Study Guide: Chemical Reactions and The Mole Concept (Chapters 7 & 8)

Chapter 7: Chemical Reactions

Components of a Chemical Equation

Chemical equations represent chemical reactions, showing the substances involved and their transformation.

• Reactants: Substances present before the reaction; found on the left side of the equation.

• Products: Substances formed as a result of the reaction; found on the right side.

• Chemical Equation Symbols: Indicate the state of substances and reaction conditions.

  • (s): solid

  • (l): liquid

  • (g): gas

  • (aq): aqueous (dissolved in water)

  • →: yields or produces

  • Δ: heat is applied

Example: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

Types of Chemical Reactions

Chemical reactions can be classified based on how reactants are transformed into products.

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

  • General form: $A + B \rightarrow AB$

• Decomposition Reaction: A single compound breaks down into two or more simpler substances.

  • General form: $AB \rightarrow A + B$

• Single Replacement Reaction: An element replaces another in a compound.

  • General form: $A + BC \rightarrow AC + B$

• Double Replacement Reaction: Two compounds exchange ions to form two new compounds.

  • General form: $AB + CD \rightarrow AD + CB$

• Neutralization Reaction: An acid reacts with a base to form a salt and water.

  • General form: $HA + BOH \rightarrow BA + H_2O$

Example: $Na_2SO_4(aq) + BaCl_2(aq) \rightarrow 2NaCl(aq) + BaSO_4(s)$ (double replacement)

Writing and Balancing Chemical Equations

Balanced equations have equal numbers of each atom on both sides, obeying the Law of Conservation of Mass.

• Identify reactants and products.

• Write correct formulas (remember diatomic molecules: $H_2, N_2, O_2, F_2, Cl_2, Br_2, I_2$).

• Balance atoms by adjusting coefficients.

Example: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Activity Series and Single Replacement Reactions

The activity series ranks elements by their ability to displace others in single replacement reactions.

• A more active metal will replace a less active metal from a compound.

• Use the activity series chart to predict if a reaction will occur.

Example: $Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$ (Zinc is more active than copper)

Solubility Rules in Chemical Equations

Solubility rules help determine if a compound is soluble (aq) or insoluble (s) in water.

• Soluble compounds are labeled (aq); insoluble compounds are labeled (s).

• Use a solubility chart to assign states in chemical equations.

Example: $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ (AgCl is insoluble)

Chapter 8: The Mole Concept

Avogadro’s Constant and Counting Particles

The mole is a counting unit in chemistry, representing $6.022 \times 10^{23}$ particles (Avogadro’s number).

• 1 mole of atoms = $6.022 \times 10^{23}$ atoms

• 1 mole of molecules = $6.022 \times 10^{23}$ molecules

• 1 mole of formula units = $6.022 \times 10^{23}$ formula units

Example: Number of molecules in 2 moles of $H_2O$: $2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$ molecules

Calculations with Moles, Mass, and Avogadro’s Constant

Conversions between mass, moles, and number of particles use molar mass and Avogadro’s constant.

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

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

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

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

Example: Find the mass of 0.5 moles of $NaCl$ ($M = 58.44$ g/mol): $0.5 \times 58.44 = 29.22$ g

Molar Volume, Molar Mass, and Density of Gases at STP

At standard temperature and pressure (STP: 0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 L.

• Molar Volume at STP: $22.4$ L/mol

• Density of a Gas at STP: $\text{density} = \frac{\text{molar mass (g/mol)}}{22.4 \text{ L}}$

Example: Density of $O_2$ at STP: $\frac{32.00}{22.4} = 1.43$ g/L

Percent Composition of Atoms in Compounds

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

• Calculate mass of each element in one mole of compound.

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

Example: Percent of C in $CO_2$: $\frac{12.01}{44.01} \times 100 = 27.3\%$

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of atoms in a molecule.

• Find moles of each element from mass or percent composition.

• Divide by smallest number of moles to get ratio.

• Molecular formula: $\text{(empirical formula)}_n$, where $n = \frac{\text{molar mass}}{\text{empirical formula mass}}$

Example: Empirical formula CH, molecular formula $C_2H_2$ (if molar mass is 26 g/mol)

Percent of Water in Hydrated Salts (Water of Crystallization)

Hydrated salts contain water molecules within their crystal structure. The percent of water is calculated as follows:

• Find mass of water in one mole of hydrate.

• Percent water: $\% = \frac{\text{mass of water}}{\text{molar mass of hydrate}} \times 100$

Example: In $CuSO_4 \cdot 5H_2O$, mass of water = $5 \times 18.02 = 90.10$ g; molar mass = $249.68$ g; $\% = \frac{90.10}{249.68} \times 100 = 36.1\%$

Common Compounds of Hydrate and Their Chemical Formulas

Hydrates are named by the salt followed by a prefix indicating the number of water molecules.

Hydrate Name	Chemical Formula
Copper(II) sulfate pentahydrate	CuSO4·5H2O
Magnesium sulfate heptahydrate	MgSO4·7H2O
Sodium carbonate decahydrate	Na2CO3·10H2O
Barium chloride dihydrate	BaCl2·2H2O
Calcium chloride dihydrate	CaCl2·2H2O

Additional info: Prefixes: mono- (1), di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6), hepta- (7), octa- (8), nona- (9), deca- (10).