Percent Yield - Video Tutorials & Practice Problems
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The Percent Yield determines how successful the product yield is in a chemical reaction.
Percent Yield
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concept
Percent Yield
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Now stoichiometry is a way of us determining theoretically on paper how much product we could make if we are given some given amount of starting material. But when we do the experiment in real life and obtain a certain amount of product, it's the percent yield that determines how successful we were. Now percent yield determines how successful the scientist was in creating their desired product. The higher the percent yield, then the higher the efficiency of a chemical reaction. So just like you want the highest possible percentages on an exam, we want the highest possible percentages in percent yield, that will show us being successful in creating our product. Now we're going to say in terms of percent yield values, we can have excellent, very good, good, and poor. Now if you have a percent yield that is equal to or greater than 90%, you could call that an excellent yield. If you have, amount that's equal to or greater than 80 percent, then that would be very good. If you were good, that means you would be equal to or greater than 70%. And then you'd have a poor yield if your percent yield was less than 40%. Now with percent yield comes the percent yield formula. And the percent yield formula equals actual yield over theoretical yield times a 100. And remember, we have our purple box here. That means that this is a formula you need to commit to memory because oftentimes it's not given on your formula sheet. Now we know that the percent yield measures how successful we are in terms of carrying this reaction out in real life. We know that theoretical yield is what we do as calculations on paper when they give us multiple amounts of given and we figure out how much product we're making. Actual yield though, actual yield is the amount of pure product actually created when the experiment is done in the laboratory. Often times, what you do on paper, you'll see that when you do the experiment in real life, things don't exactly match up. We're gonna say here that the units used in the formula are based on the units of the actual yield. So let's say that our actual yield is in grams of our product, but our theoretical yield that we calculated is in moles. You'd have to change those theoretical yield units to match the actual yield unit. So you have to change moles to grams. We're gonna say here that no chemical reaction is a 100% efficient. So you'll never ever get a percent yield that is a 100%. There's always gonna be something that happens. You're gonna spill some of your compound, you're gonna lose some randomly, outside forces will play a part. So you'll never get to a 100% yield. Because of this, that means that your actual yield is always less than your theoretical yield. So just remember, we have percent yield, we have actual yield, we have theoretical yield. Together, they give us a good insight into how efficient our chemical reaction is.
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example
Percent Yield Example 1
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4m
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Here our example question says, consider the following balanced chemical reaction. So we have 2 moles of c 6h6 reacting with 15 moles of oxygen gas to produce 12 moles of carbon dioxide gas, plus 6 moles of water as a liquid. It says, if a 2.6 gram sample of c 6h6 reacted with excess oxygen to produce 1.25 grams of water, What is the percent yield of water? Alright. They want us to determine the percent yield. We know percent yield equals actual over theoretical times a 100. If we read back on the question, they're telling me that I produced 1.25 grams of water. How can I determine if I produced something, or made something, made product? The only way I would know for sure is if I did the pro the reaction in real life and obtained that amount. That 1.25 grams represents our actual yield. So really what we have to do is calculate our theoretical yield. And remember, our theoretical yield can be determined by using stoichiometry. So if we look here, it says step 1, map out the portion of the stoichiometric chart you will use. Well, excess remember means we ignore. They're telling you that I have 2.6 grams of this reacted. That's the amount they gave to me. So that represents my grams of given. So the math that I'm going to have to work out is grams of given, convert that into moles of given, then I'm going to have to convert those moles of given to moles of unknown. What's our unknown? Well, to figure out the units that I need, remember, your theoretical yield must have the same units as your actual yield. Our actual yield is in grams per of water. So we have to go all the way to grams of water. So we're gonna have here moles of unknown which is water, and then finally end here in grams of unknown. Alright. So that's the stoichiometric chart, at least the portion of it that we need to utilize to get our theoretical yield. Alright. So we're gonna take the given quantity and change it into moles of given. We have 2.6 grams of benzene. We have 6 carbons and 6 hydrogens. When you look on the periodic table and take their atomic masses, add up the 6 carbons, add up the 6 hydrogens, you'll have a combined mass of 78.108 grams per 1 mole of c 686. Next, we're going to say here, step 3, do a mole to mole comparison to convert moles of given into moles of unknown. To do that, remember, we look at the coefficients of the balanced equation. We're gonna go from moles of c 6H6 to moles of water. And in our balanced equation, it is a 2 to 6 ratio. Then it says, if necessary, convert the moles of unknown into desired, units, into the final desired units. Since our actual yield is using grams of water, our theoretical yield needs to go to grams of water. One mole of water, taking into account the 2 hydrogens and the 1 oxygen, it has a combined mass of 18.016 grams. Then if we multiply everything out and divide by what's on the bottom, we get 1.799 grams of water. This here represents our theoretical yield. So plug that in to our answer. And then step 5, we're already doing it. We're plugging in the actual yield and the theoretical yield, and that'll help us define our percent yield. So 12.1.25 divided by 1.799 times times 100 gives me 69.48% as my percent yield for this particular question. So just remember, parts of this are pretty familiar. They're incorporating things we've learned about stoichiometry and limiting reagents, and now just tacking on actual yield. With that information, we're able to find our percent yield for this particular question.
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Problem
Problem
For the following chemical reaction 53.1 g HBrO4 is mixed with 25.8 g Li2CO3. Determine the percent yield if 7.17 g CO2 is produced.
Ammonia, NH3, can be created from the combining of H2 and N2 molecules. How many grams of ammonia are isolated when 10.0 g H2 reacts with excess N2 with a chemical reaction that has a 79.3% yield?
2 H2 + N2 → 2 NH3
A
84.4 g
B
53.0 g
C
67.0 g
D
74.5 g
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Problem
Problem
The reduction of iron (III) oxide creates the following reaction:
Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g)
If the above reaction only went to 75% completion, how many moles of Fe2O3 were require to produce 0.850 moles of Fe?