Skip to content

Calculation of Theoretical and Percent Yield

October 9, 2013

Video by Janet G. Coonce, MS

If you plan to view the video on your cell phone, consider your data plan and whether you should wait until you have a WiFi connection to avoid cellular charges.

Problem:  77 grams of glucose was allowed to ferment to form ethanol and carbon dioxide.  23.4 grams of ethanol was recovered by distillation.  What was the theoretical yield?  What was the percent yield?  In other words how efficient is my laboratory process in approaching the potential theoretical yield?


Fermentation Glucose2


When we write and balance the equation, we find that each mole of glucose, if we were able ferment 100% of 1 mole of glucose to ethanol, should yield 2 moles of ethanol and 2 moles of carbon dioxide:

C6H1206–fermented —> 2CH3CH20H + 2C02

To restate the actual problem with what we know,
77 grams   C6H120 —-fermented –> 23.4  grams of  CH3CH20H
Theoretical yield equation
To calculate the theoretical yield we first need to determine the molar mass represented by 77 grams of glucose.  We also will need to determine the molar mass of 2 moles of ethanol.
Calculate Molar Mass
First we calculate the molar mass of a mole of C6H1206 by using the periodic table to find the atomic mass of all the atoms in the glucose molecule.
Atom Atomic Mass (g/mole) Number Molecular   Mass
C 12.01 6 72.06
H 1.01 12 12.12
O 16.00 6 96.00
Total  Molar  Mass 180.18  g/mole
Therefore 1 mole C6H1206 = 180.18 grams
Now that we know how many grams in a mole of glucose we can calculate the number of moles of glucose in 77 grams.
actual moles
To calculate the molar mass of 1 mole of ethanol, referring to the formula CH3CH20H, we know there are 2 atoms of carbon, 6 of hydrogen and 1 of oxygen.  As shown in the spreadsheet below, the molecular mass of ethanol is 46.08 grams.
Atom Atomic Mass Number Molecular   Mass
C 12.01 2 24.02
H 1.01 6 6.06
O 16.00 1 16.00
 Total  Molar  Mass 46.08  g/mole
Therefore 1 mole CH3CH20H = 46.08 grams
Go back to the equation…
1 mole C6H1206 –fermented —>2 moles of CH3CH20H + 2 moles C02
The theoretical yield would be 2 moles of CH3CH20H for each mole of C6H1206
We know from our calculations above that 1 mole CH3CH20H = 46.08 grams.
theoretical yield of CH3CH20H
           = 2 x 0.43 moles x 46.08 grams/mole of CH3CH20H
           = 39.4 grams of CH3CH20H
In the video, each of these steps were linked together
Theoretical yield2
Now we can solve the next question
Percent Yield Question
Percent Yied
Percent Yield equation
                           =  59.4% yield

From → Problem Solving

Leave a Comment

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: