Chapter3-Stoichometry
Chapter 3: Stoichiometry
Page 1: Introduction to Stoichiometry
Mathematics of Chemical Reactions
Importance of algebra in stoichiometry.
Counting by weight is utilized by both airline companies and chemists.
Atomic Mass
Average atomic masses are presented on the periodic table.
Different isotopes of an element contribute to the average atomic mass.
Page 2: Counting by Weight
Average Mass Dependence
Counting by weight relies on average mass rather than individual item mass.
Example: Jelly bean analogy from the Zumdal book.
Page 3: Atomic Mass
Definition and Measurement
Atomic mass is based on the average mass of known stable isotopes.
Mass spectrometer detects mass and relative abundance of isotopes.
Periodic table reports average atomic masses.
Page 4: Example Calculation of Average Mass
Natural Copper Example
Calculation of average mass using isotopic data.
69.09% of atoms are 63Cu and 30.91% are 65Cu.
Average mass calculation:
Total mass of 100 atoms = 6355 u
Average mass = 63.55 u/atom.
Reality check confirms the answer is reasonable.
Page 5: The Mole
Definition of a Mole
A mole (mol) is a unit for counting atoms, molecules, and ions.
Defined as the number of carbon atoms in 12 grams of pure 12C.
Avogadro’s number (6.022 x 10^23) is crucial for mass calculations.
Page 6: Molar Mass
Understanding Molar Mass
Molar mass is the mass in grams of one mole of a substance.
Molecular weight and formula weight are traditional terms for molecules and ionic compounds.
Requires knowledge of the chemical formula for calculations.
Page 7: Molar Mass Calculation Example
Juglone Example
Molar mass calculation for juglone (C10H6O3).
Total molar mass = 174.1 g.
Page 8: Molar Mass of Calcium Carbonate
Calcium Carbonate Example
Molar mass calculation for CaCO3.
Total molar mass = 100.09 g.
Page 9: Molar Mass Calculation
Example Problem
Calculate the mass in micrograms for a given number of moles of pyrogallol (C6H6O3).
Page 10: Weight of Atoms
Calcium Atom Calculation
Calculate the weight of 7.81 x 10^22 atoms of calcium.
Result: 5.20 g of Ca.
Page 11: Percent Composition
Definition and Importance
Percent composition indicates the percentage of a specific element in a compound.
Requires accurate molar mass for calculations.
Page 12: Mass Percent of Glucose
Glucose Example Calculation
Molar mass of glucose (C6H12O6) = 180.156 g.
Mass percent calculations for C, H, and O.
Page 13: Practice Problems
Mass Percent Calculation Practice
Find mass percent for various compounds.
Page 14: Empirical and Molecular Formulas
Definitions
Empirical formula: simplest whole number ratio of elements.
Molecular formula: actual ratio of elements in a compound.
Example: Glucose can be represented as C6H12O6 (molecular) and CH2O (empirical).
Page 15: Empirical and Molecular Formula Calculation
Nerve Gas Example
Given mass percent analysis to determine empirical and molecular formulas.
Page 16: Empirical Formula from Combustion
Combustion Analysis Example
Determine empirical formula from combustion data of a compound containing C, H, and O.
Page 18: Chemical Equations
Basics of Chemical Equations
Simplest form: Reactant(s) → Product(s).
Importance of balancing equations for conservation of matter.
Page 19: Balancing Chemical Equations
Balancing Strategy
Example of balancing a reaction involving potassium carbonate and hydrochloric acid.
Page 20: Balancing Techniques
Steps for Balancing
Focus on balancing one atom at a time, typically starting with those present in only one reactant.
Page 21: Example Balancing Problem
Balancing NaOH and H3PO4 Reaction
Step-by-step balancing of the equation.
Page 22: Combustion Reaction Balancing
Balancing Combustion Reactions
Example of balancing a combustion reaction for C4H10.
Page 24: Limiting Reagent Concept
Understanding Limiting Reagents
Use of equations to determine grams of reactants and products.
Importance of balancing equations before calculations.
Page 25: Limiting Reagent Details
Reactant and Product Relationships
Understanding mole ratios to identify limiting reagents in reactions.