Stoichiometry Calculator
Calculate stoichiometric amounts for chemical reactions. Find theoretical yield, moles, and mass relationships.
Reaction Parameters
Reactant
Product
Example Reactions:
Theoretical Yield
89.3601 g
4.9603 moles of product
Actual Yield
89.3601 g
at 100% yield
Moles Reactant
4.9603
Mole Ratio
1.00
product/reactant
Mass Ratio
8.9360
g product / g reactant
Calculation Steps:
- Convert reactant mass to moles: 10g / 2.016 g/mol = 4.9603 mol
- Apply mole ratio: 4.9603 x (2/2) = 4.9603 mol
- Convert to mass: 4.9603 mol x 18.015 g/mol = 89.3601 g
About Stoichiometry
Stoichiometry is the study of quantitative relationships in chemical reactions. It uses the coefficients in balanced equations to calculate how much product forms from a given amount of reactant. The key steps are: convert mass to moles, apply the mole ratio from the balanced equation, then convert back to mass. Percent yield accounts for real-world inefficiencies where actual yield is less than theoretical.
What Is Stoichiometry?
Stoichiometry is the quantitative study of reactants and products in chemical reactions. It uses balanced chemical equations to calculate how much of each substance is consumed or produced, making it fundamental to chemistry, from laboratory synthesis to industrial manufacturing.
| Term | Definition | Example |
|---|---|---|
| Stoichiometric ratio | Mole ratios from balanced equation | 2H₂ + O₂ → 2H₂O (ratio 2:1:2) |
| Limiting reagent | Reactant that runs out first | Determines maximum product yield |
| Excess reagent | Reactant left over after reaction | Doesn't affect yield amount |
| Theoretical yield | Maximum possible product | Calculated from limiting reagent |
| Actual yield | Product actually obtained | Usually less than theoretical |
| Percent yield | (Actual/Theoretical) × 100 | Measures reaction efficiency |
Basic Stoichiometry Relationship
Where:
- moles= Amount of substance (mol)
- coefficient= Number in balanced equation
Balancing Chemical Equations
Before performing stoichiometric calculations, the chemical equation must be balanced to conserve mass—equal atoms on both sides.
| Step | Action | Example: Fe + O₂ → Fe₂O₃ |
|---|---|---|
| 1. Count atoms | List atoms on each side | Left: Fe(1), O(2); Right: Fe(2), O(3) |
| 2. Balance metals | Start with single elements | 4Fe + O₂ → 2Fe₂O₃ |
| 3. Balance nonmetals | Adjust oxygen, hydrogen last | 4Fe + 3O₂ → 2Fe₂O₃ |
| 4. Verify | Check all atoms balance | Fe: 4=4, O: 6=6 ✓ |
Balanced equation: 4Fe + 3O₂ → 2Fe₂O₃ (coefficients give mole ratios)
Converting Between Mass, Moles, and Particles
Stoichiometry works in moles, so you often need to convert between different units.
| From | To | Formula | Example |
|---|---|---|---|
| Mass (g) | Moles | n = mass / MW | 18 g H₂O / 18 g/mol = 1 mol |
| Moles | Mass (g) | mass = n × MW | 2 mol × 44 g/mol = 88 g CO₂ |
| Moles | Molecules | N = n × 6.022×10²³ | 0.5 mol × Nₐ = 3.01×10²³ |
| Volume (gas, STP) | Moles | n = V / 22.4 L | 11.2 L / 22.4 = 0.5 mol |
| Molarity × Volume | Moles | n = M × V(L) | 2 M × 0.5 L = 1 mol |
Key Conversion Formulas
Where:
- n= Moles
- MW= Molecular weight (g/mol)
- M= Molarity (mol/L)
- V= Volume (L)
Finding the Limiting Reagent
The limiting reagent is the reactant that gets completely consumed, determining how much product can form. To find it, compare the mole ratios of available reactants to required ratios.
| Method | Steps | Choose Limiting If... |
|---|---|---|
| Ratio comparison | Calculate (moles available) / (coefficient) for each reactant | Smallest ratio is limiting |
| Product calculation | Calculate product from each reactant assuming excess of others | Smallest product amount is actual yield |
Example: For 2H₂ + O₂ → 2H₂O with 3 mol H₂ and 2 mol O₂:
- H₂: 3 mol / 2 = 1.5
- O₂: 2 mol / 1 = 2.0
- H₂ is limiting (smaller ratio); max product = 3 mol H₂O
Percent Yield and Efficiency
Percent yield compares actual product obtained to the theoretical maximum, indicating reaction efficiency.
| Yield Type | Definition | How to Determine |
|---|---|---|
| Theoretical yield | Maximum possible product | Calculate from limiting reagent |
| Actual yield | Product actually obtained | Measure experimentally |
| Percent yield | Efficiency of reaction | (Actual / Theoretical) × 100% |
Why yield < 100%: Incomplete reactions, side reactions, loss during purification, transfer losses, equilibrium limitations.
