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0:00
[MUSIC] Lighthouse Scientific Education presents a lecture in the Stoichiometry series. The topic Stoichiometry: Problem Solving. Material in this lecture relies on an understanding of the previous lectures; Balancing Chemical Equations; The Mole as a Quantity; Molecular Weight and Conversions and Mole Ratio: Intro to Stoichiometry. In particular, this lecture should be watched and the material understood before tackling the larger stoichiometric problems.
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Still, this lecture will begin with a brief overview of stoichiometry as covered in that previous lecture. Building on that is a discussion on the different types of stoichiometry problems. From there, we develop a procedure or series of steps for solving stoichiometry problems. To help keep some order and to make sense of the steps, a visual aid or graphical guide, is developed based on a discussion of appropriate conversion factors. The steps will be employed in 3 example
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stoichiometry problem. Finally, a particular use of stoichiometry is in finding percent yields. Two additional problems will be given. What did we learn in the first stoichiometry lecture? Well, stoichiometry is the calculation of relative quantities of reactants and products in chemical reactions. It is a very powerful procedure in the study chemistry. More succinctly, stoichiometry uses the coefficients of the reactants and products in the balanced chemical equation to determine
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relative quantities. Using this stand-in equation, a series of ratios, of mole ratios can be constructed that has the coefficient and substance type of two compounds from the balanced chemical equation. Any and every combination can be made into a ratio. It can have any reactant or any product in the numerator of the ratio as well as in the denominator. A mole ratio is a conversion factor that allows amount information about one substance in the reaction
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to be converted into amount information about a different substance in the reaction. The 'Mole Ratio: Introduction to Stoichiometry' lecture covers this central concept in more depth and should be grasped fully before moving on with this more involved discussion of stoichiometry. Why is that? Stoichiometry problems are initially difficult to get a handle on because they are often solved in 3 parts. The three parts can be held together by this common refrain: get into moles;
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mole ratio; get out of moles. The previous lecture focuses on the central part, mole ratio. This lecture will add the preceding and following parts to construct a broader procedure for solving stoichiometry problems. And it is not just the fact that a stoichiometry problem is really a construct of 3 separate problems that makes this topic one of the hardest to master in the initial study of chemistry. It is also that these problems require the student to have mastered a previous
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set of skills. Including: balancing chemical equations; determining molecular or molar weights and masses; converting between weight and moles; converting between molarity and moles and determining a mole ratio which is covered in the previous lecture. Stoichiometry is a concept built on a foundation of skills. Before we get started there are a couple additional points. The first, and the most important in preventing errors, is to use the units to keep order!
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Cancel units in the solution step of the problem so that starting units are converted in to wanted or asked for units. Stoichiometry is a conversion of information and basic conversion steps are used. Stoichiometry is a broad topic and not all aspects are covered in all courses. This lecture covers the standard stoichiometry material covered in most courses. There is an additional lecture in this series that covers more advanced concepts on the topic of stoichiometry.
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If your course moves into these topics please go to the 'Stoichiometry: Advanced Concepts' lecture. In that lecture the topics discussed are: limiting reagents (or reactants); molar volume as it relates to stoichiometry and the Ideal Gas Law as it relates to stoichiometry. There are some advanced stoichiometry problems. Molar volume and the Ideal Gas Law may be covered after stoichiometry in your course. Review this material as needed. Now, onto the topic of this lecture.
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We begin with a look at the 4 types of stoichiometry problems. First there is mole to mole type of problems. This is where amount information of one substance is in number of moles and that is converted into amount information of another substance that is also in the number of moles. This problem type was addressed in the 'Mole Ratio' lecture. Then there is mass to mass or weight to weight type of problems. This is where information on the mass, or weight, of one substance is converted
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into information on the mass, or weight, of another substance in a chemical reaction. Molar masses, formula weights and atomic masses are often used in this type of problem. For example, 'How many grams of H2O can be created from 10 g O2'. That is converting the weight of water into the weight of oxygen. The third type is mass, or weight, to volume. Here weight is given for one substance and is converted into the volume, of some concentration, of another substance. Molarity is often used in
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this type of problem. Volume to mass types of problems are also under this heading. An example problem would be 'How many grams of Na2SO4 can be formed using a starting volume of 0.150 l of a 1.5 molar NaOH solution? The last type is volume to volume problems. Information on the volume of one substance is converted into information on the volume of another substance in a chemical reaction. Molarity is commonly found in these problems but volumes of gases can also be used. Gas stoichiometry
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problems are covered in the advanced concept lecture. Common problems of this type include 'How many liters of CO2 will be produced in a combustion reaction with 25 liters of O2? As can be seen stoichiometry is a pretty broad topic. We will limit our discussion in this lecture to mass and volume problems (that is liquid volumes). Some problems that fall in the third type and all gas related fourth type will be addressed in the advanced concept lecture.
