For more information, please see full course syllabus of Organic Chemistry Lab

For more information, please see full course syllabus of Organic Chemistry Lab

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### Completing the Reagent Table for Prelab

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Sample Reagent Table 0:11
- Reagent Table Overview
- Calculate Moles of 2-bromoaniline
- Calculate Molar Amounts of Each Reagent 9:20
- Calculate Mole of NaNO₂
- Calculate Moles of KI
- Identify the Limiting Reagent 11:17
- Which Reagent is the Limiting Reagent?
- Calculate Molar Equivalents 13:37
- Molar Equivalents
- Calculate Theoretical Yield 16:40
- Theoretical Yield
- Calculate Actual Yield (%Yield) 18:30
- Actual Yield (%Yield)

### Chemistry: Organic Chemistry Lab

I. Reagent Table | ||
---|---|---|

Completing the Reagent Table for Prelab | 21:09 | |

II. Melting Points | ||

Introduction to Melting Points | 16:10 | |

Melting Point Lab | 8:17 | |

III. Recrystallization | ||

Introduction to Recrystallization | 22:00 | |

Recrystallization Lab | 19:07 | |

IV. Distillation | ||

Introduction to Distillation | 25:54 | |

Distillation Lab | 24:13 | |

V. Chromatography | ||

Introduction to TLC (Thin-Layer Chromatography) | 28:51 | |

TLC Analysis Lab | 20:50 | |

VI. Extractions | ||

Introduction to Extractions | 34:25 | |

Extraction Lab | 14:49 | |

VII. Spectroscopy | ||

Infrared Spectroscopy, Part I | 1:04:00 | |

Infrared Spectroscopy, Part II | 48:34 | |

Nuclear Magnetic Resonance (NMR) Spectroscopy, Part I | 1:32:14 | |

Nuclear Magnetic Resonance (NMR) Spectroscopy, Part II | 2:03:48 | |

Mass Spectrometry | 1:28:35 |

### Transcription: Completing the Reagent Table for Prelab

*Hi and welcome back to www.educator.com.*0000

*Today, we are going to talk about how to complete the reagent table in your lab notebook,*0002

*before you do a synthesis experiment.*0007

*Here is a sample reagent table and there is a reaction taking place here.*0012

*For right now, it is not important to understand how this reaction takes place.*0018

*What you do need to know is that all of the reagents react an equal molar amount.*0022

*In other words, it takes one molecule of 2-bromoaniline and one equivalent of sodium nitrite, and HCl.*0031

*I’m sorry, not HCl, we have that in excess.*0041

*But we need the sodium nitrite and the potassium iodide, all 1 to 1 to 1 to make one product of this iodide.*0044

*As long as we know the stoichiometry of the reaction then that is all we need to do our calculations.*0055

*I put together a sample reagent table.*0062

*This is not necessarily the format that every instructor wants.*0065

*Make sure you follow the format that is required for your instructor.*0068

*What we have listed here are the reagents.*0073

*This is like our ingredient list for the recipe of the reaction that we are going to be doing.*0076

*This first item listed, I would describe as my starting material or maybe the reactant.*0081

*That is the organic species that is undergoing synthetic transformation.*0095

*These guys, I would all describe as reagents.*0101

*These are the chemicals that are causing the reaction to take place.*0106

*The water in this case is my solvent.*0110

*Almost all reactions on are run in some kind of solvent.*0113

*In this case, it is water used as an organic solvent.*0117

*Anyway, we are going to list everything that goes into the reaction at the start.*0120

*Some instructors also want included in the reagent table all of the materials used in the work up procedure.*0125

*You are kind of making sure that you know everything about everything you are using that whole day.*0131

*But in this table, all I have are the actual reagents that are used for the transformation to take place in the solvent.*0136

*That is going to be used when we mix all these together.*0144

*Very much like a recipe list, when you are going to bake some cookies.*0147

*I also have a separate row.*0154

*I separated it by a line there, I have separate row to list my product.*0157

*The product is not something that is going into the reaction.*0161

*It is something we are going to be isolating at the end of the reaction.*0164

*But this is a very nice place to include information about your product.*0166

*This is kind of one stop shopping for all of your physical properties in your information.*0170

