Theoretical Yield Calculator

Calculate theoretical yield with the Examples.com Theoretical Yield Calculator. Input reactants and product data for precise results.

How to Use the Theoretical Yield Calculator

Step 1: Understand the Formulas:

• Moles of limiting reagent: Mass of limiting reagent (g) ÷ Molecular weight of limiting reagent (g/mol)
  
• Theoretical Yield (g): Moles of Limiting Reagent × Stoichiometry × Weight of Desired Product (g/mol)
  
• Desired Product Stoichiometry: Moles of Desired Product ÷ Moles of Limiting Reagent

Step 2: Input Limiting Reagent:

• Enter the mass of the limiting reagent in grams in the “Mass” field under the “Limiting Reagent” section.
• Provide the molecular weight of the limiting reagent in grams per mole (g/mol) in the appropriate field.

Step 3: Input Desired Product:

• Enter the moles of the desired product in the “Moles” field under the “Desired Product” section.
• Provide the molecular weight of the desired product in grams per mole (g/mol) in the appropriate field.

Step 4: Stoichiometry:

• Step 5: Enter the stoichiometric ratio in the “Stoichiometry” field.

Step 6: Calculate:

• Click the “Calculate” button to compute the theoretical yield.

Step 7: Review Results:

• The theoretical yield will be displayed based on the input data. Ensure all fields are correctly filled for accurate results.

Examples of Moles of Limiting Reagent

Example 1: Combustion of Methane

Reaction: CH₄+2O₂→CO₂+2H₂O

• Mass of Limiting Reagent (Oxygen): 32 g
• Molecular Weight of Oxygen (O2): 32 g/mol

Calculation: Moles of O2=Mass of O2 (g)/Molecular weight of O2 (g/mol)

Moles of O2=32 g/32 g/mol=1 mol

Moles of Limiting Reagent (Oxygen): 1 mol

Example 2: Formation of Calcium Carbonate

Reaction: CaO+CO₂→CaCO₃

• Mass of Limiting Reagent (Calcium Oxide): 56 g
• Molecular Weight of Calcium Oxide (CaO): 56 g/mol

Calculation: Moles of CaO=Mass of CaO (g)/Molecular weight of CaO (g/mol)

Moles of CaO=56 g/56 g/mol=1 mol

Moles of Limiting Reagent (Calcium Oxide): 1 mol

Example 3: Synthesis of Ammonia

Reaction: N₂+3H₂→2NH₃

• Mass of Limiting Reagent (Hydrogen): 10 g
• Molecular Weight of Hydrogen (H2): 2 g/mol

Calculation: Moles of H2=Mass of H2 (g)/Molecular weight of H2 (g/mol)

Moles of H2=10 g/2 g/mol=5 mol

Moles of Limiting Reagent (Hydrogen): 5 mol

Examples of Desired Product Stoichiometry

Example 1: Formation of Water

Reaction: 2H₂+O₂→2H₂O

• Moles of Limiting Reagent (Oxygen): 1 mol
• Moles of Desired Product (Water): 2 mol

Calculation: Desired Product Stoichiometry=Moles of Desired Product/Moles of Limiting Reagent

Desired Product Stoichiometry=2 mol/1 mol=2

Desired Product Stoichiometry (Water): 2

Example 2: Synthesis of Ammonia

Reaction: N₂+3H₂→2NH₃

• Moles of Limiting Reagent (Hydrogen): 3 mol
• Moles of Desired Product (Ammonia): 2 mol

Calculation: Desired Product Stoichiometry=Moles of Desired Product/Moles of Limiting Reagent

Desired Product Stoichiometry=2 mol/3 mol=2/3

Desired Product Stoichiometry (Ammonia): 0.67

Example 3: Formation of Sodium Chloride

Reaction: 2Na+Cl₂→2NaCl

• Moles of Limiting Reagent (Sodium): 2 mol
• Moles of Desired Product (Sodium Chloride): 2 mol

Calculation: Desired Product Stoichiometry=Moles of Desired Product/Moles of Limiting Reagent

Desired Product Stoichiometry=2 mol/2 mol=1

Desired Product Stoichiometry (Sodium Chloride): 1

Examples of Theoretical Yield

Example 1: Formation of Water

Reaction: 2H₂+O₂→2H₂O

• Moles of Limiting Reagent (Oxygen): 1 mol
• Stoichiometry (H2 to H2O): 1
• Molecular Weight of Water (H2O): 18 g/mol

Calculation: Theoretical Yield (g)=Moles of Limiting Reagent×Stoichiometry×Weight of Desired Product (g/mol)

Theoretical Yield (g)=1 mol×1×18 g/mol=18 g

Theoretical Yield of Water: 18 g

Example 2: Formation of Sodium Chloride

Reaction: 2Na+Cl₂→2NaCl

• Moles of Limiting Reagent (Sodium): 2 mol
• Stoichiometry (Na to NaCl): 1
• Molecular Weight of Sodium Chloride (NaCl): 58.5 g/mol

Calculation: Theoretical Yield (g)=Moles of Limiting Reagent×Stoichiometry×Weight of Desired Product (g/mol)

Theoretical Yield (g)=2mol×1×58.5g/mol=117g

Theoretical Yield of Sodium Chloride: 117 g

Example 3: Synthesis of Ammonia

Reaction: N₂+3H₂→2NH₃

• Moles of Limiting Reagent (Hydrogen): 3 mol
• Stoichiometry (H2 to NH3): 2/3
• Molecular Weight of Ammonia (NH3): 17 g/mol

Calculation: Theoretical Yield (g)=Moles of Limiting Reagent×Stoichiometry×Weight of Desired Product (g/mol)

Theoretical Yield (g)=3mol×32​×17g/mol=34g

Theoretical Yield of Ammonia: 34 g

Can the actual yield be higher than the theoretical yield?

No, the actual yield cannot be higher than the theoretical yield. The actual yield is often lower due to factors such as incomplete reactions, side reactions, and losses during product recovery.

Why is the theoretical yield important?

Theoretical yield is important because it provides a benchmark to evaluate the efficiency of a chemical reaction. It helps in comparing the actual yield obtained from an experiment with the maximum possible yield.

How does the Theoretical Yield Calculator work?

The calculator requires you to input the mass and molecular weight of the limiting reagent, the stoichiometry of the reaction, and the molecular weight of the desired product.

Can I calculate the theoretical yield for any chemical reaction?

Yes, as long as you have the balanced chemical equation for the reaction, and you know the mass and molecular weight of the limiting reagent, the stoichiometry, and the molecular weight of the desired product.

How accurate is the Theoretical Yield Calculator?

The accuracy of the calculator depends on the precision of the input values (mass, molecular weights, and stoichiometry). Provided accurate data, the calculator gives a precise theoretical yield.

What are some common mistakes to avoid when using the Theoretical Yield Calculator?

Common mistakes include:

• Incorrectly identifying the limiting reagent.
• Using incorrect or inconsistent units.
• Misbalancing the chemical equation.
• Inputting wrong molecular weights.