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Fortunately for us stoichiometry problems can be summarized in a five step procedure. When starting out, it is advised that students follow these steps in a very vigorous manner. Actually, it is advised that the student always follow procedure vigorously to avoid mistakes. There is a lot of commonality in this procedure with the procedure outlined in the previous lecture including step one, write out and balance the chemical equation. The conversion of amount information that is
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stoichiometry is based on the balanced chemical equation. Step two is basic problem solving; identify the given (or current) substance in the problem and the wanted (or desired) substance. It is always beneficial to stop and consider the problem. What are the known and unknowns? What will become clearer later on is that the student also needs to verify that there is sufficient material of the other components in the reaction that will allow the reaction to proceed to completion.
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This is the advanced topic 'limiting reactant or reagent' and is discussed more in detail in the advanced concept lecture. Step 3 is to write out the mole ratio (gotten from the balanced equation) and to put this ratio in the center of the solution equation. Remember, in the refrain get into moles, mole ratio, get out of moles that the mole ratio is in the center. Not to worry, by time were done there will be something in front of and something following
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the mole ratio. Orientation of the mole ratio will have the desired substance (with its coefficient) over the current substance (with its coefficient). In this form the mole ratio is a conversion factor. Step 4 is the get into moles part of the stoichiometry solution. Convert given or current units (be they grams or ml or whatever) into moles of the given or current substance. This step uses previously developed skills such as conversion of grams into moles through molar
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mass and volume into moles using molarity. That conversion is placed in front of the mole ratio in the solution equation. At this point in the solution process the amount of the given substance should cancel out leaving moles of the wanted or desired substance. If not there might be a problem with the orientation of the mole ratio or the conversion in step 4. The last step is to convert the moles of the wanted or desired substance into asked for units. Often those units are in weight or volume.
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Place this conversion after the mole ratio in the solution equation. The moles of the wanted or desired substance should cancel out. If not, there might be an orientation problem with the conversion factor in step 5. The first three steps are the mole ratio steps. Step 4 is getting to moles and the fifth step is getting out of moles. Get into moles, mole ratio, get out of moles. These 5 steps produce that refrain. With these steps we can jump into problem solving
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but what would be really helpful at this point is to consider each aspect of that refrain in more detail. The better we understand each of the 3 parts separately the easier it will be to combined them into a final solution. Starting with a graphical representation of the five steps. The first three steps are the central component of the solution to a stoichiometric problem. The mole ratio is what allows the movement of amount information about one substance into amount
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information about another substance. But it requires that the amount of given or current substance be in moles. That often requires a conversion of its own: getting into moles of the given or current substance. If these two parts are done correctly the solution will have the correct number of moles for the desired or wanted substance. Most problems, however, do not want the value of the desired substance to be in moles. Another conversion is needed; get out of moles.
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From this graphic guide we can see that known information is converted into unknown information. This transition will take up to 3 conversions. There's a conversion factor to get the known quantity of a substance into moles of the given or current substance which is converted into moles of the desired substance with the conversion factor known as the mole ratio. The third conversion factor is sometimes needed to get into real world units like grams or ml.
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Stoichiometry problems are complex because they involve three separate conversions. The five step process separates these conversions so that the student is only focused on one conversion at a time. As mentioned the mole ratio, was covered in the previous lecture so let's look more deeply at the first and last conversion. Staring with the first conversion getting into moles. And that conversation begins with a basic set of questions. What type of units might the amount of the given
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substance be given in? Followed by what is a conversion factor that takes those given units and converts them into moles of the given substance. Well, we might get lucky in the amount of the given substance is given in moles. In that case there is no conversion. But what if the starting amount is in weight like grams? What converts grams of the substance into moles of that substance? It is the molar mass. That can be atomic weight, the formula weight or the molecular weight.
