Stoichiometry and the Mole Concept ✏ SAT Chemistry

Rucete ✏ SAT Chemistry In a Nutshell

6. Stoichiometry and the Mole Concept

Stoichiometry is the foundation of quantitative chemistry. This chapter explains the mole concept, how to convert between mass, moles, and volume, and how to apply these ideas using dimensional analysis to solve chemical equations. These skills are essential for success on the SAT Chemistry test.

The Mole Concept

• A mole is a counting unit in chemistry, just like a dozen (12 items).
  1 mole = 6.02 × 10²³ particles (Avogadro’s number)
• One mole of any element has a mass equal to its atomic mass in grams.
• This relationship allows us to convert between atoms, molecules, and grams.
• Avogadro’s number comes from the number of atoms in exactly 12 g of carbon-12.

Molar Mass

• Molar mass = mass of 1 mole of a substance (in grams).
• For monatomic elements, molar mass = atomic mass in grams (e.g., Fe = 55.8 g/mol).
• For diatomic elements (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂), add up atomic masses (e.g., O₂ = 32.0 g/mol).
• For compounds: Add atomic masses of all atoms in the formula.
  • NH₃ → 14.0 + (3 × 1.0) = 17.0 g/mol
  • CaCO₃ → 40.1 + 12.0 + (3 × 16.0) = 100.1 g/mol
• For hydrates (e.g., CuSO₄·5H₂O): 63.5 + 32.1 + (4 × 16.0) + (5 × 18.0) = 249.6 g/mol

Molar Mass and Gas Volume

• Avogadro’s Law: Equal volumes of gases contain equal numbers of particles at the same T and P.
• At STP (0°C and 1 atm), 1 mole of any gas = 22.4 L → This is the molar volume of a gas.
• Use molar mass and molar volume as conversion factors in stoichiometry problems.

Using Dimensional Analysis

• Dimensional analysis = multiply by unit fractions to convert between quantities.
• Example: To find moles of Si in 4.30 g: 4.30 g × (1 mol / 28.1 g) = 0.153 mol
• Key idea: Units cancel and the desired unit remains.

Density and Molar Mass of Gases

• For gases at STP:
  • Density = molar mass ÷ 22.4 L
  • Molar mass = density × 22.4 L
• Example: If density = 1.25 g/L → molar mass = 1.25 × 22.4 = 28.0 g/mol
• Example: O₂ (32.0 g/mol) → density = 32.0 ÷ 22.4 = 1.43 g/L

Mole-Mole Stoichiometry

• Balanced equations provide mole ratios between reactants and products.
• Example: 2NaClO → 2NaCl + O₂ → 2 mol NaClO produces 1 mol O₂
• If given moles of one substance, use mole ratio to find moles of another.
• Example: 0.750 mol O₂ × (2 mol NaClO / 1 mol O₂) = 1.50 mol NaClO

Mass-Mass Stoichiometry

• Convert mass → moles → use mole ratio → convert moles → mass.
• Example: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
  Given 2.59 g CaCO₃, find g HCl and g CO₂ produced.
• Use molar masses and mole ratios in a three-step dimensional analysis.

Volume-Volume Stoichiometry (for Gases)

• At STP, mole ratios = volume ratios for gases.
• Example: N₂ + 3H₂ → 2NH₃
  To produce 0.400 L NH₃:
  N₂ = 0.400 × (1 L N₂ / 2 L NH₃) = 0.200 L
  H₂ = 0.400 × (3 L H₂ / 2 L NH₃) = 0.600 L
• Works as long as T and P are constant, even if not at STP.

Mass-Volume and Volume-Mass Stoichiometry

• Used when a reaction involves gases and solids/liquids together.
• You need both molar mass (for mass) and molar volume (22.4 L) for gases at STP.
• Example: Mg + 2HCl → MgCl₂ + H₂
  Given: 0.250 L H₂ at STP → What mass of Mg required?
  0.250 L H₂ × (1 mol / 22.4 L) × (1 mol Mg / 1 mol H₂) × (24.3 g / 1 mol) = 0.271 g Mg
• If reaction is not at STP, use Ideal Gas Law (PV = nRT) instead of 22.4 L.

Gas Reactions at Non-STP Conditions

• Example: 2H₂O₂ → 2H₂O + O₂
  Given: 5.00 g H₂O₂ → What volume of O₂ at 70°C and 1.25 atm?
  Step 1: Convert grams H₂O₂ → moles O₂
  Step 2: Use PV = nRT to find volume.
  • Use R = 0.0821 L·atm/mol·K
  • Convert T to Kelvin

Limiting Reactant Problems

• When quantities of both reactants are given, determine which one runs out first.
• Use stoichiometry to calculate the product from both reactants.
• The smaller result = theoretical yield, and that reactant = limiting reactant.
• Example: CH₄ + 2O₂ → CO₂ + 2H₂O
  Given: 15.0 g CH₄ and 15.0 g O₂ → Calculate CO₂ produced from each reactant → O₂ is limiting reactant
• Subtract used amount to find excess reactant left over.

Percent Yield

• Real reactions often produce less than theoretical maximum.
• Percent yield = (actual yield ÷ theoretical yield) × 100
• Example: 3.89 kg Al₂O₃ → actual yield = 1.95 kg Al → Calculate theoretical yield → Use to find percent yield

In a Nutshell

This chapter introduces the core of chemical calculations: the mole concept, molar mass, and dimensional analysis. It shows how to use stoichiometry to calculate masses, moles, and volumes of reactants and products. It also covers gas law applications, limiting reactants, and reaction efficiency. Mastering these tools allows accurate and practical problem-solving in chemistry.