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# Percent Yield - Concept

###### Kendal Orenstein

###### Kendal Orenstein

**Rutger's University**

M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

Reactions don t always go to completion. In the lab, chemists usually produce less reactants than anticipated. We represent the amount we produced as **percent yield**, which represents the percent of the anticipated yield we actually produced. The formula for percent yield is *percent yield = 100 x absolute value (actual yield / predicted yield)*.

We're going to talk about percent yield. Percent yield is the ratio of actual yield to theoretical yield. Okay what is that? Actual yield is the amount of product produced in the experiment so if you're actually doing a lab, and you're doing the experiment on your own, it's the actual amount of product that you actually formed in the lab you as the student or you as a Chemist actually have produced. The theoretical yield is a maximum produced through calculations so you're not Superman, you're going to make mistakes, you're going to have product fall, you're going to have contamination whatever it may be so in theory there you're going to have something different so those you're going to get based on your stoichiometry calculations okay so it's a comparison of what we actually get in a lab versus what you should get using your stoichiometry calculations. Okay so the percent yield which is the ratio of the actual yield over theo- theoretical yield you're actual from the experiment over from your calculation times a 100 because it is a percentage. Can you have a percentage a percentage yield over a 100%? Sure you can have an actual yield that's actually higher than your theoretical yield because due to contamination and things like that you can also 00 that's lower than a 100 if you lost some products somewhere along the line so your percent yield can actually be a plus or minus a 100.

Alright let's do an ex- a problem based on percent yield. Okay so when potassium chromate K2CrO4 is added to a solution of point 500 grams of silver nitrate, AgNO3, point 455 grams of solid silver chromate, AG2CrO4 is formed alright? So what is the percent yield? So let's see here at first do we're first going to actually write a reaction of what exactly took place so we have K2CrO4 is reacting with silver nitrate, AgNO3, I'm sorry Ag yeah NO3 this is give me double replacement reaction may have two ionic compounds for producing silver chromate and potassium nitrate and we balance this we should have 2 here which means we should have 2 here alright so now this is balanced okay great we have a balanced equation so in my lab, I still actually first I was given 0.5 grams of silver nitrate great that's how much I was given so when I went back in the lab, I actually produced 0.455 grams of this, okay but how much should I produce if I did everything perfectly? Alright we're going to have to do a mass to mass stoichiometry problem of this okay so we're given 0.5 grams of AgNo3 we're going to have to figure out how much I actually can how much how many grams of AgCrO4 actually I can produce so only to do that is compare through moles so I have to change it change this to mole so I'm going to say 1 mole of this is how many grams of silver nitrate 169.9 using our periodic table okay and we say now 1 mole of AgNO3. Now looking at our formula I know that for every 2 moles of AgNO3 I produce 1 mole of Ag2CrO4 but this gets me the mole and I cross all these units out this gets me to moles I want to get to grams so I do the molar mass 1 mole of AgNO3. Oh sorry Ag2CrO4 is actually 331.8 grams of Ag2CrO4. Okay, so doing all these calculation, point 5 times 1 times 1 times 331.8 divided by 169.9 divided by 2 divided by 1 will give me 4 point 448 I'm going to write this over here, point 488 grams of Ag2CrO4 so I should have produced point 488 grams of this silver chromate but in my lab I only produced this much okay so I have to figure out the percent yield so I'm going to take the actual yield what I actually got in the lab which is point 455 I'm going to divide grams of Ag2CrO4 divided by point 488 grams which is what I should have gotten multiplied by a 100 and I get I'm sorry I'm running out of space, what is it? 92, 93.2% and this is my percent yield.

Alright so in the percent yield we actually have to figure out how much you've actually gotten in experiment how much you did in the lab and how much in theory you should have gotten and that is percent yield of the comparison of the two so hope it explains percent yield.

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###### Kendal Orenstein

M.Ed., Columbia Teachers College

Her energy is contagious. She gets you excited about learning new concepts and makes it easy to understand. Her love for chemistry comes across in all her videos.

Youre like a genius chemistry guru--thanks for helping me get that A”

My professor is about as clear as mud - thanks for explaining this concept in 7 minutes - that i haven't gotten in 6 weeks of class. Awesome - thanks you!”

You are my go-to whenever I have a chem question. You explain everything so well. THANK YOU!!”

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