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The orientation of the molar mass has to have the gram amounts in the denominator so that they cancels out with the given gram amount. If we were given 4.25 g of calcium carbonate and had to get into moles of calcium carbonate we would calculate its formula weight and insert that into the problem so that grams of calcium carbonate cancel out and we get an equivalent amount of moles of CaCO3. What conversion factor would be needed if the given amount was in volume?
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Let's just consider the volume of a solution and say that the units are liters. For such problems the molarity will also need to be given. That molarity will serve as a conversion factor with the units of liters in the denominator. If the problem has the given substance as 0.15 liters of sodium hydroxide which came in a concentration of 0.1 molar, the molarity would be oriented so that liters of NaOH cancel out and the equivalent amount of moles of NaOH is obtained.
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Returning to the graphical representation as a guide, the getting into moles section can be expanded to include the conversion factors in their correct orientation for the current substance. When the current substance is given in weight units; that will be the molar mass. and when the current substance is given in volume units; that will be the molarity. Since all conversions of this type are done the same way this guide can be directly consulted when solving stoichiometry problems.
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What about the third part of the stoichiometry solution? The getting out of moles part. A similar set of questions can be asked for this part. At the beginning of the third part the wanted or desired substance has its amount solved for in moles. A conversion factor is needed to move from that mole amount to whatever units are asked for with the desired substance. If the units are asked for in moles then no conversion is needed. If the asked for units are grams or
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some other weight then the molar mass, atomic mass or formula weight will be needed. The orientation of the molar mass needs to have moles in the denominator so that they cancel out with the moles produced from in the 'get into moles' and 'mole ratio' parts. Specifically, they will cancel out with the moles in the numerator of the mole ratio. Say coming out of step 4, the mole ratio produces 0.15 moles of water. The problem however asks for grams of water. The molecular weight of
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water is a conversion factor with moles of water in the denominator so that moles will cancel out and an equivalent weight of water is produced. This is a standard mole to weight conversion. If however the problem asked for amount of the desired substance to be given in molarity. Molarity is moles over volume. Coming out of the mole ratio step the desired substance is already in moles. To get those moles into molarity divide by the volume that contains those moles. Most solutions problems use
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dilute reactant so the volume of the products will be the same as the volume of the reactants. If not, a volume will be given. In this example, the mole ratio gives 0.44 moles of HCl. The problem however wanted HCl in molarity and was generous enough to provide access to the volume of the final solution. The molarity of the HCl will be those 0.44 moles divided by the 0.88 liters. The HCl is a 0.50 molar solution. Returning to the graphical guide and expanding the
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'get out of moles' section, when amount of the desired substance is asked for in weight the molar mass is used with moles in denominator. When the amount of desired substance is asked for in molarity then a volume is needed to move out of moles. When filled out like this the graphical guide is a very helpful tool when first attacking stoichiometry problems. There is some interesting symmetry in this representation. In the getting into moles section the moles of the conversion factor are
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always in numerator and in the getting out of moles section the moles of the conversion factors are always in the denominator. As we see the units keep the order of the stoichiometry problem. It may seem like canceling out units is an additional step but what it really does is it catches mistakes. Now we are ready take on our first stoichiometry problem. Iron can be produced from the reaction of aluminum and ferric oxide. Starting with 55 g of Fe2O3 and an excess of Al how
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much iron (in grams) can the reaction yield? This is a stoichiometry problem because we are given amount information about one substance in a chemical reaction and asked for amount information about a different substance. Before proceeding notice the wording 'an excess of'. It tell us that we do not need to know how much aluminum there is because there is more than enough to react with all of the ferric oxide. It is important for such terminology to be in a stoichiometry problem
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because it is a requirement that students 'verify that there is sufficient material' as stated in step 1 of the 5 step procedure. Keep an eye out for just such wording. Without it this might be a limiting reagent or reactant problem which is an advanced concept. Returning to the problem, it is a weight to weight or mass to mass problem. Okay, what is the first step in solving this stoichiometry problem? It is to get a balanced chemical equation for the reaction. Is this equation balanced?
