Exam 2 Review: Chapters 3 and 4

Chapter 3: Stoichiometry and the Mole Concept

Understanding Concepts of Mol, Mass, Molar Mass, and Atom
- The Mole ( $mol$ ): The mole is the SI unit for the amount of a substance. It is defined as the amount of substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly $12\,g$ of carbon-12.
- Avogadro's Number ( $N_A$ ): The number of particles in one mole of a substance, which is exactly $6.02214076 \times 10^{23}$ . For calculations, we typically use $6.022 \times 10^{23}\,mol^{-1}$ .
- Molar Mass ( $M$ ): The mass in grams of one mole of a substance. It is numerically equal to the atomic mass or molecular mass in atomic mass units ( $amu$ ) but expressed in units of $g\,mol^{-1}$ .
- Atomic Mass: The mass of an individual atom, typically found on the periodic table as a weighted average of all naturally occurring isotopes.
Calculations of Mol, Mass in grams, and Number of Atoms
- To convert between mass, moles, and number of particles, the following formulas are utilized:
- Mass to Moles: $n = \frac{m}{M}$ Where $n$ is the number of moles, $m$ is the mass in grams ( $g$ ), and $M$ is the molar mass in $g\,mol^{-1}$ .
- Moles to Number of Atoms/Particles: $N = n \times N_A$ Where $N$ is the total number of particles (atoms or molecules) and $N_A$ is Avogadro's number.
- Mass to Number of Atoms: This requires a two-step process: converting mass to moles, then moles to atoms. $N = \left( \frac{m}{M} \right) \times 6.022 \times 10^{23}$
Calculation of Molar Mass for a Given Formula
- The molar mass of a compound is the sum of the molar masses of the individual elements multiplitied by their respective subscripts in the chemical formula.
- Example: To find the molar mass of sulfuric acid ( $H_2SO_4$ ):
  - Molar mass of $H = 1.008\,g\,mol^{-1}$
  - Molar mass of $S = 32.06\,g\,mol^{-1}$
  - Molar mass of $O = 16.00\,g\,mol^{-1}$
  - $M(H_2SO_4) = (2 \times 1.008) + (1 \times 32.06) + (4 \times 16.00) = 98.08\,g\,mol^{-1}$
Concept of Percent Composition and Empirical Formula
- Percent Composition: The percent by mass of each element in a compound. $\text{Percent Composition of Element} = \left( \frac{n \times \text{molar mass of element}}{\text{molar mass of compound}} \right) \times 100\%$ Where $n$ is the number of moles of the element in 1 mole of the compound.
- Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.
  - Steps to Determine Empirical Formula:
  1. Assume a $100\,g$ sample (convert percentages directly to grams).
  2. Convert grams to moles using atomic masses ( $n = \frac{m}{M}$ ).
  3. Divide all mole values by the smallest number of moles obtained.
  4. If the results are not whole numbers, multiply all values by a small integer to obtain whole numbers.
- Molecular Formula: The actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula. $\text{Multiple} = \frac{\text{Molar Mass (experimental)}}{\text{Empirical Formula Mass}}$
Balancing Chemical Equations
- Based on the Law of Conservation of Mass, the number of atoms of each element must be the same on both sides of the equation.
- Rules:
  - Only change coefficients (the numbers in front of formulas); never change subscripts within a formula.
  - Start by balancing elements that appear in only one reactant and one product.
  - Balance polyatomic ions as single units if they appear unchanged on both sides.
  - Ensure the final coefficients are the smallest possible whole numbers.
Stoichiometry
- Stoichiometry involves using the relationships between reactants and products in a balanced chemical equation to determine quantitative data.
- Mole Ratio: A conversion factor derived from the coefficients of a balanced chemical equation.
- Calculation Flow: $\text{Mass A} \rightarrow \text{Moles A} \rightarrow \text{Moles B} \rightarrow \text{Mass B}$
  1. Convert mass of given substance to moles.
  2. Use the stoichiometric ratio from the balanced equation to find moles of the desired substance.
  3. Convert moles of the desired substance back to mass (grams).
Limiting Reagent Concept
- Limiting Reagent (Reactant): The reactant that is completely consumed in a chemical reaction and limits the amount of product formed.
- Excess Reagent: The reactant present in a quantity greater than necessary to react with the limiting reagent.
- Determining Limiting Reagent:
  1. Calculate the amount of product that could be formed by each reactant.
  2. The reactant that produces the smaller amount of product is the limiting reagent.
- Theoretical Yield: The maximum amount of product that can be produced as calculated via stoichiometry.
- Actual Yield: The amount of product actually obtained from a reaction in the laboratory.
- Percent Yield: $\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%$

