## % Yield

$c=\lambda v$

Zane Mills 1E
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### % Yield

suppose 68.5kg of CO is reacted with 8.60kg of H2....calculate the percent yield of methanol if 3.57 x 10^4g of CH3OH is ACTUALLY produced. I got everything leading up to this question I am just super confused as to what it is asking and how to get there.

Chem_Mod
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### Re: % Yield

Once you have determined limiting reagent and used stoichiometry, you will have found the THEORETICAL yield of methanol.

the experimental (actual) yield of methanol divided by the theoretical yield gives the % yield

Sina Rahmani 4A
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### Re: % Yield

Theoretical yield is the amount of a substance one has at the end of an experimental calculation. So whenever you are solving problems, the answer you find when you do all of those mole conversions is the THEORETICAL. The actual yield is the one found in the lab, and the point of the percent yield is to determine how close the lab result is to the calculation from the mole conversions. Since you are looking for a percent, the actual value (the one you find in the lab) is divided by the theoretical value, and then you multiply by 100%.

704628249
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### Re: % Yield

Just remember this simple formula: actual/theoretical * 100

Aleena_Sorf_2A
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### Re: % Yield

In this problem, we are given the masses of the reactants and the actual yield. We are asked to calculate the percent yield, which can be solved by this formula:
Percent Yield= (Actual Yield/Theoretical Yield) x 100
However, we first need to find the theoretical yield, which can be done by solving for the limiting reactant. Once found, you just plug in the values of actual yield for methanol and theoretical yield for methanol into the equation written above and you get the percent yield.

Mirian_Garcia_2G
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### Re: % Yield

So, using the information provided, you can definitely come up with a chemical equation for the rxn. This will allow you to better understand what is occurring within the reaction. The formula for % yield is:

Percent Yield= (Actual Yield/Theoretical Yield)x100%