How do you solve stoichiometry word problems step by step?

The universal 4-step gram-to-gram method: Step 1 — convert grams of given substance to moles by dividing by its molar mass. Step 2 — convert moles of given to moles of desired using the mole ratio from the balanced equation. Step 3 — convert moles of desired to grams by multiplying by its molar mass. Step 4 — check significant figures and units.

Stoichiometry Calculator — Mole Ratios, Limiting Reagent & Yields

Chemistry Tool

Stoichiometry Calculator

Q: What is stoichiometry?

Stoichiometry is the quantitative study of the relationships between reactants and products in a chemical reaction. It uses mole ratios from a balanced chemical equation to convert between amounts of different substances — grams to moles, moles to moles, and moles to grams.

Q: How do you find the limiting reagent?

To find the limiting reagent: (1) Convert mass of each reactant to moles using molar mass. (2) Divide each reactant's moles by its balanced equation coefficient. (3) The reactant with the smallest result is the limiting reagent — it runs out first and limits the amount of product formed.

Q: What is theoretical yield?

Theoretical yield is the maximum amount of product that could be formed from a given amount of limiting reagent, assuming 100% reaction completion with no side reactions or product loss. It is calculated using: moles_product = moles_limiting × (coefficient_product / coefficient_limiting), then mass = moles × molar mass.

Q: How do you calculate percent yield?

Percent yield = (actual yield / theoretical yield) × 100. The actual yield is the amount of product actually collected in the experiment. The theoretical yield is the maximum possible calculated from stoichiometry. Percent yield is always ≤ 100% due to incomplete reactions, side reactions, or product loss during collection.

Q: What is a mole ratio?

A mole ratio is the ratio of moles of one substance to moles of another substance in a balanced chemical equation, derived directly from the coefficients. For example, in 2H₂ + O₂ → 2H₂O, the mole ratio of H₂O to H₂ is 2:2 = 1, meaning 1 mole of H₂O is produced per mole of H₂ consumed. Mole ratios are the key conversion factors in stoichiometry.

Q: What is the difference between limiting reagent and excess reagent?

The limiting reagent is the reactant that is completely consumed first, limiting the amount of product that can form. The excess reagent is any other reactant present in greater amount than needed — some of it remains unreacted after the reaction is complete. The amount of excess remaining = initial moles − moles consumed by limiting reagent × (its coefficient / limiting reagent coefficient).

Solve stoichiometry problems step by step — mole ratios, limiting reagent, theoretical yield, percent yield, and gram-to-gram conversions for any balanced chemical equation.

Stoichiometry Calculator — 4 Tools

Enter a balanced equation, specify a known quantity, and find the amount of any other species.

aA + bB → cC + dD

2H₂+O₂→2H₂O | 5g H₂→H₂O

N₂+3H₂→2NH₃ | 14g N₂→NH₃

CaCO₃→CaO+CO₂ | 100g→CO₂

C₃H₈+5O₂→3CO₂+4H₂O

Balanced Chemical Equation

Use -> or → for reaction arrow. Coefficients before formulas. Use + between species.

Known Species (formula)

MW: —

Target Species (formula)

MW: —

Known Amount

Unit

Error

Step-by-Step Working

Enter balanced equation + mass/moles of EACH reactant. Identifies limiting reagent, excess, and theoretical yield.

N₂+3H₂→2NH₃ | 14g N₂, 6g H₂

2H₂+O₂→2H₂O | 4g H₂, 32g O₂

Fe+S→FeS | 56g Fe, 48g S

Balanced Chemical Equation

Enter equation above to see reactant fields

Product to Calculate Yield For

MW: —

Error

Step-by-Step Working

Calculate theoretical yield from limiting reagent moles, or enter actual yield to find percent yield.

N₂+3H₂→2NH₃ | %yield example

2H₂+O₂→2H₂O | 5 mol H₂

Moles of Limiting Reagent

Coefficient of Limiting Reagent

Coefficient of Product

Molar Mass of Product (g/mol)

Actual Yield Obtained (g) — optional for % yield

Error

Step-by-Step Working

The complete 4-step gram-to-gram stoichiometry method. Perfect for word problems.

