Stoichiometry Memory Tricks | Chemistry

⚖️ Memory Tricks

Stoichiometry — 9 Memory Tricks

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The Mole

6.022 × 10²³

Avogadro's Number — particles per mole of any substance

The Mole — Chemistry's Counting Unit

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.

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.

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

Dimensional Analysis — The Universal Strategy

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.

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.

Limiting Reagent

LESS = LIMITING

The reagent that produces LESS product is the limiting reagent

Finding 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.

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.

Percent Yield

% Yield = (Actual ÷ Theoretical) × 100

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

Percent Yield — How Efficient Was Your Reaction?

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.

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.

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

Finding Empirical & Molecular Formulas

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.

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.

Solution Concentration

M = mol ÷ L

Molarity = moles of solute ÷ liters of solution

Molarity — Concentration of Solutions

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.

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.

Gas Stoichiometry

PV = nRT

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

Ideal Gas Law — Connecting Gas Volume to Moles

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.

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.

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 — Finding a Formula from Burning

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.

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.

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

6 Concentration Units — When to Use Each

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).

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.

🎓 Common Exam Questions

Q: What does GFGW stand for and walk through a dimensional analysis problem using it?

A: GFGW = Given (what you're starting with), Find (what you need to find), Go (set up conversion factors as fractions), Work it out (cancel units). Example: how many grams of NaCl in 0.250 mol? Given: 0.250 mol NaCl. Find: grams. Go: molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol. Work: 0.250 mol × (58.44 g / 1 mol) = 14.6 g. The key is setting up conversion factors so unwanted units cancel. Chain them: grams → moles → molecules: 5.85 g NaCl × (1 mol/58.44 g) × (6.022×10²³/1 mol) = 6.02×10²² formula units.

Q: What does PGRS stand for and work through an empirical formula problem?

A: PGRS = Percent → Grams → Ratio (moles) → Simplify. A compound is 40.0% C, 6.7% H, 53.3% O. Step 1 (Percent→Grams): assume 100g sample → 40.0g C, 6.7g H, 53.3g O. Step 2 (Grams→moles): C: 40.0/12.01 = 3.33 mol; H: 6.7/1.008 = 6.65 mol; O: 53.3/16.00 = 3.33 mol. Step 3 (Ratio): divide by smallest (3.33): C=1, H=2, O=1. Step 4 (Simplify): empirical formula = CH2O. To find molecular formula: divide molar mass by empirical formula mass. If MM = 180 g/mol: 180/30 = 6 → molecular formula = C6H12O6 (glucose).

Q: Explain the limiting reagent concept and walk through a calculation.

A: The limiting reagent (LESS = LIMITING) is the reactant that runs out first, determining the maximum amount of product. Method: convert both reactants to moles of product using the stoichiometric ratio — whichever gives LESS product is limiting. Example: 5.0g H2 + 32.0g O2 → H2O. Equation: 2H2 + O2 → 2H2O. Moles: H2 = 5.0/2.016 = 2.48 mol; O2 = 32.0/32.00 = 1.00 mol. Convert to mol H2O: from H2: 2.48 mol × (2/2) = 2.48 mol H2O; from O2: 1.00 mol × (2/1) = 2.00 mol H2O. O2 gives LESS product → O2 is limiting reagent. Actual yield based on 1.00 mol O2 = 2.00 mol H2O = 36.0 g.

Q: What does MM·PP·XX stand for and when is each concentration unit used?

A: MM·PP·XX = Molarity (M), Molality (m), Parts per million (ppm), Parts per billion (ppb), Mole fraction (X), Mass percent (%). Molarity (M = mol/L): most common in lab; temperature-dependent (volume changes with temp). Molality (m = mol/kg solvent): used for colligative properties (boiling point elevation, freezing point depression) because it is temperature-independent. ppm (mg/L or mg/kg): used for trace contaminants in water/food/air. ppb: even more dilute solutions. Mole fraction (X = mol of component / total mol): used in gas mixtures (Dalton's law) and vapor pressure calculations. Mass percent (g solute/g solution × 100): used in concentrated solutions and industry.

Q: Explain CHON combustion analysis — how do you determine the molecular formula?

A: CHON: Carbon → CO2, Hydrogen → H2O, Oxygen → by difference (in original compound), Nitrogen → N2 gas. All carbon in the original sample ends up in CO2; all hydrogen ends up in H2O. Steps: (1) Measure masses of CO2 and H2O produced. (2) Find mass of C: mass CO2 × (12.011/44.010). Find mass of H: mass H2O × (2.016/18.015). (3) Find mass of O = original sample mass − mass C − mass H (− mass N if nitrogen present). (4) Convert masses to moles. (5) Find smallest whole-number ratio (empirical formula using PGRS). (6) Determine molecular formula using molar mass. This technique is fundamental to organic chemistry structure determination.

Memory tricks for moles, limiting reagents & percent yield