% Yield Calculations

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junglemonkey
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% Yield Calculations

Post by junglemonkey » Sun Nov 16, 2003 2:08 pm

Can someone please help me out with organic percentage yield calculations? I'm so lost. Even just a method or something you can write out for me, I don't really understand what I'm doing. I'm so gutted cause my organic chemistry has been top notch after 3 years of struggling to understand it!



Here's an example question...



Qu. In the oxidation of Butan-1-ol to butanone shown below, calculate the mass of butanone formed when 5.0g of butan-1-ol is reacted and the yield of the reaction is 62.5%



(sorry no subscript)



C4H9OH ---> C4H8O

butan-1-ol butanone



--------------------------------------



And this type...



The manufacture of methanol from synthesis gas can be shown by the equation:-



CO(g) + 2H2(g) ---> CH3OH(g)



In a convertor it was found that 1400kg of carbon monoxide formed 900kg of methanol. What is the percentage yield of the reaction?







Anyone!?!?!? :blink:
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Doh!-Nut
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Re: % Yield Calculations

Post by Doh!-Nut » Wed Nov 19, 2003 3:48 am

900 is what percent of 1400? ;)
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Re: % Yield Calculations

Post by junglemonkey » Wed Nov 19, 2003 1:16 pm

No, I'm talking about actual/theoretical values!! You have to work out what, in theory, would be the value and what it really is. :blink:
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Re: % Yield Calculations

Post by Doh!-Nut » Thu Nov 20, 2003 2:50 am

I know. Do you know Molar Mass?

:P
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Re: % Yield Calculations

Post by junglemonkey » Thu Nov 20, 2003 1:05 pm

mass/GFM yeah? Number of moles?
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Re: % Yield Calculations

Post by Doh!-Nut » Thu Nov 20, 2003 6:17 pm

Number of moles helps you figure out yieldage. :shifty:
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Re: % Yield Calculations

Post by junglemonkey » Thu Nov 20, 2003 6:29 pm

Yes, I know. Any chance of telling me how to do them? General method? Some sort of explanation?
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Re: % Yield Calculations

Post by Doh!-Nut » Thu Nov 20, 2003 6:32 pm

Me doing your homework for you? How silly. :P
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Re: % Yield Calculations

Post by Kristina » Thu Nov 20, 2003 10:02 pm

You first balance the equation



(which you did)



then you find the molar mass of the product





You take the mass of the actual product you got in your experiment







like a fraction





actual reaction product in molar mass

---------------------------------- = percent yeild

theroretical reaction product in molar mass





But since you have the percent yeild, you work it just like algebra



X

----------------------------------- = 62.5%

72.04 g/mol (theroretical reaction product in molar mass)





72.04 * .625 = 45.025





Now.. Check your work





Does this work out?





45.025 g/mol

--------------- = 62.5% (or .625 in decimals)

72.04 g/mol





Yesss it does.. Yay very simple





sorry I didn't get to you sooner! If you need anymore help, just ask!
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Re: % Yield Calculations

Post by junglemonkey » Sat Nov 29, 2003 1:55 pm

Me doing your homework for you? How silly.




How silly? That WASN'T my homework, this is a general point in my higher chemistry course I don't understand, therefore I asked for help. Therefore I asked for a method or an explanation.
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