Percent yield (grams or moles)?

[bdc142]

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TBR Stoichiometry, 72. If 10 g of Ca(OH)2 produces 5 grams of CaCO3, what is the percent yield for the reaction?

The TBR answer uses moles to calculate percent yield, but I thought you were supposed use mass? Which one does the MCAT want?

 

[MedPR]

TBR Stoichiometry, 72. If 10 g of Ca(OH)2 produces 5 grams of CaCO3, what is the percent yield for the reaction?

The TBR answer uses moles to calculate percent yield, but I thought you were supposed use mass? Which one does the MCAT want?

You have to write a balanced equation and convert grams to moles.

 

[bdc142]

The passage has a balanced equation (mol ratio is 1:1 here). I just don't understand why I can't do (actual product in grams / theoretical product in grams)(100%)

 

[MedPR]

The passage has a balanced equation (mol ratio is 1:1 here). I just don't understand why I can't do (actual product in grams / theoretical product in grams)(100%)

Because 1gram of 1 thing isn't equivalent to 1gram of another thing. Moles are the common unit that you have to use.

Imagine trying to convert a gram to a cubic meter without first converting grams to mL.

 

[bdc142]

http://www.800mainstreet.com/6/0006-007-percent-yield.html

Why is percent yield there using grams?

 

[Buttafuoco]

you can use grams or moles to do percent yield as they are directly related to each other, but you need to use moles at some point because of the molar ratio of product to reactant.

 

[MedPR]

http://www.800mainstreet.com/6/0006-007-percent-yield.html

Why is percent yield there using grams?

Because they already did the appropriate gram to mole calculations; they just didn't show them.

[FONT=Arial, Helvetica, sans-serif]If you burn 12 grams of carbon to make CO2, then amount of carbon dioxide expected is one mol of CO2 or 44 grams of CO2. .​

[FONT=Arial, Helvetica, sans-serif]Sadly the amount you will get will probably be less than 44 grams and more like 34 grams of CO2. The problem is a competing reaction that happens. Some carbon reacts to make CO.

.

In order to know that 12 grams of Carbon theoretically makes 44 grams of CO2, you must do a gram to mole conversion (Carbon), then a mole to mole conversion (Carbon to CO2), then a mole to gram conversion (CO2).

If you only make 34 grams, then you can do 34 grams/44 grams to get percent yield. BUT, you CANNOT know that the theoretical yield is 44 grams without converting to moles.

MCAT problems will give you an equation (probably unbalanced), tell you that the overall yield was X grams, or X moles, and then ask you the percent yield. Your task now is to find out the theoretical yield, which requires that you find the limiting reagent (which requires grams to mole conversion), then find the theoretical yield (which also requires grams to mole conversion). THEN you can do grams/grams or mole/mole or whatever else you want.

 

[DrDreams]

Hi, sorry to bring back this old thread, but for some odd reason I cant get this either. I know that 5g (CaCO3)/100g/mol makes perfect sense, but where did the 10g/74g/mol come from?

Because I was converting to moles of Ca(OH)2 and then taking this mole quantity and converting it to moles of CaCO3 and then i just began questioning what the heck I was doing because I do not recall TBR going over this theoretical over actual stuff.

 

[Czarcasm]

Hi, sorry to bring back this old thread, but for some odd reason I cant get this either. I know that 5g (CaCO3)/100g/mol makes perfect sense, but where did the 10g/74g/mol come from?

Because I was converting to moles of Ca(OH)2 and then taking this mole quantity and converting it to moles of CaCO3 and then i just began questioning what the heck I was doing because I do not recall TBR going over this theoretical over actual stuff.

% yield is actual grams/theoretical grams. The actual grams obtained is 5 grams which is less than the theoretical expected. If you write a balanced ratio, you'll find all the reactants and products are in a 1:1 to 1:1 ratio. Therefore, to find the grams of products that would be theoretically produced (if it reacted entirely), you must first covert grams to moles Ca(OH)2, since that's what you're provided, then relate that to moles CaCO3 produced, and convert moles to grams. That's essentially what they did there.

So 10g/74g represents moles Ca(OH)2, which is equal to moles CaCO3 (because they are in a 1:1 ratio), and if you multiple this times molecular weight of CaCO3, you get the grams of CaCO3 that would be produced theoretically (which as it turns out is greater than 5g we're told is produced). So essentially your % yield is: actual grams CaCO3 / theoretical grams CaCO3: (( 5g / (10g/74g) x 100g )) x 100. The 100 in the numerator and denominator cancel and this simplifies to: 5g / (10g/74g)

 

[DrDreams]

OK GREAT. Yea I knew it was a 1:1 ratio, I just got confused after I got the moles of Ca(OH)2. That makes perfect sense. Thank you very much!

 

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