# Theoretical Yield

The theoretical yield of a reaction is the amount of product that would be formed if the reaction went to completion. It is based on the stoichiometry of the reaction and ideal conditions in which starting material is completely consumed, undesired side reactions do not occur, the reverse reaction does not occur, and there no losses in the work-up procedure.

Theoretical yield calculations are carried out in the same way as they were in general chemistry: the moles of limiting reactant determines the moles of product. To calculate theoretical yield:

- Balance the reaction and determine the stoichiometry or ratios of reactants to products.
- Find the number of moles of each starting material used.
- Determine which reagent is limiting; remember that catalysts, solvents, or any compounds that are not part of the actual chemical reaction cannot be the limiting reagent.
- Calculate the moles of product expected if the yield were 100% based on the limiting reagent.
- Calculate the grams of product corresponding to the number of moles expected.

After your laboratory reaction is complete, you will isolate and measure the amount of product, then compare the **actual** yield to the theoretical yield to determine the **percent yield**:

In the laboratory, the percent yield has the practical aspect of telling you how successful was your synthesis scheme. A low percentage yield means that the conditions were not optimal and could be improved. Perhaps there are competing reactions occurring or some of the product is being lost in the purification steps.

### Example Calculations of Theoretical and Percent Yield

**Example 1**: What is the theoretical yield of ethylene in the acid-catalyzed production of ethylene from ethanol, if you start with 100 g of ethanol?

The reaction as written above is balanced, with one mole of ethanol producing one mole of ethylene, therefore the stoichiometry is 1:1. The acid, written over the arrow, is a catalyst and does not enter into the theoretical yield calculations. To calculate the theoretical yield, determine the number of moles of each reactant, in this case the sole reactant ethanol. Convert the 100 g to moles; the molecular weight of ethanol is 46 g/mole, therefore:

Since there is only one reactant, it is also the limiting reagent. The theoretical number of moles of ethylene is 2.17. Since the molecular weight of ethylene is 28 g/mole, this corresponds to 61 g from the following calculation:

The theoretical yield is therefore 61 g. Note that theoretical yield is expressed in grams.

Percent yields are expressed as percentages. For instance, if you start with 100 g of ethanol and isolate 50 g of ethylene, the actual yield is 82%:

**Example 2**: Consider the acid-catalyzed esterification of isoamyl alcohol to produce isoamyl acetate. If you begin with 10 g of isoamyl alcohol, 5 mL of acetic acid, and 1 ml of sulfuric acid, what is the theoretical yield of isoamyl acetate?

The reaction as written above is balanced, with one mole of acetic acid and one mole of isoamyl alcohol producing one mole of isoamyl acetate and one mole of water, therefore the stoichiometry is 1:1. The acid (sulfuric acid), written over the arrow, is a catalyst and does not enter into the theoretical yield calculations. Next, determine the number of moles of each reactant. To do this, first go to the Tables of Physical Constants and the Table of Acids and look up the molecular weights and densities of the reactants and product:

Compound | MW | Density | Other Information |

Isoamyl alcohol (butanol, 1-, 3-methyl) |
88.15 | 0.809 | |

Acetic acid | 60.05 | 1.05 | Conc. reagent is 17.4 N, 99.8% by weight |

Isoamyl acetate (acetic acid, 3-methylbutyl ester) |
130.19 | 0.867 |

10 g of isoamyl alcohol will produce 0.11 mole of isoamyl acetate:

5 mL of acetic acid will produce 0.087 mole of isoamyl acetate:

The limiting reagent is the acetic acid: if there are only 0.087 moles of acetic acid, the maximum number of moles of isoamyl acetate which can be produced is 0.087 moles. Convert this mole amount to grams to get 11.3 grams:

**Example 3**: If you begin with 25 mL of propionic acid, 20 mL of n-butanol, 1ml of sulfuric acid, what is the theoretical yield of n-butylpropionate in the following acid-catalyzed esterification?

