Theoretical Yield:

The prime responsibility of a chemical engineer and a chemist is to design the processes in which the known concentrations and quantities of chemical reactants can convert into the products under a given set of reaction parameters. 

The efficiency of the process can calculate via theoretical and actual yields. The maximum amount of conversion of raw materials obtained after the successful completion of a chemical or physical reaction is the theoretical yield. 

Generally, there are two types of raw materials that we use in any chemical reaction that are limiting reactants and excessive reactants. 

Limiting Reactant: 

The reactant that is completely consuming at the end of the chemical reaction are the limiting reactant. 

Excessive Reactant: 

The reactant that is still present even after the execution of the chemical reaction once the limiting reactant is fully consume is called excessive reactant. 

It is worth mentioning here that, theoretical yield depends upon the consumption of limiting reactant and hence, before calculating theoretical yield it is very important to identify the limiting reactant. Lastly, we can express theoretical yield in grams (g) or moles. 

Actual Yield: 

As stated above theoretical yield is the pre reaction calculated yield that may occur if the chemical reaction has been completed and the reactants are completely consumed and are converted into desired product. 

Hence, contrary to the above discussion, the amount of product actually produced defines actual yield. It is mostly observed and very well demonstrated that actual yield is less as compared to theoretical yield due to the fact that majority of the times, there are backward reactions, chain reactions and spilling of the products  and ultimately the amount of product obtained is in less amount as compared to the amount that has been elaborated theoretically. 

Hence, the overall percent yield, that is the ratio in between actual and theoretical yield is less than cent percent. Additionally, there are some reported cases in which the actual yield exceeds theoretical yield as the quantity of product is large due to either addition of impurities in the overall weight or due to production of extra products due to side reactions.  Hence, the overall yield is higher than 100%. 

Calculation of Theoretical Yield: 

The prime task in calculating theoretical yield is to identify the limiting reactant in any chemical equation. 

This can only achieve if you have a balanced chemical equation otherwise one has to balance it accordingly before jumping to the next step. 

When we had done to balance it, you can predict the limiting reactant by studying the mole ratio between the reactants. 

It is very important to convert the number of reactants into moles if they are given in grams. In order to get the required quantity in moles, we have to divide the given masses by their molecular weight. 

Let us understand the way of identifying the limiting reactants and calculating theoretical yield by taking a chemical reaction. 

3.45 g of methyl bromide (CH3Br) are reacted with 5.23 g of sodium hydroxide (NaOH). As a result of this reaction, sodium bromide (NaBr) will produce. Calculate the theoretical yield of the reaction. The said problem can solve by following the given steps. 

Step 1: Balanced chemical Equation

As stated earlier it is very important to either get a balanced chemical equation or in some cases where the equation is not balanced, the first step is to balance the chemical equation. The chemical equation of the up stated reaction is: 

CH3Br+ NaOH ↔ CH3OH+ NaBr

Step 2: Conversion of Moles

Weight of CH3Br: 3.45 g

Molecular Weight of CH3Br: 94.94 g/mole

Weight of NaOH: 5.23 g

Molecular Weight of NaOH: 39.9 g/mole

Similarly, NaBr: 102.8 g/mole

Molecular Weight of CH3OH: 32 g/mole

Step 3: Identification of Limiting Reactant:

Now we need to calculate how many moles of the product we get from our calculated moles of reagents. This can achieve by one step dimensional analysis

NaBr produced by CH3Br= 3.45 g CH3Br ×1 mol CH3Br 94.9 g CH3Br×1 mol NaBr1 mol CH3Br: = 0.0363 mole of NaBr

NaBr produced by NaOH= 5.33 g NaOH×1 mol of NaOH39.9 g of NaOH×1 mol  NaBr1 mol NaOH = 0.1308 mol of NaBr

It can be seen that the amount of product formed by CH3Br is less as compared to NaOH, So, methyl bromide CH3Br is the limiting reactant. 

Theoretical yield tells how much product we will end up with when our limiting reactant has completely consumed. Hence, after consuming the entire methyl bromide, we will get 0.0363 moles of NaBr and hence, this is the theoretical yield. 

As a result, while performing an experiment, the theoretical yield can use along with the actual yield to calculate the percent yield. 

With that said, if you’re focused on the theory part and don’t wanna balance it yourself, you could use an online theoretical yield calculator with steps. 

 

Easy ways to calculate theoretical and percent yield

Yield

In Chemistry or chemical engineering, yield is often quote as the measure of the number of products formed as the successful execution of the chemical reaction in relation to the quantity and composition of the reaction products. we can write yield in moles and grams. 

Percent Yield

The ratio of actual yield to theoretical yield is the percent yield. It is a dimensionless quantity that generally refers to the overall effectiveness of any chemical process. 

It is also worth mentioning here that the percentage yield can be less than or greater than 100 depending upon the reaction parameters, nature of reactants taking part in the reaction and amount of products obtain until the completion ofreaction.  

Theoretical Yield

The maximum conversion that we can obtain as a result of the successful completion of a chemical process once the limiting reactant has fully consumed refer as the theoretical yield.

