Following Zinc Through Chemical Reactions, Lab Report Example

Purpose

The purpose of this experiment is to determine the yield of zinc as it proceeds through several different chemical reactions. Through this process, zinc carbonate will be converted to zinc oxide.

Procedure

For each chemical reaction the mass of ZnCO₃ will be determined to ensure an accurate calculation of the number of moles present in the initial product. Next, the amount of moles of the final product, ZnO, will be calculated assuming a 1:1 ratio for the overall reaction. Based on this figure, the number of grams will be calculated to understand the yield of the final product. The mass of the ZnO will be considered the theoretical yield.

For the first portion of the experiment, a reaction will form from ZnCO₃ and HCl. In the second portion of the experiment, ZnCl₂ will be added to NaOH. In the third part of the experiment, Zn(OH)₂ will be heated. To isolate the ZnO that forms as a consequence of these steps, a centrifuge will be used. It will then be weighed and appropriate calculations will be performed to determine the final yield.

Data

Mass Data

Test Tube: 5.25 g

20 mL Beaker: 50.78 g

ZnCO₃: 0.24g

Weight of ZnO and Beaker: 50.87

Observations

– 2 mL of 3M HCl reacted with zinc carbonate to bubble and become cloudy

– 2.5 mL of 3M NaOH reacted with the solution to create a gelatinous substance

– As the beaker containing the precipitate heats up H₂O evaporates the solute dries up

Calculations

Weight of ZnO

Weight of Beaker: 50.78 g

Weight of ZnO and Beaker: 50.87

(Weight of ZnO and Beaker) – (Weight of Beaker) = Weight of ZnO

Weight of ZnO = 50.87 – 50.78 g = 0.09 g

Mass of ZnCO₃

5.25 g – 0.24 g = 5.01 g

Molar Mass of ZnCO₃

Molar Mass = Zn + C + 3O

Molar Mass = 65.39 + 12.01 + 16.00(3)

Molar Mass = 125.4 g

Moles of ZnCO₃

125.4 g/5.01 g = 25.03 g

25.03 g/125.4 = 0.18 moles

Theoretical Moles of ZnO

125.4 g x 0.18 moles = 22.57 g/mol

Mass of ZnO

0.09

Molar Mass of ZnO

Zn = 65.39

O = 16.00

65.39 + 16.00 = 81.39 g

Moles of ZnO

grams/molar mass = 0.09/81.39 = 0.081 g

Actual Mass of ZnO

50.87 – 50.78 g = 0.09 g

Percent Yield of ZnO

0.09 g/0.081g = 1.11 g x 100 = 111.11 % yield

Results

The data indicates that the resulting mass of ZnO was 0.09 grams, which is much less than the 5.01 g of ZnCO₃ used as a product in this reaction. This demonstrates that this reaction has a small yield of product compared to the reactant. However, the percent yield of ZnO was 111.11%, indicating that the amount of product that was expected to be retrieved was successfully isolated from the mixture of products. Little ZnO was lost during the extraction process.

Discussion Questions

• Two observations could have been whether the color change occurred or whether precipitate formed. Mass was not lost after HCl was added.
• The mass of the HCl was approximately 1 gram.
• There was approximately .1 gram excess HCl.
• Based on the percent yield of 111.11%, more than 6 grams would be needed. However, this prediction will likely not hold true in reality because the percent yield should not be greater than 100%.
• The percent yield was greater than 100% because some of the ZnCO₃ did not fully convert to ZnO and the chemical was not separated fully during filtration. This could be avoided by repeating the filtration process several times.