101 Ex. #4 ReportTemplate

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1 Report for Laboratory Exercise #4 Synthesis of Tetramethylammonium Triiodide and Tetramethylammonium Pentaiodide _ Using Microsoft Word, students are to insert responses in all yellow highlighted areas. It is recommended that the report be completed without changing font size, column width, row width, margins and highlights. The completed report must be uploaded to the 101 Brightspace site as a .pdf file by the due date posted on Brightspace. All answers must be the student’s own work without assistance from others. Only reports which are completed using the template will be marked. Name: Ton Tran Lab Section: B02 Quad 4 Date: November 2, 2021 Abstract By reacting I 2 with NMe 4 I in solution with control of the reaction stoichiometry, Me 4 NI 3 and Me 4 NI 5 were synthesised. The % yield for Me 4 NI 3 was 58.8%, and the % yield for Me 4 NI 5 was 64.2%. Data/Results Table 1. Experimental data and calculated values for the preparation of Me 4 NI 3 and Me 4 NI 5 Synthesis Me 4 NI 3 Me 4 NI 5 NMe 4 I (g) 0.141g 0.320g I 2 (g) 1.78g 0.808 actual yield (g) 0.527g 1.129g theoretical yield (g) 0.896g 1.759g % yield 58.8% 64.2%

2 Algebraic Equations (see page 12 of the lab manual) a) Balanced chemical equations for formation of each of the products: a. Me 4 NI + I 2  Me 3 NI 3 b. Me 4 NI + 2 I 2  Me 3 NI 5 b) Theoretical yield for each of the products = a. LR = Limiting Reagent b. C = Coefficient c. P = Product i. LR * (LR C / P C) c) % yield of each of the products = a. E = Experimental b. M = Mass i. E M Me 3 NI 3 / (LR * (LR C / P C)) ii. E M Me 3 NI 5 / (LR * (LR C / P C)) Discussion Respond to the following: Discuss the meaning of the % yield, do not just give a formula and comment on the % yield observed for each of the products, include the actual values (max 4 lines). The % yield is produced yielded in experiment vs the max product yield. Me 4 NI + I 2  Me 3 NI 3 could have a total yield if done 100% correct, 0.896g. Experimental Me 4 NI + I 2  Me 3 NI 3 had a total yield of 0.527g. Only 58.8% of the max 0.896g was yielded. Me 4 NI + I 2  Me 3 NI 3 can have a total yield if done 100% correct, 1.759g. Only 64.2% of the max 1.759g was yielded Provide the limiting reagent for each of the preparations. Briefly state how the limiting reagent was determined for each of the preparations . ( max. 3 lines). Find minimum number of moles of both reactants for reaction to work. Smaller mol = Limiting reagent. Me 4 NI + I 2  Me 3 NI 3 Limiting reagent was I 2 . Me 4 NI 2.49*10 -3 g/mol > I 2 1.97*10 -3 g/mol Me 4 NI + 2 I 2  Me 3 NI 5 , Limiting reagent was Me 4 NI. Me 4 NI 2.48*10 -3 g/mol < 2 I 2 2.56*10 -3 g/mol

3 Conclusions The yield % of the reaction Me 4 NI + I 2  Me 3 NI 3 was 58.8%, and the yield % for the reaction Me 4 NI + 2 I 2  Me 3 NI 5 was 64.2% References Reimer, M. et al, Laboratory Manual, Chemistry 101 , pp. 37-40. (University of Victoria: Victoria, B.C.). Fall 2019 . Feedback Summary max. Pre-lab quiz: Are all responses correct? 3 Laboratory Notebook: Have all data, observations and procedures been recorded? 1 Report: Are all sections completed accurately? Are the algebraic equations correct? 1 Participation: Did the student come prepared, was time used well in lab and was student engaged in the experiment? Did the students request the TA to check their drawers for completeness before they left the lab? 1 Performance evaluation: Did student follow the safe practice guidelines throughout the whole lab period? 1 Total mark 7

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