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Well, there is only one aluminum on the reactive side but there are two on the product side so no, no it's not. Since balancing chemical equations is a required skill at this point in our studies we will say that it can be balanced and give the coefficients (but, we will verify that it is balanced). One quick note 2 of the 4 compounds do not specifically have coefficients in front of them. They have, of course, implied 1's in front of them.
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It is not necessary to add the 1's but it is necessary to know that the 1's are there. It is also necessary to remember that the coefficients represent the number of moles of the substances in the reaction. 2 moles of aluminum and 1 mole of ferric oxide react to form 2 moles of iron and 1 mole of aluminum oxide. A quick tally shows that the three elements in the problem have the same number of atoms on the reactive side as they do on the product side allowing us
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to move on to step 2; identifying known and unknown. This is a critical step in the problem because it outlines the overall conversion and the appropriate mole ratio. Looking at the problem we see that the Fe2O3 has a weight amount attached to it. It is the current or given substance and we will write that down as our known. Returning to the problem, it is tempting to consider the aluminum as an unknown. It doesn't come with a specific value. We could add that to our
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list of known and unknowns but it is not really a factor in the problem other than verifying that it is in sufficient quantity. The iron, on the other hand, is also an unknown. The problem specifically asks how much iron is produced and it asks for it in grams. Fe does belong on our list of known and unknowns. We have the chemical equation so there is nothing else we need from the problem. That takes us to step 3, the central part of the stoichiometry solution, the mole ratio.
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The orientation of the mole ratio will have the desired substance, with its coefficient from the balanced equation, in the numerator and the current substance, with its coefficient in the balanced equation, in the denominator. What will never be appropriate is using values from the problem in the mole ratio. A clear separation needs to be made with values from the problem and coefficients from the balance equation. Values from the problem belong with the 'get into
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moles' part of the solution and values from the balanced chemical belong to the mole ratio. Perhaps it easier to just remember the only time the balanced equation is used is with the mole ratio. Looking at the balanced chemical equation we see that the desired substance, iron, has a coefficient of 2 and the known, our current substance Fe2O3, with an implied coefficient of 1. Keeping that correct orientation we have this mole ratio. For this reaction a mole ratio involving iron and
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ferric oxide will always have these coefficients. Also part of step 3 is to insert the mole ratio into the center of the solution equation. Not the most common way of setting up a mathematical equation but it will fit in well with step 4, getting into moles. That is converting the amount of the given or current substance into moles of the given or current substance. This part of the problem is a standard conversion problem. It has given units of grams that need to be converted into moles.
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If we were to consult the graphical guide covered earlier, it would say when the amount of the current substance is given in weight. Use the molar mass or molecular weight of the substance. Determining molecular weights is also a skill that the student needs to bring to stoichiometry problems solving. Make sure that the molecular weight is added with grams in the denominator so that they cancel out. Further inspection of the solution equation shows that there are moles of
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Fe2O3 in the numerator of a fraction and moles of Fe2O3 in the denominator of a fraction. These will also cancel out leaving just moles of the desired substance Fe. The canceling of units verifies that we have set up the solution equation properly. It's all too easy to flip a fraction and not realize it. At this point in the solution we should have moles of the desired substance. If that is not the case then a review of our previous work is in order.
