form aspirin (C9H8O4). 1C4H6O3 + 1C7H6O3 –>1 C2H4O2 + 1C9H8O4
When synthesizing aspirin, a student began with 3.20 mL of acetic anhydride (density = 1.08 g/mL) and 1.45 g of salicylic acid. The reaction was
allowed to run its course and 1.23 grams of aspirin was collected by the student. Determine the limiting reactant, theoretical yield of aspirin and percent yield for the reaction. Salicylic acid is the limiting reagent. The theoretical yield of aspirin is 1.89g. The percent yield is 65.0%. mole C4H6O3 = 3.20mL C4H6O3 x (1.08g C4H6O3 / mL C4H6O3) x (1 mol C4H6O3 / 102.1g C4H6O3) = 0.03385 moles C4H6O3 mole C7H6O3 = 1.45g C7H6O3 x (1 mol C7H6O3 / 138.1g C7H6O3) = 0.01050 moles C7H6O3 1 mol C7H6O3 will require 1 mole C4H6O3
so 0.01050 mol SA
requires 0.01050 mol C4H6O3
since we have 0.03385 mol C4H6O3 available, we have excess C4H6O3 this means that C7H6O3 is the limiting reagent
from the balanced equation, 1 mol C7H6O3 –> 1 mol C9H8O4 so that 0.0105 mol C7H6O3 x (1 mol C4H6O3 / 1 mol C7H6O3) = 0.0105 mol C9H8O4 theoretical mass acid 0.0105mol C9H8O4 x (180.2g C9H8O4 / mol C9H8O4) = 1.892g C9H8O4 % yield = (1.23g / 1.892g) x 100% = 65.0%