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

# Theoretical Yield Calculator

By Jack Bowater

This theoretical yield calculator will answer all the burning questions you have regarding how to calculate the theoretical yield, such as how to find theoretical yield as well as the theoretical yield definition and the theoretical yield formula. Before carrying out any kind of lab work you need to to work out what is the theoretical yield, so you know how much of your product to expect from a given amount of starting material. This allows you to work out how efficiently you carried out your reaction, which is done by calculating the percent yield. The theoretical yield equation can also be used to ensure that you react equal moles of your reactants, so no molecule is wasted.

We also have a percent yield calculator to assist you with your calculations.

IMPORTANT NOTE: Yields can only be found using the limiting reagent. If you are uncertain which of your reagents are limiting, plug in your reagents one at a time and whichever one gives you the lowest mole is the limiting reagent. Remember to hit refresh at the bottom of the calculator to reset it.

## Theoretical yield definition

What is theoretical yield? It is the amount of a product that would be formed if your reaction was 100% efficient. How to achieve 100% efficiency? Well, it would mean that every molecule reacted correctly (i.e., no side products are formed) at every step and that no molecule was lost on the sides of the glassware. As a normal reaction deals with quintillions of molecules or atoms, it should be obvious that some of these molecules will be lost. Therefore the percent yield will never be 100%, but it is still useful to know as a metric to base your efficiency of reaction off. For more on this check out our percent yield calculator (link above). Find out how to calculate theoretical yield with the theoretical yield equation below!

## Theoretical yield formula

Using the theoretical yield equation helps you in finding the theoretical yield from the mole of the limiting reagent, assuming 100% efficiency. So, to stop you from wondering how to find theoretical yield, here is the theoretical yield formula:

`mass of product = molecular weight of product * (moles of limiting reagent in reaction * stoichiometry of product)`

where:

`moles of limiting reagent in reaction = mass of limiting reagent / (molecular weight of limiting reagent * stoichiometry of limiting reagent)`

Stoichiometry is defined as the number before the chemical formula in a balanced reaction. If no number is present, then the stoichiometry is 1. The stoichiometry is needed to reflect the ratios of molecules that come together to form a product. The good thing about this calculator is that it can be used any way you like, that is to find the mass of reactants needed to produce a certain mass of your product. All this information is hidden in the moles, which can be derived from a solutions molarity or concentration.

## How to calculate theoretical yield?

Now, the theoretical yield formula may seem difficult to understand so we will show you a quick guide on how to calculate the theoretical yield. The measurements you need are the mass of the reagents, their molecular weights, the stoichiometry of the reaction (found from the balanced equation) and the molecular weight of the desired product. Look no further to know how to find the theoretical yield:

1. First, calculate the moles of your limiting reagent. This is done by using the second equation in the theoretical yield formula section (pro tip: make sure that the units of weight are the same for the correct results).
2. Select the reactant that has the lowest number of moles when stoichiometry is taken into account. This is your limiting reagent. If both have the same amount of moles, you can use either.
3. Use the first equation to find the mass of your desired product in whatever units your reactants were in.

There you go! If you are still struggling, check the examples below for a more practical approach.

## Examples of yield calculations

Time for some examples. Lets say you are doing a nucleophilic addition reaction, forming hydroxyacetonitrile from sodium cyanide and acetone. By Hazmat2 - CC BY-SA 3.0, Link

Let's ignore the solvents underneath the arrow (they will both be present in excess and therefore will not be limiting reagents), but also the sodium cation of the sodium cyanide, as it is just a spectator ion. If we react 5 g of acetone with 2 g of cyanide, what is the theoretical yield of hydroxyacetonitrile?

1. We need to work out the limiting reagent first. As the stoichiometry of both reagents is 1 (i.e., one molecule of acetone reacts with one molecule of cyanide), we can simply use the `mass = molecular weight * mole` equation to find this:

• Lets rearrange the equation to find moles. This gives: `mole = mass / molecular weight`
• Acetone has a molecular weight of 58 g / mole, so: `mole = 5 / 58 = 0.862 mol`
• Cyanide has a molecular weight of 26 g / mole, so: `mole = 2 / 26 = 0.0769 mol`
• So there are fewer moles of cyanide, meaning this is the limiting reagent.
2. Knowing the limiting reagent and its moles means that we know how many moles of the product will form. As the stoichiometry of the product is 1, 0.0769 moles will form.

3. We can once again use the `mass = molecular weight * mole` equation to determine the theoretical mass of the product. The molecular weight of hydroxyacetonitrile is 85 g / mol: `mass = 85 * 0.0769 = 6.54 g`

Now we know that if we carry out the experiment, we would expect 6.54 g of hydroxyacetonitrile. Not too bad right!

Let's say you are trying to synthesise acetone to use in the above reaction. You react 8 g of calcium carbonate (100 g / mol) with 9 g of acetic acid (60 g / mol), how much acetone is formed?

1. Once again, we need to work out which is the limiting reagent first. Let's use the `mass = molecular weight * mole` equation again:

• Let's rearrange the equation to find moles. This gives: `mole = mass / molecular weight`
• Lets find the moles of acetic acid: `mole = 9 / 60 = 0.15 mol`
• And the moles of calcium carbonate: `mole = 8 / 100 = 0.08 mol`
• It looks like calcium carbonate is the limiting reagent. But wait! We haven't considered the stoichiometry. Since we need 2 molecules of acetic acid to form one molecule of acetone, we need to divide the moles of acetic acid by 2: `mole = 0.15 / 2 = 0.075 mol`
• So it turns out that the acetic acid is the limiting reagent!
2. Now that we know the limiting reagent and its moles, we know how many moles of the product will form. As the stoichiometry of the product is 1, 0.75 moles will form.

3. Use the `mass = molecular weight * mole` equation to determine the theoretical mass of the product. The molecular weight of acetone is 58 g / mol: `mass = 58 * 0.075 = 4.35 g`

So from this reaction, we should get, theoretically speaking, 4.35 g of acetone. Nice!

Now go on and conquer the world of theoretical yield calculations, you can do it!

Jack Bowater

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