*It is a good practice to get into to include your product here.*0173

*For each of these, I have listed the molecular weights, the formula for each of the compounds.*0178

*HCL is being added as an aqueous solution.*0185

*We do not care about its molecular weight here.*0190

*We could maybe put the molarity.*0193

*I did not include that in this example because we are just using the big excess of that.*0194

*But this is maybe where we put the molarity, you have to calculate the moles of HCl, if you needed to.*0198

*Water, I did not include the molecular weight of water because that is my solvent,*0205

*that is not being used stoichiometrically in the reaction.*0209

*That is not relevant information.*0212

*Each instructor is different, they might want all of your formula weights here but that was not important to me.*0215

*And then, I have a column for density.*0220

*Density, I only put liquid compounds which are 2-bromoaniline*0222

*happens to be a liquid and their product also happens to be a liquid.*0227

*I have looked up the densities for those so that you can use the CRC handbook*0230

*or something like that to find densities or molecular weights, formula weights.*0235

*But really, you could use like Wikipedia for this, most common reagents and reactants*0240

*you are going to find organic compounds.*0249

*You are going to find Wikipedia pages for and the data that is in that is really robust, it is really reliable.*0251

*It is a great place to go to find physical properties and molecular masses and all that kind of stuff.*0257

*Do not hesitate to use that if you want to, as well.*0263

*These amounts are coming from the experimental procedure.*0265

*Read through the experimental procedure and it said something like 2ml of 2-bromoaniline are dissolved in 10ml of water.*0269

*And then, 1.9g of sodium nitrite was added to that, and so on.*0276

*These are the amounts that you need to get from the procedure.*0281

*Notice that I'm including the units here because each of these has their own units.*0286

*Now the units for molecular mass are known, those are grams per mol.*0291

*The formula would always tells us how many, if you had 1mol of this compound, how many grams would it be.*0297

*Density is always given in grams per milliliter.*0306

*If you have a hard time remembering that unit, remember the density of water is 1.*0310

*If you have 1g of water, it is going to be 1ml of or cc, or is that going to be a liter of water?*0316

*That does not sound right, a gram is a very small amount.*0324

*1cc or 1ml of water only weighs 1g.*0327

*That is how we can remember the units for density is g/ml.*0331

*Then, we have a column here and I do not have an amount for my product,*0339

*because the procedure did not tell me how much product I’m going to get.*0342

*I have to calculate that.*0348

*And then I have a column for moles; what I'm going to do is convert all of these amounts to mole,*0350

*so that we compare one species to another.*0354

*I have another column for equivalents, molar equivalents, and that is just a more simple way to express this molar amount.*0357

*We will talk about that too.*0365

*And then, I always just have this final column called remarks and that is where I put any other relevant data.*0366

*This is a place where you can put hazards.*0372

*For example, that HCl is corrosive, we want to put a note of that.*0376

*You can put melting points for solids, boiling points for liquids.*0381

*Any information here, if you are reading through the procedure and says all watch out for this or take care about this.*0385

*This is a place to put that.*0392

*Again, some instructors will always want a column for boiling point and a column for melting point, and so on.*0394

*For me, this is kind of a catch-all place to put anything else that is pertinent about each species.*0399

*Let us start doing our reagent table calculations.*0405

*One by one, the first thing we want to do is convert these amounts to moles.*0410

*We can start with our first one listed, our 2-bromoaniline is our starting material.*0416

*We need to figure out how many moles that is but we are given the volume.*0422

*We know that we are using 2ml of 2-bromoaniline.*0427

*You cannot convert directly from ml to mol.*0431

*But we could use the density to convert from volume to mass.*0435

*And then, we can use the molecular weight to go from mass to mol.*0440

*What we are going to do is, this is 2.0 ml of the 2-bromoaniline.*0444

*Do not forget to use units but then also because*0450

*we have so many species that we are dealing with in this particular reaction,*0454

*you also want to include who it is you are looking at.*0458

*This should also be included in your lab notebook.*0462

*This should be something shown in your calculations, so that you could track back if you have an error anywhere.*0464

