General Chemistry Exam 1 Study Guide: Atoms, Molecules, Stoichiometry, Solutions, and Nomenclature

Exam 1 Preparation: Study Guide

This study guide covers foundational topics in General Chemistry, including atomic structure, chemical formulas, stoichiometry, solution concentration, nomenclature, and solubility rules. Each section provides key concepts, definitions, worked examples, and essential equations to support exam preparation.

Study Checklist and Additional Practice

• Perform unit conversions and use all metric prefixes.

• Solve problems involving chemical relationships (e.g., mole calculations, empirical/molecular formulas).

• Understand the development of atomic theory and the concept of significant figures.

• Know the difference between precision and accuracy; recognize types of errors.

• Calculate atomic mass, moles, and mass relationships using chemical formulas.

• Balance chemical equations and relate products/reactants using stoichiometry.

• Determine limiting reactants, theoretical yield, and percent yield.

• Solve molarity and dilution problems for solutions.

• Apply solubility rules to predict precipitate formation.

• Name ionic and molecular compounds using IUPAC rules.

Atomic Structure and Isotopes

Atomic Theory and Isotopes

Atoms are the basic units of matter, composed of protons, neutrons, and electrons. Isotopes are atoms of the same element with different numbers of neutrons.

• Atomic Number (Z): Number of protons in the nucleus.

• Mass Number (A): Total number of protons and neutrons.

• Isotopes: Atoms with the same Z but different A.

• Average Atomic Mass: Weighted average of all isotopes' masses.

Example Calculation: For iron isotopes, the average atomic mass is calculated as:

$\text{Average mass} = \sum (\text{fractional abundance} \times \text{isotope mass})$

For boron with two isotopes, set up equations using the given atomic mass and solve for abundances.

Empirical and Molecular Formulas

Determining Empirical and Molecular Formulas

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

• Convert mass percentages to grams (assume 100g sample).

• Convert grams to moles for each element.

• Divide by the smallest number of moles to get the ratio.

• If given, use molar mass to determine the molecular formula:

$\text{Molecular formula} = (\text{Empirical formula})_n$

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

Example: A compound with 78.9% C, 7.06% H, 14.0% O has empirical and molecular formula C6H6O.

Stoichiometry

Stoichiometric Calculations

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction.

• Balancing Equations: Ensure the same number of each atom on both sides.

• Limiting Reactant: The reactant that is completely consumed first, limiting the amount of product.

• Theoretical Yield: Maximum amount of product possible from given reactants.

• Percent Yield: $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$

Example: For the reaction $\mathrm{Cl_2 + 2KI \rightarrow 2KCl + I_2}$, calculate grams of KI needed for a given amount of Cl2 using molar ratios.

Solution Concentration and Dilution

Molarity and Dilution

Molarity (M) is the number of moles of solute per liter of solution.

$M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Dilution Equation: $M_1V_1 = M_2V_2$

• Use this to calculate volumes or concentrations after dilution.

Example: To make 100.0 mL of 0.150 M HCl from 2.00 M stock:

$V_1 = \frac{M_2V_2}{M_1} = \frac{0.150 \times 100.0}{2.00} = 7.50\,\text{mL}$

Nomenclature

Naming Ionic and Molecular Compounds

Use IUPAC rules to name compounds:

• Ionic Compounds: Name cation first, then anion (e.g., NaCl: sodium chloride).

• Molecular Compounds: Use prefixes to indicate number of atoms (e.g., CO2: carbon dioxide).

• Acids: Name based on anion (e.g., HCl: hydrochloric acid).

Examples:

• NH4Br: Ammonium bromide

• KClO4: Potassium perchlorate

• H2SO4: Sulfuric acid

• Pb(C2H3O2)4: Lead(IV) acetate

Solubility and Net Ionic Equations

Solubility Rules and Precipitation Reactions

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

• Soluble Compounds: Most salts of Na+, K+, NH4+, NO3-, and CH3COO-.

• Insoluble Compounds: Most carbonates, phosphates, sulfides, except those of alkali metals and NH4+.

Net Ionic Equations: Show only the species that change during the reaction.

Example: For the reaction of Hg(C2H3O2)2 with LiI:

• Molecular: $\mathrm{Hg(C_2H_3O_2)_2(aq) + 2LiI(aq) \rightarrow 2LiC_2H_3O_2(aq) + HgI_2(s)}$

• Total Ionic: $\mathrm{Hg^{2+}(aq) + 2C_2H_3O_2^-(aq) + 2Li^+(aq) + 2I^-(aq) \rightarrow 2Li^+(aq) + 2C_2H_3O_2^-(aq) + HgI_2(s)}$

• Net Ionic: $\mathrm{Hg^{2+}(aq) + 2I^-(aq) \rightarrow HgI_2(s)}$

Sample Table: Isotopic Abundance and Atomic Mass

The following table summarizes the calculation of average atomic mass from isotopic data:

Isotope	Abundance (%)	Mass (amu)	Protons	Neutrons
Fe-54	5.82	53.940	26	28
Fe-56	91.68	55.935	26	30
Fe-57	2.12	56.935	26	31
Fe-58	0.28	57.933	26	32

Additional info: Table values inferred from example in notes.

Key Equations and Concepts

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

• Mole-Mass Conversions: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

• Percent Composition: $\frac{\text{mass of element}}{\text{total mass of compound}} \times 100\%$

• Empirical Formula Calculation: Convert % to grams, grams to moles, divide by smallest, round to nearest whole number.

• Limiting Reactant: Compare mole ratios from balanced equation.

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

• Molarity: $M = \frac{\text{mol}}{L}$

• Dilution: $M_1V_1 = M_2V_2$

Summary Table: Solution Preparation and Dilution

Problem	Equation Used	Key Steps
Prepare 0.500M KCl solution	$M = \frac{\text{mol}}{L}$	Calculate moles KCl, convert to grams
Dilute 2.00M HCl to 0.150M	$M_1V_1 = M_2V_2$	Solve for $V_1$
Serial dilution	$M_1V_1 = M_2V_2$ (twice)	Calculate each step sequentially

Additional info: This guide is based on the provided exam preparation notes and includes expanded academic context for clarity and completeness.