Theoretical Yield

Yield of a chemical reaction is the amount of the products that are produced when the reaction between two or more substances reacts with each other. Yield can be categorized as actual yield and theoretical yield. The theoretical yield of a reaction is based on the stoichiometry of the chemical reaction. Law of conservation of mass states that the mass of reactants that combine is equal to the mass of products formed but in real life, the amount of products produced is less as calculated due to various factors that affect the rate of reaction and the amount of products produced. These factors can be the conversion of some mass of reactants to heat, impurities in the products formed or impurities in the reactants, etc. The calculated amount of products is called the theoretical yield of the reaction whereas the actual amount produced is called the actual yield.

Theoretical Yield Definition

Theoretical yield can be defined as the amount of products estimated to be produced when an ideal chemical reaction takes place is called the theoretical yield of the reaction. Theoretical yield is measured using g (grams) or moles.

Theoretical yield can be considered to be the maximum amount of the products that can be produced from a given amount of reactants.

What is a Limiting Reagent?

Let us consider an example. Assume that 5 burgers require 5 buns of bread and 1kg of potatoes. Now consider that we have only 3 buns and 2 kg of potatoes. Will we be able to make 5 burgers out of the ingredients that we have? The answer is no, because although we have more potatoes than required the buns are less than required. Thus the amount of burgers produced is limited by the number of buns. Similarly in a chemical reaction, the same holds true. Even if we have a higher amount of one reactant, the yield of the reaction will be limited by the chemical which is lesser than required.

Percent Yield

It is the percent ratio of actual yield and theoretical yield. It is calculated by dividing the actual yield of the reaction by the calculated theoretical yield and then multiplying the result obtained by 100 to get the percentage. 

Percent Yield Equation

The formula or equation for percent yield is:

The value of actual yield is usually found to be lower than the theoretical yield and hence the percent yield is usually less than 100%. Percent yield helps us determine the economic feasibility of the reaction. The higher the percent yield, the more feasible is the reaction for commercial purposes because, on a commercial scale, we need to ensure that the maximum amount of products are produced from any chemical reaction.

Learn more about Percent Yield

How to Calculate Theoretical Yield?

Calculation of theoretical yield involves two steps:

Step 1: Finding the Limiting Reagent or Limiting Reactant

A limiting reagent can be defined as the reactant that determines the final amount of the product produced (as we saw buns of bread were a limiting reagent in our example). We know that elements or chemicals always react in a fixed proportion to produce products. So if the amount of one reactant is found to be lesser than the expected value needed to react with another reactant, then such reactant is called a limiting reagent or limiting reactant. In a reaction, limiting reactant sets a limit to the number of products that will be produced. Also, this limiting reagent gets fully consumed once the reaction is complete although there may be some amount of other reactants left behind.

In order to calculate the limiting reactant for the considered reaction, the following steps need to be followed:

Step 1: Calculation of molar mass of each reactant.

Molar mass of reactants can be calculated by summing up the product of the molar mass of the elements and the number of elements that make up the reactant. For example, if reactant A is of the form M_mN_n then the molar mass of A is (m * molar mass of M) + (n * molar mass of N). We may represent the molar mass of A, B, C and D as M_A, M_B, M_C, and M_D.

Step 2: Calculate the number of moles of each element. 

Now we need to calculate the number of moles of each element that we have. In order to do this, we divide the amount of reactant in grams available to us with the molar mass of each reactant. Suppose the number of moles of A and B comes out to be m_A, and m_B.

Step 3: Now we need to calculate the ratio in which the reactants combine in the reaction. In our reaction p moles of A combine with q moles of B. Thus the ratio is p:q but we need to change the denominator in the ratio to 1. Thus this ratio can be rewritten as p/q:1.

Step 4: In this, step we calculate the ratio of the moles of the reactants available with us using the values obtained in Step 2. This comes out to be m_A:m_B in this example but as we need to change the denominator in the ratio to 1, this ratio can be rewritten as m_A/m_B:1.

Step 5: Comparing the ratios.

Now we shall compare the ratios obtained in steps 3 and 4. If p/q>m_A/m_B, then it means that we have a higher amount of reactant A than required. In such a case reactant B will be the limiting reagent. If p/q<m_A/m_B, then it means that we have a lesser amount of reactant A than required to react with a given amount of B. In such a case reactant A will be the limiting reagent.

