 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.

• 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.

• 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.

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 ). 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.

No, the limiting reactant is not the theoretical yield. To find the theoretical yield, you must find the number of moles present of the limiting reagent. 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

### How do you calculate theoretical yield?

We calculate the theoretical yield by taking the given mass of the reactants and multiplying it by the molar mass of the reactants and the molar mass of the products.

#### How do you find the theoretical yield of a limiting reactant?

Example $$\PageIndex$$: Novocain – Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction \} \underset } } + \ce \nonumber \] If this reaction were carried out with 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and 15.7 g of procaine were isolated, what is the percent yield? The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water. Given : masses of reactants and product Asked for : percent yield Strategy :

1. Write the balanced chemical equation.
2. Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.
3. Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.

Solution : A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water. B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C 7 H 7 NO 2 ), 137.14 g/mol; 2-diethylaminoethanol (C 6 H 15 NO), 117.19 g/mol.

1. Thus the reaction used the following numbers of moles of reactants: \ \ The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant.
2. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine.
3. We can therefore obtain only a maximum of 0.0729 mol of procaine.

To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol. \ C The actual yield was only 15.7 g of procaine, so the percent yield (via Equation \ref ) is \ (If the product were pure and dry, this yield would indicate very good lab technique!) Lead was one of the earliest metals to be isolated in pure form.

It occurs as concentrated deposits of a distinctive ore called galena ($$\ce$$), which is easily converted to lead oxide ($$\ce$$) in 100% yield by roasting in air via the following reaction: \ The resulting $$\ce$$ is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected.

The reaction for the conversion of lead oxide to pure lead is as follows: \ If 93.3 kg of $$\ce$$ is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield? Electrolytically refined pure (99.989 %) superficially oxidized lead nodules and a high purity (99.989 %) $$1\; cm^3$$ lead cube for comparison. Figure used with permission from Wikipedia. Answer 89.2% Percent yield can range from 0% to 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%.

• This usually happens when the product is impure or is wet with a solvent such as water.
• If this is not the case, then the student must have made an error in weighing either the reactants or the products.
• The law of conservation of mass applies even to undergraduate chemistry laboratory experiments.
• A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced.
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Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair.

## What is theoretical actual yield?

Theoretical and Actual Yields – Reactants not completely used up are called excess reagents, and the reactant that completely reacts is called the limiting reagent. This concept has been illustrated for the reaction: \ Amounts of products calculated from the complete reaction of the limiting reagent are called theoretical yields, whereas the amount actually produced of a product is the actual yield.

#### Why do you calculate theoretical yield?

Before performing chemical reactions, it is helpful to know how much product will be produced with given quantities of reactants. This is known as the theoretical yield, This is a strategy to use when calculating the theoretical yield of a chemical reaction. The same strategy can be applied to determine the amount of each reagent needed to produce a desired amount of product.

## Is theoretical yield 100%?

Example Percent Yield Calculation – For example, the decomposition of magnesium carbonate forms 15 grams of magnesium oxide in an experiment. The theoretical yield is known to be 19 grams. What is the percent yield of magnesium oxide? MgCO 3 → MgO + CO 2 The calculation is simple if you know the actual and theoretical yields.

All you need to do is plug the values into the formula: percent yield = actual yield / theoretical yield x 100% percent yield = 15 g / 19 g x 100% percent yield = 79% Usually, you have to calculate the theoretical yield based on the balanced equation. In this equation, the reactant and the product have a 1:1 mole ratio, so if you know the amount of reactant, you know the theoretical yield is the same value in moles (not grams!).

You take the number of grams of reactant you have, convert it to moles, and then use this number of moles to find out how many grams of product to expect.

