CHM 420 Laboratory Report Experiment 4 PDF

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Summary

Organic chemistry is the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which include not only hydrocarbons but also compounds with any number of other elements....

Description

Experiment 4 : Stoichiometry and Theoretical Yield

Objective

To identify the limiting reactant, excess reactant and to determine the percent yield.

Introduction

Stoichiometry is a section of chemistry that involves using relationships between reactants and products in a chemical reaction to determine desired quantitative data. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions.

The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated. If the reactants are not mixed in the correct stoichiometric proportions , then one of the reactants will be entirely consumed while another will be left over.

The actual yield of desired product is simply how much is isolated, stated in either mass units or moles. The theoretical yield of the product is what must be obtained if no losses occur. When less than the theoretical yield of product is obtained, chemists generally calculate the percentage yield of product to describe how well the preparation went. The percentage yield is the actual yield calculated as a percentage of the theoretical yield. Percentage yield can be calculated by using formula :

Percentage yield =

Actual yield(g) 100% Theoretical yield(g)

Chemicals and Apparatus 0.5 M CaCl2

Gravity filter set :

Suction Filtration set :

1.5 M Na2CO3

Filter paper

Unfolded filter paper

Buret

funnel

Buchner funnel

Erlenmeyer / conical flask

or

Adapter

Watch glass

Vacuum filter flask

Oven

Rubber tube

Analytical balance

Procedure

1.

Required value of 0.5 M CaCl2 and 1.5 M Na2CO3 were dispensed(as stated in Table 4.1 below) from the buret on side bench into a clean conical flask.

2.

Reaction

0.5 M CaCl2

1.5 M Na2CO3

1

20 mL

10 mL

2

20 mL

5 mL

The flask was swirled and they were left aside for five minutes to allow precipitate to completely form.

3.

The product was suctioned filter using a Buchner funnel or gravity quarters and making a funnel. A piece of paper was folded into quarters and funnel was made. The filter paper was placed inside glass funnel. The solution was poured into the

centre of the filter paper taking care taking care not to let it got above the level of filter paper. 4.

The sides of the conical were washed with a small amount of distilled water and they were added to the filter paper.

5.

The filter paper was removed carefully and it was placed on a pre-weighed watch glass. The product was dry in the oven for half an hour.

6.

The product (CaCO3) was scraped from the filter paper onto the watch glass. The filter paper was discard and the watch glass was returned to the oven for a further ten minutes. The mass of the product was determined. The mass of product was reheated and redetermined at least three times to complete dryness and until got two weighings that were within 0.02 g of one another.

7.

The amount of CaCO3 was compared with that expected. The limiting reactant and excess reactant were identified. The percent yield was calculated.

Data

Reaction 1

2

Mass of the product, g (1st heat)

0.9031

0.4831

Mass of the product, g (after reheating)

0.9026

0.4796

Mass of the product,g (after reheating)

0.9009

0.4782

Mass of the product, g

0.9009

0.4782

Calculations

First reaction

CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq)

mol of CaCl2 = (0.02 L)(0.5 M) = 0.01 mol

mol of Na 2CO3 = (0.01 L)(1.5 M)

From equation,

From equation,

1 mol CaCl2 Ξ 1 mol CaCO3

1 mol Na2CO3 Ξ 1 mol CaCO3

0.01 mol CaCl2 = 0.01 mol CaCO3

0.015 mol Na2CO3 = 0.015 mol CaCO3

= 0.015 mol

 CaCl2 is the limiting reactant because it produce smaller mol of CaCO3 while Na2CO3 is the excess reactant because it produce bigger mol of CaCO3.

Theoretica l yield of CaCO 3 = (0.01 mol)(100.0 9gmol -1 ) = 1.0009 g

Percentage yield 0.9009 g =  100% 1.0009 g = 90.01%

Reaction 2 CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq)

mol of CaCl2 = (0.02 L)(0.5 M) = 0.01 mol

mol of Na 2CO 3 = (5 10 − 3L)(1.5M) = 7.5 10 −3 mol

From equation,

From equation,

1 mol CaCl2 Ξ 1 mol CaCO3

1 mol Na2CO3 Ξ 1 mol CaCO3

0.01 mol CaCl2 = 0.01 mol CaCO3

7.5×10-3 mol Na2CO3 = 7.5×10-3 mol CaCO3

 Na2CO3 is the limiting reactant because it produce smaller mol of CaCO 3 while CaCl2 is the excess reactant because it produce bigger mol of CaCO3.

