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For more information, please see full course syllabus of General Chemistry
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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 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
0 answers
Post by Mirriel Akoto on July 18, 2017
For sample problem 1 on stochiometryI. I do not think the calculation is right.
0 answers
Post by Evan Wang on May 18, 2017
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
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
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
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
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
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
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)?