Stoichiometry Lessons | Chemistry

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The Mole

6.022 × 10²³

Avogadro's Number — particles per mole of any substance

A mole is 6.022 × 10²³ of anything — atoms, molecules, ions, elephants. It's just a number, like a dozen = 12. The molar mass (g/mol) of any element equals its atomic mass from the periodic table. One mole of carbon = 12.011 g. One mole of water (H₂O) = 18.015 g. Everything in stoichiometry flows through moles.

📖 Full Lesson → Difficulty: Beginner

Avogadro's number

6.022 × 10²³ — the number of particles in one mole. Named for Amedeo Avogadro. Also written as Nₐ.

Molar mass

Mass of one mole in grams. For elements: read from periodic table. For compounds: sum of all atomic masses × their subscripts.

Conversions

Mass → moles: divide by molar mass. Moles → mass: multiply by molar mass. Moles → particles: multiply by 6.022×10²³.

Molar volume

At STP (0°C, 1 atm), one mole of any ideal gas occupies 22.4 L. At STP (25°C, 1 bar — new IUPAC): 24.8 L.

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🃏 The Mole

6.022 × 10²³

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🃏 Answer

Avogadro's number — 6.022 × 10²³ — the number of particles in one mole. Named for Amedeo Avogadro. Also written as Nₐ.

Molar mass — Mass of one mole in grams. For elements: read from periodic table. For compounds: sum of all atomic masses × their subscripts.

Conversions — Mass → moles: divide by molar mass. Moles → mass: multiply by molar mass. Moles → particles: multiply by 6.022×10²³.

Molar volume — At STP (0°C, 1 atm), one mole of any ideal gas occupies 22.4 L. At STP (25°C, 1 bar — new IUPAC): 24.8 L.

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Dimensional Analysis

GFGW (G=Given what you start with, F=Find what you need, G=Go set up conversion factors, W=Work it out)

Given · Find · Go (conversion factors) · Work it out

GFGW is the four-step approach to any stoichiometry problem. Given (what you're starting with) · Find (what you need) · Go (set up conversion factors so unwanted units cancel) · Work it out. Units that appear on top and bottom cancel — if your units work out, your math is right.

📖 Full Lesson → Difficulty: Beginner

Unit cancellation

Write each conversion as a fraction. Put the unit you want to cancel on the bottom. It cancels with the same unit on top in the next step.

Mole ratio

The coefficients in a balanced equation ARE the mole ratio. 2H₂ + O₂ → 2H₂O means 2 mol H₂ per 1 mol O₂ per 2 mol H₂O.

Chain approach

Mass A → moles A (÷ molar mass A) → moles B (× mole ratio) → mass B (× molar mass B). Always go through moles.

Check units

If your final units are what you were asked for — your setup is correct. Wrong units = wrong setup, not wrong math.

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🃏 Dimensional Analysis

GFGW

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🃏 Answer

GFGW (G=Given what you start with, F=Find what you need, G=Go set up conversion factors, W=Work it out)

Unit cancellation — Write each conversion as a fraction. Put the unit you want to cancel on the bottom. It cancels with the same unit on top in the next step.

Mole ratio — The coefficients in a balanced equation ARE the mole ratio. 2H₂ + O₂ → 2H₂O means 2 mol H₂ per 1 mol O₂ per 2 mol H₂O.

Chain approach — Mass A → moles A (÷ molar mass A) → moles B (× mole ratio) → mass B (× molar mass B). Always go through moles.

Check units — If your final units are what you were asked for — your setup is correct. Wrong units = wrong setup, not wrong math.

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Limiting Reagent

LESS = LIMITING

The reagent that produces LESS product is the limiting reagent

The limiting reagent is the reactant that runs out first — it determines how much product forms. Strategy: convert each reactant to moles of product using the mole ratio. The reactant that gives you LESS product is the limiting reagent. The other is in excess. Calculate actual yield from the limiting reagent only.

📖 Full Lesson → Difficulty: Intermediate

Step 1

Convert all given reactant masses to moles using their molar masses.