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The last part, getting out of moles, is all that remains in this solution. The desired substance iron is in moles of iron but the question asks for amount of iron in weight, specifically grams. If we were to consult the graphical guide and ask how to get out of moles, into a weight unit, it would say to use the molar mass (or in this case the atomic mass) of iron. Importantly, the atomic mass has to be oriented with moles in the denominator so that they cancel out with moles generated in the
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previous two parts of the solution. The remaining unit is grams and it is time to take out our calculator. It is a very busy problem and there is room for miscalculation. A common practice is to multiply all the numerators together and then divide by each denominator. Values of 1 can be omitted from the calculation. In this problem that would be 55 times 2 times 55.8 divided by 159.7 for a final value of 38 g of iron. In this reaction 55 g of Fe2O3 reacting with an excess of aluminum can produce
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38 g of iron. At least, in theory. At this point in their studies the student should recognize that in the lab almost nothing works perfectly. One last point to cover. Not all students are comfortable with combining the three parts of the problem into a single equation. That's okay. The three parts can be solved separately with the solution to one part starting the next part. The 'getting into moles' part can be solved on its own. 55 g of Fe3O2, the given substance, times it's
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molecular weight (in the correct orientation) yields 0.3444 moles of Fe3O2. The second part of the problem, the mole ratio part, uses that 0.3444 moles of the Fe3O2 and converts it into moles of desired substance Fe using the 'mole ratio'. The orientation is correct because units cancel and a value of 0.688 moles of iron is had. 55 g of Fe3O2, with an excess of aluminum, can produce 0.688 moles of iron. 'Getting out of moles' starts with that 0.688 moles of iron
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and converts it into grams of Fe using the atomic mass of iron oriented such that moles of iron cancel out yielding a final value of 38 g of iron just like in the combined equation. Yes, stoichiometry is complex because it has multiple steps and each step has to be correct for the final answer to be correct. Let's try another. The combustion of 26.2 grams of hexane requires how many grams of oxygen? This is a stoichiometry problem because amount information is given about
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one substance and amount information is asked for about a different substance. Since there are only two reactants and both of them are involved in the stoichiometry problem there is no need to consider if there is a sufficient amount of all the reactants. This is another example of the mass to mass or weight to weight stoichiometry problem. Step one in the process is to get a balanced chemical equation. Clearly this equation is not balanced and it does not balanced easy.
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A look at the tally for the three elements shows that there are the same number of atoms on the reactants side as on products side. Balancing chemical equations is a skill student bring into stoichiometry. Step two has us scouring the problem for a known and unknown. Hexane has a known quantity. Add the 26.2 g of hexane to our set up. It is our known and current substance. The only other substance mentioned in the problem is the oxygen. Its amount is unknown and ask for in grams.
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It is the desired substance. There is enough information to generate a mole ratio, step 3. We need to identify the moles of the desired substance as shown in the balanced chemical equation and the moles of the current substance as shown in the balanced chemical equation. The desired or wanted substance, oxygen, has a whopping coefficient of 19 and the given or current substance hexane has a coefficient of 2. Keeping the correct orientation gives a mole ratio of 19 moles of oxygen
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over 2 moles of hexane. Again, values from the problem like the 26.2 grams, never find their way into the mole ratio. Only the balanced chemical equation gives values for the mole ratio. Dropping that mole ratio into the center of the solution equation and moving onto step 4, getting into moles. What needs to be done here is a conversion of 26.2 grams of the current substance hexane into moles of hexane. According to our helpful guide converting weight into moles is done with a molar mass.
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Add in the molecular weight of hexane as a conversion factor. Determining the molecular weights is omitted because it is a required skill at this point in chemistry. The molecular weight is placed with an orientation of grams of hexane in the denominator so that they cancel out. The purpose of converting grams of hexane into mole of hexane is so that the moles of hexane cancel out with the moles of hexane in the mole ratio leaving moles of the oxygen
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(the desired substance). The final step is to get out of moles oxygen and into grams of oxygen. According to our guide, getting out of moles into mass or weight uses a molar or molecular weight. Moles need to be in the denominator. Placing the molecular weight of oxygen after the mole ratio and verifying that the moles of oxygen cancel out brings us to the computation. Taking our calculator and multiplying 26.2 times 19 times 32.0 and dividing that by 86.18 and 2
