Exam 2 Study Guide: Mass Relationships and Reactions in Aqueous Solution

Chapter 3: Mass Relationships in Chemical Reactions

Chemical Equations

Chemical equations represent chemical reactions, showing the substances involved and their relative quantities. Balancing equations ensures the law of conservation of mass is obeyed.

• Balanced Chemical Equation: An equation with equal numbers of each atom on both sides.

• Steps to Balance:

  1. Write correct formulas for reactants and products.

  2. Adjust coefficients to balance atoms.

  3. Check that all elements are balanced.

• Example: Balancing the combustion of methane: $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$

Molecular Weight and Molar Mass

The molecular weight (or formula mass) is the sum of atomic masses in a molecule. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

• Molar Mass Calculation: Add atomic masses from the periodic table for all atoms in the formula.

• Conversions:

  • Mass ↔ Moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

  • Moles ↔ Number of Entities: $\text{entities} = \text{moles} \times 6.022 \times 10^{23}$

• Mass Percent and Mass Fraction:

  • Mass %: $\frac{\text{mass of element}}{\text{mass of compound}} \times 100\%$

  • Mass fraction: $\frac{\text{mass of element}}{\text{mass of compound}}$

• The Mole: A counting unit equal to $6.022 \times 10^{23}$ entities (Avogadro's number).

• Example: Calculate the molar mass of $\mathrm{H_2O}$: $2 \times 1.01 + 16.00 = 18.02\ \mathrm{g/mol}$

Stoichiometry, Percent Yield, and Limiting Reactants

Stoichiometry uses balanced equations to relate quantities of reactants and products. Percent yield measures reaction efficiency, and the limiting reactant determines the maximum product formed.

• Stoichiometric Calculations: Use mole ratios from the balanced equation to convert between substances.

• Theoretical Yield: Maximum amount of product possible from given reactants.

• Percent Yield: $\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

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

• Example: If 2 mol $\mathrm{H_2}$ react with 1 mol $\mathrm{O_2}$, $\mathrm{H_2}$ is limiting in $2\ \mathrm{H_2} + O_2 \rightarrow 2\ \mathrm{H_2O}$.

Percent Composition and Empirical Formulas

Percent composition shows the mass percentage of each element in a compound. Empirical formulas give the simplest whole-number ratio of atoms, while molecular formulas show the actual number of atoms.

• Percent Composition: $\frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100\%$

• Empirical Formula: Simplest ratio of elements.

• Molecular Formula: Actual number of atoms; may be a multiple of the empirical formula.

• Finding Empirical Formula:

  1. Convert mass % to grams (assume 100 g sample).

  2. Convert grams to moles for each element.

  3. Divide by smallest number of moles to get ratio.

• Example: A compound with 40% C, 6.7% H, 53.3% O has empirical formula $\mathrm{CH_2O}$.

Chapter 4: Reactions in Aqueous Solution

Solution Concentration

Solutions are homogeneous mixtures of solute (dissolved substance) and solvent (dissolving medium). Concentration is often expressed as molarity (M).

• Solute: Substance dissolved.

• Solvent: Substance present in greater amount.

• Molarity (M): $\frac{\text{moles of solute}}{\text{liters of solution}}$

• Preparation: Calculate moles and volume needed for desired concentration.

• Example: To make 0.5 L of 1.0 M NaCl, dissolve 0.5 mol NaCl in water to make 0.5 L solution.

Dilution

Dilution reduces concentration by adding solvent. The amount of solute remains constant.

• Dilution Equation: $M_1V_1 = M_2V_2$

• Technique: Measure volume of concentrated solution, add solvent to reach final volume.

• Example: To make 100 mL of 0.2 M solution from 1.0 M stock: $V_1 = \frac{M_2V_2}{M_1} = \frac{0.2 \times 100}{1.0} = 20\ \mathrm{mL}$

Electrolytes

Electrolytes are substances that produce ions in solution. Strong electrolytes dissociate completely; weak electrolytes only partially.

• Strong Electrolytes: Ionic compounds, strong acids/bases (e.g., NaCl, HCl).

• Weak Electrolytes: Weak acids/bases (e.g., CH3COOH).

• Ion Concentrations: Calculate using stoichiometry of dissociation.

