# Theoretical Yield Calculator

Created by Davide Borchia
Last updated: Jul 01, 2022

Chemical reactions run both in a laboratory and on paper when you are preparing an experiment: with our theoretical yield calculator, you will quickly learn how to calculate the maximum yield you can expect from any reaction.

You will only have to input a few reaction parameters, and we will calculate the theoretical yield in a fraction of a second. And if you want to learn more, keep reading to find out:

• What is the theoretical yield;
• How to find the theoretical yield;
• Some practical examples of the application of the theoretical yield formula.

If you're synthesizing a new compound and want to learn something new, check how to use the Randles-Sevcik equation to find diffusion coefficient!

## Chemical reactions

A chemical reaction is a process where a certain number of substances (the reactants) combine and change their chemical bonds to form a new set of species (the products).

The rearrangement of atoms in the species is the basis of chemistry. One of the major distinctions among them is the possibility of reversing the direction of the reaction. We can identify:

• Reversible chemical reactions;
• Irreversible chemical reactions.

## What is the theoretical yield?

First, what is the yield of a chemical reaction? The yield of a chemical reaction is the quantity of a specific product obtained from a given set of reactants.

In a laboratory setting, many factors contribute to the yield:

• Procedural errors;
• Systematic errors;
• Environmental conditions.

And so on. Most of them (if not all) affect the yield negatively.

The theoretical yield is the maximum possible quantity of a given product you can obtain from a chemical reaction, assuming pure reactants and flawless execution of the experiment. This yield corresponds to a $100\%$ conversion of the reactants in the products, and perfect recovery of all the molecules of products created in the reaction. As you can easily imagine, this is not an easy task: it is actually impossible.

The theoretical yield of a reaction depends entirely on the limiting reactant: we will discover what it is and how to calculate the theoretical yield in the next section.

## How to calculate the theoretical yield?

The mass of the product (it's not important to specify which one) is determined in the theoretical yield formula by the stoichiometry of the limiting reactant . The limiting reactant is the one that participates in the reaction with the smallest number of moles — adjusted by its coefficient. The mass of the product is then:

$\footnotesize m_\text{prod} = m_{\text{mol,prod}}\cdot (n_{\text{moles,lim}}\cdot c_{\text{lim}})$

Where:

• $m_\text{prod}$ is the mass of the product;
• $m_{\text{mol,prod}}$ is the molar mass of the same product (don't mistake it with the molecular weight);
• $n_{\text{moles,lim}}$ is the number of moles of the limiting reactant in the reaction; and
• $c_{\text{lim}}$ is the stoichiometric coefficient of the limiting reactant.

You can quickly find the number of moles of the limiting reactant by dividing its mass by its stoichiometric coefficient and molar mass:

$\footnotesize n_\text{moles,lim} = \frac{m_{\text{lim}}}{m_{\text{mol,lim}}\cdot c_{\text{lim}}}$

Don't confuse the symbol $n$ with a refractive index which is, e.g., a part of the Sellmeier equation calculator. It's another method of materials characterization in chemistry by using optical absorption of light.

## How do I find the limiting reactant?

To find the limiting reactant, calculate the number of moles with which each reactant participates in the reaction. The lowest one identifies the limiting reactant.

It makes sense! If reactant $\text{A}$ participates with $4$ moles, and reactant $\text{B}$ participates with $2.5$ moles, we can expect that, at most $2.5$ moles of reactant $\text{A}$ will be used (in this examples we assumed stoichiometry 1).

## How to calculate the theoretical yield: an example

Take the synthesis of toluene from benzene and methanol.

$\footnotesize \text{C}_6\text{H}_6 + \text{CH}_3\text{OH} \rightarrow \text{C}_6\text{H}_5\text{CH}_3 + \text{H}_2\text{O}$

Assume you start with $100\ \text{g}$ of benzene and $80\ \text{g}$ of ethanol. First, we calculate the molar masses of the two molecules:

• For the benzene: $m_{\text{mol,}\text{C}_6\text{H}_6}=78.11\ \text{g}/\text{mol}$;
• For the methanol: $m_{\text{mol,}\text{CH}_3\text{OH}}=32.04\ \text{g}/\text{mol}$;

The stoichiometric coefficients in the reaction are uniformly $1$, so we don't need to adjust our calculations. Let's find the limiting reactant: divide the mass of each reactant by its molar mass.

\footnotesize \begin{align*} n_{\text{mol,}\text{C}_6\text{H}_6}=\frac{100\ \text{g}}{78.11\ \text{g}/\text{mol}}=1.28\ \text{mol}\\ \\ n_{\text{mol,}\text{CH}_3\text{OH}}=\frac{80\ \text{g}}{32.04\ \text{g}/\text{mol}}=2.5\ \text{mol} \end{align*}

Benzene is clearly our limiting reactant. Apply the formula for the theoretical yield to learn how much toluene we can expect to produce with a perfect reaction (knowing that toluene has molar mass equal to $92.14\ \text{g}/\text{mol}$).

\footnotesize \begin{align*} m_{\text{C}_6\text{H}_5\text{CH}_3}\! &=\! m_{\text{mol,}\text{C}_6\text{H}_5\text{CH}_3}\!\cdot\! (n_{\text{moles}\text{C}_6\text{H}_6}\!\cdot\! c_{\text{C}_6\text{H}_6})\\ \\ &=92.14\ \frac{\text{g}}{\text{mol}}\cdot(1.28\ \text{mol} \cdot 1)\\ \\ &= 117.96\ \text{g} \end{align*}

This quantity would correspond to a $100\%$ percent yield. However, in the lab, you're more likely to end up close to $50\%$.

Davide Borchia
Limiting reagent
Mass
g
Molecular weight
g / mol
Moles
mole(s)
Stoichiometry
Desired product
Stoichiometry
Moles
mole(s)
Molecular weight
g / mol
Theoretical yield
g
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