CHEM103 Exam 2 Study Guide: Moles, Solutions, and Chemical Reactions

Chapter 3: The Mole, Solutions, and Molarity

The Mole and Avogadro’s Number

The mole is a fundamental unit in chemistry that represents a specific quantity of particles (atoms, molecules, or ions). Avogadro’s number defines the number of particles in one mole.

• Definition of the Mole: One mole contains exactly $6.022 \times 10^{23}$ particles (Avogadro’s number).

• Avogadro’s Number: $6.022 \times 10^{23}$ particles/mol.

• Application: Used to relate mass, number of particles, and volume in chemical calculations.

• Example: 2 moles of H2O contains $2 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$ molecules of water.

Dimensional Analysis: Mass, Moles, and Particles

Dimensional analysis allows conversion between mass, moles, and number of particles using molar mass and Avogadro’s number as conversion factors.

• Molar Mass (g/mol): The mass of one mole of a substance, found by summing atomic masses from the periodic table.

• Conversions:

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

  • Moles ↔ Particles: $\text{particles} = \text{moles} \times 6.022 \times 10^{23}$

• Example: How many molecules are in 18.0 g of H2O?

  • Moles: $\frac{18.0\ \text{g}}{18.02\ \text{g/mol}} = 0.999\ \text{mol}$

  • Molecules: $0.999\ \text{mol} \times 6.022 \times 10^{23} = 6.02 \times 10^{23}$ molecules

Solutions: Solute and Solvent

A solution is a homogeneous mixture of two or more substances. The solute is the substance dissolved, and the solvent is the substance present in the greatest amount.

• Solute: The component present in a lesser amount; what is dissolved.

• Solvent: The component present in a greater amount; what does the dissolving (often water in aqueous solutions).

• Example: In a saltwater solution, NaCl is the solute and water is the solvent.

Molarity and Solution Calculations

Molarity (M) is a measure of concentration, defined as moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$

• Solving for Variables: Rearranged as needed:

  • Moles: $\text{moles} = M \times V$

  • Volume: $V = \frac{\text{moles}}{M}$

• Example: What is the molarity of a solution containing 0.5 mol NaCl in 250 mL of solution?

  • Convert 250 mL to L: $0.250$ L

  • $M = \frac{0.5}{0.250} = 2.0$ M

Calculating Molarity from Mass

To find molarity when given the mass of solute, first convert mass to moles using molar mass, then use the molarity formula.

• Example: 10.0 g NaCl in 500 mL solution.

  • Moles NaCl: $\frac{10.0\ \text{g}}{58.44\ \text{g/mol}} = 0.171\ \text{mol}$

  • Volume: $0.500$ L

  • Molarity: $M = \frac{0.171}{0.500} = 0.342$ M

Dilution Calculations

Dilution involves adding solvent to decrease the concentration of a solution. The number of moles of solute remains constant before and after dilution.

• Dilution Equation: $M_1 V_1 = M_2 V_2$

• Variables:

  • $M_1$ = initial molarity

  • $V_1$ = initial volume

  • $M_2$ = final molarity

  • $V_2$ = final volume

• Example: How much 6.0 M HCl is needed to make 250 mL of 1.0 M HCl?

  • $M_1 V_1 = M_2 V_2$

  • $6.0 \times V_1 = 1.0 \times 0.250$

  • $V_1 = \frac{0.250}{6.0} = 0.0417$ L = 41.7 mL

Chapter 4: Chemical Reactions and Stoichiometry

Writing and Balancing Chemical Equations

Chemical equations represent chemical reactions. They must be balanced to obey the law of conservation of mass.

• Steps to Balance:

  1. Write correct formulas for reactants and products.

  2. Balance atoms one element at a time using coefficients.

  3. Check that all atoms are balanced and coefficients are in lowest ratio.

• Example: Sodium sulfate solution plus silver nitrate solution react to give sodium nitrate solution and silver sulfate solid.

  • Equation: $2 \text{AgNO}_3(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow 2 \text{NaNO}_3(aq) + \text{Ag}_2\text{SO}_4(s)$

Molecular, Complete Ionic, and Net Ionic Equations

Chemical reactions in aqueous solution can be represented in three ways:

• Molecular Equation: Shows all compounds as neutral formulas.

• Complete Ionic Equation: Shows all strong electrolytes as ions.

