How To Calculate Theoretical Yield

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 ), which is done by,

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.

1. Remember to hit refresh at the bottom of the calculator to reset it.
2. What is the theoretical yield? It is the amount of a product that would be formed if your reaction was 100% efficient.
3. 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.

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

The number of moles of the limiting reagent in the reaction is equal to: n lim = m lim m mol, lim ⋅ c lim n_ } = \frac }},\text }\cdot c_ }} n lim ​ = m mol, lim ​ ⋅ c lim ​ m lim ​ ​ where:

• n lim n_ } n lim ​ — Number of moles of the limiting reagent;
• m lim m_ } m lim ​ — Mass of limiting reagent;
• m mol, lim m_,\text } m mol, lim ​ — Molecular weight of the limiting reagent; and
• c lim c_ } c 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.

1. 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 and ).
2. 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 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. Cyanohydrin acetone reaction (by Rehua – CC BY-SA 3.0, Wikipedia ). 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.

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: mole = mass / molecular weight \small\text = \text /\text mole = mass / molecular weight
   • Acetone has a molecular weight of 58 g / mole 58\ \text /\text 58 g / mole, so: mole = 5 / 58 = 0.862 mol \small\text = 5 / 58 = 0.862\ \text mole = 5/58 = 0.862 mol
   • Cyanide has a molecular weight of 26 g / mole 26\ \text /\text 26 g / mole, so: mole = 2 / 26 = 0.0769 mol \small\text = 2 / 26 = 0.0769\ \text 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 knowing how many moles the product will form. As the stoichiometry of the product is 1 1 1, 0.0769 0.0769 0.0769 moles will form.
3. We can once again use the mass = molecular weight ⋅ mole \small\text = \text \cdot\text mass = molecular weight ⋅ mole equation to determine the theoretical mass of the product. The molecular weight of hydroxyacetonitrile is 85 g / mol 85\ \text /\text 85 g / mol :

mass = 85 ⋅ 0.0769 = 6.54 g \quad\ \ \text = 85 \cdot 0.0769 = 6.54\ \text mass = 85 ⋅ 0.0769 = 6.54 g Now we know that if we carry out the experiment, we would expect 6.54 g 6.54\ \text 6.54 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 g 8\ \text 8 g of calcium carbonate ( 100 g / mol 100\ \text /\text 100 g / mol ) with 9 g 9\ \text 9 g of acetic acid ( 60 g / mol 60\ \text /\text 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 \small\text = \text \cdot \text mass = molecular weight ⋅ mole equation again:
   • Let’s rearrange the equation to find moles. This gives: mole = mass / molecular weight \small\text = \text / \text mole = mass / molecular weight
   • Let’s find the moles of acetic acid: mole = 9 / 60 = 0.15 mol \small\text = 9 / 60 = 0.15\ \text mole = 9/60 = 0.15 mol
   • And the moles of calcium carbonate: mole = 8 / 100 = 0.08 mol \small\text = 8 / 100 = 0.08\ \text 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 2 2 : mole = 0.15 / 2 = 0.075 mol \small\text = 0.15 / 2 = 0.075\ \text 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 1 1, 0.75 0.75 0.75 moles will form.
3. Use the mass = molecular weight ⋅ mole \small\text = \text \cdot\text mass = molecular weight ⋅ mole equation to determine the theoretical mass of the product. The molecular weight of acetone is 58 g / mol 58\ \text /\text 58 g / mol : mass = 58 ⋅ t i m e s 0.075 = 4.35 g \text = 58 \cdot times0.075 = 4.35\ \text mass = 58 ⋅ t im es 0.075 = 4.35 g

So from this reaction, we should get, theoretically speaking, 4.35 g 4.35\ \text 4.35 g of acetone. Nice! Now go on and conquer the world of theoretical yield calculations, you can do it! To find the theoretical yield:

1. Balance the reaction.
2. Identify the limiting reagent, which is the reagent with the fewest moles,
3. Divide the fewest number of reagent moles by the stoichiometry of the product.
4. Multiply the result of Step 3 by the molecular weight of the desired product.

The theoretical yield is the amount of product that would be formed from a reaction if it was 100% efficient, It is the maximum mass of product that the reagents can form, and you can compare your yield against it to see how successfully you carried out your reaction.

