For more information, please see full course syllabus of General Chemistry

For more information, please see full course syllabus of General Chemistry

### Stoichiometry I

- Stoichiometry uses coefficients from a balanced chemical equation as a conversion factor to relate any (2) reactants and/or products.
- Mole to mole ratios are central to any stoichiometry problem.
- The limiting reagent or reactant dictates how much product is expected to form (known as the theoretical yield).
- Molarity expresses solution concentration, and can be used as a conversion factor.
- Acid-base titrations are used to determine (or standardize) the concentration of an unknown solution.

### Stoichiometry I

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
- Lesson Overview
- Mole to Mole Ratios
- Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
- Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
- Mole to Mole Ratios Cont'd
- Mole to Mole Ratios Cont'd
- Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
- Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
- Mass to mass Conversion
- Mass to mass Conversion
- Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
- Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
- Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
- Limiting Reactants, Percent Yields
- Limiting Reactants, Percent Yields
- Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
- Percent Yield
- Example 9: How Many Grams of The Excess Reactant Remains?
- Summary
- Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide?
- Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)?
- Sample Problem 3: Part 1
- Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain?

- Intro 0:00
- Lesson Overview 0:23
- Mole to Mole Ratios 1:32
- Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
- Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
- Mole to Mole Ratios Cont'd 5:13
- Balanced Chemical Reaction
- Mole to Mole Ratios Cont'd 7:25
- Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
- Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
- Mass to mass Conversion 11:06
- Mass to mass Conversion
- Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
- Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
- Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
- Limiting Reactants, Percent Yields 20:42
- Limiting Reactants, Percent Yields
- Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
- Percent Yield
- Example 9: How Many Grams of The Excess Reactant Remains?
- Summary 29:34
- Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide? 30:47
- Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)? 33:06
- Sample Problem 3: Part 1 36:10
- Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain? 40:53

### General Chemistry Online Course

I. Basic Concepts & Measurement of Chemistry | ||
---|---|---|

Basic Concepts of Chemistry | 16:26 | |

Tools in Quantitative Chemistry | 29:22 | |

II. Atoms, Molecules, and Ions | ||

Atoms, Molecules, and Ions | 52:18 | |

III. Chemical Reactions | ||

Chemical Reactions | 43:24 | |

Chemical Reactions II | 55:40 | |

IV. Stoichiometry | ||

Stoichiometry I | 42:10 | |

Stoichiometry II | 42:38 | |

V. Thermochemistry | ||

Energy & Chemical Reactions | 55:28 | |

VI. Quantum Theory of Atoms | ||

Structure of Atoms | 42:33 | |

VII. Electron Configurations and Periodicity | ||

Periodic Trends | 38:50 | |

VIII. Molecular Geometry & Bonding Theory | ||

Bonding & Molecular Structure | 52:39 | |

Advanced Bonding Theories | 1:11:41 | |

IX. Gases, Solids, & Liquids | ||

Gases | 35:06 | |

Intermolecular Forces & Liquids | 33:47 | |

The Chemistry of Solids | 25:13 | |

X. Solutions, Rates of Reaction, & Equilibrium | ||

Solutions & Their Behavior | 38:06 | |

Chemical Kinetics | 37:45 | |

Principles of Chemical Equilibrium | 34:09 | |

XI. Acids & Bases Chemistry | ||

Acid-Base Chemistry | 43:44 | |

Applications of Aqueous Equilibria | 55:26 | |

XII. Thermodynamics & Electrochemistry | ||

Entropy & Free Energy | 36:13 | |

Electrochemistry | 41:16 | |

XIII. Transition Elements & Coordination Compounds | ||

The Chemistry of The Transition Metals | 39:03 | |

XIV. Nuclear Chemistry | ||

Nuclear Chemistry | 16:39 |

### Transcription: 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 through*0038

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

*Moreover the mole to mole ratio can be used to solve*0048

*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 encounter*0060

*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 N _{2} requires three moles of H_{2} for this reaction to occur.*0357

*Next line, we can expect two moles of ammonia to form from one mole of N _{2}.*0366

*Finally we can expect two moles of ammonia to form from three moles of H _{2}; from three moles of H_{2}.*0372

*That should be H _{2}; 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 H _{2}?*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 H _{2}.*0475

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

*3.1 moles of H _{2} 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 NH _{3} 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 H _{2}; 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 H _{2} required.*0583

*Moles of N _{2} is going to go downstairs to get cancelled.*0600

*Moles of H _{2} 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 H _{2} for every one mole of N_{2}.*0629

*3 goes in front of the moles of H _{2} and 1 goes in front of the moles of N_{2}.*0636

*When all is said and done, we are going to get 19.2 moles of H _{2} 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 H _{2}?*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 H _{2}, the first step is to go to moles of H_{2}.*0850

