Stoichiometry: Laws, Calculations, Reactions & Mole Concept

TAKEAWAYS:

• Stoichiometry connects reactants and products using balanced equations and mole ratios.
• Moles are the bridge between grams (mass) and particles (atoms/molecules).
• Empirical formulas show simplest ratios; molecular formulas show actual numbers.
• Limiting reagents determine how much product forms in real reactions.
• Percentage yield compares theoretical vs. actual product.
• Gas laws (PV=nRT) help solve stoichiometry problems involving gases.

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1. The Law of Conservation of Mass

Chemical reactions follow Lavoisier’s Law: mass is neither created nor destroyed—only rearranged.

Example: If 4g of hydrogen (H₂) reacts with 32g of oxygen (O₂), how much water (H₂O) forms?

• Step 1: Write the balanced equation: 2H₂ + O₂ → 2H₂O
• Step 2: Calculate total mass: 4g (H₂) + 32g (O₂) = 36g
• Step 3: Mass of water formed = 36g (conservation of mass).

NEB-style question: "In the reaction 2Mg + O₂ → 2MgO, 12g of Mg reacts with 8g of O₂. Calculate the mass of MgO formed and identify the limiting reagent." Answer:

• Moles of Mg = 12g / 24g/mol = 0.5 mol
• Moles of O₂ = 8g / 32g/mol = 0.25 mol
• Limiting reagent: O₂ (requires 0.5 mol Mg but only 0.25 mol O₂ is available).
• Mass of MgO = (0.25 mol O₂ × 2 × 40.3 g/mol MgO) = 20.15g.

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2. The Mole Concept

A mole (mol) is the amount of substance containing 6.022 × 10²³ particles (Avogadro’s number).

Key relationships:

Example: Calculate moles in 5g of calcium (Ca, atomic mass = 40 g/mol).

• Step 1: Use formula: moles = mass / molar mass
• Step 2: Plug in values: moles = 5g / 40 g/mol = 0.125 mol

NEB-style question: "How many molecules are in 0.5 mol of CO₂?" Answer: 0.5 mol × 6.022 × 10²³ molecules/mol = 3.011 × 10²³ molecules

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3. Empirical and Molecular Formulas

• Empirical formula: Simplest whole-number ratio of atoms (e.g., CH₂ for C₂H₄).
• Molecular formula: Actual number of atoms (e.g., C₂H₄).

Steps to find empirical formula:

1. Assume 100g sample → % = g.
2. Convert g → moles (moles = g / molar mass).
3. Divide by smallest moles → ratio.
4. Multiply by integer if needed.

Example: A compound is 40% C, 6.7% H, 53.3% O by mass.

• Step 1: Assume 100g → 40g C, 6.7g H, 53.3g O.
• Step 2: Moles:
  • C = 40g / 12 g/mol = 3.33 mol
  • H = 6.7g / 1 g/mol = 6.7 mol
  • O = 53.3g / 16 g/mol = 3.33 mol
• Step 3: Divide by smallest (3.33):
  • C: 1, H: 2, O: 1 → CH₂O (empirical formula).

Molecular formula: If molar mass = 90 g/mol, empirical mass = 30 g/mol → multiply by 3 → C₃H₆O₃.

NEB-style question: "A compound has 60% C and 40% H with a molar mass of 30 g/mol. Find its empirical and molecular formulas." Answer:

• Empirical: CH₃ (molar mass = 15 g/mol).
• Molecular: (CH₃)₂ = C₂H₆.

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4. Balancing Chemical Equations

Balanced equations show conservation of mass and law of definite proportions.

Rules:

1. Count atoms on both sides.
2. Use coefficients (never change subscripts).
3. Start with the most complex molecule.
4. Check for smallest whole numbers.

Example: Balance Fe + O₂ → Fe₂O₃.

• Step 1: Start with Fe₂O₃ → need 2 Fe and 3 O.
• Step 2: Add 2 Fe on left, but O₂ has 2 O → need 3/2 O₂.
• Step 3: Multiply all by 2 to eliminate fractions: 4Fe + 3O₂ → 2Fe₂O₃.

