Calculation of Theoretical and Percent Yield

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?

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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,

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77 grams   C₆H₁₂0₆  —-fermented –> 23.4  grams of  CH₃CH₂0H

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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.

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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

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Therefore 1 mole C₆H₁₂0₆ = 180.18 grams

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.

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Therefore 1 mole CH₃CH₂0H = 46.08 grams

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Go back to the equation…

1 mole C₆H₁₂0₆ –fermented —>2 moles of CH₃CH₂0H + 2 moles C0₂

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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.

therefore

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           = 2 x 0.43 moles x 46.08 grams/mole of CH₃CH₂0H

           = 39.4 grams of CH₃CH₂0H

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In the video, each of these steps were linked together

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Now we can solve the next question

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