*Times 1 mole of H _{2} divided by roughly 2.016 grams of H_{2}; that gets me as the moles of H_{2}.*0861

*Now from moles of H _{2}, I am going to go on to the next step to go to moles of ammonia.*0870

*That is going to be moles of NH _{3} on top and moles of H_{2} on the bottom.*0876

*That is going to be a 2:3 ratio; finally this gets me into moles of NH _{3} after everything cancels.*0883

*Now I want to go to grams of NH _{3}; 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 N _{2}, the first step is to go from grams of A to moles of A.*0956

*Times something over something; grams of N _{2} goes downstairs; 1 mole of N_{2} goes upstairs.*0965

*The molar mass of molecular nitrogen is going to be roughly 28 grams of N _{2}.*0973

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

*Times something over something; moles of H _{2} on top; moles of N_{2} 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 H _{2} on top and 1 mole of N_{2} 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 H _{2} required.*1005

*1 mole of H _{2} on the bottom; grams of H_{2} 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 prefixes*1102

*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 H _{2}?*1116

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

*1.2 kilograms of H _{2}, we want to go to grams of A or grams of H_{2}.*1128

*Times grams of H _{2} for every 1 kilogram of H_{2}.*1136

*Remember what the prefix multiplier is for kilo?--it is 10 ^{3}.*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 H _{2} for every gram of H_{2}.*1156

*That is going to be roughly 2.016 for the molar mass of H _{2}.*1162

*Now I am in moles of H _{2}, 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 H _{2}.*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 NH _{3} for every one mole of NH_{3}.*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 10 ^{6} milligrams of ammonia.*1212

*That is expected to form from 1.2 kilograms of H _{2}.*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 H _{2}?*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 H _{2} and 3.1 grams of N_{2}?*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 H _{2} times 1 mole of H_{2} divided by 2.016 grams of H_{2}.*1389

*Times 2 moles of ammonia for every 3 moles of H _{2}.*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 N _{2} times 1 mole of N_{2} divided by roughly 28 grams of N_{2}.*1432

*Now times 2 moles of ammonia for every 1 mole of N _{2}.*1441

*That is going to give me 0.22 moles of NH _{3} 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 NH _{3}.*1484

*Because of this, we conclude that N _{2} is a limiting reactant in this example.*1489

*Also H _{2} 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 H _{2} is in excess; we have too much of it and not enough of N_{2}.*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 experiment*1586

*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 H _{2} 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 N _{2}.*1650

*Basically we are going to go from 3.1 grams of N _{2} to how much grams of H_{2} required.*1655

*Once we get the grams of H _{2} required, we simply subtract that from the amount that we have to get the amount left over.*1667

*Grams of H _{2} initially have minus grams of H_{2} required is going to give you the grams of H_{2} remaining.*1675

*Let's go ahead and demonstrate; 3.1 grams of N _{2} times 1 mole of N_{2} divided by roughly 28 grams of N_{2}.*1693

*Now we go to moles of H _{2}; 3 moles of H_{2} for every 1 moles of N_{2}.*1707

*Now we go to grams of H _{2}; 2.016 grams of H_{2} for every 1 mole of H_{2}.*1715

*That is going to give us 0.67 grams of H _{2}.*1723

*What this number is, this 0.67 grams, this is the amount of H _{2}*1727

*that is going to react with the 3.1 grams of N _{2}... 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 H _{2} minus 0.67 grams of H_{2}.*1743

*That is going to say 2.43 grams of H _{2} 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 CO _{2} for every 1 mole of carbon.*1890

*And 1 mole of CO _{2} 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 10 ^{3} grams of CO_{2} for every 1 kilogram of CO_{2}.*1919

*Now that we are in grams of CO _{2}... remember grams of A.*1927

*We now go to moles of A or moles of CO _{2}; times 1 mole of CO_{2}.*1930

*The molar mass of carbon dioxide is roughly 44 grams of CO _{2}.*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 CO _{2}.*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 10 ^{4} 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 CH _{4} 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 CH _{4}.*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 CH _{4}, again we want to get to grams first.*2068

*Times 10 ^{3} grams of CH_{4} for every 1 kilogram of CH_{4}; 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 CH _{4} for every... molar mass of CH_{4} is roughly 16 grams of CH_{4}.*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 CO _{2} for every 1 mole of CH_{4}; 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 10 ^{7} milligrams of CO_{2}.*2143

*That is expected to form from 23.1 kilograms of CH _{4}.*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 CH _{4} 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 method*2219

*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 CH _{4}; we already did that.*2246

*We got roughly 10 ^{7} milligrams of CO_{2}; that is just directly from the previous example.*2259

*Now we need to determine how much CO _{2} 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 10 ^{3} 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 O _{2}.*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 CO _{2} 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 CO _{2}.*2329

*Again I want to get the grams of CO _{2}.*2342

*Now I am going to multiply this by 44 grams of carbon dioxide for every 1 mole of CO _{2}.*2345

*When all is said and done, we are going to get approximately 16,000 grams of CO _{2}.*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 CO _{2}.*2365

*What is important is that this is going to be smaller than the 10 ^{7} milligrams of CO_{2} gotten from the other answer.*2374

*Just go ahead and compare; 16,000 grams of CO _{2} is going to be...*2390

*That is going to be 16,000,000 milligrams of CO _{2} or 1.6 times 10^{7} milligrams of CO_{2}.*2396

*From the previous answer, this was I believe 6.3 times 10 ^{7} milligrams.*2409

*The 1.6 times 10 ^{7} milligrams of CO_{2} is going to be our actual answer.*2418

*We conclude that molecular oxygen is the limiting reactant.*2429

*CH _{4} 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 CH _{4}.*2450

*What that means is that we can calculate the amount of CH _{4} that is going to remain.*2454

*We started with 23.1 kilograms initially of CH _{4}.*2461

*We are going to be using roughly 16,000 grams of CH _{4} required; roughly.*2471

*That means 23,100 grams minus 16,000 grams tells me approximately 5,000 grams of CH _{4}, 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

0 answers

Post by Mirriel Akoto on July 18 at 02:21:04 PM

For sample problem 1 on stochiometryI. I do not think the calculation is right.

0 answers

Post by Evan Wang on May 18 at 10:48:53 PM

Can you give us harder examples?

I feel these are too simple.

0 answers

Post by jacob featehrstone on November 19, 2015

just noticed a mistake in the video.. 2.6 moles of H2O has 5.2 moles of H not 5.6

1 answer

Last reply by: Peter Ke

Sat Sep 12, 2015 8:30 PM

Post by Peter Ke on September 12, 2015

I am so confused in Sample Problem 3: Part 1. Please help.

Didn't the mg for CH4 converted to CO2 was 6.3 x 10^7 mg CO2 that you got in Sample Problem 2? But you put 10^7.

1 answer

Last reply by: Peter Ke

Sat Sep 12, 2015 5:14 PM

Post by Peter Ke on September 12, 2015

I was wondering when using dimensional analysis, when you get a answer with decimals don't you have to round it? Because the answer for 17.29 is 6.75 and you didnt round it up. Why is that?

2 answers

Last reply by: Parsa Abadi

Sun Apr 23, 2017 12:14 PM

Post by Muhammad Ziad on October 27, 2014

Shouldn't the answer to sample 1 be approximately 600.5 grams? I did the same work as you but my answer was different. By the way, your lecture was very helpful. Thank you so much.

2 answers

Last reply by: Professor Franklin Ow

Mon Nov 3, 2014 11:03 PM

Post by Okwudili Ezeh on October 27, 2014

The answer to sample problem 1 should be 600gms of carbon.

1 answer

Last reply by: Professor Franklin Ow

Fri Oct 17, 2014 5:34 PM

Post by Virginia Vizconde on October 17, 2014

Why 2.6 mole of H20 is 2 mole H? Thank you

1 answer

Last reply by: Professor Franklin Ow

Sun Oct 12, 2014 11:58 AM

Post by Saadman Elman on October 12, 2014

You didn't answer my Question. Can you please let me know if i am right or wrong?

Thanks.

1 answer

Last reply by: Parsa Abadi

Sun Apr 23, 2017 12:36 PM

Post by Saadman Elman on September 20, 2014

The answer of sample 3 is wrong. The question was how many of excess (CH4) will remain. You said the answer is approximately 5000 g. But the answer is 17325 gram of CH4 will remain.

1 answer

Last reply by: Professor Franklin Ow

Fri Jul 18, 2014 5:49 AM

Post by Kurt Kamena on July 15, 2014

Shouldn't the answer to Sample Question 3 (Part 2) be roughly 17,000 grams?

1 answer

Last reply by: David Millaud

Mon Jun 23, 2014 2:46 PM

Post by David Millaud on June 23, 2014

is that math correct on your conversion using the molar mass of hydrogen i thought you would only calculate 1 gram seeing how hydrogen is one gram/mole. Or is it because the element is diatomic due to the subscript is why you use 2 grams for one mole?

2 answers

Last reply by: Professor Franklin Ow

Sat Apr 26, 2014 5:19 PM

Post by Jia Cheong on March 31, 2014

2.6 moles of H2O x 2 moles of H = 5.2 not 5.6 moles of H2O...

0 answers

Post by Shane Lynch on March 31, 2014

Is that last answer about the excess CH4 correct? I got 7100g...

1 answer

Last reply by: Professor Franklin Ow

Sun Mar 16, 2014 12:21 AM

Post by Meredith Roach on March 9, 2014

Is the math correct in Sample problem #1 (33:01)?