Chapter 4: Reactions in Aqueous Solutions

Concept of Strong, Weak, and Nonelectrolytes
- Electrolyte: A substance that, when dissolved in water, results in a solution that can conduct electricity.
- Strong Electrolyte: A solute that is assumed to be $100\%$ dissociated or ionized in solution.
  - Examples: Strong acids ( $HCl$ , $HNO_3$ , $H_2SO_4$ ), strong bases ( $NaOH$ , $KOH$ ), and soluble ionic salts ( $NaCl$ , $KBr$ ).
- Weak Electrolyte: A solute that only partially ionizes in water. The solution contains both molecules and ions.
  - Examples: Weak acids ( $CH_3COOH$ (acetic acid), $HF$ ) and weak bases ( $NH_3$ (ammonia)).
- Nonelectrolyte: A substance that does not ionize when dissolved in water and does not conduct electricity.
  - Examples: Sucrose ( $C_{12}H_{22}O_{11}$ ), ethanol ( $C_2H_5OH$ ), and urea.
Writing Equations for a Precipitation Reaction
- A precipitation reaction occurs when two soluble ionic compounds in aqueous solution react to form an insoluble solid (precipitate).
- Types of Equations:
  1. Molecular Equation: Shows the complete formulas of all reactants and products as if they were molecules. $Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)$
  2. Ionic Equation: Shows all strong electrolytes as dissociated ions. $Pb^{2+}(aq) + 2NO_3^-(aq) + 2K^+(aq) + 2I^-(aq) \rightarrow PbI_2(s) + 2K^+(aq) + 2NO_3^-(aq)$
  3. Net Ionic Equation: Shows only the species involved in the chemical change, removing spectator ions (ions that appear on both sides). $Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s)$
Understanding Acids and Bases Based on Brønsted Definition
- Brønsted Acid: A substance that can donate a proton ( $H^+$ ).
- Brønsted Base: A substance that can accept a proton ( $H^+$ ).
- Examples:
  - In the reaction $HCl(aq) + NH_3(aq) \rightarrow NH_4^+(aq) + Cl^-(aq)$ , $HCl$ is the Brønsted acid and $NH_3$ is the Brønsted base.
- Monoprotic, Diprotic, and Triprotic Acids: Categorized by the number of protons they can donate (e.g., $HCl$ is monoprotic, $H_2SO_4$ is diprotic, $H_3PO_4$ is triprotic).
Identifying Products of an Acid-Base Reaction
- Neutralization Reaction: A reaction between an acid and a base.
- Standard Product: Typically, a neutralization reaction between a Brønsted acid and a metal hydroxide base produces water and a salt (an ionic compound). $\text{Acid} + \text{Base} \rightarrow \text{Salt} + H_2O$ Example: $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$
Concept of Oxidation-Reduction (Redox) Reactions
- Redox Reaction: A chemical reaction involving the transfer of electrons from one reactant to another.
- Oxidation: The loss of electrons or an increase in oxidation number.
- Reduction: The gain of electrons or a decrease in oxidation number.
- Reducing Agent: The substance that donates electrons (is oxidized).
- Oxidizing Agent: The substance that accepts electrons (is reduced).
- Mnemonic: "OIL RIG" (Oxidation Is Loss, Reduction Is Gain).
Calculation of Oxidation Number
- To track electron transfer, oxidation numbers are assigned based on specific rules:
  1. Elements in their free state (e.g., $H_2$ , $S_8$ , $Fe$ ) have an oxidation number of $0$ .
  2. For monatomic ions, the oxidation number equals the charge of the ion (e.g., $Li^+$ is $+1$ , $Fe^{3+}$ is $+3$ ).
  3. Oxygen is usually $-2$ (but $-1$ in peroxides like $H_2O_2$ ).
  4. Hydrogen is $+1$ when bonded to nonmetals and $-1$ when bonded to metals (hydrides).
  5. Fluorine is always $-1$ .
  6. The sum of oxidation numbers in a neutral compound is $0$ . In a polyatomic ion, the sum equals the charge of the ion.
Identification of Different Types of Oxidation-Reduction Reactions
- Combination Reaction: Two or more substances combine to form a single product. $A + B \rightarrow C$
- Decomposition Reaction: The breakdown of a compound into two or more substances. $C \rightarrow A + B$
- Displacement Reaction: An atom or ion in a compound is replaced by an atom or ion of another element.
  - Metal Displacement: $V_2O_5 + 5Ca \rightarrow 2V + 5CaO$
  - Hydrogen Displacement: $2Na + 2H_2O \rightarrow 2NaOH + H_2$
  - Halogen Displacement: $Cl_2 + 2KBr \rightarrow 2KCl + Br_2$
- Combustion Reaction: A substance reacts with oxygen, usually with the release of heat and light. $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
- Disproportionation Reaction: An element in one oxidation state is simultaneously oxidized and reduced. $2H_2O_2 \rightarrow 2H_2O + O_2$
Calculation of Molarity and the Use of Dilution Formula
- Molarity ( $M$ ): The number of moles of solute per liter of solution. $M = \frac{n}{V}$ Where $n$ is the number of moles and $V$ is the volume of the solution in $dm^3$ (Liters).
- Dilution: The process of preparing a less concentrated solution from a more concentrated (stock) solution.
- Dilution Formula: $M_1 V_1 = M_2 V_2$ Where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume. Note that the number of moles of solute remains constant during dilution ( $n_1 = n_2$ ).