Step 2

Use the mole ratio from the balanced equation to find how many moles of product each reactant would produce IF it were completely consumed.

Step 3

The reactant that gives LESS product is limiting. Use that smaller amount to calculate theoretical yield.

Excess reagent

To find how much excess reagent remains: calculate how much was consumed by the limiting reagent, then subtract from what you started with.

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🃏 Limiting Reagent

Limiting reagent — how do you find it?

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🃏 Answer

LESS = LIMITING

Step 1 — Convert all given reactant masses to moles using their molar masses.

Step 2 — Use the mole ratio from the balanced equation to find how many moles of product each reactant would produce IF it were completely consumed.

Step 3 — The reactant that gives LESS product is limiting. Use that smaller amount to calculate theoretical yield.

Excess reagent — To find how much excess reagent remains: calculate how much was consumed by the limiting reagent, then subtract from what you started with.

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Percent Yield

% Yield = (Actual ÷ Theoretical) × 100

Actual yield is what you got · Theoretical yield is what you calculated

Percent yield measures how close your actual result was to the theoretical prediction. % Yield = (Actual ÷ Theoretical) × 100. Actual yield = what you measured in the lab. Theoretical yield = calculated from the limiting reagent. Real reactions rarely hit 100% due to side reactions, incomplete reactions, or product lost in transfer.

📖 Full Lesson → Difficulty: Beginner

Theoretical yield

The maximum amount of product possible — calculated from the limiting reagent using stoichiometry. Assumes 100% conversion.

Actual yield

What you actually collected in the lab. Always ≤ theoretical yield. Given in the problem — you don't calculate it.

Why <100%?

Side reactions produce unwanted products · Reaction doesn't go to completion · Product lost during filtration or transfer · Measurement error.

Can't exceed 100%

If % yield > 100%, something is wrong — product is impure (contains water or other substances) or measurement error occurred.

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🃏 Percent Yield

Percent yield — the formula?

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🃏 Answer

% Yield = (Actual ÷ Theoretical) × 100

Theoretical yield — The maximum amount of product possible — calculated from the limiting reagent using stoichiometry. Assumes 100% conversion.

Actual yield — What you actually collected in the lab. Always ≤ theoretical yield. Given in the problem — you don't calculate it.

Why <100%? — Side reactions produce unwanted products · Reaction doesn't go to completion · Product lost during filtration or transfer · Measurement error.

Can't exceed 100% — If % yield > 100%, something is wrong — product is impure (contains water or other substances) or measurement error occurred.

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Empirical Formula

PGRS (P=Percent to Grams, G=Grams to moles, R=Ratio by dividing by smallest, S=Simplify to whole numbers)

Percent → Grams → Ratio (moles) → Simplify

PGRS: assume 100g sample so Percent becomes Grams → divide each by molar mass to get moles → find the Ratio by dividing all by the smallest → Simplify to whole numbers (multiply if needed). Molecular formula = empirical formula × n, where n = molar mass ÷ empirical formula mass.

📖 Full Lesson → Difficulty: Intermediate

Step 1: % → g

Assume 100 g sample. Then % becomes grams directly. 40% C = 40 g C. 6.7% H = 6.7 g H. 53.3% O = 53.3 g O.

Step 2: g → mol

Divide each mass by the element's molar mass. 40g C ÷ 12 g/mol = 3.33 mol C.

Step 3: find ratio

Divide all mole values by the smallest. If ratios aren't whole numbers, multiply all by 2, 3, 4, or 5 to clear decimals.

Molecular formula

n = (given molar mass) ÷ (empirical formula mass). Multiply empirical subscripts by n. CH₂O with n=2 → C₂H₄O₂.

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🃏 Empirical Formula

PGRS

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🃏 Answer

PGRS (P=Percent to Grams, G=Grams to moles, R=Ratio by dividing by smallest, S=Simplify to whole numbers)

Step 1: % → g — Assume 100 g sample. Then % becomes grams directly. 40% C = 40 g C. 6.7% H = 6.7 g H. 53.3% O = 53.3 g O.

