Within the field of chemistry, theoretical performance is the maximum amount of product that can create a chemical reaction. In reality, most reactions are not perfectly effective. By performing the experiment, you will get a smaller amount, which is called actual performance. If you want to express the efficiency of a reaction, you can calculate the percentage yield Using the following formula:% yield = (actual yield / theoretical yield) x 100. A 90% percent yield means that the reaction was 90% efficient, while 10% of the materials were wasted (they did not achieve a reaction or their product was not recovered).
Steps
Part 1 of 3: Finding the Limiting Reactant
Step 1. Start with a balanced chemical equation
A chemical equation describes a chemical reaction of reactants (on the left side) in order to form products (on the right side). In some problems, you will already have this equation, while in others you will have to write it yourself. Because atoms are not created or destroyed during a chemical reaction, each element must have the same number of atoms on the left and right side.

For example, oxygen and glucose can react to form carbon dioxide and oxygen: 6O2 + C6H12O6 { displaystyle 6O_ {2} + C_ {6} H_ {12} O_ {6}}
→ 6CO2+6H2O{displaystyle 6CO_{2}+6H_{2}O}
Cada lado posee exactamente 6 átomos de carbono (C), 12 átomos de hidrógeno (H) y 18 átomos de oxígeno (O). De esta manera, la ecuación queda balanceada.
 Si quieres saber cómo balancear una ecuación, lee este artículo.
Step 2. Find the molar mass of each reactant
Determine the molar mass of all the atoms in the compound, and then add them together to find the molar mass of the compound. Perform this procedure to find a single molecule of that compound.

For example, an oxygen molecule (O2 { displaystyle O_ {2}}
) contiene dos átomos de oxígeno.
 La masa molar del oxígeno es de aproximadamente 16 g/mol (puedes encontrar un valor más preciso en una tabla periódica).
 2 átomos de oxígeno x 16 g/mol por átomo = 32 g/mol de O2{displaystyle O_{2}}
 La glucosa (C6H12O6{displaystyle C_{6}H_{12}O_{6}}
), el otro reactivo, tiene una masa molar de (6 átomos C x 12 g de C/mol) + (12 átomos de H x 1 g H/mol) + (6 átomos de O x 16 g O/mol) = 180 g/mol.
Step 3. Convert the amount of each reagent from grams to moles
Now is the time to focus on the specific experiment that we must study. Write down the amounts of each reactant in grams and divide that value by the molar mass of the compound in order to convert the amount to moles.
 For example, let's say you started with 40 grams of oxygen and 25 grams of glucose.

40 g O2 { displaystyle O_ {2}}
/ (32 g/mol) = 1, 25 moles de oxígeno.
 25 g de C6H12O6{displaystyle C_{6}H_{12}O_{6}}
/ (180 g/mol) = aproximadamente 0, 139 moles de glucosa.
Step 4. Calculate the proportion of the reactions
Remember that a mole is just a large number chemists use to "count" molecules. Now you know how many molecules of each reagent you started with. Divide the moles of one reactant by the moles of the other to find the ratio of the two molecules.
In the example above, we started with 1.25 moles of oxygen and 0.139 moles of glucose. The ratio of oxygen molecules to glucose molecules is 1.25 / 0.19 = 9.0. This means that you started with 9 oxygen molecules for every glucose molecule
Step 5. Find the ideal proportion for the reaction
Go back to the balanced equation you wrote earlier, which will tell you the ideal ratio of molecules. By using this ratio, both reagents will be consumed at the same time.

The left side of the equation is 6O2 + C6H12O6 { displaystyle 6O_ {2} + C_ {6} H_ {12} O_ {6}}
. Los coeficientes te indican que hay 6 moléculas de oxígeno y 1 de glucosa. La proporción ideal para esta reacción es 6 de oxígeno / 1 de glucosa = 6, 0.
 Asegúrate de colocar los reactivos en el mismo orden en que lo hiciste para la otra proporción. Por ejemplo, si empleas la proporción de oxígeno/glucosa en un lado y glucosa/oxígeno en el otro, obtendrás un resultado incorrecto.
Step 6. Compare the proportions to find the limiting reactant
In a chemical reaction, one of the reactants is consumed before the others. This limiting reagent determines the duration of the chemical reaction. Therefore, compare both proportions that you calculated in order to identify the limiting reagent:
 If the actual proportion is greater than the ideal proportion, it means that you have more than necessary amount of the reactant located in the numerator of the fraction. The reactant located in the denominator of the fraction becomes the limiting reactant.
 If the real proportion is less than the ideal proportion, it means that the reactant located in the numerator of the fraction is insufficient, therefore it becomes the limiting reactant.
 Using the previous example, the actual ratio of oxygen and glucose (9, 0) is greater than the ideal ratio (6, 0). In conclusion, the reagent located in the denominator, glucose, must be the limiting reagent.
Part 2 of 3: Calculate Theoretical Return
Step 1. Identify the desired product
The right side of a chemical equation displays a list of the products created by a chemical reaction. Each product has a theoretical yield, that is, the amount of product that you would expect to obtain if the reaction were fully effective.

