I need help with the questions and help with calulating % yield. If I know the f
ID: 954237 • Letter: I
Question
I need help with the questions and help with calulating % yield. If I know the fomula I can plug in my results and find % yeild.
×Y, Aldehydes ×YDOak Park CexYRE The New Y 4. Ecology Lat ×Y-Experiment: × Dl > Iodination c × c https://mycourses purdue.edu/bbcswebdav/pid 6451408-dt content-rid-23467020_1 /courses/cal 65637215162/cal-65637215162 lmportedContent 201601cQ7 :: Apps Bookrmarks G Google S My PUC B Blackboard Leam ES Google Scholar y Twitter Lists Other bookmarks Psy Genetics -Sapling Learning In. lodination of Salicylamide 2016.pdf Fingerprint Infrared Absorptions for Substituted Aromatics Ring substitution pattern Expected peaks (cmSubstitution Pattern Example 770-715 (strong) 770-730 (strong) 820-760 (strong) 870-800 (strong) 1,2,-Trisubstituted 790-750 (strong) 1,2,4-Trisubstituted 850-800 (strong) 1,3,5-Trisubstituted 910-830 (strong) The numbers here apply to substitution patterns, not nomenclature. Experimental Procedure CAUTION: Use caution when handling the materials used in this experiment. Wear gloves and appropriate eye protection at all times during this procedure. Salicylamide and sodium iodide are irritants. Sodium hypochlorite and hydrochloric acid are irritants and are corrosive. Measure out 1.0 g of salicylamide and record the mass to the nearest 0.001 g. Place the ethanol, warming the flask with your hand to speed up the dissolution. reaction mixture, stirring with a glass stirring rod until the solution is homogeneous. about thermal contact!! Ice bath means ice AND water.) salicylamide into a 100-mL round-bottom flask, Dissolve the salicylamide in 20 mL of absolute Once the salicylamade is oompletely dissolved, add 1.2 g of sodium iodideExplanation / Answer
mass of salicyclalamide taken = 1.0 g
molecular mass of salicyclalamide = 137.136 g/mol
Hence moles of salicyclalamide = 1.0 g / 137.136 g/mol = 0.00729 mol
mass of sodium iodide taken = 1.2 g
molecular mass of NaI = 149.9 g/mol
Hence moles of NaI = 1.2 g /149.9 g/mol = 0.008005 mol
Volume of 6%(w/v) NaOCl taken = 9.2 mL
100 mL of the solution contains 6.0g NaOCl.
=> 9.2 mL of the solution that would contain the mass of NaOCl = (6.0 g / 100 mL) x 9.2 mL = 0.552 g
molecular mass of NaOCl = 74.44 g/mol
Hence moles of NaOCl = 0.552 g / 74.44 g/mol = 0.007415 mol
The balanced equation for the iodination reaction is
C6H4(OH)(CONH2) + NaI + NaOCl ------ > C6H3(OH)(CONH2)(I) + Na2O
1 mol --------------------- 1 mol, 1 mol ----------- 1 mol
Equal mol of each of C6H4(OH)(CONH2), NaI and NaOCl react with each other.
Since salicyclalamide, C6H4(OH)(CONH2) contains least number of moles( 0.00729 mol) it is the limiting reactant and decides the theoritical mass of product formd
Hence theoritical moles of the product, C6H3(OH)(CONH2)(I) formed = 0.00729 mol
molecular mass of C6H3(OH)(CONH2)(I) = 263.03 g/mol
Hence the theoritical mass of C6H3(OH)(CONH2)(I) formed = 0.00729 mol x 263.03 g/mol = 1.9175 g
Hence % yield = [(Actual mass of crystal foromed) / 1.9175] x 100 (formulae)
By putting the value of Actual mass of crystal foromed in the above formulae we can calculate the % yield