Recording-2025-02-27T22:09:16.744Z

Moles and Molar Mass

• One mole is equal to 6.022 x 10²³ entities, known as Avogadro's number.

• When converting grams to moles for compounds, use molar mass (molecular mass), which is expressed in grams per mole (g/mol).

• To find the molar mass of a compound, add up the atomic masses of all elements in the compound:

  • Example: For water (H₂O), the calculation includes:

    • 2 hydrogen (H) atoms × atomic mass of H (1.008 g/mol)

    • 1 oxygen (O) atom × atomic mass of O (16.00 g/mol)

• To convert from moles to molecules, use Avogadro's number.

• Questions may involve conversions between grams, moles, and molecules in relation to compounds.

Percent by Mass

• Percent by mass is calculated using the formula:

  • Percent by mass = (mass of part / total mass) × 100%

• Example Calculation:

  • For a compound containing carbon and chlorine, to find percent of chlorine:

    • Find mass of chlorine (35.453 g/mol), multiply by the number of chlorine atoms for total mass.

    • Divide that by the total mass of the compound.

  • Result: The example found 58.64% chlorine.

• Percent by mass can be converted into a conversion factor:

  • 58.64 g Cl / 100 g compound or vice versa.

Using Percent by Mass as a Conversion Factor

• Using percent by mass allows quick calculations in problems involving compounds.

• Example Practice Problem:

  • Given 79.2% carbon in benzaldehyde, it can be expressed as 79.2 g of carbon per 100 g of benzaldehyde.

• From sodium chloride, calculate grams of sodium in a given mass:

  • Start by calculating percent by mass of sodium in sodium chloride.

  • Use molecular masses:

    • Sodium (Na): 22.99 g/mol

    • Chloride (Cl): 35.453 g/mol

    • Sodium chloride (NaCl) = 22.99 + 35.453 = 58.44 g/mol

Empirical and Molecular Formulas

Determining Empirical Formulas

• Steps to find an empirical formula using percent composition:

  1. Assume a 100 g sample to simplify percent to grams.

  2. Convert grams to moles using atomic masses.

  3. Divide by the smallest mole value to simplify to ratios.

  4. If ratios are not whole numbers, multiply by a factor to obtain whole number ratios.

• Example with Stannous Fluoride:

  • Given 75.7% tin, calculate amounts for other element (fluoride).

Molecular Formulas

• The molecular formula is the actual formula of the compound.

• To derive the molecular formula:

  1. Calculate the empirical formula mass.

  2. Divide the molar mass by the empirical formula mass to find a factor.

  3. Multiply the subscripts in the empirical formula by this factor.

• Example:

  • Empirical formula yields C₆H₁₂O₅ with molar mass derived from the calculation leading to a final molecular formula.

Combustion Analysis

• Although it wasn't tested in class, combustion analysis methods often involve determining the empirical formula but is noted to be complex.

Naming Compounds

• Transition to Chapter 5:

  • Emphasis on naming compounds to understand next steps in drawing Lewis structures and understanding bond polarity.

  • Importance of knowing compounds to answer related questions throughout quizzes and tests.