General Chemistry Stoichiometry I

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be another...0003

It is one of those central topics that you are going to be using for the rest of your general chemistry career.0007

That is going to be the concept of stoichiometry.0014

This is going to be the first of two lectures on stoichiometry.0017

Let's go over the lesson overview.0024

What is very core to stoichiometry problems is what is called a mole to mole ratio.0027

This is what we are going to introduce right from the start.0033

After we introduce the mole to mole ratio, we are going to go through0038

some traditional types of problems and sample calculations you can encounter.0041

Moreover the mole to mole ratio can be used to solve0048

a very common type of problem which is called the mass to mass conversion.0053

Right after the mass to mass conversion, we are going to encounter0060

another unique type of problem which is called a limiting reactant.0064

In addition to that, we are going to be discussing what is called a percent yield.0069

I am going to not only show you how to do it.0073

But more importantly, how do you know when you are dealing with a limiting reactant problem?0075

Because it involves a different strategy of its own.0079

We are going to finish up with a brief summary as always.0083

Then we will go ahead and tackle some sample problems together.0087

What exactly is a mole to mole ratio?0094

In chemical formulas, that is the first place where we are going to look.0098

In chemical formulas, the subscripts can actually be interpreted as the following.0102

The moles of that element for every one mole of compound.0108

For example, in one mole of water, there are two moles of hydrogen and there is going to be one mole of oxygen.0113

We literally can translate the subscripts in terms of moles.0128

That is going to make calculations a lot easier to interpret later on.0135

We not only have to deal with one mole of a compound.0141

For example, let's go ahead and look at the next question.0144

It is the same question: how many moles are there of each element?0147

But not in one mole of water but this time in two moles of water.0150

Remember the concept of dimensional analysis.0156

We saw that dimensional analysis used what was called a conversion factor.0160

In this whole lesson, conversion factors are key.0166

We are going to be doing dimensional analysis so much in this chapter.0175

But the nice news about this is that you already know how to do dimensional analysis.0182

You already know how to use conversion factors.0186

It is the same mathematical tools that we were introduced to several lectures ago.0189

Basically 2.6 moles of water is given to me; 2.6 moles of water.0195

Let's go ahead and get the moles of hydrogen.0204

2.6 moles of water times something over something, that is going to give me my moles of hydrogen.0208

Let's go ahead and enter the conversion factor.0218

I want moles of water to cancel; so moles of water goes downstairs.0220

I want moles of hydrogen to go upstairs to get carried through to the final answer.0224

Where do I get the numbers for the conversion factor?0231

You get it from the subscripts in the chemical formula.0233

For every one mole of water, there is going to be two moles of hydrogen.0237

We get an answer of 5.6 moles of hydrogen; that is it; it is that simple.0242

Let's go ahead and finish up the problem and now calculate the moles of oxygen that are contained in 2.6 moles of water.0250

2.6 moles of water times something over something equals to the moles of oxygen.0258

We are going to be using the same tools as before.0272

Moles of water is going to get cancelled; that goes downstairs.0276

Moles of oxygen now is going to the numerator.0280

That is going to get carried through to the final answer.0286

My mole to mole ratio is just 1:1, very conveniently.0289

In 2.6 moles of water, there are 2.6 moles of oxygen.0294

What we just did was we used and we came up with mole to mole ratios from a chemical formula.0300

Mole to mole ratios from a chemical formula.0311

The nice thing about mole to mole ratios is that they easily carry over from the chemical formula to a balanced chemical equation.0317

For example, in the following equation, nitrogen plus three hydrogens goes on to form two ammonias.0329

The coefficients in the balanced chemical equation, they are actually interpreted and treated as moles; we treat the coefficients as moles.0341

For example, let's go ahead and translate this.0354

One mole of N₂ requires three moles of H₂ for this reaction to occur.0357

Next line, we can expect two moles of ammonia to form from one mole of N₂.0366

Finally we can expect two moles of ammonia to form from three moles of H₂; from three moles of H₂.0372

That should be H₂; my apologies.0380

It is that simple; again we use the coefficients in terms of moles now.0386

This now leads us to our definition of stoichiometry.0394

If you look at everything in this phrase here, every sentence is relating one compound from the chemical reaction to another one.0399