1. No, the limiting reactant is not the theoretical yield.
2. To find the theoretical yield, you must find the number of moles present of the limiting reagent.
3. You can then multiply this number by the stoichiometry of the desired product to find the number of moles formed, then use this to derive the theoretical yield.

The theoretical yield of CO 2 depends on the reaction taking place and the amount of reagents. To find the theoretical yield, you can follow the steps below:

1. Find the moles of the limiting reagent.
2. Multiply the moles of the limiting reagent by the stoichiometry of carbon dioxide in the reaction to give the moles of CO 2 produced.
3. Multiply the moles of CO 2 produced by 44, the molecular weight of CO 2, to get the theoretical yield of your reaction.

: Theoretical Yield Calculator

What is theoretical yield equal to?

Percent yield – As we just learned, the theoretical yield is the maximum amount of product that can be formed in a chemical reaction based on the amount of limiting reactant. In practice, however, the actual yield of product—the amount of product that is actually obtained—is almost always lower than the theoretical yield.

This can be due to a number of factors, including side reactions (secondary reactions that form undesired products) or purification steps that lower the amount of product isolated after the reaction. The actual yield of a reaction is typically reported as a percent yield, or the percentage of the theoretical yield that was actually obtained.

The percent yield is calculated as follows: start text, P, e, r, c, e, n, t, space, y, i, e, l, d, end text, equals, start fraction, start text, a, c, t, u, a, l, space, y, i, e, l, d, end text, divided by, start text, t, h, e, o, r, e, t, i, c, a, l, space, y, i, e, l, d, end text, end fraction, times, 100, percent Based on this definition, we would expect a percent yield to have a value between 0% and 100%.

How is theoretical value calculated?

Theoretical Nil Paid Price – If an investor chooses to sell his or her right outright in the market, or if they choose to let the right lapse, which may result in a minimal administrative charge, they will receive the theoretical nil paid price of the right.

What is the formula for yield?

What is Current Yield? – The Current Yield measures the expected annual return of a bond and is calculated by dividing the annual coupon by the current market price.

What is theoretical probability and how do you calculate it?

Theoretical probability is a method to express the likelihood that something will occur. It is calculated by dividing the number of favorable outcomes by the total possible outcomes.

How do you find theoretical yield in grams?

How to Calculate Theoretical Yield: 12 Steps (with Pictures)

1. 1 Start with a balanced, A chemical equation is like a recipe. It shows the reactants (on the left side) reacting to form products (on the right side). A properly will show the same number of atoms going into the equation as reactants as you have coming out in the form of products.
   • For example, consider the simple equation H 2 + O 2 +O_ } → H 2 O O}, There are two of hydrogen on both the left and right. But there are two atoms of oxygen going in as a reactant and only one atom in the product on the right.
   • To balance, double the product, to get H 2 + O 2 +O_ } → 2 H 2 O O},
   • Check the balance. This change has corrected the oxygen, which now has two atoms on both sides. But you now have two atoms of hydrogen on the left with four atoms of hydrogen on the right.
   • Double the hydrogen in the reactant. This will adjust the equation to 2 H 2 + O 2 +O_ } → 2 H 2 O O}, This change now has 4 atoms of hydrogen on both sides, and two atoms of oxygen. The equation is balanced.
   • As a more complicated example, oxygen and glucose can react to form carbon dioxide and water: 6 O 2 + C 6 H 12 O 6 +C_ H_ O_ } → 6 C O 2 + 6 H 2 O +6H_ O} In this equation, each side has exactly 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 18 oxygen (O) atoms. The equation is balanced.
2. 2 Calculate the of each reactant. Using the or some other reference, look up the molar mass of each atom in each compound. Add them together to find the molar mass of each compound of reactant. Do this for a single molecule of the compound. Consider again the equation of converting oxygen and glucose into carbon dioxide and water: 6 O 2 + C 6 H 12 O 6 +C_ H_ O_ } → 6 C O 2 + 6 H 2 O +6H_ O}
   • For this example, one molecule of oxygen ( O 2 } ) contains two oxygen atoms.
   • The molar mass of one atom of oxygen is about 16 g/mol. If necessary, you can find more precise values.)
   • 2 oxygen atoms x 16 g/mol per atom = 32 g/mol of O 2 },
   • The other reactant, glucose ( C 6 H 12 O 6 H_ O_ } ) has a molar mass of (6 atoms C x 12 g C/mol) + (12 atoms H x 1 g H/mol) + (6 atoms O x 16 g O/mol) = 180 g/mol.

