Chemistry 2045 Exam 2 Study Guide: Chemical Reactions, Stoichiometry, and Solutions

Chemical Reactions and Reaction Stoichiometry

Definitions and Key Terms

Understanding fundamental definitions is essential for mastering chemical reactions and stoichiometry. Below are important terms:

• Limiting Reactant: The reactant that is completely consumed first in a chemical reaction, thus determining the maximum amount of product formed.

• Percent Yield: The ratio of actual yield to theoretical yield, expressed as a percentage.

• Solution: A homogeneous mixture of two or more substances.

• Solvent: The substance present in the greatest amount in a solution; it dissolves the solute.

• Solute: The substance dissolved in the solvent.

• Electrolyte: A substance that produces ions when dissolved in water, thus conducting electricity.

Balancing Chemical Equations and Conservation of Mass

Balancing chemical equations ensures the Law of Conservation of Mass is obeyed, meaning the total mass of reactants equals the total mass of products.

• Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction.

• Balancing Equations: Adjust coefficients to ensure equal numbers of each atom on both sides of the equation.

Example:

Balance the equation for the combustion of methane:

$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$

Patterns of Reactivity

Chemical reactions can be classified by their patterns:

• Combination (Synthesis): Two or more substances combine to form one product.

• Decomposition: A single compound breaks down into two or more products.

• Combustion: A substance reacts with oxygen, often producing heat and light.

Example:

• Combination: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$

• Decomposition: $\mathrm{2H_2O \rightarrow 2H_2 + O_2}$

• Combustion: $\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$

Atoms, Molecules, and Ions

Molecular and Formula Mass

The mass of a compound can be determined from its chemical formula:

• Molecular Mass: The sum of the atomic masses of all atoms in a molecule.

• Formula Mass: The sum of atomic masses in an ionic compound's formula unit.

Example:

Calculate the molecular mass of water ($\mathrm{H_2O}$):

$\mathrm{(2 \times 1.01) + (1 \times 16.00) = 18.02\ g/mol}$

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole ($g/mol$).

Example:

Molar mass of sodium chloride ($\mathrm{NaCl}$):

$\mathrm{22.99\ g/mol\ (Na) + 35.45\ g/mol\ (Cl) = 58.44\ g/mol}$

Percent Composition by Mass

Percent composition indicates the mass percentage of each element in a compound.

• Formula: $\mathrm{\%\ by\ mass = \frac{\text{mass of element in 1 mole}}{\text{molar mass of compound}} \times 100\%}$

Example:

Percent composition of hydrogen in water:

$\mathrm{\frac{2.02}{18.02} \times 100\% = 11.2\%}$

Chemical Reactions and Reaction Stoichiometry

The Mole Concept and Avogadro's Number

The mole is a fundamental unit for counting particles in chemistry.

• Mole: $6.022 \times 10^{23}$ particles (atoms, molecules, or ions).

• Avogadro's Number: $6.022 \times 10^{23}$

Example:

1 mole of $\mathrm{H_2O}$ contains $6.022 \times 10^{23}$ molecules.

Conversions: Grams, Moles, and Particles

Use molar mass and Avogadro's number to convert between grams, moles, and number of particles.

• Grams to Moles: $\mathrm{\text{moles} = \frac{\text{grams}}{\text{molar mass}}}$

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

Example:

Convert 18.02 g of water to molecules:

$\mathrm{\frac{18.02\ g}{18.02\ g/mol} = 1\ mol}$

$\mathrm{1\ mol \times 6.022 \times 10^{23} = 6.022 \times 10^{23}\ molecules}$

Stoichiometric Calculations

Stoichiometry involves calculations based on balanced chemical equations.

• Use mole ratios from the equation to relate reactants and products.

Example:

Given $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$, 4 moles of $\mathrm{H_2}$ produce 4 moles of $\mathrm{H_2O}$.

Empirical and Molecular Formulas

Empirical formula shows the simplest ratio of elements; molecular formula shows the actual number of atoms.

• Empirical Formula: Simplest whole-number ratio.

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

• Formula: $\mathrm{\text{Molecular formula} = n \times \text{Empirical formula}}$

• n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Example:

Empirical formula: $\mathrm{CH_2}$; Molar mass: 28 g/mol; Empirical formula mass: 14 g/mol; $n = 2$; Molecular formula: $\mathrm{C_2H_4}$

Combustion Analysis

Combustion analysis determines empirical formulas by measuring the amounts of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ produced.

• Calculate moles of C and H from $\mathrm{CO_2}$ and $\mathrm{H_2O}$.

• Determine the simplest ratio for the empirical formula.

Limiting Reactant and Percent Yield

Identify the limiting reactant to determine the maximum product yield.

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

Example:

If theoretical yield is 10 g and actual yield is 8 g:

$\mathrm{\frac{8}{10} \times 100\% = 80\%}$

Reactions in Aqueous Solution

Solubility Rules and Precipitation Reactions

Solubility rules help predict whether a compound will dissolve in water or form a precipitate.

• Use rules to determine if a reaction forms an insoluble product (precipitate).

Example:

Mixing $\mathrm{NaCl}$ and $\mathrm{AgNO_3}$ forms $\mathrm{AgCl}$ precipitate.

Net Ionic Equations

Net ionic equations show only the species that participate in the reaction.

• Remove spectator ions.

Example:

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

Acid-Base Equilibria

Acids and Bases

Acids donate protons ($\mathrm{H^+}$); bases accept protons.

• Strong Acids/Bases: Completely dissociate in water.

• Weak Acids/Bases: Partially dissociate.

• Common Acids: Hydrochloric acid ($\mathrm{HCl}$), sulfuric acid ($\mathrm{H_2SO_4}$), nitric acid ($\mathrm{HNO_3}$).

• Common Bases: Sodium hydroxide ($\mathrm{NaOH}$), potassium hydroxide ($\mathrm{KOH}$).

Neutralization Reactions

Acids and bases react to form water and a salt.

Example:

$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$

Properties of Solutions

Molarity (Molar Concentration)

Molarity is the number of moles of solute per liter of solution.

• Formula: $\mathrm{M = \frac{\text{moles of solute}}{\text{liters of solution}}}$

Example:

2 moles of $\mathrm{NaCl}$ in 1 L solution: $\mathrm{M = 2\ mol/L}$

Dilution Calculations

To dilute a solution, use the equation:

• $\mathrm{M_1V_1 = M_2V_2}$

Example:

To dilute 1.0 L of 2.0 M solution to 0.5 M:

$\mathrm{(2.0)(1.0) = (0.5)(V_2) \Rightarrow V_2 = 4.0\ L}$

Lab Techniques and Procedures

Significant Figures and Conversions

All answers must use the correct number of significant figures. Basic conversions are essential for calculations.

• Temperature Conversions: $\mathrm{K = ^\circ C + 273.15}$

• Metric Prefixes: 1 kg = 1000 g; 1 mg = 0.001 g

Example:

Convert 250 mg to grams: $\mathrm{250\ mg \times \frac{1\ g}{1000\ mg} = 0.25\ g}$

Additional info: Review all conversions and significant figure rules from previous material for comprehensive exam preparation.