Using Stoichiometry and Percent Yield to Create a Given Amount of Substance

The lab was an example of a case where these subjects would be used. In order to predict the amount of reactants needed, stoichiometry problems needed to be used. This shows practical applications to somewhat abstract concepts. The result of this lab was an 84% yield of actual yield to predicted yield. This shows that there is a difference between what is done on paper and what happens in the lab.

1. Pre-Lab

   Determine the identity of the product

   made when solutions of your two reactants are mixed in a double replacement reaction. Consult figure 17-9, “Solubility Rules for Ionic compound” found on page 574 of you textbook.
   
   The product is MgCO3.

2. Write the balanced chemical equation for this precipitation reaction including state symbols (aq, s, l, g).

   Note that each pair of reactants contains at least one hydrate, so water will also be a product of the reaction.

   MgSO4 · 7H20(aq) + Na2CO3(s)  MgCO3(s) + Na2SO4(aq) + 7H20(l)

3. Write a complete word equation for the reaction including the number of moles of each reactant and product.

   1 mol Magnesium sulfate and 1 mol Sodium carbonate form 1 mol Magnesium carbonate, 1 mol Sodium sulfate, and 7 mol water.

4. Calculate the moles of product in 2.00 g of product.

   2.00g MgCO3 ( 1 mol MgCO3 ) = 0.0237 mol MgCO3

   (84.313g MgCO3)

5. Calculate the moles of each reactant required to produce 2.00g of product.

   2.00g MgCO3 ( 1 mol MgCO3 )(1 mol MgSO4 · 7H2O) = 0.0237 mol MgSO4 · 7H2O

   (84.313g MgCO3)( 1 mol MgCO3 )

   2.00g MgCO3 ( 1 mol MgCO3 )(1 mol Na2CO3) = 0.0237 mol Na2CO3

   (84.313g MgCO3)(1 mol MgCO3)

6. Calculate the mass of each reactant required to produce 2.00g of product.

   2.00g MgCO3 ( 1 mol MgCO3 )(1 mol MgSO4 · 7H2O)(246.472g MgSO4 · 7H2O) = 5.847g MgSO4 · 7H2O

   (84.313g MgCO3)( 1 mol MgCO3 )( 1 mol MgSO4 · 7H2O )

   2.00g MgCO3 ( 1 mol MgCO3 )(1 mol Na2CO3)(105.988g Na2CO3) = 2.514g Na2CO3

   (84.313g MgCO3)(1 mol MgCO3)( 1 mol Na2CO3 )

Procedure:

1. Mass empty beaker and filter paper.
2. Measure out 50 mL of water.
3. Add 5.847g MgSO4 · 7H2O to the water.
4. Add 2.524g Na2CO3 to the water.
5. Wait for reaction to take place.
6. Filter liquid with filter paper.
7. Mass product in filter paper.
8. If needed use chemical oven to evaporate trapped water.

Results

The actual yield from this lab was 1.67g MgCO3 and the percent yield was 84%. There are several possible reasons for this low of a yield to have taken place. When the reaction was complete, and filter paper dry, the product weighed 3.09g. Since this value was extremely far from the expected yield, we placed the product in a beaker by scraping all of the residue off the filter paper and placed it into a chemical oven. After the beaker had been in the chemical oven, the weight was 1.67g. The loss in mass was attributed to trapped water evaporating. Some extra product could have been lost in process of scraping it off of the filter paper. An amount of product was could have been stuck on the filter paper and was not removed. This could have accounted for the .33g of MgCO3 that was missing. Another reason for the low actual yield could have been inaccurate measurements. There could also have been some issues with equations where a small mistake was made which would disrupt the rest of the lab.

What You Learned

I learned several things in this lab. I learned that the equations we learned in class can be used to find the amount of substances needed to make reactions take place in the laboratory. The percent yield can be used to determine exactly how close or how far from the predicted yield your actual yield was. From this lab, I gained a better perception of how the stoichiometry problems which we have been dealing with work in a lab where the problems are applied to real reactions. This lab has helped me gain a better understanding of what the equations are used for and how to solve them.