Anytime you relate any two elements or molecules in a chemical reaction, this is what is called stoichiometry.0410

Once again this is what is called stoichiometry.0419

Another name to be specific for these coefficients, sometimes you will hear this, these are also called stoichiometric coefficients.0423

Again it is just a fancy name for what you already know how to do.0435

That is to use whole numbers to balance a chemical reaction.0439

Let's go ahead and now apply this to some problems.0447

We are going to take the following, the same reaction, the ammonia formation.0450

The question is the following: how many moles of ammonia can form if you have 3.1 moles of H₂?0456

I want you to get into habit; what type of problem is this?0465

This is stoichiometry because we are asked to relate one compound of the chemical reaction to another.0467

In other words, we are asked to relate ammonia with H₂.0475

We are going to use dimensional analysis again to do our problem.0481

3.1 moles of H₂ times something over something is going to give me the moles of ammonia that we can expect to form.0485

The moles of hydrogen goes downstairs to get cancelled.0500

The moles NH₃ goes upstairs to get carried through to the final answer.0508

What is going to make the numbers for the conversion factor?--the coefficients; that is it.0515

When I look here, it is basically a 3:2 ratio of hydrogen to ammonia.0523

We put the coefficient in front of ammonia there and you put the coefficient in front of H₂; just like that.0533

When all is said and done, we are going to get an answer of 2.1 moles of ammonia.0542

Again all we are doing is we are using the coefficients in a balanced chemical equation directly into our conversion factor.0550

It is that simple; let's go on to another illustration of this.0557

Same reaction, now how many moles of hydrogen gas are required to react with 6.4 moles of nitrogen gas?0563

Once again this is stoichiometry all the way through because we are asked to relate one compound of the balanced chemical equation with another.0572

6.4 moles of nitrogen times something over something is going to give me the moles of H₂ required.0583

Moles of N₂ is going to go downstairs to get cancelled.0600

Moles of H₂ is going to go upstairs to get carried through to the final answer.0606

Once again where do we get the numbers from?--we get it from the coefficients.0611

When I go ahead and look at the balanced chemical equation... that should be a three; this should be a two.0618

It is going to be three moles of H₂ for every one mole of N₂.0629

3 goes in front of the moles of H₂ and 1 goes in front of the moles of N₂.0636

When all is said and done, we are going to get 19.2 moles of H₂ required.0644

Once again what we just did was we were able to derive a conversion factor from the coefficients in the balanced chemical equation.0655

It is always more practical to work not in terms of moles but in grams.0668

Because when you go into the laboratory, that is what you are actually weighing.0672

So what we are going to do now is a mass to mass conversion.0675

What we previously did was we converted from moles of one compound to moles of another.0680

All we are going to do next is now convert from mass of one compound to mass of another.0686

That is usually of course going to be grams.0692

So we need a conversion factor that relates moles to grams; but we already know that.0700

Do you remember what the conversion factor is called that relates moles of one entity to grams of the same entity?0705

It is called molar mass; once again we are using nothing new.0715

We are using material that we have learned previously.0720

Remember, let's just go ahead and do a refresher.0724

If we start with the grams of one compound and we want to go to moles of that compound.0727

Grams goes downstairs and 1 mole goes upstairs; so we are essentially going to divide by molar mass.0736

If I start with moles of the compound and I want to go to grams of the compound.0744

1 mole goes downstairs and grams goes upstairs; I am essentially multiplying by molar mass.0750

Let's go ahead and apply this to a question now.0759

I am going to now work with the same identical question.0764

Except that you notice that I changed the units of moles to grams now.0768

How many grams of ammonia can form if you have 3.1 grams of H₂?0773

Before what we did was we did moles of A to moles of B.0781

But now we want to go from grams of A to grams of B.0790

All we are going to do, we are going to retain this very important step.0794

All we are going to do is put two steps on the outside now.0799

We are going to start with grams of A and then go to moles of A.0803

Once we are in moles of A, we are going to go to moles of B.0808

Then once we are in moles of B, we are going to finish up the problem and go on to grams of B.0811

We already know how to do this.0820

This is just a mole to mole ratio from the balanced chemical equation; that is what we just did.0822

If I want to go from grams of A to moles of A, I am going to divide by molar mass of A.0829

Here if I want to go from moles of B to grams of B, I am going to multiply by molar mass of B.0838

