Che130 L8

Important Reminders for Chemistry 130

  • Review Practice Problems from today’s lecture
  • No reading assigned
  • Homework 4 is due on Monday, September 29
  • Exam 1 is scheduled for Wednesday, October 1
  • Monday before the exam will be an Exam 1 Review Day
  • A worksheet will be provided for practice during this review.

Exam 1 Details

  • Format:
    • 10 Multiple Choice questions
    • 2-3 Short Answer questions
    • 5-8 Problem solving questions

Note: The number of questions in each category may vary

  • Exam Instructions:
    • Show ALL work clearly with units.
    • Demonstrate all conversions, and equations used.
    • Clear presentation enhances chances for partial credit.
    • If unsure whether to show a step, it is advisable to go ahead and show it.

Extra Examples and Calculations

Empirical and Molecular Formula Calculation

  • Example Problem: Determine the empirical formula of a compound with the following mass percentages:
    • 40.0% C
    • 6.71% H
    • 53.28% O
Step 1: Convert mass percentages to moles assuming 100 g sample.
  • Moles of Carbon (C):
    • Calculation: 40.0 \text{ g C} \times \frac{1 \text{ mol C}}{12.011 \text{ g C}} = 3.3303 \text{ mol C}
  • Moles of Hydrogen (H):
    • Calculation: 6.71 \text{ g H} \times \frac{1 \text{ mol H}}{1.008 \text{ g H}} = 6.6567 \text{ mol H}
  • Moles of Oxygen (O):
    • Calculation: 53.28 \text{ g O} \times \frac{1 \text{ mol O}}{15.999 \text{ g O}} = 3.3302 \text{ mol O}
Step 2: Normalize the mole ratio by dividing by the smallest number of moles (mol of O).
  • For Carbon:
    • Ratio: \frac{3.3303 \text{ mol C}}{3.3302 \text{ mol O}} \rightarrow 1 \text{ mole C per mole O}
  • For Hydrogen:
    • Ratio: \frac{6.6567 \text{ mol H}}{3.3302 \text{ mol O}} \rightarrow 2 \text{ moles H per mole O}
  • For Oxygen:
    • Ratio: \frac{3.3302 \text{ mol O}}{3.3302 \text{ mol O}} \rightarrow 1 \text{ mole O}
Step 3: Determine empirical formula and calculate empirical formula mass (emf).
  • Empirical Formula: CH₂O
  • Empirical Formula Mass:
    • Calculation: 30.0261 \text{ g/mol}
Step 4: Calculate n for the molecular formula using molecular mass.
  • Given Molar Mass: 90.08 g/mol
  • Calculation of n:
    • Formula: n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}
    • Result:
    • Calculation: n = \frac{90.08 \text{ g/mol}}{30.0261 \text{ g/mol}} = 3
Step 5: Determine molecular formula.
  • Molecular Formula:
    • Calculation based on n:
    • \text{Molecular Formula} = \text{Empirical Formula} \times n = \text{CH}2\text{O} \times 3 = \text{C}3\text{H}6\text{O}3

Molarity Calculation Example

  • Example Problem: What volume of a 1.50 M Potassium Bromide solution contains 66.0 g?
  • Conversion Process:
    • We need to convert grams to moles to liters:
    • Calculation: 66.0 \text{ g} \times \frac{1 \text{ mol}}{119.002 \text{ g}} \times \frac{1 \text{ L}}{1.50 \text{ mol}} = 0.370 \text{ L}

Concentration Units in Solutions

Alternative Units for Solution Concentration

  • Molarity (M) is the most widely used unit for evaluating the concentration of solutions:
    • Example notations for concentration:
    • [HCl] = 12 M
    • [NH₃] = 2.5 M
Mass Percentage
  • The mass percentage of a solution component is defined as the ratio of the component's mass to the solution's total mass, expressed as a percentage.
  • Commonly used to compute the mass percentages of solutes, but can apply to solvents as well.
  • Terminology: Percent mass, percent weight, weight/weight percent.
  • Notable for consumer products, e.g., 7.4% bleach.
Example Calculation of Percent by Mass
  • Context: A bottle of tile cleaner contains 135 g of HCl and 775 g of water.
  • Calculation of percent by mass of HCl:
    • Formula: \text{percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100
    • Total mass of solution: 135 g + 775 g = 910 g.
    • Percent by mass calculation:
    • \text{Percent by mass of HCl} = \frac{135 \text{ g}}{910 \text{ g}} \times 100
Molarity Calculation for Concentrated HCl
  • Problem: Concentrated HCl is 37.2% HCl by mass. The density of this solution is 1.19 g/mL. A) What is the molarity of this solution?
  • To find molarity, first calculate mass of HCl in 1 L of solution using its volume and density. Then, find the number of moles of HCl. B) What volume of concentrated HCl solution contains 125 g of HCl?
  • Use total mass and molarity to find the required volume.
Volume Percent
  • The concentration of a solution formed by dissolving a liquid solute in a liquid solvent can be expressed in volume percentage terms.
Mass-Volume Percentage
  • Definition: A mass-volume percent is a ratio of a solute’s mass to the solution’s volume, expressed as a percentage.
  • Variable specific units for solute mass and solution volume apply depending on the solution context.
  • Example: Physiological saline solution has a concentration of 0.9% mass/volume (m/v). This indicates that the solution contains 0.9 g of solute per 100 mL of solution.
Parts per Million (ppm) and Parts per Billion (ppb)
  • These units are often used to express very low solute concentrations.
  • Mass-based definitions of ppm and ppb are particularly useful in reporting the concentrations of pollutants and trace contaminants in water.
  • Example: Maximum level of fluoride ion: 4 ppm.
  • Density factors are generally required for calculations regarding ppm and ppb, base density of water is typically 1.00 g/mL.
Extra Practice Problems
  • Problem Example 1: Calculate the concentration of phosphate ions in 345 mL of a 0.535 M calcium phosphate solution.
  • Problem Example 2: The density of a 2.95 M solution of HNO₃ is 1.094 g/mL. What is the percent by mass of HNO₃ in this solution?
  • Problem Example 3: To what volume should you dilute 50.0 mL of a 12 M stock solution of nitric acid to obtain a 0.100 M solution?