Reaction Stoichiometry and Limiting Reactants

Review of Concepts

• Goal: Count particles to determine molecular formulas and do chemical reaction calculations.
• Method: Count moles of particles.
  • Mole: A specific number of particles, equal to the number of atoms in 12 grams of carbon.
  • Analogy: Like a dozen or a gross, a mole represents a specific quantity.
  • Equal moles imply equal particle numbers: One mole of carbon contains the same number of atoms as one mole of helium.

Molar Mass

• Definition: The mass of one mole of a substance.
• Carbon: 12 grams (by definition).
• Helium: 4 grams, based on the mass ratio of carbon to helium atoms (12:4).
• Molar mass in grams is numerically equal to the mass of a single atom/molecule in atomic mass units (amu).

Calculating Moles

• Formula: $\text{Number of moles} = \frac{\text{Mass of sample}}{\text{Mass of one mole}}$
• Significance: By weighing a sample, we can determine the number of particles it contains.

Importance of Mole Calculations

• Determining molecular formulas.
• Performing calculations for chemical reactions.
• Determining concentrations of solutions.

Example: Burning Acetylene

• Reaction: Acetylene (CxHy) + Oxygen → Carbon Dioxide + Water
• Goal: Determine the empirical formula of acetylene.

Counting Carbon Atoms

• Measure the mass of carbon dioxide produced.
• Calculate moles of carbon dioxide: $\text{Moles of } CO<em>2 = \frac{\text{Mass of } CO</em>2}{\text{Molar mass of } CO_2}$
  • Molar mass of CO2 = 12 (C) + 2 * 16 (O) = 44 grams/mole.
• Given data: 33.8 grams of $CO_2$ produced.
• Calculation: $\text{Moles of } CO_2 = \frac{33.8 \text{ grams}}{44 \text{ grams/mole}} = 0.76 \text{ moles}$
• Since each CO2 molecule contains one carbon atom, the number of moles of carbon equals the number of moles of carbon dioxide (0.76 moles).

Counting Hydrogen Atoms

• Measure the mass of water produced.
• Calculate moles of water: $\text{Moles of } H<em>2O = \frac{\text{Mass of } H</em>2O}{\text{Molar mass of } H_2O}$
  • Molar mass of H2O = 16 (O) + 2 * 1 (H) = 18 grams/mole.
• Given data: 6.9 grams of H2O produced.
• Calculation: $\text{Moles of } H_2O = \frac{6.9 \text{ grams}}{18 \text{ grams/mole}} = 0.38 \text{ moles}$
• Since each water molecule contains two hydrogen atoms, the number of moles of hydrogen is double the number of moles of water.
• Moles of hydrogen = 2 * 0.38 moles = 0.76 moles.

Determining Empirical Formula

• The molar ratio of carbon to hydrogen is 0.76:0.76, which simplifies to 1:1.
• Empirical formula of acetylene is CH, while the actual formula is C2H2.

Balanced Chemical Equations

• Atomic Molecular Theory: Atoms are conserved in a chemical reaction.
• The number and type of atoms in the reactants must equal the number and type of atoms in the products.

Example: Burning Propane

• Unbalanced equation: $C<em>3H</em>8 + O<em>2 \rightarrow CO</em>2 + H_2O$
• Balancing the equation:
  • Balance carbon: $C<em>3H</em>8 + O<em>2 \rightarrow 3CO</em>2 + H_2O$
  • Balance hydrogen: $C<em>3H</em>8 + O<em>2 \rightarrow 3CO</em>2 + 4H_2O$
  • Balance oxygen: $C<em>3H</em>8 + 5O<em>2 \rightarrow 3CO</em>2 + 4H_2O$
• Balanced equation: $C<em>3H</em>8 + 5O<em>2 \rightarrow 3CO</em>2 + 4H_2O$

Example: Copper and Nitric Acid

• Unbalanced reaction: $Cu + HNO<em>3 \rightarrow Cu(NO</em>3)<em>2 + NO</em>2 + H_2O$
• Balancing the equation:
  • $Cu + 4HNO<em>3 \rightarrow Cu(NO</em>3)<em>2 + 2NO</em>2 + 2H_2O$
• Verification: Count the number of each type of atom on both sides to ensure they are equal.

Chemical Stoichiometry (Chemical Algebra)

• Definition: Using balanced chemical equations to perform calculations involving the amounts of substances in a reaction.
• Example: Relating the mass of copper to the mass of nitrogen dioxide produced.

Stoichiometry Calculation

• Given: 10 grams of copper.
• Goal: Determine the mass of nitrogen dioxide produced.
• Steps:
  • Calculate moles of copper: $\text{Moles of } Cu = \frac{10 \text{ grams}}{63.55 \text{ grams/mole}} = 0.157 \text{ moles}$
  • Use the balanced equation to determine the mole ratio of copper to nitrogen dioxide (1:2).
  • Calculate moles of nitrogen dioxide produced: $\text{Moles of } NO_2 = 2 \times 0.157 \text{ moles} = 0.314 \text{ moles}$
  • Calculate the mass of nitrogen dioxide: $\text{Mass of } NO_2 = 0.314 \text{ moles} \times 46 \text{ grams/mole} = 14.44 \text{ grams}$

Limiting Reactant

• Scenario: Reactants may not be present in stoichiometric quantities.
• Definition: The reactant that is completely consumed first, limiting the amount of product formed.

Example: Copper and Limited Nitric Acid

• Given: 0.2 moles of nitric acid and 10 grams of copper (0.157 moles).
• Determine if nitric acid is the limiting reactant.
• Moles of nitric acid needed to react with all copper: $0.157 \text{ moles of } Cu \times 4 = 0.628 \text{ moles}$
• Since we only have 0.2 moles of nitric acid, it is the limiting reactant.

Calculating Product with Limiting Reactant

• Use the moles of the limiting reactant (nitric acid) to determine the moles of product (nitrogen dioxide).
• Mole ratio of nitric acid to nitrogen dioxide is 4:2 (or 2:1).
• Moles of nitrogen dioxide produced: $\text{Moles of } NO<em>2 = 0.2 \text{ moles of } HNO</em>3 \times \frac{1}{2} = 0.1 \text{ moles}$
• Calculate the mass of nitrogen dioxide produced: $\text{Mass of } NO_2 = 0.1 \text{ moles} \times 46 \text{ grams/mole} = 4.6 \text{ grams}$
• Conclusion: The amount of nitrogen dioxide produced is limited by the amount of nitric acid available.