Stoichiometry

CHAPTER 3: STOICHIOMETRY

Calculations with Chemical Formulas and Equations

Understanding Stoichiometry

  • Definition: Stoichiometry involves calculations based upon the understanding of atomic masses, chemical formulas, and the law of conservation of mass.

  • Historical Figures: Key figures in the foundational understanding of stoichiometry include Mme. and Antoine Lavoisier.

I. Chemical Equations

  • The basis for stoichiometry starts with chemical equations, often requiring students to predict products of chemical reactions.

II. Simple Patterns of Chemical Reactivity

  • Net Ionic Equations (NIE):

    • Essential for predicting chemical reactions, especially on the AP Chemistry Examination, appearing in various formats post-2014 revision.

    • A NIE demonstrates the chemical process focusing on species that change identity by excluding spectator ions.

    • Scoring Criteria:

    • 1 point for correct reactants,

    • 2 points for correct products,

    • 1 point for a balanced equation,

    • 1 point for answering any follow-up questions correctly.

  • Example Reaction:

    • Molecular Equation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

    • Typical NIE: Ag+ + Cl− → AgCl

    • Evidence of Reaction: Formation of a white precipitate (ppt).

III. Types of Reactions

  1. Double Replacement Reactions:

    • Involves exchanging of ions. Types include acid/base and precipitation reactions.

    • Example:

      • NH3 + HF → NH4+ + F−

      • Ca(OH)2 + H2SO4 → CaSO4 + 2H2O

  2. Redox/Single Replacement Reactions:

    • Explain oxidation and reduction:

      • Oxidation: Loss of electrons, gains higher positive charge.

      • Reduction: Gain of electrons, lowers positive charge.

    • Example:

      • K + H2O → K+ + OH− + H2

  3. Combustion Reactions:

    • Reaction with oxygen that typically produces CO2 and H2O. Example reactions must assume complete combustion unless specified otherwise.

    • Example: C3H8 + O2 → CO2 + H2O.

  4. Complex Ion Formation:

    • A complex ion consists of a central metal ion bonded covalently to surrounding ligands.

    • Example of ligands: NH3, H2O, CO, CN−, Cl−, and OH−.

    • Coordination compounds exhibit a wide range of colors.

  5. Synthesis and Decomposition Reactions:

    • Synthesis Reaction: Combining reactants to form a single product; e.g., 2 K + Br2 → 2 KBr.

    • Decomposition Reaction: Splitting a compound into two or more products; e.g., 2 Al2O3 → 4 Al + 3 O2.

IV. Formula Weights

  • Formula Weight (FW): The sum of atomic weights of atoms in a compound’s formula, also known as Molecular Weight (MW).

  • Percent Composition:

    • Formula:
      ext{Percent Composition} = rac{ ext{Number of atoms} imes ext{Atomic weight of element}}{ ext{Formula weight of compound}} imes 100 ext{%}

    • Example: Calculate percent composition of nitrogen in ammonium persulfate

    • Molecular Weight of (NH4)2SO5 = 148.1 g:
      ext{Percent of N} = rac{(2 imes 14)}{148.1} imes 100 ext{%} = 18.9 ext{%}

V. Avogadro’s Number and the Mole

  • Concept: Relates quantities of substances to moles. List of substances for reference includes 1 mol of common compounds and elements.

VI. Empirical Formulas

  • Determining Molar Ratios: Derived from laboratory analyses offering empirical formulas based on percent composition.

  • Example Calculation: Given a compound’s mass percentages (C, H, Cl) to derive the molecular formula when a molecular weight is provided.

  • General Steps:

    1. Assume a 100 g sample to convert percentages to grams.

    2. Convert grams to moles.

    3. Determine the simplest ratios to yield empirical or molecular formulas.

Example Problem:

  • Given 24.27% carbon, 4.07% hydrogen, and 71.65% chlorine, and a molecular weight of 98.96 amu, calculate as follows:

    1. 24.27 g C → 2.021 mol C

    2. 4.07 g H → 4.04 mol H

    3. 71.65 g Cl → 2.021 mol Cl

    4. Empirical formula derived (C, H, Cl ratios) yields C2H4Cl2.

VII. Quantitative Information from Balanced Equations

  • Conversion factors relate different amounts of substances through stoichiometric coefficients in a balanced equation.

  • Common conversions encompass mass to moles, moles to particles, mol gas to volume, and volume of solution to moles of solute. Examples illustrate these concepts.

Limiting Reactants

  • Definition: The reactant that is consumed first in a chemical reaction, determining the amount of product generated.

  • Experimental Context: Limiting reactants are often more expensive or complicated to handle compared to others.

  • Methodology:

    • Identify limiting reactants by analyzing amounts given in stoichiometry problems.

    • Calculate theoretical versus actual yield using:
      ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100 ext{%}

Example Yield Calculation

  • Methanol Synthesis:
    2extH<em>2(g)+extCO(g)ightarrowextCH</em>3extOH(l)2 ext{H}<em>2(g) + ext{CO}(g) ightarrow ext{CH}</em>3 ext{OH}(l)

  • Calculate theoretical yield and percent yield. Exemplar calculations through mass conversions reveal:

  • Results: 68.5 kg CO and 8.60 kg H2 yield a theoretical yield of 68,200 g (actual yield of 35,700 g), yielding a percent yield of 52.3%.