Lab Report of Experiment 4 (Stoichiometry and Theoretical Yield) PDF

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EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

Name

SITI NUR AFIQAH BINTI MAHAZAN

Student ID

2020897786

Course Code

CHM420

Group

AS2531A1

Title of Experiment

EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

Lecturer’s Name

DR. NURUNAJAH BINTI AB. GHANI

Date

30 OCTOBER 2020

CHM420 (LAB)

EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

JOTTER OF THE EXPERIMENT

:

CHM420 (LAB)

EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

DATA Reactions

Mass of the products (g)

1

2

0.888g

0.542g

ABSTRACT The aim of the experiment is to identify the limiting reactant, excess reactant and to determine the percent yield. Stoichiometry technique can be used to calculate masses, number of moles and percent of yield within a balanced chemical equation. The method used to conduct this experiment is through the mixture of two aqueous solutions for Reaction 1 and Reaction 2 by dispensing the required volume of each solution from the burette into a clean conical flask. Once the two aqueous solutions are mixed together, a white insoluble solid will form precipitate out of the solution. Then, the product was filtered from the solution with the aid of filter paper and the remaining amount of product in the flask was washed with a small amount of distilled water. The filter paper with the product attached to it was dried in the oven to dry off the water from the filter paper. The product was scraped from the filter paper onto the watch glass and it was reheated in the oven. The mass of the product was determined after the reheating process. Again, the product was reheated for at least three times to complete dryness and the mass of the product was redetermined until two weighings that are within 0.02 g of one another were obtained. The amount of product obtained was compared between Reaction 1 and Reaction 2 to identify the limiting reactant, excess reactant and the calculated percent yield. From the reaction between calcium chloride, CaCl 2 and sodium carbonate, Na2 CO 3 , the limiting reactant, excess reactant and the calculated percent yield can be identified. Based on the balanced chemical equation CaCl 2 was determined as the limiting reactant in Reaction 1. This is due to the number of moles required for CaCl 2 (0.015 mol) is more than the number of moles available for CaCl 2 (0.01 mol) in the reaction, which automatically makes Na2 CO 3 the excess reactant. On the other hand, Na2 CO 3 was determined as the limiting reactant in Reaction 2 as the number of moles required for Na2 CO3 (0.01 mol) is more than the number of moles available for Na2 CO 3 (0.0075 mol) in the reaction. This will make CaCl 2 the excess reactant. Next, the theoretical mass of calcium carbonate (product) can be determined from the stoichiometry between the number of moles available for the limiting reactant in the reaction with the number of moles of calcium carbonate based on the balanced chemical equation for both reactions. The theoretical yield obtained from Reaction 1 and Reaction 2 are 1.001g and 0.75075g respectively. After that, the percent yield of the product can be obtained from the calculation of the actual mass divided by the theoretical mass of the product times by 100 percent to obtain the yield in percentage. The percent yield obtained from the calculation for CHM420 (LAB)

EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

Reaction 1 and Reaction 2 are 88.71% and 72.19% respectively. In conclusion, stoichiometry technique can be used to calculate masses, number of moles and percent of yield within a balanced chemical equation. QUESTIONS 1) For each of the two reactions : a) Write a balanced chemical equation CaCl 2 ( aq ) +Na2 CO 3 ( aq) →Ca CO3 ( s ) +2 NaCl (aq) b) Determine the limiting reactant Reaction 1: nCa Cl = MV

n Na CO =MV

2

2

3

¿ ( 0.5 M ) ( 20 ×10 L ) ¿ 0.01 mol ( available) ¿ 0.015 mol ( available) From the balanced chemical equation, −3

¿ ( 1.5 M ) (10 × 10−3 L)

1 mol of CaCl 2 ≡1 mol of Na2 CO 3

0.01 mol of CaCl 2 ≡

1 ×0.01 mol of Na2 CO3 1

¿ 0.01 mol of Na2 CO3 (required)

thus

Na2 CO3

Since the number of moles for available

∴

CaCl 2 is the limiting reactant and

Na2 CO 3

Number of moles for excess reactant ( Na2 CO 3 )

is more than required,

is the excess reactant. ¿ navailable −nrequired

¿ 0.015 mol−0.01 mol ¿ 0.005 mol Reaction 2: nCa Cl = MV

n Na CO =MV

2

2

¿ ( 0.5 M ) ( 20 ×10 L ) ¿ 0.01 mol ( available) ¿ 0.0075 mol ( available) From the balanced chemical equation, −3

1 mol of Na2 CO3 ≡ 1 mol of CaCl 2

0.0075 mol of Na2 CO 3 ≡

1 ×0.0075 mol of CaCl 2 1

¿ 0.0075 mol of CaCl 2 (required) CHM420 (LAB)

3

−3 ¿ ( 1.5 M ) (5 × 10 L)

EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

∴

Since the number of moles for available

Na2 CO 3

is the limiting reactant and

CaCl 2

is more than required, thus

CaCl 2 is the excess reactant.

Number of moles for excess reactant ( CaCl 2 ) ¿ navailable −nrequired ¿ 0.01 mol−0.0075 mol ¿ 0.0025 mol c) Determine the theoretical yield Reaction 1: From the balanced chemical equation, 1 mol of CaCl 2 ≡ 1 mol of Ca CO3

0.01 mol of CaCl 2 ≡

1 ×0.01 mol of CaCO 3 1

¿ 0.01 mol of CaCO3 Mass of CaCO 3 : nCa CO = 3

mass of Ca CO3 molar mass of CaCO 3

Mass of CaCO3=nCa CO ×molar mass of Ca CO3 3

¿ 0.01 mol ×100.1 g /mol ¿ 1.001 g

Reaction 2: From the balanced chemical equation, 1 mol of Na2 CO3 ≡ 1 mol of CaCO 3

0.0075 mol of Na2 CO 3 ≡

1 ×0.0075 mol of CaCO3 1 ¿ 0.0075 mol of CaCO3

Mass of CaCO 3 : nCa CO = 3

mass of Ca CO3 molar mass of CaCO 3

Mass of CaCO3=nCa CO ×molar mass of Ca CO3 3

¿ 0.0075 mol ×100.1 g /mol ¿ 0.75075 g d) Determine the percent yield of the product

CHM420 (LAB)

EXPERIMENT 4: STOICHIOMETRY AND THEORETICAL YIELD

Reaction 1: % yield = =

Actual mass of Ca CO3 × 100 % Theoretical mass of Ca CO3 0.888 g 1.001 g

×100 %

= 88.71% Reaction 2: % yield =

=

Actual mass of Ca CO3 × 100 % Theoretical mass of Ca CO3

0.542 g 0.75075 g

×100 %

= 72.19% 2) Was the yield obtained satisfactory? Justify your answer.

The actual mass of the product

(CaCO3)

obtained from the experiment for Reaction 1

and Reaction 2 are 0.888g and 0.542g respectively. Based on this data, the theoretical mass of the product also can be obtained in order to calculate the percent yield. It can be determined from the stoichiometry between the number of moles available for the limiting reactant in the reaction with the number of moles of product based on the balanced chemical equation for both reactions. The percent yield of product calculated in Reaction 1 and Reaction 2 are 88.71% and 72.19% respectively. The yield obtained for both reactions are satisfactory as stoichiometry is used to measure the quantitative proportions or mass ratios in which chemical elements relate to one another.

CHM420 (LAB)...