Step 2: g → mol — Divide each mass by the element's molar mass. 40g C ÷ 12 g/mol = 3.33 mol C.

Step 3: find ratio — Divide all mole values by the smallest. If ratios aren't whole numbers, multiply all by 2, 3, 4, or 5 to clear decimals.

Molecular formula — n = (given molar mass) ÷ (empirical formula mass). Multiply empirical subscripts by n. CH₂O with n=2 → C₂H₄O₂.

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Solution Concentration

M = mol ÷ L

Molarity = moles of solute ÷ liters of solution

Molarity (M) is the most common concentration unit in chemistry. M = moles of solute ÷ liters of solution. A 2M NaCl solution has 2 moles of NaCl per liter. Key formula triangle: moles = M × L · M = mol ÷ L · L = mol ÷ M. For dilutions: M₁V₁ = M₂V₂ — moles of solute are conserved.

📖 Full Lesson → Difficulty: Intermediate

Molarity triangle

Cover what you need: moles = M × L. M = mol ÷ L. L = mol ÷ M. Always use LITERS, not mL (convert first).

Dilution formula

M₁V₁ = M₂V₂. Concentration × volume before = concentration × volume after. Moles of solute don't change when you add water.

Making solutions

Calculate moles needed → convert to grams → dissolve in less than final volume of water → bring to final volume in volumetric flask.

Solution stoichiometry

moles = M × L. Use this to get moles from solution problems, then proceed with normal stoichiometry through the mole ratio.

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🃏 Solution Concentration

Molarity — the formula?

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🃏 Answer

Molarity triangle — Cover what you need: moles = M × L. M = mol ÷ L. L = mol ÷ M. Always use LITERS, not mL (convert first).

Dilution formula — M₁V₁ = M₂V₂. Concentration × volume before = concentration × volume after. Moles of solute don't change when you add water.

Making solutions — Calculate moles needed → convert to grams → dissolve in less than final volume of water → bring to final volume in volumetric flask.

Solution stoichiometry — moles = M × L. Use this to get moles from solution problems, then proceed with normal stoichiometry through the mole ratio.

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Gas Stoichiometry

PV = nRT

Pressure · Volume = moles · R (gas constant) · Temperature

PV = nRT links all gas properties in one equation. Rearrange to find moles: n = PV ÷ RT. Use R = 0.08206 L·atm/mol·K. Always convert temperature to Kelvin (K = °C + 273.15) and pressure to atm. Once you have moles, proceed with normal stoichiometry using the balanced equation's mole ratios.

📖 Full Lesson → Difficulty: Intermediate

Variables

P = pressure (atm) · V = volume (L) · n = moles · R = 0.08206 L·atm/mol·K · T = temperature (K, not °C).

STP shortcut

At STP (0°C, 1 atm): 1 mole of ideal gas = 22.4 L. Skip PV=nRT and use 22.4 L/mol directly as a conversion factor.

Molar mass from gas

M = mRT/PV or M = dRT/P where d = density in g/L. Useful for identifying unknown gases from density measurements.

Dalton's law

Total pressure = sum of partial pressures. Pₜₒₜₐₗ = P₁ + P₂ + P₃... Each gas behaves independently in a mixture.

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🃏 Gas Stoichiometry

PV = nRT — what does each term mean?

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🃏 Answer

Variables — P = pressure (atm) · V = volume (L) · n = moles · R = 0.08206 L·atm/mol·K · T = temperature (K, not °C).

STP shortcut — At STP (0°C, 1 atm): 1 mole of ideal gas = 22.4 L. Skip PV=nRT and use 22.4 L/mol directly as a conversion factor.

Molar mass from gas — M = mRT/PV or M = dRT/P where d = density in g/L. Useful for identifying unknown gases from density measurements.

Dalton's law — Total pressure = sum of partial pressures. Pₜₒₜₐₗ = P₁ + P₂ + P₃... Each gas behaves independently in a mixture.

