#10 SCH3U Lesson

docx

School

Lakehead University *

*We aren’t endorsed by this school

Course

SCH3U

Subject

Chemistry

Date

Feb 20, 2024

Type

docx

Pages

5

Uploaded by ProfessorIronTapir18

Report
Percentage Yield January 12, 2016 SCH3U: University Preparation Chemistry Lesson # 10
Percentage Yield January 12, 2016 Percentage Yield Lab Lakehead Adult Education SCH3U: University Preparation Chemistry
Percentage Yield January 12, 2016 Lab Determining the Percentage Yield of a Reaction Purpose - Determine the percentage yield of a precipitate. Chemicals Lead(II) nitrate Potassium iodide Results Item Mass (g)   Trial 1 Trial 2 Filter paper 1.36 1.29 Filter paper + PbI 2 5.35 5.50 PbI 2   3.99   4.21 Conclusion Based on the results from this lab it can be concluded that when lead (II) nitrate and potassium iodide are combine in solution a precipitate forms. This precipitate is lead (II) iodide. When this lab is completed with accuracy it is possible to get, or get very close to 100% yield of precipitate.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Percentage Yield January 12, 2016 Questions 1. What was your average yield of precipitate? Pb(NO 3 ) + 2Kl PBl 2 + 2KNO 3 Pb(NO 3 ) 2 Mol = __g___ mm = 3.31g 331.2g/mol = 0.0099mol Kl Mol = __g___ mm = 2g 166g/mol = 0.012mol Pb(NO 3 ) 2 Pbl 2 1 : 1 0.0099 X X = 0.0099mol Pbl 2 Theoretical Yield: 0.0099 mol *461.01 g/mol = 4.56g PbI 2 Trial#1 Trial#2 3.99 4.56 x100 = 87.5% 4.21 4.56 x100 = 92% Average Yield: 87.5 + 92 2 = 89.8% 2. Calculate the percentage yield of your reaction. Use the mass of lead(II) nitrate   to calculate the theoretical yield. Pb(NO 3 ) + 2Kl PBl 2 + 2KNO 3 Pb(NO 3 ) 2 Mol = __g___ mm = 3.31g 331.2g/mol = 0.0099mol
Percentage Yield January 12, 2016 Kl Mol = __g___ mm = 2g 166g/mol = 0.012mol Pb(NO 3 ) 2 Pbl 2 1 : 1 0.0099 X X = 0.0099mol Pbl 2 Theoretical Yield: 0.0099 mol *461.01 g/mol = 4.56g PbI 2 Trial#1 Trial#2 3.99 4.56 x100 = 87.5% 4.21 4.56 x100 = 92% 3 . Give two reasons why the yield is below 100%. When transferring the solution from the beaker to the filter paper some of the precipitate could have been left behind causing a percent yield less than 100% When transferring the filter paper from the beaker to the hotplate or from the hotplate to the scale some of the precipitate could have been lost, thus causing a percent yield less than 100% 4 . State one way to improve upon the technique, and increase the percentage yield. When the residue went through the paper filter it would have affected the percent yield because not all the residue would have been captured in the filter and measured If I Could have gotten more residue off the beaker I would have measured more of the residue in the filter.