Chemistry: An Introduction to General, Organic, and Biological Chemistry - Chapter 7 Notes

Understand Concepts in Chemical Quantities
- Use Avogadro’s number to determine the number of particles in moles.
- Calculate molar mass from chemical formulas.
- Convert between grams and moles using molar mass.
- Write balanced chemical equations based on reactant and product formulas.
- Identify reactants and products in terms of chemical reaction types: combination, decomposition, single replacement, double replacement, and combustion.
- Define oxidation and reduction; identify oxidized and reduced reactants.
- Utilize mole-mole factors from balanced equations for calculations involving moles of substances in reactions.
- Calculate mass in grams for substances involved in reactions based on given mass in grams for another substance.
- Describe exothermic and endothermic reactions and recognize factors affecting reaction rates.

Definition: Avogadro’s number is the number of particles in one mole of a substance, which is approximately $6.02 imes 10^{23}$ particles.
Historical Note: This constant is named after Amedeo Avogadro (1776–1856), an Italian physicist.
Equality: 1 mole = $6.02 imes 10^{23}$ items (particles, atoms, ions, or molecules).

Application of Avogadro’s Number: Used for converting moles of a substance into molecules.
- Example Problem: To determine the number of CO2 molecules in 0.50 mole of CO2.
- Steps:
  1. Identify given (0.50 mole of CO2) and target quantities (molecules).
  2. Establish conversion (1 mole of CO2 = $6.02 imes 10^{23}$ molecules of CO2).
  3. Set up calculation to find number of molecules.

Subscript Representation: In a chemical formula, subscripts indicate:
- Relationship of atoms in the formula.
- The number of moles of each element present in one mole of compound.
Example: Aspirin (C9H8O4)
- 1 molecule of aspirin contains:
- 9 atoms of Carbon (C)
- 8 atoms of Hydrogen (H)
- 4 atoms of Oxygen (O)
- Hence, in 1 mole of aspirin:
- 9 moles of C
- 8 moles of H
- 4 moles of O

Molar Mass: The mass of one mole of an element or compound, expressed in grams.
- Example: For Carbon (C), 1 mole = 12.01 g.
Calculating Molar Mass of Li2CO3:
- Steps:
1. Obtain molar mass of each element from the periodic table.
2. Multiply each by the number of moles (from the subscripts in the formula).
3. Add the products to find total molar mass:
  - $2 ext{ moles of Li} = 13.88 g \ \ 1 ext{ mole of C} = 12.01 g \ \ 3 ext{ moles of O} = 48.00 g \ \ 1 ext{ mole of Li2CO3} = 73.89 g$

Key Symbols in Equations:
- Arrow separates reactants from products.
- Reactants on the left, products on the right, separated by a plus sign (+).
- Physical states indicated as: (s) solid, (l) liquid, (g) gas, (aq) aqueous.
Balancing Chemical Equations: Ensure the number of atoms in reactants equals that in products.
- Example: Balancing the reaction of aluminum and sulfur to form aluminum sulfide:
- Unbalanced: $ext{Al(s)} + ext{S(s)} ightarrow ext{Al2S3(s)}$
- Balanced: $2 ext{Al(s)} + 3 ext{S(s)} ightarrow ext{Al2S3(s)}$

Combination Reactions: Two or more reactants form one product.
- Example: $2 ext{Mg(s)} + ext{O2(g)} ightarrow 2 ext{MgO(s)}$
Decomposition Reactions: One substance splits into two or more simpler substances.
- Example: $2 ext{HgO(s)} ightarrow 2 ext{Hg(l)} + ext{O2(g)}$
Single Replacement Reactions: One element replaces another in a compound.
- Example: $ext{Zn(s)} + 2 ext{HCl(aq)} ightarrow ext{ZnCl2(aq)} + ext{H2(g)}$
Double Replacement Reactions: Positive ions in two compounds swap places.
- Example: $ext{AgNO3(aq)} + ext{NaCl(aq)} ightarrow ext{AgCl(s)} + ext{NaNO3(aq)}$
Combustion Reactions: Carbon-containing compound burns in oxygen to produce CO2 and H2O, releasing energy.
- Example: $ext{CH4(g)} + 2 ext{O2(g)} ightarrow ext{CO2(g)} + 2 ext{H2O(g)} + ext{energy}$

Definition: Reactions involving the transfer of electrons.
- Oxidation is described as the loss of electrons.
- Reduction is described as the gain of electrons.
- Mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain).
Example of Oxidation: $ext{2Cu(s)} ightarrow ext{2Cu}^{2+}(s) + 4e^{-}$ (loss of electrons).
Example of Reduction: $ext{O2(g)} + 4e^{-} ightarrow ext{2O}^{2-}(s)$ (gain of electrons).

Law of Conservation of Mass: Matter is neither created nor destroyed. The mass of products is equal to the mass of reactants.
Example Calculation: For $4 ext{Fe(s)} + 3 ext{O2(g)} ightarrow 2 ext{Fe2O3(s)}$ , calculate how many moles of Fe2O3 can form from 6.0 moles of O2.
- Mole-Mole relationship: $3 ext{ moles of O2} ightarrow 2 ext{ moles of Fe2O3}$ .

Exothermic Reactions:
- Heat is released; the energy of the products is less than that of the reactants.
- Example: $ext{C(s)} + 2 ext{H2(g)} ightarrow ext{CH4(g)} + 18 ext{ kcal}$ .
Endothermic Reactions:
- Heat is absorbed; the energy of the products is greater than that of the reactants.
- Example: $ext{N2(g)} + ext{O2(g)} + 43.3 ext{ kcal} ightarrow 2 ext{NO(g)}$ .

Collision Theory: Reactants must collide with sufficient energy and proper orientation to react.
Factors Increasing Reaction Rate:
- Increase in temperature increases the speed of molecules.
- Higher concentration of reactants results in more frequent collisions.
- Presence of a catalyst lowers activation energy and increases the rate of reaction without being consumed.

Given Problem 1: Determine how many atoms are in 2.0 moles of Aluminum (Al).
- $ext{Answer: } 3.0 imes 10^{23}$ atoms.
Given Problem 2: Balance a reaction involving solid Fe3O4 and hydrogen gas to produce iron and water.
- Balanced: $ext{Fe3O4(s)} + 4 ext{H2(g)} ightarrow 3 ext{Fe(s)} + 4 ext{H2O(l)}$ .

Understanding chemical quantities and reactions is fundamental to chemistry. The principles outlined in this chapter, from Avogadro's number to the various types of chemical reactions, provide a solid foundation for further study in chemistry and its applications in various scientific fields.