We are looking essentially at a minimum of three steps, three conversion factors.0844

Let's go ahead and do this; 3.1 grams of H₂, the first step is to go to moles of H₂.0850

Times 1 mole of H₂ divided by roughly 2.016 grams of H₂; that gets me as the moles of H₂.0861

Now from moles of H₂, I am going to go on to the next step to go to moles of ammonia.0870

That is going to be moles of NH₃ on top and moles of H₂ on the bottom.0876

That is going to be a 2:3 ratio; finally this gets me into moles of NH₃ after everything cancels.0883

Now I want to go to grams of NH₃; times something over something gives me grams of ammonia.0891

1 mole of ammonia goes downstairs and grams of ammonia goes upstairs.0898

This is going to be roughly 17 grams for the molar mass.0904

After following this three step process, we are going to get an answer of 17.4 grams of ammonia.0910

What we just did again is called a mass to mass conversion.0924

Grams of one compound to grams of another; it is a basic three step process.0929

Let's go ahead and do one more problem.0936

How many grams of hydrogen gas are required to react with 6.4 grams of nitrogen gas?0939

It is the same repetitive machinery; we are just going to go through the work now.0949

6.4 grams of N₂, the first step is to go from grams of A to moles of A.0956

Times something over something; grams of N₂ goes downstairs; 1 mole of N₂ goes upstairs.0965

The molar mass of molecular nitrogen is going to be roughly 28 grams of N₂.0973

After I am in moles of A, I then go on to moles of B.0980

Times something over something; moles of H₂ on top; moles of N₂ on the bottom.0983

Remember I get this mole to mole ratio from the balanced chemical equation, the coefficient.0992

That is going to be 3 moles of H₂ on top and 1 mole of N₂ on the bottom.0996

I am now in units of moles of B; let's go ahead and finish up now and go to grams of B.1000

Times something over something is going to give me grams of H₂ required.1005

1 mole of H₂ on the bottom; grams of H₂ on top.1012

The molar mass of molecular hydrogen is roughly 2.016.1016

When all is said and done, we are going to get an answer of 1.4 grams of hydrogen gas that is required.1021

Remember it is a basic three step process.1033

You go from grams of A to moles of A; moles of A to moles of B.1036

Then on to moles of B to grams of B; this is called a mass to mass conversion.1043

Because we are working with mass, that does not mean we have to end in grams.1053

Any of the mass units, you should be able to get.1058

Let me go ahead and redraw what we just did.1062

What we did was we went from grams of A to moles of A.1065

Then to moles of B; then to grams of B.1071

But remember we know the prefixes for a multiplier, such as kilo, such as milli; this is always important.1075

From grams of A, you can also work with milligrams, kilograms, etc.1086

Similarly for the mass of the product, you can also work with milligrams, kilograms, etc.1094

This is a nice cumulative problem; it really requires you to know those prefixes1102

that you are undoubtedly going to have to memorize; especially milli and kilo.1107

Let's go ahead and do an example.1113

How many milligrams of ammonia can form if you have 1.2 kilograms of H₂?1116

We are just going to follow this basic flow chart.1121

1.2 kilograms of H₂, we want to go to grams of A or grams of H₂.1128

Times grams of H₂ for every 1 kilogram of H₂.1136

Remember what the prefix multiplier is for kilo?--it is 10³.1140

Remember you always put the multiplier with the prefix-less unit; times.1144

Now I want to go from grams of A to moles of A.1152

Times moles of H₂ for every gram of H₂.1156

That is going to be roughly 2.016 for the molar mass of H₂.1162

Now I am in moles of H₂, moles of A.1167

Now I want to go to moles of B which is going to be moles of ammonia.1170

Times moles of ammonia for every mole of H₂.1174

Remember I get the mole to mole ratio from the balanced chemical equation.1180

That is going to be 2 here and 3 there.1183

Now I can go from moles of ammonia to grams of ammonia.1187

Times roughly 17 grams of NH₃ for every one mole of NH₃.1191

The question is asking for milligrams; so now I am going to multiply this by 1 mg over 10^-3 grams.1198

When all is said and done, I am going to get 6.7 times 10⁶ milligrams of ammonia.1212

