Sample Results:
Mass (g)
Trial 1 Trial 2
Filter paper 1.36 1.29
Filter paper + PbI2 3.20 3.09
PbI2 ? ?
Calculations and Discussion Questions: ( show the step by step)
1. Calculate the mass of precipitate collected in Trial 1 and Trial 2. Include these values in your data table.
2. Calculate the average yield of precipitate in this experiment.
3. Calculate the theoretical yield of precipitate that was expected based on the initial concentration of the reactants. Note: Use the moles of KI when calculating the theoretical yield of precipitate possible, as potassium iodide is the limiting reactant.
4. Calculate the percentage yield of precipitate in this experiment.
5. State two reasons why the yield is below 100%.
6. State one way to improve upon the experimental technique that could increase the percentage yield.
1. To calculate the mass of precipitate collected in Trial 1 and Trial 2, subtract the mass of the filter paper from the mass of filter paper with PbI2.
Trial 1:
Mass of precipitate = Mass of filter paper + PbI2 - Mass of filter paper
= 3.20g - 1.36g
= 1.84g
Trial 2:
Mass of precipitate = Mass of filter paper + PbI2 - Mass of filter paper
= 3.09g - 1.29g
= 1.80g
2. To calculate the average yield of precipitate, add the mass of precipitate from Trial 1 and Trial 2, and divide by 2.
Average yield of precipitate = (Mass of precipitate in Trial 1 + Mass of precipitate in Trial 2) / 2
= (1.84g + 1.80g) / 2
= 1.82g
3. To calculate the theoretical yield of precipitate, you need to calculate the moles of KI present in the reaction. Since KI is the limiting reactant, the moles of KI will determine the maximum amount of precipitate that can be formed.
First, convert the mass of KI used to moles using the molar mass of KI.
Molar mass of KI = 39.10g/mol (potassium) + 126.90g/mol (iodine)
= 166.00g/mol
Theoretical moles of KI = Mass of KI used / Molar mass of KI
= Mass of KI used / 166.00g/mol
Theoretical yield of precipitate = Theoretical moles of KI * Molar mass of PbI2
= Theoretical moles of KI * (Molar mass of Pb + Molar mass of I2)
= Theoretical moles of KI * (207.20g/mol + 253.80g/mol)
= Theoretical moles of KI * 461.00g/mol
4. To calculate the percentage yield of precipitate, divide the actual yield (average mass of precipitate) by the theoretical yield, and multiply by 100.
Percentage yield of precipitate = (Average yield of precipitate / Theoretical yield of precipitate) * 100
= (1.82g / Theoretical yield of precipitate) * 100
5. Two reasons why the yield is below 100% could be:
a) Incomplete reaction: The reaction may not have gone to completion due to factors such as side reactions or impurities in the reactants.
b) Loss of product: Some of the precipitate may have been lost during the filtration or transfer process, leading to a lower yield.
6. One way to improve upon the experimental technique and increase the percentage yield could be to ensure thorough mixing of the reactants to promote a more complete reaction. Additionally, minimizing any possible loss of product during filtration or transfer can also help improve the yield.
To answer these questions, we need to perform a series of calculations based on the given data. Let's go through each step:
1. To calculate the mass of precipitate collected in Trial 1 and Trial 2, we subtract the mass of the filter paper from the mass of the filter paper + PbI2.
- In Trial 1: Mass of precipitate = Mass of filter paper + PbI2 - Mass of filter paper = 3.20g - 1.36g = 1.84g
- In Trial 2: Mass of precipitate = Mass of filter paper + PbI2 - Mass of filter paper = 3.09g - 1.29g = 1.80g
Data Table Update:
Mass (g)
Trial 1 Trial 2
Filter paper 1.36 1.29
Filter paper + PbI2 3.20 3.09
PbI2 1.84 1.80
2. To calculate the average yield of precipitate in this experiment, we sum up the masses of precipitate from Trial 1 and Trial 2, and then divide by the number of trials.
- Average yield = (Mass of precipitate in Trial 1 + Mass of precipitate in Trial 2) / 2 = (1.84g + 1.80g) / 2 = 1.82g
3. To calculate the theoretical yield of precipitate, we need to know the mole ratio between PbI2 and KI in the reaction and the initial concentration of KI. Once we have these values, we can convert the moles of KI to PbI2 and calculate the mass.
- For example, if the balanced chemical equation is 2KI + Pb(NO3)2 -> 2KNO3 + PbI2, we know that the mole ratio between PbI2 and KI is 1:2.
- Let's say the initial concentration of KI is given as 0.1 M. To find moles of KI, we multiply the concentration by the volume used (in liters).
- Theoretical moles of KI = 0.1 M x [volume of KI solution (in L)]
- From the balanced equation, we know that 2 moles of KI give 1 mole of PbI2, so the theoretical moles of PbI2 = Theoretical moles of KI / 2
- Theoretical mass of PbI2 = Theoretical moles of PbI2 x molar mass of PbI2
This step requires additional information that is not provided in the question. The concentration of KI and the volume used need to be known to proceed with this calculation.
4. To calculate the percentage yield of precipitate, we divide the actual yield (average yield from step 2) by the theoretical yield (step 3) and multiply by 100.
- Percentage yield = (Average yield / Theoretical yield) x 100
This step also requires the theoretical yield (step 3) to be known.
5. Two reasons why the yield is below 100% could be:
a) Incomplete reaction: Not all of the reactants may have reacted to form the desired product. Side reactions might have occurred or there could be impurities in the reactants.
b) Losses during the experiment: Some of the precipitate could have been lost during filtration or transfer, leading to a lower yield.
6. One way to improve upon the experimental technique and increase the percentage yield could be:
- Better mixing: Ensuring thorough mixing of reactants can promote complete reaction and improve yield.
- Using a catalyst: Adding a catalyst can increase the rate of reaction and make it more efficient.
- Optimization of reaction conditions: Adjusting factors such as temperature, pressure, or concentration can sometimes increase the yield of the desired product.
In this investigation, you will study the reaction between lead(II) nitrate, Pb(NO3)2, and potassium iodide, KI, to determine the percentage yield of lead(II) iodide. This reaction occurs based on the following equation:
This image shows Pb(NO sub 3 ) sub 2(aq) + 2KI sub (aq), reaction arrow, Pbl sub 2(s) + 2KNO sub 3(aq).
Purpose:
To determine the percentage yield of lead(II) iodide precipitate produced in a reaction between lead(II) nitrate and potassium iodide.
Materials:
Chemicals:
1.0 mol/L lead(II) nitrate solution
1.0 mol/L potassium iodide solution
Apparatus:
pipettes (10 mL), pipette bulb, beaker (250 mL), Erlenmeyer flask (250 mL), funnel, filter paper, electronic balance, wash bottle with distilled water
Safety:
Safety goggles and lab aprons should always be worn when working with chemicals.
Procedure:
1. Use a pipette to measure 10.0 mL of the each of the two solutions and add to the beaker. Use a different pipette for each solution to avoid contaminating your stock solutions.
2. Swirl gently.
3. Determine the mass of the filter paper. An accurate measurement is very important.
4. Fold the filter paper in four and place it in the funnel supported by a flask.
5. Slowly pour the mixture into the funnel. Be careful not to overflow the filter paper. Use a wash bottle of distilled water to wash any remaining solid out of the beaker and onto the filter paper.
6. After filtering, let the paper dry overnight.
7. When it is completely dry, determine it's mass. Once again, accuracy is important. Be careful not to remove any of the solid product on the filter paper.
8. If time permits, repeat the entire procedure for a second time for accuracy.