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3. 3 Convert the amount of each reactant from, For an actual experiment, you will know the mass in grams of each reactant that you are using. Divide this value by that compound’s molar mass to convert the amount to moles.
   • For example, suppose you begin with 40 grams of oxygen and 25 grams of glucose.
   • 40 g O 2 } / (32 g/mol) = 1.25 moles of oxygen.
   • 25g C 6 H 12 O 6 H_ O_ } / (180 g/mol) = about 0.139 moles of glucose.
4. 4 Determine the molar ratio of the reactants. A mole is a tool used in chemistry to count molecules, based on their mass. By determining the number of moles of both oxygen and glucose, you know how many molecules of each you are starting with. To find the ratio between the two, divide the number of moles of one reactant by the number of moles of the other.
   • In this example, you are starting with 1.25 moles of oxygen and 0.139 moles of glucose. Thus, the ratio of oxygen to glucose molecules is 1.25 / 0.139 = 9.0. This ratio means that you have 9 times as many molecules of oxygen as you have of glucose.
5. 5 Find the ideal ratio for the reaction. Look at the balanced equation for the reaction. The coefficients in front of each molecule tell you the ratio of the molecules that you need for the reaction to occur. If you use exactly the ratio given by the formula, then both reactants should be used equally.
   • For this reaction, the reactants are given as 6 O 2 + C 6 H 12 O 6 +C_ H_ O_ }, The coefficients indicate that you need 6 oxygen molecules for every 1 glucose molecule. The ideal ratio for this reaction is 6 oxygen / 1 glucose = 6.0.
6. 6 Compare the ratios to find the limiting reactant. In most, one of the reactants will be used up before the others. The one that gets used up first is called the limiting reactant. This limiting reactant determines how long the chemical reaction can take place and the theoretical yield you can expect. Compare the two ratios you calculated to identify the limiting reactant:
   • In this example, you are beginning with 9 times as much oxygen as glucose, when measured by number of moles. The formula tells you that your ideal ratio is 6 times as much oxygen as glucose. Therefore, you have more oxygen than required. Thus, the other reactant, glucose in this case, is the limiting reactant.

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1. 1 Review the reaction to find the desired product. The right side of a chemical equation shows the products created by the reaction. The coefficients of each product, if the reaction is balanced, tells you the amount to expect, in molecular ratios. Each product has a theoretical yield, meaning the amount of product you would expect to get if the reaction is perfectly efficient.
   • Continuing the example above, you are analyzing the reaction 6 O 2 + C 6 H 12 O 6 +C_ H_ O_ } → 6 C O 2 + 6 H 2 O +6H_ O}, The two products shown on the right are carbon dioxide and water.
   • You can begin with either product to calculate theoretical, In some cases, you may be concerned only with one product or the other. If so, that is the one you would start with.
2. 2 Write down the number of moles of your limiting reactant. You must always compare moles of reactant to moles of product. If you try to compare the mass of each, you will not reach the correct results.
   • In the example above, glucose is the limiting reactant. The molar mass calculations found that the initial 25g of glucose are equal to 0.139 moles of glucose.
3. 3 Compare the ratio of molecules in product and reactant. Return to the balanced equation. Divide the number of molecules of your desired product by the number of molecules of your limiting reactant.
   • The balanced equation for this example is 6 O 2 + C 6 H 12 O 6 +C_ H_ O_ } → 6 C O 2 + 6 H 2 O +6H_ O}, This equation tells you that you expect 6 molecules of the desired product, carbon dioxide ( C O 2 } ), compared to 1 molecule of glucose ( C 6 H 12 O 6 H_ O_ } ).
   • The ratio of carbon dioxide to glucose is 6/1 = 6. In other words, this reaction can produce 6 molecules of carbon dioxide from one molecule of glucose.
4. 4 the ratio by the limiting reactant’s quantity in moles. The answer is the theoretical yield, in moles, of the desired product.
   • In this example, the 25g of glucose equate to 0.139 moles of glucose. The ratio of carbon dioxide to glucose is 6:1. You expect to create six times as many moles of carbon dioxide as you have of glucose to begin with.
   • The theoretical yield of carbon dioxide is (0.139 moles glucose) x (6 moles carbon dioxide / mole glucose) = 0.834 moles carbon dioxide.
5. 5 the result to grams. This is the reverse of your earlier step of calculating the number of moles or reactant. When you know the number of moles that you expect, you will multiply by the molar mass of the product to find the theoretical yield in grams.
   • In this example, the molar mass of CO 2 is about 44 g/mol. (Carbon’s molar mass is ~12 g/mol and oxygen’s is ~16 g/mol, so the total is 12 + 16 + 16 = 44.)
   • Multiply 0.834 moles CO 2 x 44 g/mol CO 2 = ~36.7 grams. The theoretical yield of the experiment is 36.7 grams of CO 2,
6. 6 Repeat the calculation for the other product if desired. In many experiments, you may only be concerned with the yield of one product. If you wish to find the theoretical yield of both products, just repeat the process.
   • In this example, the second product is water, H 2 O O}, According to the balanced equation, you expect 6 molecules of water to come from 1 molecule of glucose. This is a ratio of 6:1. Therefore, beginning with 0.139 moles of glucose should result in 0.834 moles of water.
   • Multiply the number of moles of water by the molar mass of water. The molar mass is 2 + 16 = 18 g/mol. Multiplying by the product, this results in 0.834 moles H 2 O x 18 g/mol H 2 O = ~15 grams. The theoretical yield of water for this experiment is 15 grams.