That is expected to form from 1.2 kilograms of H₂.1219

Just be on the lookout for that again.1225

Just because the most common unit of mass we work with is grams, it doesn't mean we have to stop there or start there.1229

It is any of the mass units we can work with using stoichiometry and mole to mole ratios.1235

We now move on to a very specific type of stoichiometry problem.1244

This is what is called the limiting reactant and percent yields.1248

What we have done is the following before.1253

How many grams of ammonia can form if you have 3.1 grams of H₂?1256

You notice one thing is that only one reactant amount is specified; that is of hydrogen gas.1260

When you encounter these types of problems where only one reactant amount is specified, you are making several assumptions.1268

First, you have enough of the other reactant.1276

Second, there are no side reactions or unexpected occurrences.1280

Finally, third, say you are doing this in a lab.1286

You are assuming that everything goes perfect; there are no experimental errors at all.1290

If all goes well, you expect the maximum amount possible of product to form.1297

This is what we call the theoretical yield.1306

In practice, you never ever get the theoretical yield because errors happen all the time.1312

There are going to be inefficiencies in the system; you may get side reactions, etc.1318

Again the theoretical yield is never obtained; in practice never obtained.1324

Remember there is no such thing as a 100 percent efficient process.1335

Let's look at the following question.1346

How many grams of ammonia can form if you have 3.1 grams of H₂ and 3.1 grams of N₂?1348

This is the first time we have encounter this type of problem, when both reactant amounts is specified.1355

This is what we call a limiting reactant problem.1361

Because one of these reactant amounts is going to dictate, is going to limit how much product we can get.1364

To solve a limiting reactant problem, we are going to be using what is called the smaller amount reacted.1371

Let me go ahead and demonstrate.1377

We are going to first determine the amount of product that can form from each of the reactants; let's go ahead and do both.1380

3.1 grams of H₂ times 1 mole of H₂ divided by 2.016 grams of H₂.1389

Times 2 moles of ammonia for every 3 moles of H₂.1401

That is going to give me 1.03 moles of ammonia expected; remember that is the key word, expected.1415

What we are going to do now, we are going to repeat the process.1426

But for the other reactant amount, the 3.1 grams of nitrogen gas.1428

3.1 grams of N₂ times 1 mole of N₂ divided by roughly 28 grams of N₂.1432

Now times 2 moles of ammonia for every 1 mole of N₂.1441

That is going to give me 0.22 moles of NH₃ expected.1449

This is again called the smaller amount method for the following reason.1459

The route that forms the smaller amount of product is going to be what is called the limiting reactant.1464

It is going to be what dictates the amount of product formed.1476

We are not going to get the 1.03 moles of ammonia expected.1481

But instead we expect the smaller amount, the 0.22 moles of NH₃.1484

Because of this, we conclude that N₂ is a limiting reactant in this example.1489

Also H₂ therefore is something we have plenty of; we have more than enough.1504

The nitrogen gas, the limiting reactant, we don't have enough of it; that is why it is limiting.1511

That is why it is going to dictate how much product we are going to get.1516

We say that H₂ is in excess; we have too much of it and not enough of N₂.1519

The percent yield, the percent yield is the following.1528

This is what is called the actual yield divided by the theoretical yield times 100 percent.1534

The actual yield is what you physically get in the lab; this is from practice.1547

You are usually told this in the question; given.1558

The theoretical yield is what you always calculate from a mole to mole ratio.1562

In other words, it is the 0.22 moles; from a calculation.1568

In practice of course we want the percent yield to be as close to 100 percent as possible.1577

But that is just never ever going to happen.1581

What your percent yield tells you is pretty much how well of an experiment1586

you were able to pull off doing this type of synthesis problem.1593

What we just did was called the smaller amount method.1602

It is used to solve when two different reactants are given.1604

Another common type of question you can encounter is the following.1611

How many grams of the excess reactant is going to remain?1614

In other words, how much H₂ is going to be left over?1620

This is another type of common question you can be asked.1630

I have seen it asked many many times which is why I want to bring it up.1633

Basically what we are going to do is we are going to once again use stoichiometry mole to mole ratios.1637

Let's start with what we know for sure; we know for sure we are going to consume all of the limiting reactant.1646

We don't have enough of it, the 3.1 grams of N₂.1650

Basically we are going to go from 3.1 grams of N₂ to how much grams of H₂ required.1655