5g H₂ → g H₂O

100g CaCO₃ → g CO₂

44g C₃H₈ → g CO₂

27g Al → g AlCl₃

Balanced Chemical Equation

Given Compound (formula)

MW: —

Mass of Given Compound (g)

Desired Compound (formula)

MW: —

Error

Step-by-Step Working (4-Step Gram-to-Gram Method)

Common Molar Masses — Reference Table

Common Elements

Symbol	Element	Molar Mass (g/mol)
H	Hydrogen	1.008
C	Carbon	12.011
N	Nitrogen	14.007
O	Oxygen	15.999
Na	Sodium	22.990
Mg	Magnesium	24.305
Al	Aluminium	26.982
S	Sulfur	32.060
Cl	Chlorine	35.453
K	Potassium	39.098
Ca	Calcium	40.078
Fe	Iron	55.845
Cu	Copper	63.546
Zn	Zinc	65.380
Br	Bromine	79.904
Ag	Silver	107.868
I	Iodine	126.904
Au	Gold	196.967
Pb	Lead	207.200

Common Compounds

Formula	Name	Molar Mass (g/mol)
H₂O	Water	18.015
H₂	Hydrogen gas	2.016
O₂	Oxygen gas	31.998
N₂	Nitrogen gas	28.014
CO₂	Carbon dioxide	44.009
NH₃	Ammonia	17.031
NaCl	Sodium chloride	58.443
HCl	Hydrochloric acid	36.461
H₂SO₄	Sulfuric acid	98.072
CaCO₃	Calcium carbonate	100.086
CaO	Calcium oxide	56.077
AlCl₃	Aluminium chloride	133.341
C₃H₈	Propane	44.094
FeS	Iron(II) sulfide	87.905
Cl₂	Chlorine gas	70.906

What Is Stoichiometry? — The Mole Ratio Method

This stoichiometry calculator solves all types of stoichiometry problems step by step — including mole ratio calculations, limiting reagent identification, theoretical yield, percent yield, and complete gram-to-gram stoichiometry word problems for any balanced chemical equation.

Stoichiometry is the quantitative study of the relationships between reactants and products in a balanced chemical equation. The word comes from the Greek stoicheion (element) and metron (measure). Every stoichiometry calculation depends on one key tool: the mole ratio.

aA + bB → cC + dD Mole ratio: c moles of C produced per a moles of A consumed

The coefficients in a balanced equation (a, b, c, d) give the exact mole ratios between all species. The mole ratio is the conversion factor that makes stoichiometry work — it links the amount of one substance to the amount of any other.

Why Must the Equation Be Balanced?

The Law of Conservation of Mass requires that atoms are neither created nor destroyed — the same number of each atom type must appear on both sides. An unbalanced equation gives wrong mole ratios, leading to incorrect stoichiometry results. Always balance before calculating.

The Universal 4-Step Gram-to-Gram Method

Every stoichiometry word problem — no matter how complex — can be solved with this four-step process:

Step 1: Convert grams of given substance → moles of given (divide by molar mass MW_A)
Step 2: Convert moles of given → moles of wanted (multiply by mole ratio b/a from balanced equation)
Step 3: Convert moles of wanted → grams of wanted (multiply by molar mass MW_B)
Step 4: Check significant figures and units

g_B = g_A × (1/MW_A) × (b/a) × MW_B Combined gram-to-gram stoichiometry formula — all 4 steps in one line

How to Solve Stoichiometry Problems — Step-by-Step

The 4-step gram-to-gram method works for every type of stoichiometry problem. Here are three fully worked examples.

Example 1 — Simple Gram-to-Gram: How many grams of H₂O form from 5g H₂?

Balanced equation: 2H₂ + O₂ → 2H₂O

Step 1 — g H₂ → mol H₂: 5g ÷ 2.016 g/mol = 2.480 mol H₂
Step 2 — mol H₂ → mol H₂O using mole ratio: mole ratio = 2H₂O / 2H₂ = 1. So 2.480 mol H₂ × 1 = 2.480 mol H₂O
Step 3 — mol H₂O → g H₂O: 2.480 × 18.015 = 44.68 g H₂O
Step 4 — Significant figures: 3 sig figs → 44.7 g H₂O

Example 2 — Polynomial Stoichiometry: CaCO₃ → CaO + CO₂ | 100g CaCO₃ → g CO₂?