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gives a final answer of 92.4 g of oxygen. 26.2 g of hexane require at least 92.4 g of oxygen to burn completely. While some students feel comfortable combining the 3 parts of the stoichiometry solution into one equation others prefer addressing this in a step-by-step manner. 26.2 g of hexane, the current or given substance, is converted into moles of hexane in the 'get into moles' part by multiplying it by the molecular weight of hexane. Grams of hexane will cancel out with the correct
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orientation of the molecular weight giving an equivalent mole value of 0.3040 mole of hexane. These moles of hexane are converted into moles of oxygen using the mole ratio. If it is oriented correctly moles of hexane cancel out yielding 2.888 moles oxygen. Once the amount is in moles of oxygen the problem solves by getting out those moles of oxygen and into grams of oxygen using the molecular weight of oxygen oriented such that moles of oxygen cancel out
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leaving 92.4 g of oxygen. Just like in the the single equation. As long as the units are oriented correctly the problem will solve to the same value. The third problem is an acid-base reaction. It wants to know what is the maximum number of grams of sodium sulfate that can be formed using a starting volume of 0.150 l of 1.5 molar sodium hydroxide solution. This is a stoichiometry problem because it gives amount information about one substance in the reaction and asks for information about a
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different substance in the reaction. It is also the third type of stoichiometry problem: a mass to volume conversion. In this problem it will actually be a volume to mass conversion. Like all stoichiometry problems this one begins with a balanced chemical equation. Since there are no coefficients attached to these compounds it is probably reasonable guess that the equation is not balanced. It actually balances rather easily which can be verified with a look at the
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tally of elements. There are equal number atoms on the reactant and product sides. With that complete is time to return to the problem and find a known and unknown. For the known, we are given the amount of the reactant sodium hydroxide. It is given as a volume but that volume needs to be given with a particular concentration. Solving stoichiometry problems that has molarity does require an understanding of volume and concentration. The other substance mentioned
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in the problem is the sodium sulfate. It is a product of this neutralization reaction. The problem asks for its mass or weight in grams. Sodium hydroxide is the known or current substance and sodium sulfate is the unknown or desired substance. Now to the central part of the stoichiometry solution, the mole ratio. The orientation of the mole ratio has the desired substance with its coefficient in the numerator and the current substance with its coefficient in the denominator.
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The mole ratio is 1 to 2 moles of sodium sulfate to sodium hydroxide. The mole ratio holds the center of the stoichiometry solution. Before we can use the mole ratio we need to get the current substance into moles. Our current substance sodium hydroxide has its amount value given in volume. The volume has a molarity. How do we go from volume into moles? Bringing up the graphical guide we see that in the getting into moles panel that when volume is the given unit the molarity is the
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conversion factor. Be sure to have liters in the denominator. The given molarity of the sodium hydroxide is 1.5 molar. That is 1.5 moles of sodium hydroxide for every 1 liter. In this orientation liters cancel as do moles of sodium hydroxide. Volume information about sodium hydroxide has been converted into mole information about a sodium sulfate. But the problem did not ask for moles of sodium sulfate, it asked for grams. One more step is needed and that is to get out of moles
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and into the requested unit of grams. What converts moles and grams? The same thing as in the previous two example problems; molar mass. This time it is the formula weight since sodium sulfate is an ionic compound. Be sure the orientation is such that units of moles of sodium sulfate cancel out leaving grams of sodium sulfate. Taking out our calculator and ignoring the 1's: multiplying 0.150 times 1.5 times 142 divided by 2 gives a final answer of 16.0 g of sodium sulfate.
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0.15 liters of a 1.5 molar NaOH solution can at most produce 16.0 g of the product Na2SO4. The three parts of the stoichiometry problem combine into a final solution using a single equation or can be solved in a sequential manner. 0.150 l of the given substance NaOH times the morality of the sodium hydroxide, oriented so that liters cancel out, produces 0.225 moles of NaOH. There are 0.225 moles of NaOH in 0.150 l of a 1.5 molar NaOH solution. Take that value and put it
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before the mole ratio which has the current substance in the denominator so that moles of NaOH cancel out leaving 0.1125 moles of Na2SO4. That value is used as the starting amount in the getting out of moles (into grams) part. The formula weight of sodium sulfate is a conversion factor oriented so that moles of sodium sulfate cancel out leaving that same 16.0 g of sodium sulfate. There is an additional stoichiometry problem in the next section.