Once we get the grams of H₂ required, we simply subtract that from the amount that we have to get the amount left over.1667

Grams of H₂ initially have minus grams of H₂ required is going to give you the grams of H₂ remaining.1675

Let's go ahead and demonstrate; 3.1 grams of N₂ times 1 mole of N₂ divided by roughly 28 grams of N₂.1693

Now we go to moles of H₂; 3 moles of H₂ for every 1 moles of N₂.1707

Now we go to grams of H₂; 2.016 grams of H₂ for every 1 mole of H₂.1715

That is going to give us 0.67 grams of H₂.1723

What this number is, this 0.67 grams, this is the amount of H₂1727

that is going to react with the 3.1 grams of N₂... will react.1731

But the question is asking for now how much is going to react or how much is going to remain.1738

Now 3.1 grams of H₂ minus 0.67 grams of H₂.1743

That is going to say 2.43 grams of H₂ left over.1750

Once again we are not reinventing the wheel at all today.1757

We are simply using tools that we know already, dimensional analysis.1760

We are really using the concept of the mole to mole ratio incredibly heavily.1765

Let's go ahead and summarize before we get into the sample problems.1776

The mole to mole ratio is incredibly fundamental; I cannot underscore this enough.1780

We derive the mole to mole ratio from two sources today.1786

Number one was the chemical formula; number two was from a balanced chemical equation.1789

That is why you always have to make sure to balance your chemical equation.1794

Always make sure it is balanced if not done so for you already.1798

After introducing the mole to mole ratio, we then define stoichiometry.1802

Once again stoichiometry refers to relating the amounts of one compound to that of another in a chemical formula or a reaction.1807

Finally we introduced a very specific commonly asked stoichiometry problem, the limiting reactant.1821

How do you know you are doing a limiting reactant?1828

Because you are going to be specified two reactant amounts.1830

That is our summary for this first presentation of stoichiometry.1836

Let's now spend some time on doing some calculations and representative problems.1841

How many grams of carbon are in 2.2 kilograms of carbon dioxide?1848

How many grams of carbon are in so much carbon dioxide?1854

You don't see any chemical equation or there is no mention of any chemical equation.1858

This is using a mole to mole ratio from a chemical formula... mole to mole ratio from a chemical formula.1863

For carbon dioxide, it is basically 1 mole of carbon dioxide contains 1 mole of carbon.1878

And 1 mole of carbon dioxide contains 2 moles of oxygen.1886

So 1 mole of CO₂ for every 1 mole of carbon.1890

And 1 mole of CO₂ for every 2 moles of oxygen.1894

Remember we use the subscripts directly as moles from a chemical formula.1900

Now that we have our conversion factors that we can possibly use, let's go ahead and solve using dimensional analysis.1906

2.2 kilograms of carbon dioxide, remember we want to get this to grams first.1912

Times 10³ grams of CO₂ for every 1 kilogram of CO₂.1919

Now that we are in grams of CO₂... remember grams of A.1927

We now go to moles of A or moles of CO₂; times 1 mole of CO₂.1930

The molar mass of carbon dioxide is roughly 44 grams of CO₂.1937

Now we are in moles of A, moles of carbon dioxide.1943

We want to go to moles of B which is going to be the moles of carbon.1950

That is going to be 1 mole of carbon for every 1 mole of CO₂.1954

Finally moles of B goes to grams of B; times 12.01 grams of carbon for every 1 mole of carbon.1960

When all is said and done, we are going to get 2.6 times 10⁴ grams of carbon.1971

Again this is a representative problem of using a mole to mole ratio directly from a chemical formula.1979

Right away, sample problem two, you see a chemical equation immediately.1989

You know this is going to be using a mole to mole ratio from the coefficients of the balanced chemical equation.1994

The very first thing you want to do is to balance or make sure it is balanced.2001

This is an ordinary combustion problem that we have looked at before.2005

We are going to need in the end two waters and two oxygens.2009

Now the question: how many milligrams of carbon dioxide can form from so many kilograms of CH₄ gas?2017

You see milligrams and you see kilograms; therefore what type of problem is this?2027

That is right; this is a mass to mass conversion.2032

In addition, do you see amounts of both reactants specified?2039

We don't; we only see the amount of one reactant specified, the CH₄.2045

So we know that this is not limiting reactant at all... not limiting reactant.2052