Step 2: Estimating the Theoretical Yield

To estimate the theoretical yield, we need to follow the following sub-steps:

Step 1: Now we need to find the number of moles of C or D produced with 1 mole of the limiting reagent.

Step 2: The number of moles of C or D produced is multiplied by the given amount of moles of the limiting reagent to get the moles of C or D produced.

Step 3: In order to get the value in grams, this value is further multiplied with the molar mass of the respective product i.e. moles of C are multiplied with a molar mass of C and similarly for D.

Example: Let us understand the calculation of theoretical yield with an example. Consider the following balanced chemical equation

We are provided with 50g of O₂ and 30g of H₂. We need to determine the amount of H₂O produced.

Solution:

The first step is finding the limiting reagent.

Step 1:

Molar mass of H₂ = 2 * molar mass of H = 2 * 1 = 2 g/mol

Molar mass of O₂ = 2 * molar mass of O = 2 * 16 = 32 g/mol

Step 2:

We are given with 50g of O₂ and 30g of H₂. Thus,

Moles in 50g of O₂ = given weight/molar mass of O₂ = 50/32

= 1.5625 moles

Moles in 30g of H₂ = given weight/molar mass of H₂ = 30/2

= 15 moles

Step 3:

The ratio of H₂ and O₂ that combine together = 2:1 because the coefficients of H₂ and O₂ are 2 and 1 respectively.

Step 4:

The ratio of moles of H₂ and O₂ available with us = 15:1.5625 = 9.6:1

Step 5:

As we have greater amount of Oxygen available as 9.6 > 2, O₂ becomes the limiting reagent in this case because the equation shows that 2 moles of hydrogen combine with one mole of oxygen. No doubt that we have 9.6 moles of hydrogen for every 1 mole of oxygen but still the reaction will proceed only with 2 moles of hydrogen for every 1 mole of oxygen and the remaining moles of hydrogen will be left unused as the oxygen available with us will react completely with hydrogen to produce water. Thus the amount of water produced will be limited by the amount of oxygen available. If we were to utilise the hydrogen available (15 moles) completely, we would have needed 7.5 moles of oxygen but we have only 1.5625 moles available.

Step 6:

From the equation we can see that 1 mole of O₂ produces 2 moles of H₂O as the coefficient with O₂ is 1 and that with H₂O is 2.

Step 7:

Thus, moles of H₂O produced = 2 * moles of oxygen available = 2 * 1.5625 = 3.125

Molar mass of H₂O = 2 * molar mass of H + 1 * molar mass of O = 2 * 1 + 16 = 18 g/mol

Weight of H₂O produced in grams = 6.25 * molar mass of H₂O = 3.125 * 18 = 25 g

Note: If we are given the mass of actual yield and percent yield of a reaction and need to find theoretical yield, it can be calculated as:

Solved Examples of Theoretical Yield Formula

Problem 1: Calculate the theoretical yield of a reaction if the actual yield is 30 g and the percent yield is 90%.

Solution:

Given Actual Yield = 30g

Percent Yield = 90%

Using, 

Theoretical Yield = 3000/90 = 33.33 g

Problem 2: Calculate the theoretical yield of a reaction if the actual yield is 100 g and the percent yield is 50%.

Solution:

Given Actual Yield = 100g

Percent Yield = 50%

Using, 

Theoretical Yield = 10000/50 = 200 g

Problem 3: Calculate the percent yield of a reaction if the actual yield is 105 g and the theoretical yield is 95.6 g.

Solution:

Given Actual Yield = 105g

Percent Yield = ?

Theoretical Yield = 95.6 g

Using, 

Percent Yield = 10500/95.6 = 109.8%

Problem 4: Consider the following balanced chemical equation

We are provided with 96g of O₂ and 10g of H₂. Determine the amount of H₂O produced.

Solution:

The first step is finding the limiting reagent.