The reaction as written above is balanced, with one mole of propionic acid and one mole of n-butanol producing one mole of n-butylpropionate and one mole of water, therefore the stoichiometry is 1:1. The mineral acid (sulfuric acid), written over the arrow, is a catalyst and does not enter into the theoretical yield calculations. Next, determine the number of moles of each reactant. To do this, consult the Tables of Physical Constants:

Compound | MW | Density |

Propionic acid | 74.08 | 0.993 |

n-Butanol | 74.12 | 0.8098 |

n-Butylpropionate | 130.19 | 0.875 |

25 mL of propionic acid is will produce 0.335 mole of n-butylpropionate:

20 mL of n-butanol will produce 0.218 mole of n-butylpropionate:

Therefore, the n-butanol is the limiting reagent: only 0.218 mole of n-butylpropionate can be produced using 0.218 moles of n-butanol and 0.335 moles of propionic acid. The theoretical yield is 28.4 gram:

**Example 4**:

Three moles of 1-propanol reacts with one mole of phosphorus trichloride to produce 1-chloropropane. What is the theoretical yield if you begin with 75 g of 1-propanol and 75 g of phosphorus trichloride?

The stoichiometry of this reaction is 1:1 for 1-propanol to 1-chloropropane, but 1:3 for phosphorus trichloride to 1-chloropropane. These ratios must be incorporated into the theoretical yield calculations.

The physical data (from the CRC) is:

Compound | MW | Density |

1-Propanol | 60.10 | 0.803 |

Phosphorus trichloride | 137.33 | |

1-Chloropropane | 78.54 | 0.891 |

75 g of 1-propanol will produce 1.25 mole of 1-chloropropane:

75 g of phosphorus trichloride will produce 1.64 mole of 1-chloropropane:

Therefore, the limiting reagent is 1-propanol, and the theoretical yield is 98.1 g:

### Helpful Nomenclature and Relationships

The calculation of the number of moles of acetic acid (Example 2) brings up several questions which students often ask when encountering physical data involving acids. For acetic acid, you find in the table the MW, normality of concentrated reagent, the percent by weight, and the specific gravity.

**Normality** is the number of equivalents per liter. For an acid, an equivalent is the mass of acid that can furnish exactly 1 mole of H^{+} ions. The relationship between molarity (M) and normality (N) for several acids is summarized in the table below.

Acid | Name | N:M |

HCl | Hydrochloric acid | N = M |

H_{2}SO_{4} |
Sulfuric acid | N = 2M |

H_{3}PO_{4} |
Phosphoric acid | N = 3M |

CH_{3}CO_{2}H |
Acetic acid | N = M |

**Specific gravity** is the ratio of the density (in g/mL) of the compound at standard conditions relative to the density of water at the same conditions. Practically, you can assume that specific gravity equals density, therefore if concentrated acetic acid has a specific gravity of 1.05, it has a density of 1.05 g/mL.

**Percent by weight** is the number of grams of the compound in a particular mass of the concentrated solution. If you have 100 g of concentrated acetic acid, you have 99.8 g of acetic acid. From this, you can calculate the number of moles using the molecular weight.

Which values should you use to calculate the number of moles of the acid in question? Whichever values you find more convenient, depending on which values you are given in the experimental procedure or measured in your experiment. If you are given a volume measurement, it is easiest to use the normality as in example 2 above. You could also use the specific gravity and the molecular weight to calculate moles from a volume measurement. If you are given a number of grams, you can calculate the moles from the % by weight and the molecular weight.

How do you calculate the number of grams when given the number of milliliters? The answer is to convert using the following relationship of mass, volume, and density:

mass = volume x density

Density values are listed as grams per milliliter. For example, if you have 25 mL of a compound and its density is 0.97 g/mL, the number of grams is found by the following calculation:

mass = 25 mL x 0.97 g/mL = 24.2 g

Back to Lab Notebook