PERCENT YIELD YES I'M TYPING IN CAPS.

Hi there.
Imagine if you had a bag of chips. Yes I just made you hungry. I bet 50% of you just went into the kitchen to check if you had chips. Anyways, back to my point. Imagine the bag is big, and by big, I don't mean those baby bags you find in vending machines. By big, I mean momma sized chips, like them big ones you get in Costco. Appealing, I know. Would you be able to finish them? Perhaps some of you could, but most of you most likely can't without getting heartburn. This is the same with chemical reactions; sometimes not all of the reactants or products get used up; they can only do so much. What if we wanted to know how much of the compound was used up? (or how much of the bag of chips we finished) Thus now, ladies and germs, I present to you the PERCENT YIELD.


And yes I'm aware the chip explanation was uncalled for. But it's a good analogy no?

Percent Yield can be generally calculated using this formula:
Percent Yield= mass of product actually formed
                        --------------------------------------------------     x 100%
                        mass of product expected to be formed
This formula is actually not necessary when you use logic. It is like calculating a test mark. When you calculate a test mark, you divide what you got, by what you possibly could've gotten. Ahhhh logic feels good when you use it, doesn't it?

Here are a few examples:
EX/ What is the percent yield for a reaction if you predicted the formation of 21. grams of C₆H₁₂ and actually recovered only 3.8 grams?
-21g is expected, but you only got 3.8 grams. Bummer.
You would figure it out like so:    (3.8g/21g)x 100% = 18% (don't forget sig figs)


Now that was easy, wasn't it?
Now here, I will throw a curve ball at your way.


EX/ Consider the reaction: U+3Br2--> UBr6. What is my actual yield of uranium hexabromide if I start with 100 grams of uranium and get a percent yield of 83%?
First, we would use basic stoichiometry calculations to figure the amount of uranium hexbromide that is expected. Like so:
100g U x  1 mol Mg  x   1 mol H2   x 2.0g 
                --------------     -------------     ------------   = 301.4g (don't round yet! This is NOT your final ans)
                24.3g Mg       1 mol Mg     1 mol H2 


Since we know 2 variables (percent yield and expected amount), Let x= actual produced amount.
   x/301.4g=0.83. Therefore, x=250g=200g (S.F.)


DID THAT BEND YOUR MIND OR WHAT?!
Yes I know I'm an enthusiastic person when it comes to chemistry.


So long, young disciples.

^GG is all I have to say to that.