### How do you calculate theoretical yield in Excel?

When assessing the profitability of bonds, analysts use a concept called yield to determine the amount of income an investment expects to generate each year. Yield is prospective and should not be confused with the rate of return, which refers to realized gains,

To calculate the current yield of a bond in Microsoft Excel, enter the bond value, the coupon rate, and the bond price into adjacent cells (e.g., A1 through A3). In cell A4, enter the formula “= A1 * A2 / A3” to render the current yield of the bond. However, as a bond’s price changes over time, its current yield varies.

Analysts often use a much more complex calculation called yield to maturity (YTM) to determine the bonds’ total anticipated yield, including any capital gains or losses due to price fluctuation.

#### What is the theoretical yield in grams?

The theoretical yield is the amount of the product in g formed from the limiting reagent. From the moles of limiting reagent available, calculate the grams of product that is theoretically possible (same as Step 4 above). The actual yield is the amount of the product in g actually formed in the laboratory.

#### What is limiting reactant and theoretical yield in stoichiometry?

Stoichiometry, Product Yield, and Limiting Reactants Chemical equations represent how a chemical reaction proceeds from reactants to products through physical or chemical change using chemical formulas. Stoichiometry is a term that describes the relative quantities of reactants and products in a chemical reaction.

It is based on the Law of Conservation of Mass, which is a fundamental law that states that matter is neither created nor destroyed. Quite simply, the number and identity of reactant atoms must equal the number and identity of product atoms. Reactions rearrange atoms but do not create or destroy them.

This requires that a proposed reaction must be balanced, meaning that the number of atoms for each element are equal on the reactant and product sides. For example, in the chemical equation below, the left side (the reactants) includes one copper atom, one hydrogen atom, one nitrogen atom, and three oxygen atoms.

1. Cu + HNO 3 → Cu(NO 3 ) 2 + 4 H 2 O + NO On the right side, notice the water product has a number preceding it.
2. This is a coefficient and represents the number of molecules in the reaction.
3. Using this information, we can tally the number of atoms on the product side.
4. There is one copper atom and eight hydrogen atoms (4 x 2).

Tallying the nitrogen and oxygen atoms requires a bit more math. There are two nitrogen atoms in the first product and one nitrogen atom in the third product, which gives a total of three nitrogen atoms. For oxygen, there are six oxygen atoms in the first product, four oxygen atoms in the second product, and one oxygen atom in the third product for a total of 11 oxygen atoms.

If left in this form, the reaction wouldn’t be feasible because it defies the Law of Conservation of Masses. There are more hydrogen, nitrogen, and oxygen atoms on the product side. Therefore, the equation needs to be balanced. Balancing an equation is an iterative process that requires adding coefficients to each side until the numbers become equal.

There are several approaches to balance a chemical equation. One approach uses a table to visualize the numbers and a bit of trial and error.

 # of atoms on the reactant side # of atoms on the product side Copper Hydrogen Nitrogen Oxygen Copper Hydrogen Nitrogen Oxygen 1 1 1 3 1 8 3 11

Since there is only one hydrogen atom on the reactant side but eight hydrogen atoms on the product side, multiplying the compound containing the nitrogen on the reactant side by eight would balance hydrogen. The second row in the new table reflects this change to the number of atoms.

 # of atoms on the reactant side # of atoms on the product side Copper Hydrogen Nitrogen Oxygen Copper Hydrogen Nitrogen Oxygen 1 1 1 3 1 8 3 11 8 HNO 3 1 8 8 24 1 8 3 11

Next, to increase the number of nitrogens on the product side, multiplying the product Cu(NO 3 ) 2 by three would raise the number of nitrogen atoms from three to seven. The number of nitrogen atoms is not balanced yet, but there are still coefficients for one reactant and one product to consider.

 # of atoms on the reactant side # of atoms on the product side Copper Hydrogen Nitrogen Oxygen Copper Hydrogen Nitrogen Oxygen 1 1 1 3 1 8 3 11 8 HNO 3 1 8 8 24 1 8 3 11 3 Cu(NO 3 ) 2 1 8 8 24 3 8 7 23

If there are two molecules of NO produced, this adds one more nitrogen atom and one more oxygen atom to the product side, balancing these two species with the reactant side.

