To find the percent yield, things are actually pretty easy!
All you have to do is take the actual yield divided by the theoretical yield and multiply that by 100.
I remember this because anything actual comes over anything theoretical in real science.
You just have to remember that your percent yield will always be less than 100 due to error, whether it be human error, machine error, or unknown error.
Finding percent yield can be pretty disappointing, too. It's like baking a batch of cookies and starting with all of this:
http://noentreecakes.com/wp-content/uploads/2015/03/Baking-Ingredients.jpg
Only to end up with this:
http://upload.wikimedia.org/wikipedia/commons/thumb/6/61/Single_chocolate_chip_cookie.jpg/300px-Single_chocolate_chip_cookie.jpg
Here are some websites that'll help you if you're struggling:
https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/mass-relationships-and-chemical-equations-3/reaction-stoichiometry-44/calculating-theoretical-and-percent-yield-234-4704/
http://www.sparknotes.com/chemistry/stoichiometry/realworldreactions/section2.rhtml
Marie, I really liked the memory hook you came up with for percent yield!! It will definitely help me to remember that actual yield goes over theoretical in the equation. Thanks for sharing!
ReplyDeleteI agree with Brianna about your memory hook, it's really helpful! Also, your comparison that you made with finding percent yield and making cookies was really funny. Thanks for sharing this information!
ReplyDeleteMarie, I like how our first explained the process to find percent yield and shared how you remember the equation. In addition, the way you compared percent yield to baking cookies was very unique and helpful. I also found the website you posted very educational with its examples, explanations, and bozeman video.
ReplyDeleteAgreed with Carly and Bri! That is super interesting and I never thought of it in that way, let alone a real life example at all. I also really like the website you included because Bozeman is my best friend; he helps me every unit! This is a really good post that helps understand the concept better.
ReplyDeleteThe analogy you used was really helpful to me and your explanation was super through. Thank you so much for the great example and for helping me better understand this subject!
ReplyDelete