Step 1 — 100g CaCO₃ ÷ 100.086 g/mol = 0.9991 mol CaCO₃
Step 2 — Mole ratio CO₂:CaCO₃ = 1:1. So 0.9991 mol CO₂
Step 3 — 0.9991 mol × 44.009 g/mol = 43.97 g CO₂

Example 3 — Combustion Stoichiometry: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O | 44g C₃H₈ → g CO₂?

Step 1 — 44g ÷ 44.094 g/mol = 0.9979 mol C₃H₈
Step 2 — Mole ratio = 3CO₂ / 1C₃H₈ = 3. So 0.9979 × 3 = 2.994 mol CO₂
Step 3 — 2.994 × 44.009 = 131.8 g CO₂

Limiting Reagent — How to Find It

The limiting reagent (limiting reactant) is the reactant that is completely consumed first in a chemical reaction. It limits the amount of product that can form. Even if other reactants are still available, the reaction stops when the limiting reagent runs out.

The excess reagent is any reactant present in greater quantity than needed — some remains unreacted after the reaction is complete.

Method: The Mole-to-Coefficient Ratio Test

For reaction aA + bB → products:

Convert mass of each reactant to moles: n = m/M
Divide each reactant's moles by its balanced equation coefficient
The reactant with the smallest ratio (mol/coefficient) is the limiting reagent
Calculate excess remaining: mol_excess_remaining = mol_excess − (limiting_ratio × coefficient_excess)

Worked Example: N₂ + 3H₂ → 2NH₃ | 14g N₂ and 6g H₂

N₂: 14g ÷ 28.014 g/mol = 0.4998 mol N₂. Ratio = 0.4998/1 = 0.4998
H₂: 6g ÷ 2.016 g/mol = 2.976 mol H₂. Ratio = 2.976/3 = 0.9921
N₂ has the smaller ratio (0.4998 < 0.9921) → N₂ is the limiting reagent
H₂ consumed = 0.4998 × 3 = 1.499 mol. H₂ remaining = 2.976 − 1.499 = 1.477 mol H₂ excess
NH₃ produced: 0.4998 × (2/1) = 0.9996 mol × 17.031 g/mol = 17.03 g NH₃ theoretical yield

Key insight: You cannot identify the limiting reagent just by comparing masses or moles — you must divide by the stoichiometric coefficient. A reactant with more moles can still be limiting if its coefficient is large.

Theoretical Yield, Actual Yield, and Percent Yield

Theoretical yield is the maximum amount of product that could form from the limiting reagent, assuming the reaction goes to completion with no losses. It is calculated from stoichiometry using mole ratios.

Actual yield is the amount of product actually collected in the laboratory experiment. It is always less than or equal to the theoretical yield.

Percent yield measures reaction efficiency:

% Yield = (Actual Yield / Theoretical Yield) × 100 Always ≤ 100% — actual yield can never exceed maximum possible

Why Is Actual Yield Less Than Theoretical?

Incomplete reaction — equilibrium, slow kinetics
Side reactions consuming reactants or products
Product loss during filtration, transfer, or crystallization
Impure reactants — effective amount is less than measured mass
Measurement errors — weighing inaccuracies

Worked Example: N₂ + 3H₂ → 2NH₃

Limiting reagent: N₂ (0.4998 mol from 14g)
Theoretical yield: 0.4998 mol N₂ × (2 mol NH₃/1 mol N₂) = 0.9996 mol NH₃ × 17.031 g/mol = 17.03 g theoretical yield
Suppose actual yield collected = 8.5g
Percent yield = (8.5/17.03) × 100 = 49.9%

Stoichiometry Word Problems — Examples and Solutions

These six complete stoichiometry word problem solutions cover every type encountered in general chemistry courses.

Word Problem 1 — Simple Gram-to-Gram (4-step method)

"How many grams of water are produced when 36g of propane (C₃H₈) combusts?"

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

36g C₃H₈ ÷ 44.094 g/mol = 0.8163 mol C₃H₈
Mole ratio H₂O:C₃H₈ = 4:1. So 0.8163 × 4 = 3.265 mol H₂O
3.265 × 18.015 = 58.8g H₂O

Word Problem 2 — Limiting Reagent

"56g Fe + 48g S react: Fe + S → FeS. Find limiting reagent and yield."

