Chapter 3 - Composition of Substances and Solutions

Composition of Substances and Solutions

1. Chapter 3: Composition of Substances and Solutions Summary, Examples, and Key Exercises | Chemistry 2e

2. Introduction: Why Chemistry of Substances Matters Real-world example: Swimming pool maintenance. Chemicals like Ca²⁺ are added based on measured composition. This chapter explores formula masses, moles, and solution concentrations.

3. 3.1 Formula Mass and the Mole Concept Formula Mass = sum of atomic masses (amu). Mole = 6.022 × 10²³ entities (Avogadro’s number). Molar Mass = g/mol (numeric match with formula mass).

4. Sample Calculations – Formula Mass & Mole Example: Ibuprofen C₁₃H₁₈O₂ → 206.29 amu. Convert 5g Cu to atoms: mol = g ÷ g/mol → atoms = mol × Avogadro's number. 0.443 mol N₂H₄ = 14.2 g.

5. 3.2 Empirical and Molecular Formulas Percent Composition → Empirical Formula. Empirical Formula → simplified atom ratio. Molecular Formula = multiple of empirical formula.

6. Empirical Formula Steps – Sample Steps: 1. Convert grams to moles. 2. Divide all by smallest mole value. 3. Adjust to whole numbers. Example: Fe (34.97g), O (15.03g) → Fe₂O₃

7. 3.3 Molarity – Calculating Concentration M = mol of solute ÷ L of solution. Used to measure strength of solutions. Dilution: M₁V₁ = M₂V₂

8. Sample Molarity Calculations 355 mL with 0.133 mol sucrose → 0.375 M. 25.2 g CH₃COOH in 0.5 L → 0.839 M. 0.850 L of 5.00 M Cu(NO₃)₂ → diluted to 1.80 L → 2.36 M.

9. 3.4 Other Units of Concentration Mass %, Volume %, Mass/Volume % ppm = mg/L, ppb = µg/L Used in medicine, industry, environmental testing

10. Practice Problems 1. Determine empirical formula from: 34.97 g Fe and 15.03 g O. 2. Calculate formula mass of glucose (C₆H₁₂O₆). 3. 6.52 g CoCl₂ in 75.0 mL → what is M? 4. 0.48 mg Hg in 50.0 g solution → ppm & ppb?