CHEM 100 – Exam 3 Study Guide: Core Concepts and Practice Problems

1. Reaction Prediction and Balancing

Introduction

Chemical reactions can be classified by their type and must be balanced to obey the law of conservation of mass. Predicting products and balancing equations are foundational skills in general chemistry.

• Reaction Types:

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

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

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

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

  • Combustion: A substance reacts with oxygen, often producing CO2 and H2O.

• Balancing Equations:

  • Ensure the same number of each atom on both sides of the equation.

  • Use coefficients to balance atoms.

• Example: Calcium hydroxide reacts with hydrochloric acid: $\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow 2\text{H}_2\text{O} + \text{CaCl}_2$

• Practice:

  • Potassium chlorate heated: $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$ (decomposition)

  • Zinc added to copper(II) nitrate: $\text{Zn} + \text{Cu(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Cu}$ (single replacement)

  • Sodium reacts with water: $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$ (single replacement)

• Hint: Use the activity series to predict if single replacement occurs.

2. Mole-Particle Conversions

Introduction

The mole is a fundamental unit in chemistry, relating the number of particles to a measurable amount of substance. Avogadro's number is used to convert between moles and particles.

• Key Terms:

  • Mole (mol): The amount of substance containing $6.022 \times 10^{23}$ particles.

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

• Conversion Formula: $\text{Number of moles} = \frac{\text{Number of particles}}{6.022 \times 10^{23}}$

• Example: How many moles of nitrogen gas are formed when $4.56 \times 10^{22}$ molecules of ammonia decompose? Balanced equation: $2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2$ $\frac{4.56 \times 10^{22}}{6.022 \times 10^{23}} = 0.0757$ mol NH3 Moles of N2: $0.0757 \text{ mol NH}_3 \times \frac{1 \text{ mol N}_2}{2 \text{ mol NH}_3} = 0.0379$ mol N2

3. Stoichiometry and Mole Calculations

Introduction

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction. Calculations often require balanced equations and molar ratios.

• Steps for Stoichiometric Calculations:

  1. Write and balance the chemical equation.

  2. Use molar ratios from the balanced equation.

  3. Convert moles to grams using molar mass.

• Example: How many grams of oxygen gas are produced from decomposing 15.0 moles of hydrogen peroxide (H2O2)? Balanced equation: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$ $15.0 \text{ mol H}_2\text{O}_2 \times \frac{1 \text{ mol O}_2}{2 \text{ mol H}_2\text{O}_2} \times 32.00 \text{ g O}_2 = 240.0 \text{ g O}_2$

4. Limiting Reactant and Percent Yield

Introduction

The limiting reactant determines the maximum amount of product formed in a reaction. Percent yield compares actual yield to theoretical yield.

• Limiting Reactant:

  • Calculate moles of each reactant.

  • Determine which reactant produces less product; this is the limiting reactant.

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

• Example: A mixture of 25.0 g methane (CH4) reacts with 80.0 g oxygen (O2): Balanced equation: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

  • Calculate moles: $80.0 \text{ g O}_2 \times \frac{1 \text{ mol O}_2}{32.00 \text{ g O}_2} = 2.50 \text{ mol O}_2$ $25.0 \text{ g CH}_4 \times \frac{1 \text{ mol CH}_4}{16.04 \text{ g CH}_4} = 1.56 \text{ mol CH}_4$

  • Limiting reactant: O2 (produces less CO2)

  • Theoretical yield of CO2: $2.50 \text{ mol O}_2 \times \frac{1 \text{ mol CO}_2}{2 \text{ mol O}_2} \times 44.01 \text{ g CO}_2 = 55.0 \text{ g CO}_2$

  • Percent yield if 60.0 g CO2 is produced: $\frac{60.0}{55.0} \times 100\% = 109\%$

5. Empirical Formula

Introduction

The empirical formula represents the simplest whole-number ratio of elements in a compound. It is determined from percent composition data.

• Steps to Determine Empirical Formula:

  1. Assume a 100 g sample.

  2. Convert percentages to grams.

  3. Convert grams to moles using atomic masses.

  4. Divide by the smallest mole value to get ratios.

  5. Write the formula using whole-number ratios.

• Example: Compound contains 40.0% C, 6.7% H, 53.3% O.

  • C: $\frac{40.0}{12.01} = 3.33$ mol

  • H: $\frac{6.7}{1.008} = 6.65$ mol

  • O: $\frac{53.3}{16.00} = 3.33$ mol

  • Divide by 3.33: C = 1, H = 2, O = 1

  • Empirical formula: CH2O

6. Titration and Molarity

Introduction

Titration is a technique to determine the concentration of a solution using a reaction with a standard solution. Molarity is the number of moles of solute per liter of solution.

• Key Terms:

  • Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

  • Equivalence Point: The point at which stoichiometric amounts of reactants have reacted.

• Neutralization Equation: $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

• Titration Formula: $M_1V_1 = M_2V_2$ (adjust for stoichiometric ratio)

• Example: 25.00 mL NaOH titrated with 0.200 M H2SO4, requiring 18.75 mL acid. $M_1V_1 = 2M_2V_2$ $M_1 \times 25.00 = 2 \times 0.200 \times 18.75$ $M_1 = \frac{2 \times 0.200 \times 18.75}{25.00} = 0.300 \text{ M}$

Additional info: All calculations and equations have been expanded for clarity and completeness. Practice problems are based on standard general chemistry curriculum.