Theoretica l yield of CaCO 3 -3

-1

= (7.5 10 mol )(100.09gm ol ) = 0.7507g

Question

1. For each of the two reactions:

Percentage yield 0.4782g =  100% 0.7507 g = 63.70%

(a) Write balanced chemical equation. CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq)

(b) Determine the limiting reactant Reaction 1

Calcium chloride, CaCl2

Reaction 2

Sodium carbonate, Na2CO3

(c) Determine the theoretical yield Reaction 1

Theoretica l yield of CaCO 3 = (0.01 mol)(100.0 9gmol -1 ) = 1.0009 g

Reaction 2

Theoretica l yield of CaCO 3 = (7.5 10 -3 mol )(100.09gm ol-1 ) = 0.7507g

(d) Determine the percent yield of the product. Reaction 1

Percentage yield 0.9009 g =  100% 1.0009 g = 90.01%

Reaction 2

Percentage yield 0.4782g =  100% 0.7507 g = 63.70%

2. Was the yield you obtained satisfactory? Justify your answer.

Yes because generally, less than 100% yields are obtained and there is always some small loss. In reaction 1, the loss amount is 18.82% while in reaction 2, the loss amount is 39.84%. Discussions

In reaction 1,

CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq)

Given the concentration : 0.5M CaCl2 1.5M Na2CO3 and the volume 20mL CaCl2 and 10 mL Na2CO3. To determine the number of moles of reactants present, the number of moles is calculated by multiplying concentration and volume. The number of mole of CaCl2 is 0.01 mol while the number of mole of Na2CO3 is 0.015 mol. The number of mole of each reactants is compared with the product ( CaCO3 ) to find mole ratio based on chemical equations. From the calculation above, in this reaction mixture, CaCl2 is called the limiting reactant because it limits the amount of product (CaCO3) that forms.

Theoretical yield is the amount of products created by a chemical reaction, provided the reaction was fully completed. The theoretical yield that we get is 1.0009g while the actual yield is 0.9009g. Generally the percentage yield of product will be calculated to describe how well the preparation went. It is calculated to be the experimental yield divided by theoretical yield multiplied by 100% . The percentage yield is 90.01%. Usually, percent yield is lower than 100% because the actual yield is often less than the theoretical value. This is because of incomplete or competing reactions and loss of sample during recovery.

In reaction 2,

CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq)

Given the concentration : 0.5M CaCl2 1.5M Na2CO3 and the volume 20mL CaCl2 and 5 mL Na2CO3. To determine the number of moles of reactants present, the number of moles is calculated by multiplying concentration and volume. The number of mole of CaCl2 is 0.01 mol while the number of mole of Na2CO3 is 7.5×10-3 mol. The number of mole of each reactants is compared with the product ( CaCO3 ) to find mole ratio based on chemical equations. From the calculation above, in this reaction mixture, Na2CO3 is called the limiting reactant because it produce smaller amount of product (CaCO3) that forms.

Theoretical yield is the amount of products created by a chemical reaction which helps determine a reaction's efficiency. Generally the percentage yield of product will be calculated to describe how well the preparation went. It is calculated to be the experimental yield divided by theoretical yield multiplied by 100% . The theoretical yield that we get is 0.7507g while the actual yield is 0.4782g. Thus, the percentage yield is 63.70%. Usually, percent yield is lower than 100% because loss of product often occurs during purification or isolation steps.

There are factors that can cause the skewed result.Firstly, the precipitate which was left to dry could have accumulated with dust and reacted with other substances in the air Thus, it weigh more than it should. Incomplete drying also will cause the addition of weigh since the solid is wet. In addition, if the limiting reactant was not completely decomposed, this also would add the weight of the procut which affect the percentage yield.

There are some precaution steps to avoid this problem. First all, method of heating to constant could be used to avoid this error. Moreover, the wet filter sheets must be placed in a closed environment. The solutions also should be mixed properly

to ensure that all of the solid particles will fully decomposed. Then, use the accurate equipment to get an accurate readings of measurement.

Conclusion

From the Reaction 1, the limiting reactant is Calcium chloride, CaCl2 and the excess reactant is Sodium carbonate, Na2CO3. The percentage yield of Reaction 1 is 90.01%. From the Reaction 2, the limiting reactant is Sodium carbonate, Na2CO3 and the excess reactant is Calcium chloride, CaCl2 . The percentage yield of Reaction 2 is 63.70%.

References Neil D. Jespersen & Alison Hyslop. (2014). Chemistry: The Molecular Nature of Matter, 7th Edition. John Wiley and Sons, Incorporated.

Joseph Nijmeh & Mark Tye. (2019). Stoichiometry and Balancing Reactions. Retrieved October 12, 2019 from https://chem.libretexts.org/Bookshelves/ Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Chemical_ Reactions/Stoichiometry_and_Balancing_Reactions

Sarick Shah. (2019). Limiting Reagents. Retrieved October 12, 2019 from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Mod ules_(Inorganic_Chemistry)/Chemical_Reactions/Limiting_Reagents

Chung (Peter) Chieh. (2019). Thereotical and Actual Yields. Retrieved October 12, 2019 from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/ Supplemental_Modules_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiomet ry/Theoretical_and_Actual_Yields...