 # of atoms on the reactant side # of atoms on the product side Copper Hydrogen Nitrogen Oxygen Copper Hydrogen Nitrogen Oxygen 1 1 1 3 1 8 3 11 8 HNO 3 1 8 8 24 1 8 3 11 3 Cu(NO 3 ) 2 1 8 8 24 3 8 7 23 2 NO 1 8 8 24 3 8 8 24

The only atom left unbalanced now is copper. Increasing the number to three copper atoms on the reactant side balances the equation.

 # of atomson the reactant side # of atoms on the product side Copper Hydrogen Nitrogen Oxygen Copper Hydrogen Nitrogen Oxygen 1 1 1 3 1 8 3 11 8 HNO 3 1 8 8 24 1 8 3 11 3 Cu(NO 3 ) 2 1 8 8 24 3 8 7 23 2 NO 1 8 8 24 3 8 8 24 3 Cu 3 8 8 24 3 8 8 24

The balanced equation is written as follows: 3 Cu + 8 HNO 3 → 3 Cu(NO 3 ) 2 + 4 H 2 O + 2 NO Balancing the equation is also essential for determining the limiting reactant because the coefficient of the compounds is used to calculate how much product is produced by each reactant (product yield).

From this quantity, the reactant producing the least amount of product is considered the limiting reactant-which is completely consumed in the reaction and therefore limits the total amount of product generated. This calculated quantity also represents the theoretical yield of the reaction, which is needed to calculate the percent yield.

The balanced equation is more than a simple accounting of atoms. The coefficients describe the molar relationship between products and reactants, i.e., how much product is produced by each reactant. The number of moles of reactant is used to calculate the number of moles of another product or reactant.

1. The reactant that produces the least amount of product is considered the limiting reactant.
2. The limiting reactant is completely consumed in the reaction and therefore limits the total amount of product generated.
3. Once the limiting reactant is entirely consumed, no more product will form.
4. The possible amount of product that could be formed based on the limiting reactant is the theoretical yield of the reaction.

The actual yield is compared to the theoretical yield, resulting in the ‘percent yield’. A percent yield of 100% means that, based on the reactants used, the maximum possible amount of product was produced. Percent yields less than 100% are common and indicate that there was some product loss during the reaction.

#### Does the theoretical yield depend on the limiting reactant?

The theoretical yield is based on the limiting reactant, not the excess reactant. The limiting reactant is the one that will be completely consumed in the reaction.

## What is an example of a theoretical yield?

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.

#### Is theoretical yield the same as percent yield?

Key Takeaways –

• Theoretical yield is what you calculate the yield will be using the balanced chemical reaction.
• Actual yield is what you actually get in a chemical reaction.
• Percent yield is a comparison of the actual yield with the theoretical yield.

## Is theoretical yield always the same?

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### Why isn’t the yield 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 calculate theoretical yield in Excel?

When assessing the profitability of bonds, analysts use a concept called yield to determine the amount of income an investment expects to generate each year. Yield is prospective and should not be confused with the rate of return, which refers to realized gains,

1. To calculate the current yield of a bond in Microsoft Excel, enter the bond value, the coupon rate, and the bond price into adjacent cells (e.g., A1 through A3).
2. In cell A4, enter the formula “= A1 * A2 / A3” to render the current yield of the bond.
3. However, as a bond’s price changes over time, its current yield varies.

Analysts often use a much more complex calculation called yield to maturity (YTM) to determine the bonds’ total anticipated yield, including any capital gains or losses due to price fluctuation.

### What is the theoretical formula?

TL;DR (Too Long; Didn’t Read) – To calculate the theoretical percentage of an element in a compound, divide the molar mass of the element by the mass of the compound and multiply by 100. In a chemical reaction, the percent yield of a product is its actual yield divided by its theoretical yield and multiplied by 100.

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## What is the theoretical yield of an experiment?