Fe: 56/55.845 = 1.003 mol. Ratio = 1.003/1 = 1.003
S: 48/32.060 = 1.497 mol. Ratio = 1.497/1 = 1.497
Fe is limiting (1.003 < 1.497)
FeS produced: 1.003 × 1 = 1.003 mol × 87.905 = 88.2g FeS
S remaining: 1.497 − 1.003 = 0.494 mol × 32.060 = 15.8g S excess

Word Problem 3 — Percent Yield

"In the synthesis of aspirin C₉H₈O₄, theoretical yield = 14.2g, actual = 11.9g. Find %yield."

%yield = (11.9/14.2) × 100 = 83.8%
% error = 100 − 83.8 = 16.2% loss

Word Problem 4 — Gas Volume at STP (22.4 L/mol)

"How many liters of CO₂ at STP form from 100g CaCO₃?"

CaCO₃ → CaO + CO₂

100g ÷ 100.086 = 0.9991 mol CaCO₃
Mole ratio CO₂:CaCO₃ = 1:1 → 0.9991 mol CO₂
Volume at STP = 0.9991 × 22.4 L/mol = 22.4 L CO₂

Word Problem 5 — Solution Stoichiometry (Molarity)

"250 mL of 2.0 M HCl reacts with excess NaOH: HCl + NaOH → NaCl + H₂O. Find g NaCl."

mol HCl = M × V = 2.0 × 0.250 = 0.500 mol HCl
Mole ratio NaCl:HCl = 1:1 → 0.500 mol NaCl
0.500 × 58.443 = 29.2g NaCl

Word Problem 6 — Multi-Step (Limiting Reagent + Percent Yield)

"28g N₂ + 9g H₂ → NH₃ (Haber process). Actual yield = 30g. Find theoretical and %yield."

N₂ + 3H₂ → 2NH₃

N₂: 28/28.014 = 0.9995 mol. Ratio = 0.9995/1 = 0.9995
H₂: 9/2.016 = 4.464 mol. Ratio = 4.464/3 = 1.488
N₂ is limiting (0.9995 < 1.488)
Theoretical: 0.9995 × (2/1) = 1.999 mol × 17.031 = 34.04g NH₃
%yield = (30/34.04) × 100 = 88.1%

Frequently Asked Questions

What is stoichiometry?

Stoichiometry is the quantitative study of the relationships between reactants and products in a chemical reaction. It uses mole ratios from balanced chemical equations to convert between amounts of different substances. The word stoichiometry comes from the Greek for "element measure." Every stoichiometry calculation relies on the mole concept and balanced equations.

How do you find the limiting reagent?

To find the limiting reagent: (1) Convert each reactant mass to moles using n = m/M. (2) Divide each reactant's moles by its stoichiometric coefficient from the balanced equation. (3) The reactant with the smallest mol/coefficient ratio is the limiting reagent — it runs out first and determines the maximum yield.

What is theoretical yield?

Theoretical yield is the maximum mass of product that can be formed from the limiting reagent, assuming 100% complete reaction with no losses. Calculate it using: moles_product = moles_limiting × (coefficient_product / coefficient_limiting), then mass = moles × molar mass. Theoretical yield is always greater than or equal to actual yield.

How do you calculate percent yield?

Percent yield = (actual yield / theoretical yield) × 100. The actual yield is the mass of product collected in the experiment. The theoretical yield is the maximum calculated from stoichiometry. Percent yield is always ≤ 100% because actual experiments involve incomplete reactions, side reactions, and product losses during collection and purification.

What is a mole ratio and how is it used?

A mole ratio is the ratio of stoichiometric coefficients of two species in a balanced equation. For 2H₂ + O₂ → 2H₂O, the mole ratio of H₂O to H₂ is 2:2 = 1 — meaning exactly 1 mole of H₂O forms per mole of H₂ consumed. Mole ratios are used as conversion factors in Step 2 of every stoichiometry calculation: mol_wanted = mol_given × (coefficient_wanted / coefficient_given).

What are the 4 steps of gram-to-gram stoichiometry?

Step 1: Convert grams of given → moles of given (divide by molar mass). Step 2: Convert moles of given → moles of desired (multiply by mole ratio from balanced equation). Step 3: Convert moles of desired → grams of desired (multiply by molar mass). Step 4: Check significant figures. Combined: g_B = g_A × (1/MW_A) × (coefficient_B/coefficient_A) × MW_B.

What is the difference between limiting reagent and excess reagent?