Percent Yield Formula
Where:
- Actual Yield= Product obtained experimentally
- Theoretical Yield= Maximum calculated from stoichiometry
Stoichiometry with Solutions
For reactions in solution, use molarity to find moles: n = M × V (where V is in liters).
| Application | Approach | Example |
|---|---|---|
| Titration | MₐVₐ = MᵦVᵦ (for 1:1 reactions) | Find unknown acid concentration |
| Precipitation | Find limiting reagent in solution | How much precipitate forms? |
| Dilution for reaction | Calculate moles, then concentration | Product concentration after mixing |
Remember: For non-1:1 ratios, use the full stoichiometric relationship with coefficients.
Gas Stoichiometry
For gas-phase reactions, stoichiometry can work with volumes directly (at same T and P) or through ideal gas law calculations.
| Condition | Approach | Key Relationship |
|---|---|---|
| Same T and P | Volume ratios = mole ratios | 2 L H₂ + 1 L O₂ → 2 L H₂O (gas) |
| At STP | 1 mol = 22.4 L | n = V / 22.4 |
| Any T and P | Use PV = nRT | n = PV / RT |
| Density given | MW = dRT / P | Identify gas from density |
Worked Examples
Mass-to-Mass Calculation
Problem:
How many grams of CO₂ are produced when 50 g of CH₄ burns completely? CH₄ + 2O₂ → CO₂ + 2H₂O
Solution Steps:
- 1Calculate moles of CH₄: MW = 12 + 4(1) = 16 g/mol; n = 50/16 = 3.125 mol
- 2Use mole ratio: 1 mol CH₄ → 1 mol CO₂, so 3.125 mol CO₂
- 3Convert to grams: MW(CO₂) = 12 + 2(16) = 44 g/mol
- 4mass = 3.125 mol × 44 g/mol = 137.5 g
Result:
50 g of CH₄ produces 137.5 g of CO₂. Note the mass increases because oxygen from air is incorporated into the product.
Limiting Reagent Problem
Problem:
10 g of Mg reacts with 10 g of O₂. How much MgO forms? (2Mg + O₂ → 2MgO)
Solution Steps:
- 1Convert to moles: Mg = 10/24.3 = 0.412 mol; O₂ = 10/32 = 0.313 mol
- 2Check limiting reagent: Mg: 0.412/2 = 0.206; O₂: 0.313/1 = 0.313
- 3Mg is limiting (smaller ratio)
- 4From 0.412 mol Mg: 0.412 mol MgO (1:1 ratio with 2Mg)
- 5Mass MgO = 0.412 × 40.3 = 16.6 g
Result:
16.6 g MgO forms. Excess O₂ remaining: 0.313 - 0.206 = 0.107 mol = 3.4 g O₂ unreacted.
Percent Yield Calculation
Problem:
A reaction's theoretical yield is 25 g, but you obtained 21 g. What is the percent yield?
Solution Steps:
- 1Identify values: Actual yield = 21 g, Theoretical yield = 25 g
- 2Apply formula: % Yield = (Actual / Theoretical) × 100
- 3Calculate: % Yield = (21 / 25) × 100 = 84%
Result:
Percent yield = 84%. This means 84% of the theoretical maximum was recovered. The remaining 16% was lost to incomplete reaction, side reactions, or purification losses.
Tips & Best Practices
- ✓Always balance the equation first—unbalanced equations give wrong stoichiometric ratios.
- ✓Convert everything to moles before using mole ratios, then convert back to desired units.
- ✓For limiting reagent: divide moles by coefficient for each reactant; smallest = limiting.
- ✓At STP, 1 mole of any gas occupies 22.4 L—useful for gas stoichiometry.
- ✓Percent yield = (actual/theoretical) × 100; values over 100% indicate measurement error.
- ✓Keep track of units throughout calculations—dimensional analysis prevents errors.
- ✓For solution reactions, moles = Molarity × Volume (in liters).
Frequently Asked Questions
Sources & References
- Chemistry LibreTexts - Stoichiometry (2024)
- OpenStax Chemistry 2e (2023)
- Khan Academy - Stoichiometry (2024)
- NIST Chemistry WebBook (2024)
Last updated: 2026-01-22