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Percent Yield Lab Report
Instructions: For this investigative phenomenon, you will need to determine the percent yield of magnesium oxide from the given reaction to determine if it is a useful commercial process. Record your data and calculations in the lab report below. You will submit your completed report.
Title:
Objective(s):
Hypothesis:
No hypothesis needed for this lab. Your theoretical yield calculation serves as your prediction for what you expect the lab to produce, and that will be determined later in the lab.
Procedure:
Access the virtual lab. Because this lab is virtual, summarize the steps used to collect your data. In addition, list and explain your controlled variables, independent variable, and dependent variable for this lab.
Materials:
Percent Yield Virtual Lab
Variables:
Remember, controlled variables are factors that remain the same throughout the experiment. An independent (test) variable changes so that the experimenter can see the effect on other variables. The dependent (outcome) variable will change in response to the test variable.
Controlled variables:
Independent Variable:
Dependent Variable:
Summary of Steps:
Data:
Type the data in the data table below. Don’t forget to record measurements with the correct number of significant figures. Hint: Using the same instrument, you should have the same number of digits to the right of the decimal.
Data
Trial 1
Trial 2
Mass of empty crucible with lid
26.698(g) 26.691(g)
Mass of Mg metal, crucible, and lid
27.040(g) 27.099(g)
Mass of MgO, crucible, and lid
27.198(g) 27.361(g)
Calculations:
Show your calculations for each of the following. Remember, calculations should follow rules for significant figures.
Write the balanced chemical equation for the reaction you are performing.
Subtract the mass of the crucible and lid (row 1 in the chart) from the total mass of Mg, crucible, and lid (row 2 in the chart) to find the mass of magnesium for each trial.
Trial 1:
Trial 2:
Subtract the mass of the crucible and lid (row 1 in the chart) from the total mass of MgO, crucible, and lid (row 3 in the chart) to find the mass of magnesium oxide for each trial. This is the actual yield of magnesium oxide for each trial.
Trial 1:
Trial 2:
Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial.
Trial 1:
Trial 2:
Determine the percent yield of MgO for your experiment for each trial.
Trial 1:
Trial 2:
Determine the average percent yield of MgO for the two trials.
Conclusion:
Write a conclusion statement that addresses the following questions:
Explain why the product had a higher mass than the reactant, and how this relates to conservation of matter.
What sources of error may have contributed to the percent yield not being 100 percent? (Think about things that may have led to inaccurate measurements or where mass of the product could have been lost if this experiment was conducted in a physical laboratory.)
How do you think the investigation can be explored further?
Post-Lab Reflection Questions
Answer the reflection questions using what you have learned from the lesson and your experimental data. It will be helpful to refer to your chemistry journal notes. Answer questions in complete sentences.
When conducting this experiment, some procedures call for heating the substance several times and recording the mass after each heating, continuing until the mass values are constant. Explain the purpose of this process and how it might reduce errors.
Your company currently uses a process with a similar cost of materials that has an average percent yield of 91 percent. If the average percent yield of this process is higher than that, this could save the company money. What is your recommendation to the company? Please support your recommendation using your data, calculations, and understanding of stoichiometry gathered from this lab.

Sagot :

Based on the data obtained from the reaction, the following conclusions can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

What is the percent yield of the reaction?

Equation of the reaction is given below:

  • 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

Moles of Mg used

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

Moles of MgO expected

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

  • molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

Percent yield

  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

Moles of Mg used

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

Moles of MgO expected

Based on the equation of reaction;

moles of MgO expected = 0.0170

  • moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

Percent yield

  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: https://brainly.com/question/1824546

Answer:

90.3

Explanation:

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