Che130 L7

Extra Practice Problem: Carbon Tetrachloride

• Problem: Determine the number of chlorine atoms present in a 25.00 µL sample of carbon tetrachloride (CCl₄) given that its density is 1.59 g/mL.

• Steps:

  1. Convert µL to mL:
     25.00 \text{ µL} \times 10^{-6} \text{ L/µL} \times 1 \text{ mL}/10^{-3} \text{ L} = 2.5 \times 10^{-2} \text{ mL}

  2. Calculate mass of CCl₄:
     2.5 \times 10^{-2} \text{ mL} \times 1.59 \text{ g/mL} = 0.03975 \text{ g}

  3. Convert grams to moles:
     \frac{0.03975 \text{ g}}{153.82 \text{ g/mol}} \approx 2.58 \times 10^{-4} \text{ mol CCl}_4

  4. Determine moles of Cl:
     2.58 \times 10^{-4} \text{ mol CCl}4 \times 4 \text{ mol Cl/mol CCl}4 = 1.03 \times 10^{-3} \text{ mol Cl}

  5. Convert moles of Cl to atoms:
     1.03 \times 10^{-3} \text{ mol Cl} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 6.22 \times 10^{20} \text{ atoms Cl}

Practice Problem: PO₄³⁻ Ions in Ca₃(PO₄)₂

• Problem: How many PO₄³⁻ ions are in 23.25 g of Ca₃(PO₄)₂?

• Steps:

  1. Convert grams to moles:

  • Molar mass of Ca₃(PO₄)₂ is calculated:
    3 \times 40.08 \text{ g/mol (Ca)} + 2 \times (30.97 \text{ g/mol (P)} + 4 \times 16.00 \text{ g/mol (O)}) \approx 310.17672 \text{ g/mol}

  • Moles:
    23.25 \text{ g} \times \frac{1 \text{ mol Ca}3(\text{PO}4)2}{310.17672 \text{ g}} \approx 0.075 \text{ mol Ca}3(\text{PO}4)2

  1. Determine moles of PO₄³⁻ ions:

  • Each molecule of Ca₃(PO₄)₂ produces 2 mol of PO₄³⁻ ions:
    0.075 \text{ mol Ca}3(\text{PO}4)2 \times 2 \text{ mol PO}4^{3-}/\text{mol Ca}3(\text{PO}4)2 = 0.15 \text{ mol PO}4^{3-}

  1. Convert to number of ions:
     0.15 \text{ mol PO}4^{3-} \times 6.022 \times 10^{23} \text{ ions/mol} \approx 9.028 \times 10^{22} \text{ ions PO}4^{3-}

Percent Composition of C₂H₈O₂

• Problem: Determine the % composition of H in C₂H₈O₂.

• Steps:

  1. Calculate molar mass:

  • C: 12.01 g/mol \times 2 = 24.02 g/mol

  • H: 1.008 g/mol \times 8 = 8.064 g/mol

  • O: 16.00 g/mol \times 2 = 32.00 g/mol

  • Total molar mass = 24.02 + 8.064 + 32.00 = 64.084 g/mol.

  1. Calculate mass percent of H:
     \frac{8.064 \text{ g}}{64.084 \text{ g}} \times 100\% \approx 12.59\%

Chapter 3.2: Determining Molecular and Empirical Formulas

Determination of Empirical Formula

1. Calculate mass of A and mass of X atoms.

2. Divide the mass of each element by its molar mass.

3. Find the number of moles for A and X.

4. Divide by the smallest number of moles to find the mole ratio.

5. Convert the ratio to the lowest whole numbers to establish the empirical formula.

Determination of Empirical Formula from Percent Composition

1. Assume a 100 g sample of the compound to convert percent composition to masses of elements.

2. Derive number of moles from the mass of each element.

3. Divide each molar amount by the smallest molar amount.

4. Multiply coefficients by an integer for whole number ratios.

Example Problem: Finding an Empirical Formula

• Problem: What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?