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

• State Symbols: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous

• Example: For the reaction above:

  • Molecular: $2 \text{AgNO}_3(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow 2 \text{NaNO}_3(aq) + \text{Ag}_2\text{SO}_4(s)$

  • Complete Ionic: $2 \text{Ag}^+(aq) + 2 \text{NO}_3^-(aq) + 2 \text{Na}^+(aq) + \text{SO}_4^{2-}(aq) \rightarrow 2 \text{Na}^+(aq) + 2 \text{NO}_3^-(aq) + \text{Ag}_2\text{SO}_4(s)$

  • Net Ionic: $2 \text{Ag}^+(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{Ag}_2\text{SO}_4(s)$

Precipitation Reactions and Solubility Rules

Precipitation reactions occur when two aqueous solutions form an insoluble solid (precipitate). Solubility rules help predict product solubility.

• Recognizing Precipitation: Formation of a solid from two solutions.

• Solubility Table: Used to determine if a compound is soluble or insoluble.

• Example: Mixing AgNO3(aq) and NaCl(aq) forms AgCl(s) precipitate.

Acids, Bases, and Neutralization Reactions

Acids produce H+ in solution; bases produce OH-. A neutralization reaction occurs when an acid reacts with a base to form water and a salt.

• Strong Acids/Bases: Completely ionize in water (e.g., HCl, NaOH).

• Neutralization Equation: $\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}$

• Example: $\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$

Oxidation-Reduction (Redox) Reactions

Redox reactions involve the transfer of electrons. Oxidation is loss of electrons; reduction is gain of electrons. Oxidation numbers help identify what is oxidized and reduced.

• Single Displacement: One element replaces another in a compound.

• Assigning Oxidation Numbers: Use rules (element = 0, group 1 = +1, group 2 = +2, F = -1, O = -2, H = +1 with nonmetals, etc.).

• Identifying Agents:

  • Oxidizing agent: Causes oxidation, is reduced.

  • Reducing agent: Causes reduction, is oxidized.

• Example: In $2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)$, Na is oxidized, H2O is reduced.

Stoichiometry and Limiting Reactants

Stoichiometry uses balanced equations to relate quantities of reactants and products. The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Stoichiometric Factors: Coefficients from balanced equations used as conversion factors.

• General Steps:

  1. Convert given quantities to moles.

  2. Use mole ratios to find moles of desired substance.

  3. Convert moles to grams if needed.

• Limiting Reactant: Compare mole ratios to determine which reactant produces less product.

• Theoretical Yield: Maximum amount of product possible from limiting reactant.

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

Stoichiometry with Solutions

When reactants are in solution, use molarity and volume to find moles: $\text{moles} = M \times V$.

• Example: Calculate the mass of Ca(OH)2 required to react with 25.0 mL of 10.0 M HC2H3O2.

  1. Write balanced equation: $2\text{HC}_2\text{H}_3\text{O}_2(aq) + \text{Ca(OH)}_2(s) \rightarrow \text{Ca(C}_2\text{H}_3\text{O}_2)_2(aq) + 2\text{H}_2\text{O}(l)$

  2. Find moles of HC2H3O2: $M \times V = 10.0\ \text{mol/L} \times 0.0250\ \text{L} = 0.250\ \text{mol}$

  3. Use stoichiometry: $2\ \text{mol HC}_2\text{H}_3\text{O}_2 : 1\ \text{mol Ca(OH)}_2$

  4. Moles Ca(OH)2: $0.250\ \text{mol} \times \frac{1}{2} = 0.125\ \text{mol}$

  5. Mass Ca(OH)2: $0.125\ \text{mol} \times 74.09\ \text{g/mol} = 9.26\ \text{g}$

Key Equations

Equation	Purpose
$M = \frac{\text{moles}}{\text{volume (L)}}$	Calculate molarity
$M_1 V_1 = M_2 V_2$	Dilution calculations
$\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$	Calculate percent yield

Summary Table: Types of Chemical Equations

Type	Description	Example
Molecular	All species as compounds	$\text{Na}_2\text{SO}_4(aq) + 2\text{AgNO}_3(aq) \rightarrow 2\text{NaNO}_3(aq) + \text{Ag}_2\text{SO}_4(s)$
Complete Ionic	All strong electrolytes as ions	$2\text{Na}^+(aq) + \text{SO}_4^{2-}(aq) + 2\text{Ag}^+(aq) + 2\text{NO}_3^-(aq) \rightarrow 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq) + \text{Ag}_2\text{SO}_4(s)$
Net Ionic	Only species that change	$2\text{Ag}^+(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{Ag}_2\text{SO}_4(s)$

Study Tips

• Practice converting between mass, moles, and particles using dimensional analysis.

• Be able to write and balance chemical equations from word descriptions.

• Familiarize yourself with solubility rules and strong acids/bases.

• Work through stoichiometry problems, including limiting reactant and percent yield calculations.

• Understand the rationale behind each answer, not just the procedure.

Additional info: This guide expands on the exam review list by providing definitions, worked examples, and key equations for each topic. For more practice, refer to your course slides, notes, and assigned problems.