Stoichiometry and Mole Calculations

Stoichiometry

• Stoichiometry is the quantitative study of reactants and products in a chemical reaction.
• "Stochio" means element in Greek, and "metry" means measurement.
• Stoichiometric coefficients in a chemical equation represent the number of moles of each substance.

Mole Method

• Stoichiometric coefficients in a chemical equation represent the number of moles of each substance.
• One mole of any pure substance contains Avogadro's number of particles.
• Avogadro's number (N_A) = 6.0221415 x 10²³

Mole

• A mole (mol) is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in 12 grams of carbon-12.
• Molar mass is the mass (in grams or kilograms) of one mole of a substance.
• Atomic mass (or atomic weight) is the mass of an atom, expressed in atomic mass units (amu).

Example Calculation

• Calculate the moles of Zn in 23.3 g of Zn:
• Molar mass of Zn = 65.39 g/mol
• Moles of Zn = 23.3 g / 65.39 g/mol = 0.356 mol

Example Calculation 2

• Calculate the number of atoms in 16.3 g of S:
• Molar mass of S = 32.07 g/mol
• Moles of S = 16.3 g / 32.07 g/mol = 0.508 mol
• Number of atoms = 0.508 mol x 6.022 x 10²³ atoms/mol = 3.06 x 10²³ atoms

Molecular Mass

• Molecular mass is the sum of the atomic masses of all atoms in a molecule.
• Calculate the molecular mass of SO₂:
• Molecular mass of SO₂ = (32.07 amu) + 2(16.00 amu) = 64.07 amu
• Calculate the molar mass using the molecular mass:
• The molar mass of a compound is numerically equal to its molecular mass in amu. For example, the molar mass of SO2 is 64.07 g/mol.

Molar Mass Calculation Example

• Calculate the moles of CH4 in 6.07 g of CH4:
• Molar mass of CH4 = 16.04 g/mol
• Moles of CH4 = 6.07g / 16.04 g/mol = 0.378 mol

Percent Composition

• The percentage composition of an element in a compound is the mass percent of that element in the compound.
• Calculate the percent composition of H in H₃PO₄:
• Percent composition of H = (3 x 1.008 g/mol) / (97.99 g/mol) x 100% = 3.09%

Stoichiometry of Amounts of Reactants and Products

• Coefficients in a balanced chemical reaction represent the mole ratio of reactants and products.
• Example:
• In the reaction 4NH₃ + 5O₂ → 6H₂O + 4NO, the mole ratio between NH₃ and H₂O is 4:6, which can simplify to 2:3 .

Limiting Reactants

• Limiting reagent: The reactant that is completely consumed in a chemical reaction. Determines the maximum amount of product.
• Excess reagents: The reactants that are present in greater quantity than is required to react completely with the limiting reagent.

Sample Problem Calculation (Limiting Reagent)

• Given 8 moles of NO and 7 moles of O₂, determine the limiting reagent for the reaction 2NO + O₂ → 2NO₂
• Calculate moles of NO₂ from NO: 8mol NO x (2 mol NO₂/2 mol NO) = 8 mol NO₂
• Calculate moles of NO₂ from O₂: 7 mol O₂ x (2 mol NO₂/1 mol O₂) = 14 mol NO₂
• The smaller value is 8, meaning NO is the limiting reagent.

Reaction Yield

• Theoretical yield: The maximum amount of product that can be produced given a complete reaction of the limiting reagent.
• Actual yield: The amount of product that is actually isolated or obtained from a reaction often less than the theoretical yield.

Percent Yield

• % yield = (actual yield / theoretical yield) x 100%

Stoichiometry Problems Examples

• The given examples include problems involving calculating moles, grams, substances present in a chemical reaction, using different amounts of reagents present, and calculating the percent yield.

Additional Concepts

• Stoichiometry calculations often involve converting between moles, mass, and number of particles.