 ## GOB Chemistry

Learn the toughest concepts covered in your GOB - General, Organic, and Biological Chemistry class with step-by-step video tutorials and practice problems.

Chemical Reactions & Quantities

# Percent Yield

The Percent Yield determines how successful the product yield is in a chemical reaction.

Percent Yield
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concept

## Percent Yield 3m
Play a video:
now stocky geometry is a way of us to turning, theoretically on paper how much product we could make if we're 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 the desired product. The higher the percent yield than 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 gonna 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% then that would be very good. If you were good, that means you'd be equal to or greater than 70% and then you have a poor yield. If you're 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 100 remember, we have our purple box here. That means that this is a formula you need to commit to memory because often times it's not given on your formula shoot. 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. Oftentimes 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 isn't moles. You'd have to change those theoretical yield units to match the actual yield units. So you have to change moles to grants. We're going to say here that no chemical reaction is 100% efficient, so you'll never, ever get a percent yield. That is 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 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. 2
example

## Percent Yield Example 1 4m
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Here are example, Question says Consider the following balanced chemical reaction. So we have two moles of C six h six reacting with moles of oxygen gas to produce 12 moles of carbon dioxide gas, plus six moles of water as a liquid. It says. If a 2.6 g sample off see 686 reacted with excess oxygen to produce 1.25 g of water. What is the percent yield of water? All right, they want us to determine the percent yield. We know percent yield equals actual over theoretical times 100. If we read back on the question, they're telling me that I produced to 5 g 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 reaction in real life and obtain that amount. That 1.25 g represents our actual yield. So, really, what we have to do is calculate our theoretical yield. Remember, our theoretical yield can be determined by using store geometry, So if we look here, it says, step one map out the portion of the stoke geometric chart you will use well, excess, remember, Means we ignore. They're telling me that I have to 0.6 g of this reacted. That's the amount that gave to me. So that represents my grams off given. So the map that I'm gonna have to work out is grams of given Convert that into moles of given. Then I'm gonna have to convert. Those moles are given two moles of unknown. What's are 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. Yeah. All right, so that's the story. Metric chart. At least a portion of it that we need to utilize to get our theoretical yield. All right, so we're gonna take the given quantity and change it into moles of given. We have to 6 g of benzene. We have six carbons and six hydrogen. When you look on the periodic table and take their atomic masses out of the six carbons out of the six hydrogen, you'll have a combined mass of 78.108 Gramps per one mole off. See 686 Next we're going to say here, Step three joy. Multiple 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 for moles of C six h 62 moles of water and in our balance equation, it is a 2 to 6 ratio. Then it says, if necessary, convert the moles of unknown into desired units into the final design units. Since our actually yielded using grams of water are theoretical, yield needs to go to grams of water. One mole of water taken into account the two hydrogen and one oxygen. It has a combined mass of 18.16 g. Then if we multiply everything out and divide the what's on the bottom, we get 1.799 g of water. This year represents our theoretical yield. So plug that in tow. Our answer and then stop five We're already doing it. We're plugging in the actual yield and the theoretical yield, and that will help us to find our percent yield. So 1 to 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 stoke geometry and limiting re agents. And now just tacking on actual yield with that information were able to find our percent yield for this particular question.
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Problem

What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO4?

Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)

4
Problem

Ammonia, NH3, reacts with hypochlorite ion, OCl, to produce hydrazine, N2H4. How many grams of hydrazine are produced from 115.0 g NH3 if the reaction has a 81.5% yield?

2 NH3 + OCl → N24 + Cl + H2O 