# Calculation of Theoretical and Percent Yield

Video by Janet G. Coonce, MS

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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:

C_{6}H_{12}0_{6}–fermented —> 2CH_{3}CH_{2}0H + 2C0_{2 }

_{6}H

_{12}0

_{6 }—-fermented –> 23.4 grams of CH

_{3}CH

_{2}0H

**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 the molar mass of a mole of C**by using the periodic table to find the atomic mass of all the atoms in the glucose molecule.

_{6}H_{12}0_{6}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 |

**1 mole C**

_{6}H_{12}0_{6}= 180.18 grams**.**

**moles of glucose in 77 grams.**

**calculate the molar mass of 1 mole of ethanol,**referring to the formula CH

_{3}CH

_{2}0H, 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 |

**1 mole CH**

_{3}CH_{2}0H = 46.08 grams_{6}H

_{12}0

_{6}–fermented —>2 moles of CH

_{3}CH

_{2}0H + 2 moles C0

_{2 }

_{3}CH

_{2}0H for each mole of C

_{6}H

_{12}0

_{6}

**1 mole CH**

_{3}CH_{2}0H = 46.08 grams.**theoretical yield of CH**

_{3}CH_{2}0H_{3}CH

_{2}0H

**39.4 grams of CH**

_{3}CH_{2}0H