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Combustion Analysis

CHON (C=Carbon→CO₂, H=Hydrogen→H₂O, O=Oxygen by difference, N=Nitrogen→N₂) → CO₂ + H₂O

Carbon → CO₂ · Hydrogen → H₂O · Oxygen → by difference · Nitrogen → N₂

Combustion analysis burns an organic compound and measures the CO₂ and H₂O produced. CHON: all Carbon ends up in CO₂ · all Hydrogen ends up in H₂O · Oxygen in the original compound is found by difference · Nitrogen forms N₂. From CO₂ get moles C; from H₂O get moles H; then find empirical formula.

📖 Full Lesson → Difficulty: Advanced

Carbon

Moles C = moles CO₂ (1:1 ratio). Mass C = moles C × 12.011 g/mol.

Hydrogen

Moles H = 2 × moles H₂O (each water has 2 H). Mass H = moles H × 1.008 g/mol.

Oxygen by difference

Mass O = mass of sample − mass C − mass H. Then moles O = mass O ÷ 16.00. Only if oxygen is present in the compound.

Then use PGRS

Once you have moles of each element, apply the PGRS method to find the empirical formula from the mole ratios.

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🃏 Combustion Analysis

CHON

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🃏 Answer

CHON (C=Carbon→CO₂, H=Hydrogen→H₂O, O=Oxygen by difference, N=Nitrogen→N₂) → CO₂ + H₂O

Carbon — Moles C = moles CO₂ (1:1 ratio). Mass C = moles C × 12.011 g/mol.

Hydrogen — Moles H = 2 × moles H₂O (each water has 2 H). Mass H = moles H × 1.008 g/mol.

Oxygen by difference — Mass O = mass of sample − mass C − mass H. Then moles O = mass O ÷ 16.00. Only if oxygen is present in the compound.

Then use PGRS — Once you have moles of each element, apply the PGRS method to find the empirical formula from the mole ratios.

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Concentration Units

MM·PP·XX (M=Molarity, M=Molality, P=Parts per million, P=Parts per billion, X=mole fraction X, X=mass percent %)

Molarity · Molality · Parts per million · Parts per billion · Mole fraction · Mass percent

MM·PP·XX: Molarity (M = mol/L — most common) · Molality (m = mol/kg solvent — used for colligative properties) · Parts per million (mg/L — used for very dilute solutions like drinking water) · Parts per billion · Mole fraction (Xₐ = molₐ/moltotal — used in gas laws) · Mass percent (g solute/g solution × 100).

📖 Full Lesson → Difficulty: Advanced

Molarity (M)

mol solute / L solution. Changes with temperature (volume expands). Most used in stoichiometry and titrations.

Molality (m)

mol solute / kg solvent. Temperature-independent (mass doesn't change). Used for boiling point elevation and freezing point depression.

Mole fraction (X)

Xₐ = moles A / total moles. Dimensionless. Xₐ + X_b + ... = 1. Used in Raoult's law for vapor pressure of solutions.

ppm and ppb

ppm = mg solute / L solution (for water). ppb = μg/L. Used for trace contaminants in water, air, and environmental analysis.

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🃏 Concentration Units

Concentration units — MM·PP·XX

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🃏 Answer

MM·PP·XX (M=Molarity, M=Molality, P=Parts per million, P=Parts per billion, X=mole fraction X, X=mass percent %)

Molarity (M) — mol solute / L solution. Changes with temperature (volume expands). Most used in stoichiometry and titrations.

Molality (m) — mol solute / kg solvent. Temperature-independent (mass doesn't change). Used for boiling point elevation and freezing point depression.

Mole fraction (X) — Xₐ = moles A / total moles. Dimensionless. Xₐ + X_b + ... = 1. Used in Raoult's law for vapor pressure of solutions.

ppm and ppb — ppm = mg solute / L solution (for water). ppb = μg/L. Used for trace contaminants in water, air, and environmental analysis.

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🎓 Common Exam Questions

Memory tricks for VSEPR, polarity & intermolecular forces

Memory Tricks