*How do we go from ml to g?*0469

*I know it is 1.56g for every 1ml.*0473

*Every 1ml weighs 1.56g.*0479

*Remember, it is g/ml.*0481

*I want to put it like this so that the ml cancels.*0484

*In other words, I’m going to multiply my volume in ml × my density to now being g.*0488

*Now I want to go from g to mol.*0496

*I know it is 172g for every mol.*0498

*I’m going to put the 172g in the denominator so that those will cancel.*0501

*Every 1mol has 172g of 2-bromoaniline.*0506

*Now my grams cancel.*0511

*And then, when we do that math, we get 0.018 mol of 2-bromoaniline.*0515

*It looks like there is two sig figs through each of this.*0524

*It is reasonable why I have two sig figs here, maybe an extra one.*0528

*Again, each instructor is a little different about how much they care about this.*0531

*This is also the same as saying 18mmol.*0534

*A lot of times our reactions are done at very small scale.*0538

*It makes sense to talk in terms of mmol instead of mol.*0542

*If you want to change the title of this column to be mmol, you can do that if you want it.*0545

*We will keep it as mol, I will say 0.018 is our mol of 2-bromoaniline.*0551

*Let us move on to our next one.*0558

*Now let us calculate the mol of NaNO₂.*0562

*Now we are given a mass of NaNO₂.*0566

*We are starting with 1.9g of NaNO₂.*0569

*So we can use our formula weight to immediately convert from grams to mol.*0573

*It is 69g of NaNO₂, for every 1 mol.*0580

*Again, in effect what we are doing is dividing by the formula weight.*0586

*If you always do the work and you can see that your units cancel,*0589

*then you do not have to memorize whether you are multiplying or dividing.*0594

*You just always put the numbers in the right place so that our units are canceling.*0597

*Let us see when we do that math, we get 0.028 mol of NaNO₂ or 28mmol of NaNO₂.*0602

*We are just going to fill out our chart, our reagent table like this.*0612

*Like I said, the HCl is just in excess, I already put that in there because it was not significant.*0616

*But if you had your molarity here, you could calculate your mol, and so on.*0622

*But sometimes, things are in catalytic amounts or huge excess amounts.*0626

*You are not required to calculate those.*0631

*How about finally mol of potassium iodide, our last reagent?*0635

*We start with 3.6g of KI.*0639

*Every 1mol of KI, we can put in our species here to weigh 166g.*0644

*Our grams cancel and we are going to do 3.6 divided by 166, we get 0.022 mol of potassium iodide.*0654

*We put that in our table as well.*0665

*One by one, we are going to calculate the mol of every one of our reagents that are added.*0667

*I do not care how many mol of solvent there are, that is not relevant.*0672

*It is not involved in the reaction.*0675

*The next thing we need to do is we need to figure out which of these reagents is the limiting reagent.*0679

*What is the limiting reagent mean?*0685

*It is the reagent with the smallest number of mol.*0687

*Now we have to be careful because sometimes a reagent is not used stoichiometrically in the reaction.*0690

*Sometimes an acid catalyst is needed or a base catalyst, perhaps.*0697

*We only need a catalytic amount meaning we do not need a full equivalent of that,*0701

*in order to convert 1mol of starting material of the product, we do not need a whole mol of the catalyst.*0707

*Even though we have a very small amount of that, that is not going to limit the amount of product that we can get.*0712

*Because remember, a catalyst is not consumed in the reactions.*0719

*Even though, I have a very tiny amount of it, it will get used but then it will get regenerated.*0722

*I can keep using that tiny amount and convert that whole amount of the starting material to product.*0729

*Keeping in mind that we are going to ignore the catalyst and*0735

*this is where we do need to know for this mystery reaction that we have never seen before, that each of these is used.*0737

*The 2-bromoaniline and the NaNO₂ and the KI are all used stoichiometrically in a 1 to 1 ratio.*0746

*In that case, we are looking for which of these have the smallest number of mol.*0752

*That is what is going to determine the mol of product, we call that the theoretical yield.*0758