Theoretical yield is the amount of product that should be produced in a chemical reaction based on stoichiometric calculation, if the reaction goes to completion and no errors are made. The actual yield is the amount of product measured at the end of the reaction.

### How do you calculate the yield of a product?

12 Yield Testing Yield in culinary terms refers to how much you will have of a finished or processed product. Professional recipes should always state a yield; for example, a tomato soup recipe may yield 15 L, and a muffin recipe may yield 24 muffins.

Yield can also refer to the amount of usable product after it has been processed (peeled, cooked, butchered, etc.) For example, you may be preparing a recipe for carrot soup. The recipe requires 1 kg of carrots, which you purchase. However, once you have peeled them and removed the tops and tips, you may only have 800 grams of carrots left to use.

In order to do accurate costing, yield testing must be carried out on all ingredients and recipes. When looking at yields, you must always consider the losses and waste involved in preparation and cooking. There is always a dollar value that is attached to vegetable peel, meat and fish trim, and packaging like brines and syrups.

• Any waste or loss has been paid for and is still money that has been spent.
• This cost must always be included in the menu price.
• Note: Sometimes, this “waste” can be used as a by-product.
• Bones from meat and fish can be turned into stocks.
• Trimmings from vegetables can be added to those stocks or, if there is enough, made into soup.

All products must be measured and yield tested before costing a menu. Ideally, every item on a menu should be yield tested before being processed. Most big establishments will have this information on file, and there are many books that can also be used as reference for yields, such as

1. Record the original weight/volume of your item. This is your raw weight or,
1. Whole tenderloin – 2.5 kg
2. Whole sockeye salmon – 7.75 kg
3. Canned tuna flakes in brine – 750 mL
2. Process your product accordingly, measure and record the waste or trim weight.
1. Tenderloin fat, sinew, chain, etc. – 750 g tenderloin trim
2. Salmon head, bones, skin, etc. – 2.75 kg salmon trim
3. Brine – 300 mL canned tuna waste
3. Subtract the amount of trim weight from the AP weight and you will have what is referred to as your processed or edible product (EP) weight, The formula is: AP weight – waste = EP weight.
1. 2500 g − 750 g = 1750 g processed tenderloin
2. 7750 g − 2750 g = 5000 g processed salmon
3. 750 mL − 300 mL = 400 mL processed canned tuna
4. Get your yield percentage by converting the edible product weight into a percentage. The formula is EP weight ÷ AP weight × 100 = yield %.
1. (1750 ÷ 2500) × 100 = 70% for the tenderloin
2. (5000 ÷ 7750) × 100 = 64.51% for the salmon
3. (400 ÷ 750) × 100 = 53.33% for the canned tuna

Yield percentage is important because it tells you several things: how much usable product you will have after processing; how much raw product to actually order; and the actual cost of the product per dollar spent. Once you have your yield percentage, you can translate this information into monetary units.

1. Record the AP cost, what you paid for the item:
1. Whole tenderloin – $23.00/kg 2. Whole sockeye salmon –$5.00/kg
3. Canned tuna flakes in brine – $5.50/750 mL can 2. Obtain your factor. This factor converts all your calculations into percentages. The formula is: 1. 100 ÷ yield % = factor 2. 100 ÷ 70 tenderloin = 1.42 3. 100 ÷ 64.51 salmon = 1.55 4. 100 ÷ 53.33 canned tuna = 1.875 3. Once the factor has been determined, it is now an easy process to determine your EP cost. The formula is: factor × as purchased cost per (unit) = edible product cost per (unit) 1. Tenderloin$23.00 × 1.42 = $32.66/kg 2. Salmon$5.00 × 1.55 = $7.75/kg 3. Canned tuna$5.50 × 1.875 = \$10.78/750 mL

There could be a considerable difference in costs between the raw product and the processed product, which is why it is important to go through all these steps. Once the EP cost is determined, the menu price can be set.

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