👩🏽‍🔬Honors Chemistry Unit 5 – Chemical Quantities & Stoichiometry

Key Concepts and Definitions

• Stoichiometry studies the quantitative relationships between reactants and products in a chemical reaction
• Mole is the SI unit for measuring the amount of a substance, equal to $6.022 \times 10^{23}$ particles (atoms, molecules, or formula units)
• Avogadro's number represents the number of particles in one mole of a substance ($6.022 \times 10^{23}$)
• Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol)
  • Calculated by adding the atomic masses of all atoms in a compound
• Empirical formula represents the simplest whole-number ratio of atoms in a compound
• Molecular formula shows the actual number of atoms of each element in a molecule
• Limiting reactant is the reactant that is completely consumed first, determining the amount of product formed
• Theoretical yield is the maximum amount of product that can be obtained based on the balanced chemical equation and the limiting reactant
• Percent yield compares the actual yield to the theoretical yield, expressed as a percentage

The Mole and Avogadro's Number

• The mole is a convenient unit for measuring large quantities of particles (atoms, molecules, or formula units)
• One mole contains Avogadro's number of particles ($6.022 \times 10^{23}$)
  • This number was chosen to relate the mass of a substance to the number of particles it contains
• Avogadro's number is named after the Italian scientist Amedeo Avogadro, who proposed the concept of the mole
• The mole allows chemists to easily convert between the number of particles, mass, and volume of a substance
• Examples:
  • One mole of carbon atoms ($6.022 \times 10^{23}$ atoms) has a mass of 12 grams
  • One mole of water molecules ($6.022 \times 10^{23}$ molecules) has a mass of 18 grams
• The mole concept simplifies stoichiometric calculations by relating the number of particles to measurable quantities like mass and volume

Molar Mass and Molecular Weight

• Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol)
• Molecular weight is the sum of the atomic masses of all atoms in a molecule
  • For compounds, molar mass and molecular weight are numerically equivalent
• To calculate molar mass, add the atomic masses of all atoms in the chemical formula
  • Atomic masses can be found on the periodic table
• Examples:
  • The molar mass of water (H2O) is 18.02 g/mol (2 × 1.01 g/mol for H + 16.00 g/mol for O)
  • The molar mass of glucose (C6H12O6) is 180.16 g/mol (6 × 12.01 g/mol for C + 12 × 1.01 g/mol for H + 6 × 16.00 g/mol for O)
• Molar mass is used to convert between the mass and the number of moles of a substance
• Knowing the molar mass allows chemists to determine the number of moles in a given mass of a substance, or vice versa

Percent Composition and Empirical Formulas

• Percent composition is the mass percentage of each element in a compound
  • Calculated by dividing the mass of each element by the total mass of the compound and multiplying by 100%
• Empirical formula represents the simplest whole-number ratio of atoms in a compound
  • Determined by calculating the mole ratio of each element in the compound
• To find the empirical formula:
  1. Calculate the mass percentage of each element
  2. Convert the mass percentages to moles using molar masses
  3. Divide each mole value by the smallest mole value to obtain whole-number ratios
  4. If necessary, multiply the ratios by a common factor to obtain whole numbers
• Examples:
  • A compound containing 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen has an empirical formula of CH2O
  • The molecular formula of glucose (C6H12O6) can be derived from its empirical formula (CH2O) by multiplying each subscript by 6
• Percent composition and empirical formulas are useful for determining the relative amounts of elements in a compound and for comparing compounds with different molecular formulas but the same empirical formula

Balancing Chemical Equations

• A balanced chemical equation has equal numbers of each type of atom on both sides of the arrow
• To balance an equation:
  1. Identify the reactants and products
  2. Count the number of each type of atom on both sides
  3. Adjust the coefficients (not subscripts) to make the number of each type of atom equal on both sides
  4. Verify that the equation is balanced
• Examples:
  • The unbalanced equation

    CH4 + O2 → CO2 + H2O

    becomes balanced as

    CH4 + 2O2 → CO2 + 2H2O

  • The unbalanced equation

    Fe + O2 → Fe2O3

    becomes balanced as

    4Fe + 3O2 → 2Fe2O3

• Balancing equations is crucial for stoichiometric calculations, as it ensures that the law of conservation of mass is followed
• A balanced equation provides the relative numbers of moles of reactants and products, which is essential for determining the quantities of substances involved in a reaction

Stoichiometric Calculations

• Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction using the balanced equation
• Mole ratios from the balanced equation are used to convert between moles of reactants and products
• Steps for stoichiometric calculations:
  1. Write and balance the chemical equation
  2. Convert given quantities to moles using molar mass or other conversion factors
  3. Use mole ratios from the balanced equation to calculate the moles of the desired substance
  4. Convert moles of the desired substance to the requested unit (mass, volume, or particles)
• Examples:
  • For the reaction

    2H2 + O2 → 2H2O

    , if 4 moles of H2 react, 2 moles of H2O will be produced
  • If 5 grams of H2 react, the mass of H2O produced can be calculated using the molar mass of H2O (18.02 g/mol)
• Dimensional analysis is often used in stoichiometric calculations to ensure that units cancel out properly and the desired unit is obtained
• Stoichiometric calculations are essential for determining the amounts of reactants needed or products formed in a chemical reaction, which is crucial for industrial processes and laboratory experiments

Limiting Reactants and Percent Yield

• The limiting reactant is the reactant that is completely consumed first, limiting the amount of product that can be formed
• Excess reactants are the reactants that remain after the limiting reactant is depleted
• To determine the limiting reactant:
  1. Calculate the moles of each reactant
  2. Determine the mole ratio of the reactants from the balanced equation
  3. Divide the moles of each reactant by its coefficient in the balanced equation
  4. The reactant with the smallest mole ratio is the limiting reactant
• Theoretical yield is the maximum amount of product that can be obtained based on the balanced equation and the limiting reactant
• Actual yield is the amount of product actually obtained in a reaction
• Percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage
  • Calculated as: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
• Examples:
  • In the reaction

    2Al + 3Cl2 → 2AlCl3

    , if 4 moles of Al and 5 moles of Cl2 are used, Al is the limiting reactant
  • If the theoretical yield of AlCl3 is 10 grams and the actual yield is 8 grams, the percent yield is 80%
• Identifying the limiting reactant and calculating percent yield are important for optimizing reaction conditions and assessing the efficiency of a chemical process

Real-World Applications and Examples

• Stoichiometry has numerous applications in various fields, such as chemical engineering, pharmaceuticals, and environmental science
• In the production of ammonia (Haber-Bosch process), stoichiometry is used to determine the optimal ratio of nitrogen and hydrogen gases to maximize yield and minimize waste
• In the synthesis of pharmaceuticals, stoichiometric calculations ensure that the correct amounts of reactants are used to produce the desired active ingredients
• Environmental scientists use stoichiometry to study the effects of pollutants on air and water quality, such as calculating the amount of oxygen consumed by the decomposition of organic matter in water (biochemical oxygen demand)
• In the combustion of fuels, stoichiometry helps determine the amount of air needed for complete combustion and the composition of the exhaust gases
• Examples:
  • The production of 1 ton of ammonia (NH3) requires 0.82 tons of nitrogen (N2) and 0.18 tons of hydrogen (H2)
  • The complete combustion of 1 mole of octane (C8H18) requires 12.5 moles of oxygen (O2) and produces 8 moles of carbon dioxide (CO2) and 9 moles of water (H2O)
• Understanding stoichiometry enables chemists and engineers to optimize processes, minimize environmental impact, and produce desired products efficiently and cost-effectively