Organic Chemistry Lab 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-bromoaniline0222

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 handbook0230

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 reactants0240

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 because0450

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 and0735

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 product0793

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 numbers0962

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 reaction0981

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 product1223

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 it1259

and successfully calculate the numbers you need to calculate.1264

Good luck with your reactions.1267