• Example: 0.1 M Na2SO4 yields 0.2 M Na+ and 0.1 M SO42−.

Aqueous Reactions and Net Ionic Equations

Reactions in solution can be classified by patterns and written as molecular, total ionic, and net ionic equations. Spectator ions do not participate in the reaction.

• Five Patterns of Reactions: Combination, decomposition, single replacement, double replacement, combustion.

• Types: Precipitation, acid-base, redox.

• Net Ionic Equation: Shows only species that change during the reaction.

• Spectator Ions: Ions unchanged on both sides.

• Example: For $\mathrm{AgNO_3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}$:

  • Molecular: $\mathrm{AgNO_3} (aq) + \mathrm{NaCl} (aq) \rightarrow \mathrm{AgCl} (s) + \mathrm{NaNO_3} (aq)$

  • Total ionic: $\mathrm{Ag^+} + \mathrm{NO_3^-} + \mathrm{Na^+} + \mathrm{Cl^-} \rightarrow \mathrm{AgCl} (s) + \mathrm{Na^+} + \mathrm{NO_3^-}$

  • Net ionic: $\mathrm{Ag^+} (aq) + \mathrm{Cl^-} (aq) \rightarrow \mathrm{AgCl} (s)$

Precipitation Reactions

Precipitation occurs when an insoluble solid forms from mixing solutions. Solubility rules predict precipitate formation.

• Solubility Rules: Guidelines for predicting if a compound is soluble or forms a precipitate.

• Double Replacement: Two ionic compounds exchange ions.

• Example: Mixing BaCl2 and Na2SO4 forms BaSO4 (s), a precipitate.

Acid-Base Neutralization Reactions

Acid-base reactions involve transfer of protons (H+). Neutralization produces water and a salt.

• Acid: Proton donor; Base: Proton acceptor.

• Strong vs. Weak: Strong acids/bases dissociate completely; weak only partially.

• Neutralization: $\mathrm{HCl} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{NaCl} (aq) + \mathrm{H_2O} (l)$

• Ionic Equation: $\mathrm{H^+} (aq) + \mathrm{OH^-} (aq) \rightarrow \mathrm{H_2O} (l)$

Solution Stoichiometry and Titrations

Solution stoichiometry relates volumes and concentrations in reactions. Titration determines unknown concentrations using a reaction with known stoichiometry.

• Mole-Volume Conversion: $\text{moles} = M \times V$ (V in liters)

• Titration: Gradual addition of one solution to another until reaction is complete (equivalence point).

• Example: 25.0 mL of 0.100 M HCl requires 25.0 mL of 0.100 M NaOH for neutralization.

Redox Chemistry

Redox (oxidation-reduction) reactions involve electron transfer. Oxidation numbers help identify what is oxidized and reduced.

• Oxidation: Loss of electrons; Reduction: Gain of electrons.

• Oxidizing Agent: Causes oxidation (is reduced); Reducing Agent: Causes reduction (is oxidized).

• Oxidation Number: Assigned to atoms to track electron transfer.

• Activity Series: Ranks metals by reactivity; predicts if a single replacement reaction occurs.

• Redox Titration: Used to determine unknown concentrations via redox reactions.

• Example: In $\mathrm{Zn} + \mathrm{CuSO_4} \rightarrow \mathrm{ZnSO_4} + \mathrm{Cu}$, Zn is oxidized, Cu2+ is reduced.

Summary Table: Key Concepts and Formulas

Concept	Definition/Formula	Example
Molarity	$M = \frac{\text{moles}}{\text{liters}}$	1.0 mol NaCl in 1.0 L = 1.0 M
Percent Yield	$\frac{\text{actual}}{\text{theoretical}} \times 100\%$	8.0 g actual / 10.0 g theoretical = 80%
Empirical Formula	Smallest whole-number ratio	CH2O for glucose
Dilution	$M_1V_1 = M_2V_2$	20 mL of 1.0 M to 100 mL = 0.2 M
Redox	Oxidation: loss e-; Reduction: gain e-	Zn + Cu2+ → Zn2+ + Cu

Additional info: Academic context and examples have been added to expand on the brief points in the original study guide, ensuring the notes are self-contained and suitable for exam preparation.