Stoichiometry Calculator
Calculate the mass of product formed from a given mass of reactant using stoichiometric relationships. Enter the reactant mass, molar masses, mole ratio, and optional percent yield to find theoretical and actual product masses. Essential for lab preparation, industrial chemistry, and academic problem-solving. See also our Theoretical Yield Calculator and Molar Mass Calculator.
Leave at 100% for theoretical yield
How to Solve Stoichiometry Problems
Stoichiometry is the quantitative study of reactants and products in chemical reactions. It uses the balanced chemical equation as a recipe that tells you the exact proportions in which substances react and form products. The word comes from the Greek "stoicheion" (element) and "metron" (measure), literally meaning "measuring elements."
Every stoichiometry problem follows the same general approach: convert the given quantity to moles, use the mole ratio from the balanced equation to find moles of the desired substance, then convert moles to the requested unit (grams, liters, molecules, etc.). This "mole bridge" method works for any stoichiometric calculation regardless of complexity.
- Write and balance the chemical equation
- Convert the given mass of reactant to moles (divide by molar mass)
- Use the mole ratio from the balanced equation to find moles of product
- Convert moles of product to grams (multiply by molar mass)
- If percent yield is given, multiply theoretical yield by (yield/100)
Formula
Mass-to-Mass Stoichiometry:
Product mass = (Reactant mass / MW_reactant) × (ratio_product / ratio_reactant) × MW_product
With Percent Yield:
Actual mass = Theoretical mass × (percent yield / 100)
Step-by-step:
1. Moles of reactant = mass (g) / molar mass (g/mol)
2. Moles of product = moles of reactant × (product coefficient / reactant coefficient)
3. Mass of product = moles of product × molar mass of product (g/mol)
Limiting Reagent:
The reactant that produces the least amount of product is the limiting reagent.
Excess = moles available - moles consumed
Example Calculation
Problem: How many grams of CO₂ are produced from burning 10 g of methane (CH₄)?
Equation: CH₄ + 2O₂ → CO₂ + 2H₂O
Given: mass(CH₄) = 10 g, MW(CH₄) = 16 g/mol, MW(CO₂) = 44 g/mol, ratio = 1:1
Solution:
Moles CH₄ = 10 / 16 = 0.625 mol
Moles CO₂ = 0.625 × (1/1) = 0.625 mol
Mass CO₂ = 0.625 × 44 = 27.5 g
With 85% yield:
Actual mass = 27.5 × 0.85 = 23.375 g CO₂
Stoichiometry Reference Table
| Reaction | Mole Ratio | 10g Reactant → Product |
|---|---|---|
| 2H₂ + O₂ → 2H₂O | 2:1:2 | 10g H₂ → 89.4g H₂O |
| CH₄ + 2O₂ → CO₂ + 2H₂O | 1:2:1:2 | 10g CH₄ → 27.5g CO₂ |
| 2Na + Cl₂ → 2NaCl | 2:1:2 | 10g Na → 25.4g NaCl |
| CaCO₃ → CaO + CO₂ | 1:1:1 | 10g CaCO₃ → 5.6g CaO |
| Fe₂O₃ + 3CO → 2Fe + 3CO₂ | 1:3:2:3 | 10g Fe₂O₃ → 7.0g Fe |
| N₂ + 3H₂ → 2NH₃ | 1:3:2 | 10g N₂ → 12.1g NH₃ |
| 2Al + 3Cl₂ → 2AlCl₃ | 2:3:2 | 10g Al → 49.4g AlCl₃ |
| C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O | 1:6:6:6 | 10g glucose → 14.7g CO₂ |
Limiting Reagent and Excess
In most real reactions, reactants are not present in exact stoichiometric proportions. The limiting reagent is the reactant that is completely consumed first, determining the maximum amount of product that can form. The other reactant(s) are in excess — some will remain unreacted after the reaction is complete.
To identify the limiting reagent: (1) calculate the moles of each reactant, (2) divide each by its stoichiometric coefficient, (3) the reactant with the smallest value is the limiting reagent. All product calculations must be based on the limiting reagent, not the excess reactant.
Understanding limiting reagents is crucial in industrial chemistry where raw materials have different costs. Manufacturers often use an excess of the cheaper reactant to ensure complete consumption of the more expensive one, maximizing economic efficiency while accepting some waste of the cheaper material.
Percent Yield
Percent yield measures the efficiency of a reaction by comparing the actual product obtained to the theoretical maximum: % yield = (actual yield / theoretical yield) × 100. Yields less than 100% result from incomplete reactions, side reactions, mechanical losses during purification, and equilibrium limitations.
In organic synthesis, multi-step reactions compound yield losses. If each step has 90% yield, a 5-step synthesis gives only 0.9⁵ = 59% overall yield. This is why synthetic chemists strive for high-yielding reactions and minimize the number of steps. Industrial processes are optimized to achieve yields as close to 100% as economically feasible.
Frequently Asked Questions
What is stoichiometry?
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It uses balanced equations and molar masses to calculate how much of each substance is consumed or produced, enabling predictions of reaction outcomes.
What is a mole ratio?
A mole ratio is the ratio of moles of one substance to moles of another in a balanced chemical equation, determined by the coefficients. In 2H₂ + O₂ → 2H₂O, the mole ratio of H₂ to O₂ is 2:1, and H₂ to H₂O is 2:2 (or 1:1). Mole ratios are the conversion factors for stoichiometric calculations.
What is the limiting reagent?
The limiting reagent is the reactant that is completely consumed first in a reaction, limiting the amount of product that can form. It is identified by comparing the mole-to-coefficient ratios of all reactants — the one with the smallest ratio is limiting. All yield calculations are based on the limiting reagent.
How do I convert grams to moles?
Divide the mass in grams by the molar mass (g/mol): moles = mass / molar mass. For example, 10 g of NaOH (MW = 40 g/mol) = 10/40 = 0.25 mol. The molar mass is found by summing the atomic masses of all atoms in the formula from the periodic table.
Why is my actual yield less than theoretical?
Actual yield is less than theoretical due to: incomplete reactions (equilibrium), side reactions producing unwanted products, mechanical losses during transfer and purification, impure reagents, and measurement errors. Percent yield = (actual/theoretical) × 100 quantifies this difference.
Can stoichiometry be used for solutions?
Yes. For solutions, use molarity × volume (in liters) to get moles: n = M × V. Then apply the mole ratio as usual. This is the basis of titration calculations. For gases at STP, use the molar volume (22.4 L/mol) to convert between volume and moles.