The limiting reagent is completely consumed first and determines the maximum yield of product. The excess reagent is present in greater quantity than required by the mole ratio — some remains unreacted when the limiting reagent is exhausted. To find excess remaining: subtract moles consumed (= limiting_ratio × coefficient_excess) from initial moles, then convert to grams.

Related Calculators

Key Formulas

n = m / M Moles = mass ÷ molar mass

mole ratio = coeff_B / coeff_A From balanced equation

mol_B = mol_A × (b/a) Mole ratio conversion

LR: min(mol_i / coeff_i) Limiting reagent test

%yield = (actual/theoretical)×100 Percent yield formula

STP: 1 mol gas = 22.4 L At standard conditions

g_B = g_A/MW_A × (b/a) × MW_B Gram-to-gram combined

Quick Examples

2H₂+O₂→2H₂O | 5g H₂→H₂O

N₂+3H₂→2NH₃ | 14g N₂

CaCO₃→CO₂ | 100g

Limiting: N₂+3H₂→2NH₃

Gram→Gram: H₂→H₂O

Common MW Values

H₂2.016
O₂31.998
N₂28.014
H₂O18.015
NH₃17.031
CO₂44.009
NaCl58.443
CaCO₃100.086

Related Tools

Stoichiometry Calculator — Mole Ratios, Limiting Reagent & Yields

Stoichiometry Calculator

What Is Stoichiometry? — The Mole Ratio Method

Why Must the Equation Be Balanced?

The Universal 4-Step Gram-to-Gram Method

How to Solve Stoichiometry Problems — Step-by-Step

Example 1 — Simple Gram-to-Gram: How many grams of H₂O form from 5g H₂?

Example 2 — Polynomial Stoichiometry: CaCO₃ → CaO + CO₂ | 100g CaCO₃ → g CO₂?

Example 3 — Combustion Stoichiometry: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O | 44g C₃H₈ → g CO₂?

Limiting Reagent — How to Find It

Method: The Mole-to-Coefficient Ratio Test

Worked Example: N₂ + 3H₂ → 2NH₃ | 14g N₂ and 6g H₂

Theoretical Yield, Actual Yield, and Percent Yield

Why Is Actual Yield Less Than Theoretical?

Worked Example: N₂ + 3H₂ → 2NH₃

Stoichiometry Word Problems — Examples and Solutions

Word Problem 1 — Simple Gram-to-Gram (4-step method)

Word Problem 2 — Limiting Reagent

Word Problem 3 — Percent Yield

Word Problem 4 — Gas Volume at STP (22.4 L/mol)

Word Problem 5 — Solution Stoichiometry (Molarity)

Word Problem 6 — Multi-Step (Limiting Reagent + Percent Yield)

Frequently Asked Questions

Related Calculators

Pages

Newsletter

Stoichiometry Calculator — Mole Ratios, Limiting Reagent & Yields

Stoichiometry Calculator

What Is Stoichiometry? — The Mole Ratio Method

Why Must the Equation Be Balanced?

The Universal 4-Step Gram-to-Gram Method

How to Solve Stoichiometry Problems — Step-by-Step

Example 1 — Simple Gram-to-Gram: How many grams of H₂O form from 5g H₂?

Example 2 — Polynomial Stoichiometry: CaCO₃ → CaO + CO₂ | 100g CaCO₃ → g CO₂?

Example 3 — Combustion Stoichiometry: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O | 44g C₃H₈ → g CO₂?

Limiting Reagent — How to Find It

Method: The Mole-to-Coefficient Ratio Test

Worked Example: N₂ + 3H₂ → 2NH₃ | 14g N₂ and 6g H₂

Theoretical Yield, Actual Yield, and Percent Yield

Why Is Actual Yield Less Than Theoretical?

Worked Example: N₂ + 3H₂ → 2NH₃

Stoichiometry Word Problems — Examples and Solutions

Word Problem 1 — Simple Gram-to-Gram (4-step method)

Word Problem 2 — Limiting Reagent

Word Problem 3 — Percent Yield

Word Problem 4 — Gas Volume at STP (22.4 L/mol)

Word Problem 5 — Solution Stoichiometry (Molarity)

Word Problem 6 — Multi-Step (Limiting Reagent + Percent Yield)

Frequently Asked Questions

Related Calculators

Share This Tool

Pages

Newsletter