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However, at some point the student needs to sit down and work through enough problems so that they feel comfortable with the procedure. There are no shortcuts concerning stoichiometry. The final topic of the lecture is yields. A yield, in the context of a chemistry course, is the amount of product obtained in a chemical reaction. But no reaction is complete. There is always some reactant that didn't get consumed or they were additional chemical reactions that occurred
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outside of the expected reaction. Some reactions also go in the reverse direction. A common step in doing a chemical reaction is finding out just how complete the reaction was. In considering the level of completion there are a couple terms that need to be introduced. The first is the theoretical yield. It is the maximum amount of product as calculated from the balanced equation. There is an upper limit on how much matter can come from a chemical reaction
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and it is set by the amount of matter of the reactants. Theoretical yields are calculated on paper. Theoretical yields are exactly what we did in the previous 3 stoichiometry problems. A second important term is the actual yield. It is the amount of product actually recovered (and measured) when the reaction is carried out (generally in a lab setting). Often this measurement is in weight or volume. In a lab setting it is measured but in a homework problem it has to be given.
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Taken together, they can produce a percent yield. It is the ratio of actual yield to theoretical yield. The ratios is multiplied by 100 to give percent. Percent yield reveals how close the experiment was to 100% complete. Since no reaction is 100% complete the actual yield should be lower than the theoretical maximum yield. In practice the actual yield is lower than the theoretical yield because mistakes get made or the reactants are not completely pure or the reaction goes
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in both directions. We're not going to go into the hows and whys right now rather we are going to focus on making the percent yield calculation. Starting with how the different yields are found. As mentioned, the theoretical yield is gotten from the balance chemical equation and stoichiometry. The actual yield can be given (like it would be in a problem) or it is measured in a lab setting. The actual and the theoretical yield have to be in the same units. That is if the measurement
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is in weight they have to both be in grams or both have to be in kilograms. If the measurement is volume they both have to be in milliliters or they both have to be in liters. So, getting a percent yield has us getting two pieces of information and putting them into ratio and multiplying by 100. With this understanding we can proceed into the process of determining percent yields. We will look at 2 problems. The first problem will focus on the process of getting the percent yield without doing
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the stoichiometry and the second one will incorporate stoichiometry on the way to getting percent yield. Our first problem says 55g of Fe2O3, iron (III) oxide, and an excess of Al produce 26 g of Fe. What is the percent yield? If this were a straight out stoichiometry problem it would offer the 55 g of iron (III) oxide and ask for grams of iron. While it doesn't explicitly say how the 26 g of iron was obtained it implies that it was measured. To get the percent yield we need
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to have the actual yield and the theoretical yield. First to the actual yield. Where can we get that? Well since we did not perform the experiment it has to be given. Actual yields cannot be calculated, they have to be measured. So in percent problems like this, that value will be in the problem. Half of our ratio is done with the 26 g of iron. What about the theoretical yield? It has to be calculated or given. Since this is our first percent yield problem let's say that we
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ran through the steps of solving a stoichiometry problem and had the given value of 55 gram of iron(III) oxide, turn into moles, and used a mole ratio to get into moles of iron (the desired substance) and then used an atomic mass of iron to get out of moles and came up with 38 grams of iron. That is a theoretical maximum that can be produced from 55 grams of iron(III) oxide. Clearly, it is more than the actual yield. Onto the percent yield equation. It has actual yield of 26 g of Fe over
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theoretical yield of 38 grams. Both yields given in the same gram units. The ratio is multiplied by 100 to give percent. The percent yield of the reaction is 68.4%. The actual yield is 68.4% of the maximum yield. Other than the stoichiometry part not too bad of a calculation. For a final example of percent yield, using stoichiometry, a problem says that 5.0 grams of ammonia was produced from 1.0 g H2 with an excess of N2. What is the percent yield? According to the percent
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yield equation we need to get an actual yield and a theoretical yield. Looking at the problem shows that ammonia, a product, is measured to give an actual yield of 5.0 g. In a word problem actual yields are given. That leaves just the theoretical yield. Sometimes that value is provided and sometimes it is not. This time it is not. The only way it can get a theoretical yield is to calculate it. Percent yield questions are generally stoichiometry questions.
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So the theoretical yield of ammonia is gotten by using the five steps of the stoichiometry procedure. The first step is already completed for us. There is a balanced chemical equation The next step is to find the known and unknown. From the problem the hydrogen gas is found as a known quantity. There is 1.0 g of H2. That is our current unit or the given substance. And we know we are trying to calculate the theoretical maximum amount of ammonia.