*I only had 18 mmol of the 2-bromoaniline.*0765

*I have 28 mmol and then 22 mmol.*0768

*These guys are in excess. We are not going to run out of these before we ran out of the 2-bromoaniline.*0772

*This is the number, we are going to take this.*0781

*This is our limiting reagent.*0784

*This is our limiting reagent because it is the smallest number.*0790

*We are going to take that and we are going to copy that number down here to the mol of product*0793

*because that is the maximum amount of product I can get.*0800

*If every single molecule of the 2-bromoaniline gets converted to product, the most I can get is 18 mmol.*0803

*Now we know how many mol of product we can get.*0812

*We can use that to calculate the amount that we have.*0818

*Before we get there, I want to talk about this last column, the molar equivalents.*0825

*What we are going to do, now that we have identified the limiting reagent,*0830

*we are going to describe that reagent as being 1.0 equivalent.*0833

*We are going to describe all the other reagents with respect to that.*0840

*You know, compared to the 1.0 as a molar ratio.*0846

*What we are going to do is we are going to divide all the molar amounts by that limiting reagent molar amount to see what we have.*0849

*First of all, we can see that if we have exactly one equivalent of the 2-bromoaniline,*0856

*that will give us exactly one equivalent of our product.*0861

*But we see that we have an excess of the NaNO₂.*0864

*If we take 0.028 and we divide that by the limiting reagent amount 0.018,*0867

*then that comes out to 1.6 equivalent of NaNO₂.*0876

*In other words, there is a 60% excess of NaNO₂ is used.*0884

*We wanted to make sure that we used up all the 2-bromoaniline possible,*0894

*so we threw an extra amount of the sodium nitrite, and we use a 60% excess.*0899

*We can put here 1.6.*0904

*You want to make sure you have at least 1 to 2 decimal points here.*0907

*Because sometimes you have just maybe a 10% excess or maybe even a just a 5% excess.*0911

*You want to go out as many decimal points as possible,*0917

*to see whether not you have exactly one equivalent or maybe you do have a little in excess.*0920

*We could do the same thing for the potassium iodide.*0926

*If we divide the 22mmol by the 18mmol, we get 1.2.*0930

*We have 1.2 equivalents of KI, that has a smaller excess, smaller amount of that.*0937

*But still an excess amount, there are some KI leftover but not quite as much.*0944

*We just say a 20% excess of KI is used.*0948

*What is nice about these equivalents is you can look very quickly at this table.*0955

*Rather than looking at all these complicated molar numbers which are very tiny numbers*0962

*and try to figure out what is the ratio of these numbers,*0967

*now you can look at this very quickly and say I used twice as much of the sodium nitrite.*0971

*Maybe I used 10 equivalent, maybe I use a very slight excess 1.2 equivalent.*0976

*This column is very nice, specially if you want to scale up the reaction*0981

*or someone else is going to do the reaction at a different scale.*0986

*Then, they can just use this to say however much 2-bromoaniline,*0989

*I need 1.6 times that many mol of the NaNO₂, and so on.*0993

*The last thing that we had figure out here is our theoretical yield.*1002

*That is how much product can we expect, what is the maximum amount of product we can expect?*1006

*We already know that number in terms of mol.*1011

*But in the end of the day, we are going to have our product in our round bottom flask and we are going to weigh it.*1014

*We are going to find out how many grams we have of that.*1019

*Or maybe by volume, we will find out what volume we have of that.*1020

*We need to calculate that.*1024

*We are going to use that molar amount to calculate this number.*1026

*We are going to have to include units in that because, remember, this amounts column,*1030

*just putting in number 6 here does not mean anything as that 6.*1036

*Kilograms is that 6.*1040

*Liters is that 6, ml, include the units there.*1042

*How do we do that?*1046

*We know we are expecting 0.018 mol of the 2-iodobromobenzene.*1047

*That is the amount that we can make and we know that every 1mol of the 2-iodobromobenzene weighs 263g.*1063

*I’m sorry, 283g.*1075

*Our mol cancel, when we do that math we get 5.1g of product is our theoretically yield.*1078