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• Question Doesn’t one molecule of glucose produce six molecules of water, not one? Yes. One molecule of glucose plus six molecules of oxygen = six molecules of water plus six molecules of carbon dioxide.
• Question What should I do if there is more than one reactant? Find out which of the reactants is the “limiting” reactant and use that to calculate the theoretical yield. This can be done using Part 1 of this article.
• Question What should I do if the reactants have the same number of moles? That’s not a problem! It only means that the molar ratio of your reactants is 1. In the next step, you need to compare it to the ideal molar ratio from your chemical equation to find the limiting reactant and continue as described in the article.

Ask a Question Advertisement This article was co-authored by, Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group.

• Co-authors: 15
• Updated: May 19, 2023
• Views: 997,853

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“This explained it better than my actual chemistry teacher!”

: How to Calculate Theoretical Yield: 12 Steps (with Pictures)

What is theoretical yield simple?

Theoretical Yield – Definition, Formula and Examples The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction. The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield. A reaction yield is reported as the percentage of the theoretical amount.

• The formula for percentage yield is given by
• Percentage yield= (Actual yield/theoretical yield )x100
• Rearrange the above formula to obtain theoretical yield formula
• Example 1

Determine the theoretical yield of the formation of geranyl formate from 375 g of geraniol. A chemist making geranyl formate uses 375 g of starting material and collects 417g of purified product. Percentage yield is given as 94.1%

1. Solution
2. The actual yield is 417 g which is the quantity of the desired product.
3. Percentage yield is 94.1%
4. Therefore,
5. Theoretical yield=(Actual yield/percentage yield) x 100
6. = (417 / 94.1)100
7. = 443g

Where is the theoretical yield?

How to Calculate Theoretical Yield – Theoretical yield is found by identifying the limiting reactant of a balanced chemical equation. In order to find it, the first step is to balance the equation, if it’s unbalanced. The next step is to identify the limiting reactant.

If the quantity of reactants is given in moles, convert the values to grams.Divide the mass of the reactant in grams by its molecular weight in grams per mole.Alternatively, for a liquid solution, you can multiply the amount of a reactant solution in milliliters by its density in grams per milliliter. Then, divide the resulting value by the reactant’s molar mass.Multiply the mass obtained using either method by the number of moles of reactant in the balanced equation.Now you know the moles of each reactant. Compare this to the molar ratio of the reactants to decide which is available in excess and which will get used up first (the limiting reactant).

Once you identify the limiting reactant, multiply the moles of limiting reaction times the ratio between moles of limiting reactant and product from the balanced equation. This gives you the number of moles of each product. To get the grams of product, multiply the moles of each product by its molecular weight.