It is worth noting here that the conversion of limiting reactants affects the theoretical yield. Moreover, the actual conversion obtained of raw materials into useful products is termed as actual yield. The values of theoretical and actual yield are different due to reaction parameters and the effectiveness and efficiency of the chemical process. 

Steps to Calculate Theoretical Yield

The following steps should always be keep in mind while calculating theoretical and percentage yield. The overall calculations can calculate if we follow these steps. 

• Determination of balanced chemical equations having reactants on the left side and products on the right side. 
• Calculation of molar masses in grams and moles of the chemical reagents taking part in the reaction and products. 
• The identification of limiting reactants.
• The calculation of theoretical yield and actual yield. 
• The calculation of percentage yield.

Ways to calculate Theoretical and Percentage Yield

The process of calculating the theoretical and percentage yield follows the same steps that have explained above. 

If we explore the relevant literature and ways to calculate the yields then we conclude that there might be two ways to calculate the theoretical and percentage yield.

One of them is by taking the molar masses of the reagents and the products and the second one is to convert the molar masses to moles. 

In both ways, the schematic procedure remains the same. The difference comes in the units of calculations of theoretical and actual yields. 

You can also use a percent yield of chemical reactions calculator to do that job. 

Calculation of Yields via Molecular masses 

Let us consider a chemical reaction in which potassium chlorate KClO3 decomposes into potassium chloride KCl and oxygen O2 in the presence of catalyst and heat. In the reaction, around 50 grams of KClO3 is heated until it is decomposed and the mass of oxygen gas produced as the by-products is 14.9 grams. Calculate:

• Theoretical Yield
• Percentage Yield

Solution:

The list of known parameters includes the mass of KClO3, and the mass of oxygen produced. Secondly, both theoretical and percentage yields need to calculate. 

Determination of balanced chemical equation:

2 KClO3→ 2KCl + 3O2

Calculation of Molar Masses or Moles of Chemical Reagents and Products

Mass of KClO3 = 122 g/mole

KCl= 74 g/mole

O2= 32 g/mole

Identification of Limiting Reactant

It can be seen in the chemical equation that it is a simple decomposition reaction in which KClO3 is decomposed into KCl and oxygen under heat and catalyst. 

There is only one chemical reagent that is KClO3 and hence, the completion of the reaction will be judged once KClO3 is completely utilized. Hence, we may say KClO3 as the limiting reactant.

Calculation of Theoretical Yield

The theoretical yield is: 

Theoretical Yield= 50 g of KClO3 × 1 mole of KClO3 122 g ×3 moles of O22 moles of KClO3 ×32 g of O21 mole of O2

= 19.67 grams. 

Hence, the theoretical yield calculated is 19.67 grams of oxygen. 

Calculation of percentage Yield

Percentage Yield=Actual YieldTheoretical Yield×100

Actual Yield= 14.9 grams

Theoretical Yield= 19.67 gram percentage Yield=14.919.67×100

Percentage Yield=75%

Henceforth, the percentage yield of the above decomposition reaction is  75% and the whole calculations had done in molar masses. 

Related: Also Read The Chemistry Behind the Undesirable Redox Reaction.

Calculation of Yields via Moles

Let us consider another example in which carbon sulfide (CS2) and oxygen (O2) react to form carbon dioxide (CO2) and sulfur oxide (SO2). 

What is the percentage yield if 25 grams of CS2 reacts with 40 grams of O2 and 22.4 grams of SO2 is produced? Also, calculate the theoretical yield and identify the limiting reactant of the process. 

Solution: 

The question can solve by following the steps elaborated above. 

Determination of balanced chemical equation:

CS2 + 3O2 → CO2 + 2 SO2

Calculation of Molar Masses or Moles of Chemical Reagents and Products

Mass of CS2 = 76 g/mole

O2= 32 g/mole

CO2= 44 g/mole

SO2= 64 g/mole

Identification of Limiting Reactant

The identification of limiting reactants is very important. 

Amount of SO2 produced by CS2= 25 grams ×1 mole of CS276 grams×2 moles of SO2I mole of CS2= 0.65 moles of SO2

Similarly by O2= 40 grams × 1 mole of O232 grams×2 moles of SO23 moles of O2= 0.83 moles of SO2

It can be seen in the calculation above, that the amount of SO2 produced by CS2 is less as compared to O2 and hence, CS2 is the limiting reactant.

Calculation of theoretical Yield

Theoretical Yield= 0.65 moles

Calculation of Percentage Yield

Percentage Yield=Actual YieldTheoretical Yield×100

Theoretical Yield= 0.65 moles

Actual Yield= 22.4 grams of SO2= 22.4/64= 0.35 moles of SO2 

Percentage Yield=0.350.65×100

=54%

Conclusion: 

It can conclude from the above discussion that the calculation of theoretical and percentage yields is a very simple process that can accomplish by following the very simple steps elaborate here 

Among all the steps the identification of the limiting reactant is very important. Yields can calculate via both molar masses and moles but the main idea is to keep the overall units consistent.