Theoretical Yield Calculator

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 work out what is the theoretical yield so you know how much of your product, be it a molecule or lattice, you can expect from a given amount of starting material. This allows you to work out how efficiently you carried out your reaction (the quantity you can find at the actual yield calculator), which is done by calculating the percent yield. You can also use the theoretical yield equation to ensure that you react with equal moles of your reactants so no molecule is wasted.

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 number of moles is the limiting reagent. Remember to hit refresh at the bottom of the calculator to reset it.

What is the 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 the reaction. For more on this, check out our percent yield calculator (link above). Find out how to calculate theoretical yield with the equation below!

Using the equation below helps you find the theoretical yield from the moles of the limiting reagent, assuming 100% efficiency. This is the formula:

where:

• $m_{\text{product}}$ — Mass of product;
• $m_{\text{mol},\text{product}}$ — Molecular weight of the desired product;
• $n_{\text{lim}}$ — Moles of the limiting reagent; and
• $c$ — The stoichiometry of the desired product.

The number of moles of the limiting reagent in the reaction is equal to:

where:

• $n_{\text{lim}}$ — Number of moles of the limiting reagent;
• $m_{\text{lim}}$ — Mass of limiting reagent;
• $m_{\text{mol},\text{lim}}$ — Molecular weight of the limiting reagent; and
• $c_{\text{lim}}$ — Stoichiometry of the 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 solution's molarity or concentration (you can learn how to do so with our molarity calculator and concentration calculator).

Now, the theoretical yield formula may seem challenging 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. We do this 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: you can use the weight converter if you need help with the factors).

2. Select the reactant with 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.

Time for some examples. Let's say you are doing a nucleophilic addition reaction, forming hydroxyacetonitrile from sodium cyanide and acetone.

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 $5\ \text{g}$ of acetone with $2\ \text{g}$ 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:

   • Let's rearrange the equation to find moles. This gives:
     $\small\text{mole} = \text{mass}/\text{molecular weight}$

   • Acetone has a molecular weight of $58\ \text{g}/\text{mole}$, so:
     $\small\text{mole} = 5 / 58 = 0.862\ \text{mol}$

   • Cyanide has a molecular weight of $26\ \text{g}/\text{mole}$, so:
     $\small\text{mole} = 2 / 26 = 0.0769\ \text{mol}$

   • So there are fewer moles of cyanide, meaning this is the limiting reagent.

2. Knowing the limiting reagent and its moles means knowing how many moles the product will form. As the stoichiometry of the product is $1$, $0.0769$ moles will form.

3. We can once again use the $\small\text{mass} = \text{molecular weight} \cdot\text{mole}$ equation to determine the theoretical mass of the product. The molecular weight of hydroxyacetonitrile is $85\ \text{g}/\text{mol}$:

Now we know that if we carry out the experiment, we would expect $6.54\ \text{g}$ of hydroxyacetonitrile. Not too bad, right?

Let's say you are trying to synthesize acetone to use in the above reaction.

You react $8\ \text{g}$ of calcium carbonate ($100\ \text{g}/\text{mol}$) with $9\ \text{g}$ of acetic acid ($60\ \text{g}/\text{mol}$), how much acetone is formed?

1. Once again, we need to work out which is the limiting reagent first. Let's use the $\small\text{mass} = \text{molecular weight}\cdot \text{mole}$ equation again:

   • Let's rearrange the equation to find moles. This gives:
     $\small\text{mole} = \text{mass} / \text{molecular weight}$

   • Let's find the moles of acetic acid:
     $\small\text{mole} = 9 / 60 = 0.15\ \text{mol}$

   • And the moles of calcium carbonate:
     $\small\text{mole} = 8 / 100 = 0.08\ \text{mol}$

   • 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 $2$:
     $\small\text{mole} = 0.15 / 2 = 0.075\ \text{mol}$

   • 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 $\small\text{mass} = \text{molecular weight}\cdot\text{mole}$ equation to determine the theoretical mass of the product. The molecular weight of acetone is $58\ \text{g}/\text{mol}$:

   $\text{mass} = 58 \cdot 0.075 = 4.35\ \text{g}$

So from this reaction, we should get, theoretically speaking, $4.35\ \text{g}$ of acetone. Nice!

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