• For example, in an experiment in which you prepare acetylsalicylic acid (aspirin) from salicylic acid, you know from the balanced equation for aspirin synthesis that the mole ratio between the limiting reactant (salicylic acid) and the product (acetylsalicylic acid) is 1:1.
• If you have 0.00153 moles of salicylic acid, the theoretical yield is: Theoretical yield = 0.00153 mol salicylic acid x (1 mol acetylsalicylic acid / 1 mol salicylic acid) x (180.2 g acetylsalicylic acid / 1 mole acetylsalicylic acid Theoretical yield = 0.276 grams acetylsalicylic acid Of course, when preparing aspirin, you’ll never get that amount.

If you get too much, you probably have excess solvent or else your product is impure. More likely, you’ll get much less because the reaction won’t proceed 100 percent and you’ll lose some product trying to recover it (usually on a filter).

Why is percentage yield never 100?

There are a few reasons why percentage yield will never be 100%. This could be because other, unexpected reactions occur which don’t produce the desired product, not all of the reactants are used in the reaction, or perhaps when the product was removed from the reaction vessel it was not all collected.

How do you convert theoretical yield to actual yield?

The formula to determine actual yield is simple: you multiply the percentage and theoretical yield together.

Why is percent yield over 100?

It is only possible to get a percentage yield greater than 100 percent if the product is contaminated with impurities or if all the solvent from the reaction mixture has not been dried off.

What is theoretical vs actual value?

Percent Error Formula When you calculate results that are aiming for known values, the percent error formula is useful tool for determining the precision of your calculations. The formula is given by: The experimental value is your calculated value, and the actual value is the known value (sometimes called the accepted or theoretical value).

• A percentage very close to zero means you are very close to your targeted value, which is good.
• It is always necessary to understand the cause of the error, such as whether it is due to the imprecision of your equipment, your own estimations, or a mistake in your experiment.
• The 17th century Danish astronomer, Ole Rømer, observed that the length of the eclipses of Jupiter by its satellites would appear to fluctuate depending on the direction Earth was traveling relative to Jupiter at the time of the eclipse.

If Earth was traveling toward Jupiter, the eclipes of Jupiter by, say, Io, would last for a shorter amount of time, while if Earth was traveling away from Jupiter, the eclipses would appear to be longer. In 1676, he determined that this phenomenon was due to the fact that the speed of light was finite, and subsequently estimated its velocity to be approximately 220,000 km/s.

Experimental value = 220,000 km/s = 2.2 x 10 8 m/s Actual value = 299,800 km/s = 2.998 x 10 8 m/sSo Rømer was quite a bit off by our standards today, but considering he came up with this estimate at a time when a majority of respected astronomers, like Cassini, still believed that the speed of light was infinite, his conclusion was an outstanding contribution to the field of Astronomy.

: Percent Error Formula

What is the theoretical value?

Theoretical Value. The theoretical value of a security or a derivative is the price as determined by a quantitative model. A theoretical price is a good guide as to whether something is overpriced or under-priced.

What is the formula for theoretical mass percent?

To calculate the mass percent of an element in a compound, we divide the mass of the element in 1 mole of the compound by the compound’s molar mass and multiply the result by 100.

How do you calculate the percent yield?

The percent yield is the actual yield divided by the theoretical yield and multiplied by 100%. Percent yield = actual yield / theoretical yield x 100%.

How do you calculate the percent yield of a reaction?

To express the efficiency of a reaction, you can calculate the percent yield using this formula: %yield = (actual yield/theoretical yield) x 100. A percent yield of 90% means the reaction was 90% efficient, and 10% of the materials were wasted (they failed to react, or their products were not captured).

What is the formula for percentage yield in economics?

Percentage Yield Formula Yield is a percentage word for the earnings created and realised on an investment over a certain period. The percentage is calculated using the amount invested, the current market, or the investment security’s face value. Interest or dividends obtained from owning certain securities for a specified period are included in yield.

1. However, financial gains are not taken into account.
2. Yields are characterised as known or expected depending on their nature and valuation (whether constant or fluctuating).
3. The ratio of actual yield to theoretical yield is the percentage yield formula.
4. Yield is a measure of the number of moles of a product generated in a chemical reaction in proportion to the amount of reactant consumed, generally given as a percentage.

The percentage yield is the difference between the actual amount of product produced and the maximum calculated yield. The experimental yield divided by the theoretical yield multiplied by 100 is the percentage yield formula. The percent yield is 100 percent if the actual and theoretical yields are equal.