Compound Stoichiometry and Reaction Stoichiometry: Key Concepts and Practice

Compound Stoichiometry

Key Conversion Factors and Relationships

Compound stoichiometry involves quantitative relationships between elements in compounds and the conversion between different units (atoms, molecules, moles, and mass). Understanding these relationships is essential for solving chemical problems and interpreting laboratory data.

• Compound Formula: Shows the ratio of elements in a compound (e.g., H2O has a 2:1 ratio of H to O).

• Avogadro's Number: $6.022 \times 10^{23}$ particles/mole, used to convert between moles and number of particles.

• Molar Mass: The mass of one mole of a substance, expressed in g/mole.

• Percent Composition / Mass Fraction: The percentage by mass of each element in a compound.

Micro vs. Macro View

Stoichiometry can be approached from both the micro (atoms, ions) and macro (molecules, formula units) perspectives. The following conversions are commonly used:

• Atoms or Ions: Use Avogadro's number and compound formula ratios.

• Molecules or Formula Units: Use molar mass and mass fraction for conversions.

Finding the Molecular Formula

To determine the molecular formula, use the ratio of molar masses and empirical formula:

• Molecular Formula: Actual number of atoms of each element in a molecule.

• Empirical Formula: Simplest whole-number ratio of elements.

• Molar Mass: Used to relate empirical and molecular formulas.

• Percent Composition: Used to determine empirical formula from mass data.

Example Table: Steps to Find Molecular Formula

Step	Description
1	Calculate moles of each element from mass or percent composition.
2	Find simplest whole-number ratio (empirical formula).
3	Compare empirical formula mass to molar mass to find molecular formula.

Reaction Stoichiometry

Balanced Chemical Equations

Reaction stoichiometry uses balanced equations to relate amounts of reactants and products. The coefficients in a balanced equation indicate the mole ratios needed for calculations.

• Balanced Equation Example: $1 \text{C}_2\text{H}_5\text{OH} + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} + 200 \text{kJ}$

• Mole Ratio: Used to convert between moles of different substances in a reaction.

• Mass-Mole-Particle Relationships: Mass (g) → Moles (mol) → Particles (atoms, molecules).

Example Table: Stoichiometric Relationships

Quantity	Conversion Factor
Mass (g)	Molar Mass (g/mol)
Moles	Avogadro's Number ($6.022 \times 10^{23}$)
Particles	Compound Formula Ratio

Key Calculations in Reaction Stoichiometry

• Limiting Reactant: The reactant that is completely consumed first, limiting the amount of product formed.

• Theoretical Yield: Maximum amount of product that can be formed from given reactants.

• Percent Yield: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

Example Calculation:

• Given: 6.38 g C2H5OH, find moles of CO2 produced.

• Step 1: Convert mass to moles using molar mass.

• Step 2: Use mole ratio from balanced equation ($1 \text{mol C}_2\text{H}_5\text{OH} : 3 \text{mol CO}_2$).

Stoichiometry Practice

Balancing Equations and Limiting Reactant

Practice problems often involve balancing chemical equations and determining the limiting reactant based on initial amounts.

• Balance the Equation: Ensure the same number of atoms of each element on both sides.

• Limiting Reactant Table: Track initial, change, and final amounts of each reactant and product.

Example Table: Limiting Reactant Calculation

Substance	Initial Amount (g)	Change (g)	Final Amount (g)
C2H5O2	45	?	?
O2	152	?	?
CO2	0	?	?
H2O	0	?	?

Finding a Compound Formula from Percent Composition

Empirical Formula Calculation

Given mass percentages, the empirical formula can be determined by converting percentages to grams (assuming 100 g sample), then to moles, and finding the simplest ratio.

• Step 1: Convert percent to grams.

• Step 2: Convert grams to moles using atomic masses.

• Step 3: Divide by smallest number of moles to get whole-number ratio.

Example:

• Given: 52.1% C, 6.08% H, 41.7% O

• Assume 100 g sample: 52.1 g C, 6.08 g H, 41.7 g O

• Convert to moles: $\text{moles C} = \frac{52.1}{12.01}$, $\text{moles H} = \frac{6.08}{1.008}$, $\text{moles O} = \frac{41.7}{16.00}$

• Find simplest ratio for empirical formula.

Additional info: These notes cover core concepts from Ch. 5 (Chemical Quantities and Stoichiometry) and Ch. 3 (Compounds and Chemical Formulas) in General Chemistry.