Derivation of Molecular Formula

• A molecular formula is established from its empirical formula and its molecular or molar mass.

• To find the molecular formula, multiply each subscript in the empirical formula by an integer n, where:

  • n = \frac{\text{molar mass}}{\text{molar mass of empirical formula}}

Chapter 3.3: Molarity

Introduction to Solutions and Molarity

• Solutions are homogeneous mixtures of two or more substances.

• Mixtures are more common than pure substances in nature.

• The effectiveness of a solution, ex: medicine, depends on the concentration of its active ingredient.

Components of a Solution

• Solvent: Major component, typically with a higher concentration.

• Solute: Minor component, typically present at a lower concentration.

• An aqueous solution is one where water is the solvent.

Examples of Solutions

• Kool-Aid in water: solute = Kool-Aid, solvent = water.

• Saltwater: solute = salt, solvent = water.

• Air: gas/gas solution.

• Bronze: solid solution of copper and zinc (Zn/Cu).

• Vodka: solution of ethanol and water.

Making a Solution

1. Weigh the solute.

2. Transfer to a volumetric flask.

3. Fill the flask to the mark with distilled or deionized water.

Concentration Terms: Concentrated vs. Dilute

• Concentrated: high solute concentration.

• Dilute: low solute concentration.

• Note: Water molecules are not visible in the solution.

Molarity Definition

• Molarity (M): Number of moles of solute per liter of solution. It is expressed as:

  • M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}

Practice Problem: Calculating Molarity

• Problem: Calculate molarity of 6.52 g of CoCl₂ (molar mass = 128.9 g/mol) in 75.0 mL of solution.

• Conversion to mol:

  • \text{Moles} = \frac{6.52 \text{ g}}{128.9 \text{ g/mol}} = 0.0507 \text{ mol}

• Molarity calculation:

  • Volume in L: 75.0 mL = 0.075 L.

  • M = \frac{0.0507 \text{ mol}}{0.075 \text{ L}} = 0.676 \text{ M}

Dilutions

Understanding Dilutions

• A dilution lowers the concentration of a solution by adding solvent.

• Moles of solute remain constant during dilution:

  • \text{moles solute in concentrated solution} = \text{moles solute in dilute solution}

• The dilution relation is given by:

  • M1V1 = M2V2

• M₁ = molarity of concentrated solution, V₁ = volume of concentrated solution,

• M₂ = molarity of diluted solution, V₂ = volume of diluted solution.

Dilution Process

• Measure desired volume (V₁) of stock solution.

• Transfer to a second volumetric flask.

• Add solvent to achieve the final desired volume.

Solutions Concentration Units

Other Concentration Units

• Molarity is widely used but there are other important concentration units.

Mass Percentage

• Defined as the ratio of the component’s mass to the total solution’s mass expressed as a percentage.

• Often used for solutes and can also compute solvent percentages.

• Example: 7.4% bleach.

Practice Problem: Percent Mass Calculation

• Question: A bottle of tile cleaner contains 135 g HCl in 775 g of water. Calculate mass percentage of HCl:

  • \text{Percent mass HCl} = \frac{135 \text{ g}}{135 \text{ g} + 775 \text{ g}} \times 100\% \approx 14.83 \%

Volume Percent

• Concentration of liquid solute in a liquid solvent expressed as volume percentage.

Mass-Volume Percentage

• The ratio of a solute’s mass to the solution’s volume as a percentage.

• Example: 0.9% (m/v) for physiological saline solution indicates 0.9 g of solute per 100 mL of solution.

Parts per Million and Billion

• Used for very low concentrations, especially in environmental science.

• Definitions:

  • ppm: mg solute per L of solution.

  • ppb: µg solute per L of solution.

• Example Application: Reporting pollutant concentrations (max level of fluoride ion: 4 ppm).