Let's go ahead and jump into the problem now.2063

23.1 kilograms of CH₄, again we want to get to grams first.2068

Times 10³ grams of CH₄ for every 1 kilogram of CH₄; that gets us into grams of A.2074

Now from grams of A, I want to go to moles of A.2083

Times 1 mole of CH₄ for every... molar mass of CH₄ is roughly 16 grams of CH₄.2086

That gives me moles of A; now from moles of A, I want to go to moles of B.2096

That is going to be carbon dioxide; it is conveniently a 1:1 ratio.2102

1 mole of CO₂ for every 1 mole of CH₄; moles of B.2111

Now from moles of B, I want to go to grams of B.2120

Carbon dioxide is roughly 44 grams for every 1 mole.2125

The question is asking for milligrams of carbon dioxide; times 1 milligram divided by 10^-3 grams.2129

After we do this calculation, we are going to get 6.3 times 10⁷ milligrams of CO₂.2143

That is expected to form from 23.1 kilograms of CH₄.2150

Again this is a mole to mole ratio from the balanced equation and not a limiting reactant problem.2162

Sample problem number three; once you see a chemical equation, make sure it is balanced.2174

Again it is going to be two of these and two of those.2183

How many milligrams of carbon dioxide can form when 23.1 kilograms of CH₄ are combined with 23.1 kilograms of oxygen gas?2189

First of all, you see that both of the reactant amounts are specified.2205

You know right away, 100 percent, there is no question, it is limiting reactant.2211

Because it is a limiting reactant problem, we are going to use our method2219

that we call the smaller quantity approach or the smaller quantity method... smaller quantity method.2223

Remember we are going to now determine the theoretical amount of product that can form from each reactant and compare.2235

The smaller one is going to be the answer.2242

Now the 23.1 kilograms of CH₄; we already did that.2246

We got roughly 10⁷ milligrams of CO₂; that is just directly from the previous example.2259

Now we need to determine how much CO₂ is going to form from 23.1 kilograms of oxygen gas.2271

23.1 kilograms of oxygen, I want to get to grams first.2280

Times 10³ grams for every 1 kilogram of oxygen; that is grams of A.2288

Now from grams of A, I go to moles of A.2296

Times 1 mole of oxygen for every... the molar mass of molecular oxygen is going to be 32 grams of O₂.2298

That is moles of A; now from moles of A, I go to moles of B.2314

That is going to be 1 mole of CO₂ for every 2 moles of oxygen.2318

Remember I am getting this from the balanced chemical equation.2324

Now from this I want to go ahead and get the grams of CO₂.2329

Again I want to get the grams of CO₂.2342

Now I am going to multiply this by 44 grams of carbon dioxide for every 1 mole of CO₂.2345

When all is said and done, we are going to get approximately 16,000 grams of CO₂.2355

I apologize; I don't have that number in front of me.2363

But it is going to be roughly that number; roughly 16,000 grams of CO₂.2365

What is important is that this is going to be smaller than the 10⁷ milligrams of CO₂ gotten from the other answer.2374

Just go ahead and compare; 16,000 grams of CO₂ is going to be...2390

That is going to be 16,000,000 milligrams of CO₂ or 1.6 times 10⁷ milligrams of CO₂.2396

From the previous answer, this was I believe 6.3 times 10⁷ milligrams.2409

The 1.6 times 10⁷ milligrams of CO₂ is going to be our actual answer.2418

We conclude that molecular oxygen is the limiting reactant.2429

CH₄ in this case is going to be in excess.2440

We don't have enough oxygen; that is why it is limiting.2448

But we have plenty of CH₄.2450

What that means is that we can calculate the amount of CH₄ that is going to remain.2454

We started with 23.1 kilograms initially of CH₄.2461

We are going to be using roughly 16,000 grams of CH₄ required; roughly.2471

That means 23,100 grams minus 16,000 grams tells me approximately 5,000 grams of CH₄, the excess amount, will remain unused.2489

What we just did again for sample problem three was a limiting reactant problem.2511

We not only determined the limiting reactant.2516

We also determined how much of the excess reactant is going to remain.2519

Thank you for using Educator.com; it was great to see you guys again.2525

I will see you all next time.2529