Returning to the previous example, it is necessary to analyze the following reaction: 6O2 + C6H12O6 { displaystyle 6O_ {2} + C_ {6} H_ {12} O_ {6}}
→ 6CO2+6H2O{displaystyle 6CO_{2}+6H_{2}O}
. En el lado derecho de la ecuación, se muestran dos productos, dióxido de carbono y agua. Ahora debemos calcular el rendimiento del dióxido de carbono (CO2{displaystyle CO_{2}}
).
Step 2. Write the number of moles of the limiting reagent
The theoretical yield of an experiment is the amount of product created under perfect conditions. To calculate this value, you must start with the number of moles of the limiting reagent (this process is detailed earlier in the instructions for finding the limiting reagent).
In the example above, you found that glucose was the limiting reagent, and you also calculated that you started with 0.139 moles of glucose
Step 3. Find the ratio of molecules in the product and the reactant
To do this, take up the balanced equation. Then divide the number of molecules of the desired product by the number of molecules of the limiting reagent.

The balanced equation is as follows: 6O2 + C6H12O6 { displaystyle 6O_ {2} + C_ {6} H_ {12} O_ {6}}
→ 6CO2+6H2O{displaystyle 6CO_{2}+6H_{2}O}
. Hay
Paso 6. moléculas del producto deseado, el dióxido de carbono (CO2{displaystyle CO_{2}}
), y
Paso 1. molécula del reactivo limitante, la glucosa (C6H12O6{displaystyle C_{6}H_{12}O_{6}}
).
 La proporción de dióxido de carbono y glucosa es de 6/1 = 6. En otras palabras, esta reacción puede producir 6 moléculas de dióxido de carbono por cada molécula de glucosa.
Step 4. Multiply the proportion by the amount in moles of the reactant
The result will be the theoretical yield of the desired product expressed in moles.
The initial amount of glucose was 0.139 moles, while the ratio of carbon dioxide to glucose is 6. The theoretical yield of carbon dioxide is (0.139 moles of glucose) x (6 moles of carbon dioxide / 1 mole of glucose) = 0.834 moles of carbon dioxide
Step 5. Convert the result obtained to grams
Multiply the result in moles by the molar mass of that compound to find the theoretical yield in grams. In most experiments, it is best to use the gram.
 For example, the molar mass of CO_{2} is about 44 g / mol (the molar mass of carbon is about 12 g / mol and oxygen is about 16 g / mol, so the total amount is 12 + 16 + 16 = 44).
 Multiply 0.834 moles of CO_{2} x 44 g / mol CO_{2} to get as a result about 36.7 grams. Therefore, the theoretical yield of the experiment is 36.7 grams of CO_{2}.
Part 3 of 3: Calculate Percentage Yield
Step 1. You have a good understanding of the meaning of percentage yield
The theoretical yield you calculated assumes that the entire equation occurred under perfect conditions, which never happens in a real experiment, since contaminants and other unpredictable problems can cause some of the reactants to fail to become the product. This is why some chemists use three different concepts to refer to performance:
 The theoretical yield is the maximum amount of product that the experiment could generate.
 Actual yield is the actual amount you created, measured directly on a scale.

The formula for the percentage return is the following Actual Return Theoretical Return × 100% { displaystyle { frac { text {Actual Return}} { text {Theoretical Return}}} times 100 \%}
Por ejemplo, un rendimiento porcentual del 50 % indica que, al finalizar una reacción química, obtienes como máximo el 50 % de la cantidad esperada.
Step 2. Write the actual performance of the experiment
If you did the experiment on your own, collect the purified product from the reaction and weigh it on a scale to calculate its mass. On the other hand, if it is an exercise or task, you should already have the information of the actual performance.
 Suppose the actual reaction is 29 grams of CO_{2}.
Step 3. Divide the actual return by the theoretical return
Make sure to use the same unit for both values (usually grams). The result will be a ratio less than unity.

The actual yield was 29 grams, while the theoretical yield was 36.7 grams. Therefore, if we do the calculation, we will obtain the following result: 29g36, 7g = 0.79 { displaystyle { frac {29g} {36, 7g}} = 0.79}
Step 4. Multiply the result obtained by 100 to convert it to a percentage
The answer will be the percentage yield.
 0.79 x 100 = 79, so the percent yield for the experiment is 79%. This means that you created 79% of the maximum possible amount of CO._{2}.
Advice

Some students confuse the term percentage performance (the result obtained from the maximum possible amount) with the percentage error (the difference between an experimental result and the expected result). The correct formula for the percentage return is Actual Return Theoretical Return × 100 { displaystyle { frac { text {Actual Return}} { text {Theoretical Return}}} times 100}
si por el contrario restas ambos rendimientos, estarás utilizando la fórmula para hallar el error porcentual.
 si obtienes resultados ampliamente diferentes, verifica las unidades. si el rendimiento real difiere del rendimiento teórico por un orden de magnitud o más, probablemente empleaste las unidades incorrectas en algún punto del cálculo. repite los procedimientos y lleva un registro de las unidades en cada paso de la ecuación.
 si el rendimiento porcentual es mayor al 100 % (y sabes a ciencia cierta de que los cálculos son correctos), significa que hay otras sustancias que han contaminado al producto. por consiguiente, deberás purificarlo (por ejemplo, secándolo o filtrándolo) y pesarlo nuevamente.