Step 1:

Molar mass of H₂ = 2 * molar mass of H = 2 * 1 = 2 g/mol

Molar mass of O₂ = 2 * molar mass of O = 2 * 16 = 32 g/mol

Step 2:

We are given with 96g of O₂ and 10g of H₂. Thus,

Moles in 96g of O₂ = given weight/molar mass of O₂ = 96/32

= 3 moles

Moles in 10g of H₂ = given weight/molar mass of H₂ = 10/2

= 5 moles

Step 3:

The ratio of H₂ and O₂ that combine together = 2:1

Step 4:

The ratio of moles of H₂ and O₂ available with us = 5:3 = 1.6:1

Step 5:

As we have lesser amount of Oxygen available as 1.6 < 2, H₂ becomes the limiting reagent.

Step 6:

From the equation we can see that 2 moles of H₂ produces 2 moles of H₂O which means 1 mole of H₂ produces 1 mole of H₂O.

Step 7:

Thus, moles of H₂O produced = 2 * moles of hydrogen available = 1 * 5 = 5

Weight of H₂O produced in grams = 5 * molar mass of H₂O = 5 * 18 = 90 g

Problem 5: Consider the following balanced chemical equation

We are provided with 160g of CO₂ and 100g of H₂O. Determine the amount of O₂ produced.

Solution:

The first step is finding the limiting reagent.

Step 1:

Molar mass of H₂O = 2 * molar mass of H + 1 * molar mass of O = 2 * 1 + 1 * 16 = 18 g/mol

Molar mass of CO₂ = 1 * molar mass of C + 2 * molar mass of O = 1 * 12 + 2 * 16 = 44 g/mol

Step 2:

We are given with 160g of CO₂ and 100g of H₂O. Thus,

Moles in 160g of CO₂ = given weight/molar mass of CO₂ = 160/44

= 3.63 moles

Moles in 100g of H₂O = given weight/molar mass of H₂O = 100/18

= 5.55 moles

Step 3:

The ratio of CO₂ and H₂O that combine together = 6:6 = 1:1

Step 4:

The ratio of moles of CO₂ and H₂O available with us = 3.63:5.55 = 0.654:1

Step 5:

As we have lesser amount of CO₂ available as 0.654 < 1, CO₂ becomes the limiting reagent.

Step 6:

From the equation we can see that 6 moles of CO₂ produces 6 moles of O₂ which means 1 mole of CO₂ produces 1 mole of O₂.

Step 7:

Thus, moles of O₂ produced = 1 * moles of CO₂ available = 1 * 0.654 = 0.654

Weight of O₂ produced in grams = 0.654 * molar mass of O₂ = 0.654 * 32 = 20.928 g

FAQs of Theoretical Yield

Q1: What is Theoretical Yield?

Answer:

Theoretical yield is the theoretical maximum amount of products that can be produced in any chemical reaction.

Q2: How to Find the Theoretical Yield?

Answer:

To find the theoretical yield:

• Balance the chemical equation.
• Determine the stoichiometry (relationship between reactants and products).
• Identify the limiting reactant (the one that is completely used up first).
• Calculate the moles of the limiting reactant.
• Use stoichiometry to find the moles of the product.
• Convert moles of product to grams.
• The result is the theoretical yield, representing the maximum possible amount of product.

Q3: How is Theoretical Yield Different from Actual Yield?

Answer:

Theoretical yield refers to the expected amount of the products whereas the actual yield refers to the amount of products actually produced during a chemical reaction.

Q4: Define Limiting Reactant.

Answer:

The reactant which determines the final amount of the products produced and is present in lesser quantity than required is called limiting reagent.

Q5: What is the Effect of Physical Conditions such as Pressure, Temperature, etc. on the Theoretical Yield?

Answer:

According to Chatelier’s principle, the equilibrium shifts either forward or backward depending upon the physical conditions of the reaction. Thus a change in physical conditions of reaction may either lead to increase or decrease in the amount of actual yield.

Q6: Why is it Necessary to use a Balanced Equation to Calculate Theoretical Yield?

Answer:

It is necessary to use a balanced chemical equation to calculate theoretical yield because according to law of fixed proportion every element combines with another element in a fixed proportion by mass and it is necessary to take into account this proportion while calculating theoretical yield.

Q7: Can the Theoretical Yield be Lesser than the Actual Yield?

Answer:

Yes, theoretical yield can be lesser than the actual yield in situations where the products produced have certain impurities such as high water content, precipitated salts, etc.