Calculate the theoretical yield of a reaction given the limiting reagent and desired product using the theoretical yield calculator. It figures out the maximum yield you can expect from any limiting reaction.

## What Is Theoretical Yield?

The amount of product that would have resulted from a chemical process at 100% efficiency is known as the theoretical yield.

It describes the maximum product that a chemical reaction can yield.

## Theoretical Yield Calculator Formula:

$$ m_{product} = m_{mol,product} \times (n_{moles,lim} \cdot c_{lim}) $$

Where,

- \(m_{product}\) (Mass of the product you want to find)
- \(m_{mol,product}\) (Molar mass of the same product)
- \(n_{moles,lim}\) (Number of moles of the limiting reactant)
- \(c_{lim}\) (Stoichiometric coefficient of the limiting reactant)

If you want to find the number of moles in the limiting reactant, divide its mass by its stoichiometry coefficient and molar mass, such as:

$$ n_{moles, lim}=\dfrac{m_{lim}}{m_{mol, lim} \cdot c_{lim}} $$

## How do you Calculate the Theoretical Yield?

The calculating process is simple, the following steps show how the maximum quantity of moles of a product yield in a chemical reaction.

Let’s elaborate on how maximum a product can be yielded! Follow the steps below:

- A chemical equation must be a
__balanced equation__ - Analyze the mole ratios of the reactants and products
- Now calculate the theoretical yield by the formula

### Example:

Let’s say 3.45g of CH3B is reacted with 5.23g NaOH. Calculate the theoretical yield of the NaBr.

### Solution:

The formula for the theoretical yield helps to gain the maximum product using the balancing chemical equation.

$$ CH_3Br\;+\;NaOH\;\rightarrow\; $$

$$ CH_3OH\;+\;NaBr $$

Now, figure out how many moles of each reagent are present so that we can find the limiting reagent.

First, convert the reagents listed from grams to moles.

Next, find how many moles of the product we can form with our calculated moles of reagent.

$$ 3.45g\;\;CH_3Br\;*\; $$

$$ \frac{1\;mole\;CH_3Br}{94.94g\;CH_3Br}\;*\;\frac{1\;mole\;NaBr}{1\;mole\;CH_3Br} $$

$$ =0.0363\;\text{moles}\;NaBr $$

Now we will do the same for sodium hydroxide.

$$ 5.23g\;\;NaOH\;*\; $$

$$ \frac{1\;mole\;NaOH}{39.99g\;NaOH}\;*\;\frac{1\;mole\;NaBr}{1\;mole\;NaOH} $$

$$ =0.1308\;\text{moles}\;NaBr $$

$$ 0.0363\;\;<\;\;0.1308 $$

We can see that 3.45 grams of methyl bromide will produce fewer moles of bromine than hydroxide.

Thus, all methyl bromide will react completely. Therefore, methyl bromide is our limiting reagent.

This theoretical yield tells us how much product we should end up with if our reaction runs to completion. So, after using up all of the methyl bromide, we are left with 0.0363 moles of bromine. This is our theoretical yield.

Now, let’s figure out what the theoretical yield would be in terms of grams.

$$ 0.0363\;\text{moles}\;NaBr\;*\; $$

$$ 39.99\;\frac{g}{mol}\;=\;1.452g\;NaBr $$

We see there is expected to be 1.452 grams of sodium bromide product. Set Apart these lengthy calculations, and let our theoretical yield calculator show the exact numbers that formed during the chemical reactions you performed.

To calculate the __percent yield__, you simply take the actual yield of 1.099 grams of sodium bromide divided by the theoretical yield of 1.452 grams of sodium bromide.

$$ \frac{1.099g\;NaBr}{1.452g\;NaBr}\;*\;100\;=\;75.69% yield $$

## Steps To Use This Theoretical Yield Calculator:

**Step 1:**First, select the unit of measurement**Step 2:**Put the value of the mass, moles, and molecular weight in their respective boxes**Step 3:**Click Calculate**Step 4:**This tool provides you with the theoretical yield of a balanced chemical equation and the number of moles of the balanced equation along with step-by-step calculations

## Comparison between Theoretical Yield vs. Actual Yield, and Percent Yield:

Feature | Theoretical Yield | Actual Yield | Percent Yield |
---|---|---|---|

Definition | Maximum amount of product expected based on the balanced chemical equation and complete reaction of the limiting reagent. | Amount of product actually obtained from the experiment. | Ratio of the actual yield to the theoretical yield, expressed as a percentage. |

Calculation | Based on stoichiometry and reactant quantities. | Measured mass or moles of the product. | (Actual Yield / Theoretical Yield) * 100% |

Units | Moles or grams | Moles or grams | Percentage (%) |

Value | Always equal to or greater than the actual yield. | Always equal to or less than the theoretical yield. | Can range from 0% (no product formation) to 100% (complete reaction), and may even exceed 100% due to side reactions. |

Significance | Ideal benchmark for efficiency and optimization. | Shows the real outcome of the experiment and potential differences. | Indicates the efficiency of the reaction and success in maximizing product formation. |

Limitations | Assumes no side reactions, complete consumption of limiting reagent, and perfect reaction conditions. | May be affected by factors like incomplete reactions, side reactions, loss of product during purification, and measurement errors. | Sensitive to both theoretical and actual yield values. |

### What Are Limiting Reactants?

In a chemical reaction, the reactants that are consumed first and therefore limit how much product can be formed. This is known as the limiting reactants.

### Is The Amount of Product Limited By The Limiting Reactant?

Yes, the amount of product is limited by the limiting reactant. From the available reactants, it gives the smallest yield of product. This smallest yield is known as the theoretical yield.

**Wikipedia: **Yield (chemistry), Definitions, Theoretical, actual, and percent yields, Example, Purification of products, Internal standard yield, Reporting of yields.

**Khan Academy:** Limiting reactant and theoretical yield, calculating theoretical yield, Stoichiometry, Calculating the amount of product formed from a limiting reactant.