Stoichiometric Relationships (The Mole)

Seven Molecular Elements (Diatomic):

  • Hydrogen (H2)
  • Nitrogen (N2)
  • Oxygen (O2)
  • Fluorine (F2)
  • Chlorine (Cl2)
  • Iodine (I2)
  • Bromine (Br2)

Seven Complex Ions:

Introduction to Chemical Change

Types of Chemical Reactions:

Synthesis: 2 or more reactants combine and form one product (A + B → AB)

Decomposition: 1 reactant breaks down into 2 or more products (AB → A+B)

Single Displacement Reaction: More reactive element replaces less reactive element in a compound

(A + XY → AY + X)

Double Displacement Reaction: 2 compounds react to form 2 new compounds (usually involves aqueous solutions and 1 product is a precipitate)

(AB + XY → XB + AY)

The Mole Concept

Mole: 6.02 x 10^23

Avogadro’s Number (NA) or L = 6.02 x 10^23

Number of Fundamental Particles (N)

Number of Moles (mol) Present (n)

Formula: N = nNA

  1. To calculate # of molecules when mol is given, multiply mol by 6.02 x 10^23 (N = nNA)
  2. To calculate # of atoms when mol and molecules are given, multiply number of atoms by N of previous number

Molar Mass (g/mol)

Mass of atoms relative to carbon (12 amu)

Mole: Number of atoms in 12 grams of Carbon

Refer to Periodic Table

Example: Molar Mass of H2O

Relating Mass and Moles

Calculate mass from the moles, and moles from the mass using MOLAR MASS

Problem Type #1

Problem Type #2:

Percentage Composition of a Compound

Problem Type #3:

Empirical and Molecular Formula

Empirical Formula: Simplest Ratio

Example: Benzene (C6H6 → CH)

Molecular Formula: Actual number of atoms of each element making up molecule

Problem Type #4:

Finding Empirical Formula

  1. Convert percentage to mass (assuming 100g of compound)
  2. Find # of moles (n) using mass (m) and molar mass M [n = m/M]
  3. Compare ratios

Molecular Formula:

Let x be (EF)x

x = Actual Molar Mass/Empirical Molar Mass

Multiply EF formula (subscripts) by x

Gravimetric Stoichiometry

  • Mass and mole relationships

    1. Balance Chemical Reaction (find missing precipitates, combustions, subscripts)
    2. Find mol by dividing mass/molar mass (n=m/M)
    3. Find mass by multiplying mol by molar mass (m=nM)
    4. Follow coefficients (don’t include them in molar mass) and multiply/divide accordingly to the ratios to get mols of other elements/compounds

  • Both sides should be balanced

  • Hydrocarbons always produce CO2 and H2O (carbon dioxide and water vapor)

Percentage Yield

% Yield = Actual Yield/Theoretical Yield x 100%

  • Actual Yield is less due to incomplete reaction, side reaction, loss of product, etc.

Limiting Reagents

Limiting Reagent: Reactant that is COMPLETELY consumed (controls how much product produced)

Excess Reagent: Reactant NOT completely consumed

  1. Find LR (smaller number)
  2. Find ER (bigger number)
  3. Find amount of product produced (Compare LR to ratio of Product)
  4. Find excess reagent remaining (Compare LR to ratio of ER a and subtract original ER to a)