*That is the maximum amount of product we can get.*1085

*We are going to put that in this table here.*1088

*If we wanted, we can use the density to convert that to volume and get that in volume,*1093

*if you are going to maybe measure it with a graduated cylinder or something.*1102

*Mass is usually the way that we measure how much we have of something.*1105

*Once the reaction is over, we can come back to our reagent table and we can find out what our actual yield is.*1115

*We call that the percent yield of the reaction.*1121

*Now that I have run the reaction, I have isolated my 2-iodobromobenzene.*1124

*I have purified it, I now have it in my flask.*1129

*Now I can report how successful is my reaction.*1133

*How much of my product did I get, compared to the theoretically yield that I was expecting, the maximum amount.*1137

*Let us say we ran the reaction, let us say we isolated 4.3g of product.*1143

*If we isolated 4.3g of product, what is the percent yield that we will report for that reaction?*1156

*We very simply take the actual yield in grams and divide it by the theoretical yield, and we convert it to a percentage.*1161

*It is quite a simple description there.*1170

*4.3 g is the actual, 5.1g was the theoretical.*1175

*When I do that math, I get 0.843 as my decimal number.*1180

*By multiplying that by 100%, it converts that to a percentage, in a yield form.*1187

*We get 84.3.*1194

*Now again, significant figures, to go beyond, you need the three here.*1196

*Just saying 84% yield is probably reasonable but certainly not saying 84.3215.*1201

*Those are not significant numbers.*1207

*This will tell you I was expecting the most that I could have gotten was 5g, but again, at 84% yield.*1210

*Most reactions do not go to 100%.*1216

*Even if they do, by the time you isolate and purify your product, some is lost along the way.*1218

*It is very difficult to get a reaction where you actually get a quantitative yield of the product*1223

*and get 100% yield of the product.*1227

*Whatever you get is what you report.*1231

*You do not want to try and make up a number thinking that this does not look so good.*1233

*A lot of times we will find, a typical student yield for a given reactions, maybe 60%.*1237

*It is okay if you get a number like that.*1243

*You just want to make sure you are reporting the actual yield of whatever your actual yield is.*1245

*Hopefully this helps, you get an idea of how to calculate your reagent table.*1249

*Definitely pay attention to how your particular instructor wants the reagent table set up.*1254

*But given that, hopefully, you have gone through the calculations that will help you successfully navigate it*1259

*and successfully calculate the numbers you need to calculate.*1264

*Good luck with your reactions.*1267

1 answer

Last reply by: Professor Starkey

Sat Aug 25, 2018 7:01 PM

Post by Dakla de Souza Lima on August 15 at 11:41:14 AM

Professor, if my ratio it isn't 1:1, how that I do to find the molar equivalent?

1 answer

Last reply by: Professor Starkey

Sun Apr 15, 2018 12:45 PM

Post by Peter Garney on April 14 at 11:18:42 AM

Is this the only lecture on ogarnic lab

1 answer

Last reply by: Professor Starkey

Fri Oct 21, 2016 12:46 AM

Post by Hector Flores on October 21, 2016

This question is not concerning this lecture...you dont have any lab videos on the grignard lab? :(

1 answer

Last reply by: Professor Starkey

Sat Sep 5, 2015 2:43 PM

Post by David Steele on September 5, 2015

are you going to talk about this. https://en.m.wikipedia.org/wiki/Enzyme_kinetics

where can i learn about this.

5 answers

Last reply by: Professor Starkey

Thu Jul 23, 2015 3:07 PM

Post by Akilah Futch on July 22, 2015

Professor, is it possible to calculate Moles if mass(g) is not given, but molarity is given?

1 answer

Last reply by: Professor Starkey

Fri May 1, 2015 1:57 PM

Post by Rene Whitaker on May 1, 2015

Wow, I wish this had been available when I started organic chemistry! I have been watching your other o-chem videos religiously to get me through the lecture class (with A's so far...orgo II final is in 3 days)! Just want to say I am forever grateful to you for the help!!! Unfortunately, I still have less than stellar synthesis skills...sigh.