NEB-style question: "Balance: Al + HCl → AlCl₃ + H₂." Answer: 2Al + 6HCl → 2AlCl₃ + 3H₂

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5. Stoichiometry: Mole Ratios and Calculations

Use coefficients in balanced equations as mole ratios.

Example: How many grams of H₂O form from 8g of O₂ in 2H₂ + O₂ → 2H₂O?

• Step 1: Moles of O₂ = 8g / 32 g/mol = 0.25 mol.
• Step 2: Mole ratio (O₂:H₂O) = 1:2 → 0.25 mol O₂ produces 0.5 mol H₂O.
• Step 3: Mass of H₂O = 0.5 mol × 18 g/mol = 9g.

Limiting Reagent Example: In N₂ + 3H₂ → 2NH₃, 10g N₂ and 5g H₂ react.

• Moles N₂ = 10g / 28 g/mol = 0.357 mol.
• Moles H₂ = 5g / 2 g/mol = 2.5 mol.
• Limiting reagent: N₂ (requires 1.071 mol H₂, but only 2.5 mol available).
• NH₃ produced: 0.357 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 0.714 mol → 12.85g NH₃.

NEB-style question: "In 2SO₂ + O₂ → 2SO₃, 32g SO₂ reacts with 16g O₂. Find the limiting reagent and mass of SO₃ formed." Answer:

• Limiting reagent: O₂ (0.5 mol O₂ needs 1 mol SO₂, but only 0.5 mol O₂ is available).
• Mass of SO₃ = 0.5 mol O₂ × (2 mol SO₃ / 1 mol O₂) × 80 g/mol = 80g SO₃.

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6. Percentage Yield

Theoretical yield: Maximum product possible (from stoichiometry). Actual yield: What you get in the lab. Percentage yield = (Actual / Theoretical) × 100%.

Example: In C + O₂ → CO₂, 12g C produces 30g CO₂ theoretically but only 25g in lab.

• Percentage yield = (25g / 30g) × 100% = 83.3%.

NEB-style question: "In 2Mg + O₂ → 2MgO, 24g Mg produces 40g MgO. Calculate % yield." Answer:

• Theoretical yield: 24g Mg × (80.6 g/mol MgO / 24 g/mol Mg) = 80.6g MgO.
• % yield = (40g / 80.6g) × 100% = 49.6%.

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7. Gas Stoichiometry (Using PV = nRT)

For gases, use molar volume (22.4 L/mol at STP) or ideal gas law.

Example: What volume of O₂ (at STP) reacts with 5.6 L H₂ in 2H₂ + O₂ → 2H₂O?

• Step 1: Mole ratio (H₂:O₂) = 2:1 → 5.6 L H₂ needs 2.8 L O₂.
• Answer: 2.8 L O₂.

Using PV = nRT: At 273K and 1 atm, 1 mol gas = 22.4 L.

• Example: Find moles of CO₂ in 11.2 L at STP. n = PV/RT = (1 atm × 11.2 L) / (0.0821 L·atm·K⁻¹·mol⁻¹ × 273 K) = 0.5 mol.

NEB-style question: "What volume of CO₂ (at STP) is produced from 2 mol C in C + O₂ → CO₂?" Answer:

• 2 mol C → 2 mol CO₂ → 2 × 22.4 L = 44.8 L CO₂.

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8. Applications of Stoichiometry

Real-world example:

• Fertilizer production: N₂ + 3H₂ → 2NH₃ (Haber process).
  • 1 ton N₂ + 3 tons H₂ → 2 tons NH₃ (ammonia fertilizer).

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Exam Tip

1. Always balance equations before calculations.
2. Show all steps—examiners reward method marks.
3. Unit consistency: Convert grams → moles → moles → grams.
4. Limiting reagent: Compare mole ratios, not masses.
5. Gas problems: Use STP (22.4 L/mol) or PV = nRT.
6. Percentage yield: Theoretical yield comes from stoichiometry.

Common mistakes to avoid:

• Forgetting to balance equations.
• Mixing up empirical/molecular formulas.
• Ignoring significant figures in answers.
• Misapplying mole ratios (e.g., using coefficients incorrectly).

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