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It would be a mistake to put that 5 g into the known and unknown setup. That 5 g belongs to the actual yield and not the theoretical yield. Only the 1.0 g of H2 are relevant in getting the theoretical yield. Probably the single most important distinction that needs to be made in calculating a percent yield is that the theoretical amount of product an unknown Even if there is an amount value attached to that substance in the problem. Two amount values for the product are needed.
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The actual is known and the theoretical yield is not known. Moving onto a mole ratio and the balanced equation. The desired substance ammonia has and a coefficient of 2 and the given our current substance has a coefficient of 3. The mole ratio for this conversion is 2 moles of ammonia over 3 mole of hydrogen. Put that ratio in the center of the solution equation and move on to the get into moles part. What converts grams of hydrogen into moles of hydrogen?
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It is the molecular weight of hydrogen with grams of hydrogen in the denominator so they cancel out with grams of the starting amount. Moles of hydrogen also cancel out leaving moles of ammonia. The third part of the stoichiometry solution is getting out of moles. That is getting out of moles of ammonia into grams of ammonia. What is the conversion factor for that transition? According to the graphical guide it is the molar mass or molecular weight. The molecular weight of
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ammonia is 17.0 grams per mole. Be sure that moles of ammonia are in the denominator so they cancel out with moles of ammonia in the mole ratio. The math produces a yield of 5.6 g of ammonia. That is our theoretical yield from stoichiometry. Now we are ready to determine percent yield. In the numerator, the actual yield is the 5.0 g of ammonia. And in the denominator the theoretical yield is that 5.6 g of ammonia. Multiplied by 100 to get percent and the 5 g of ammonia is an 89% yield.
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That is a pretty good yield and it concludes the material of this lecture. Recapping the lecture. Stoichiometry is the calculation of relative quantities of reactants and products in chemical reaction. At the heart of stoichiometry is the mole ratio. Coefficients of the reactants and/or products in the balanced chemical equation can be used to determine the relative quantities. There are 4 types of stoichiometry problems. The first type is where moles are given and moles
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are asked for. This type is addressed in the 'Mole Ratio' lecture. The second type has amount of the given substance in mass or weight and the amount of the asked for substance also in mass or weight. The third type has amount of the given substance in mass or weight and the amount of the asked for substance in volume. The opposite arrangement also falls under this heading. The last type has the amount of the given substance in volume and the amount of the asked for substance
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in volume. This type is discussed in more detail in the 'Advanced Concept' lecture. Stoichiometry problems are complex because they are often solved in 3 parts: The refrain we used for the 3 parts is to get into moles; mole ratio; get out of moles. The central component of the solution to a stoichiometry problem is the mole ratio. Since most measurements are not in moles there is a part that converts given units into moles. When the given unit is in mass the molar mass is the
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conversion factor (grams are in the denominator). When the given unit is in volume of a solution, molarity is the conversion factor (liters are in the denominator). Since most problems ask for real world units like grams and molarity there is a third part to the solution. If the problem wants the final units to be in weight the molar mass is the conversion factor. If the problem wants the final units to be in molarity then the moles from parts 1 and 2 are divided by the final volume.
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A 5 step procedure for solving most stoichiometry problems was offered. It begins by writing out and balancing the chemical equation. out and balancing the chemical equation. Then it has the student digging through the problem to identify the given or current substance and the wanted or desired substance. Find a known and an unknown. From steps 1 and 2 write out the mole ratio and put it in the center of the solution equation. Now get into moles by converting the units of the
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given substance into moles. Place that conversion in front of mole ratio in the solution equation. Get out of moles by converting the mole units of the wanted substance into asked for units (g, ml or whatever). Place this conversion after the mole ratio in the solution equation. Cancel units and solve. The mole ratio; central part. Get into moles. Get out of moles. The last topic was yields. The theoretical yield is the maximum amount of product as calculated from
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the balanced chemical equation. This is done on paper and is a stoichiometry problem. Contrast that with the actual yield: It is the amount of product actually recovered (and measured) when the reaction is carried out in a lab. It should be less than the theoretical yield. These two yields come together in the percent yield: It is the ratio of the actual yield to theoretical yield (times 100). It is a marker for how efficient the reaction was. And that completes our lecture.
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The single best way to prevent errors in stoichiometry is to keep track of the units! [MUSIC]
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