Stoichiometry of a Precipitation Reaction Essay

Custom Student Mr. Teacher ENG 1001-04 1 June 2016

Stoichiometry of a Precipitation Reaction

Abstract: The purpose of the lab, Stoichiometry of a Precipitation Reaction, is to be able to calculate the amount of a second reactant we need to react with the reactant one. You must calculate the amount of the second reactant using stoichiometry to figure out what amount is needed. After the amount is calculated, you then can add it to the first reactant and it will give you a full reaction. To figure out what you need you have to use stoichiometry. My calculation for the second reactant was: 1.0g CaCl2*2H2O(1mol CaCl2*2H­2O/147g CaCl2*2H2O)(1mol Na2CO3/1mol CaCl2*2H2O)(106g Na2CO3/1mol Na2CO3) = 0.72g Na2CO3. The final, when its dried in the paper filtered weighed at 1.6 grams.

Experiment and Observations: As I performed the lab, I had to retrieve all my equipment’s from my labpaq. As I started to setup, I had to weigh out 1.0 gram of CaCl2*2H2O on the digital scale. After it was weighed, it was added to the 100 mL beaker and 25 mL of distilled was added and stirred. Next, I had to figure out the amount of the second reactant, so that it can be added to the solution. I had to perform stoichiometry to figure out the amount of Na2CO3 is needed to make a precipitate of calcium carbonate.

After the calculations, I arrived at .72 grams of Na2CO3 and rounded to the nearest tenth to mix with 25 mL of distilled water. After both of the reactants were mixed separately with distilled water, they both were clear color. As soon as you added the solution of Na2CO3 to CaCl2*2H2O the color of solution changed to a cloud white color. When both solutions are mixed, the paper filter that was provided with the lab must be weighed and recorded. Then you pour the solution into the paper filter to filter out the precipitate and dry the filter to weigh the mass of the precipitate when it dries. After it dried, it was weighed and stoichiometry was performed to see what the theoretical yield, actual yield and percent yield was.

Calculations and Errors: During the lab there were some errors that I found that might affect the final solution when trying to find the theoretical, actual and percent yield. Some of the error could be the amount of distilled water because reading the amount in a graduated cylinder is hard to see if you are exactly below or above 25 ml. Also rounding the amount of Na2CO3 ­ to the nearest tenth. Having to round it off gram needed because the digital scale does not read to the hundredth place. The only way was to read it to the nearest tenth, so it can be measured.

Discussion and Conclusion: Stoichiometry of a Precipitation Reaction was different from the previous two labs. This lab required more mathematic arithmetic to figure what was needed to make a full reaction with another compound. Although it had calculations to the lab, it was completely different and required more thinking of figuring out how to solve the equations, but similar to mathematics formulas. The similar part about stoichiometry is the dividing, the converting and the canceling of the same units. By doing the lab and seeing the reaction between the two compounds, it was amazing to see what the reaction would be.

At first I didn’t think that anything would happen because the two solutions were clear and when two clear mix together it does do much as I saw in the first lab I did. Guess I was wrong and it produced something amazing that I thought I would not see. It produced something that looked like clouds, which made it neat to watch as it started to change. This experiment as taught me that given the right formula regardless of chemical or life always expect the unexpected.


A) From your balanced equation what is the theoretical yield of your product?

From my balanced equation my theoretical yield of the product is .6808 g of CaCO3.

B) According to your data table, what is the actual yield of the product? 
 The actual yield is .6 gram
C) What is the percent yield?

The percent yield is 88 percent.

D) A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error. When performing the lab, I measured out the compound to the amount that was needed and added to the amount of water it needed. Poured into the paper filter as suggested and waited for it to dry. Even if I measured everything correctly, I rounded to the nearest tenth and measured the compound when calculated. It was calculated to be .72 grams and I measured .7 grams.

The reason I measured it that way because the digital scale does not read to the hundredth place. If it read out to the hundredth, my result may be closer to 100 percent yield. Also when pouring the solution through the filter, it was hard to keep it from tilting and using the beaker to mix the solution. Having the beaker as a mixing container, it left residue on the side of the glass and was hard to get it complete off into the filter. E) How could these errors be reduced in the future?

I think the error could be reduced in the future is if we have a digital scale that can read to the hundredth. If we can weigh to the compound to the exact calculation I think that the percent yield would be closer to 100 percent. Also we can use a different type of container to mix the compounds. I would suggest using a funnel, where the bottom can be plugged and unplugged, so that you can get everything off the side of the funnel without losing anything.

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.

There would be produce .68 grams of CaCO3.

Na2CO3+CaCl2*2H2O —> CaCO3+2NaCl+2H2O
Molecular mass of Na2CO3+CaCl2*2H2O = 147.01
Moles =1/147.01 which equals 6.8*10-3 mol
Molecular mass of Na2CO3 = 105.99 g/mol
Moles = 1/105.99 which equals 9.43*10­-3 mol
CaCO3 Produced
6.8 * 10-3 * 100 = .68 grams

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 grams of excess reagent remain is .28 grams.

(9.43*10-3 – 6.8*10-3) * 105.99 = .28 grams of Na2CO3 left over. 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).

C4H6O3 + C7H6O3 –> C2H4O2­ + C9H8O4

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.

The limiting reactant is salicylic, theoretical yield of aspirin is 1.89 grams and the percent yield for the reaction is 65.1%.

Free Stoichiometry of a Precipitation Reaction Essay Sample


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  • University/College: University of Chicago

  • Type of paper: Thesis/Dissertation Chapter

  • Date: 1 June 2016

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