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Chemistry Review Notes

Key Definitions

  • Mole: A quantity of 6.02 x 10²³ of anything.

  • Molar Mass: The atomic mass expressed in grams, equivalent to the mass of one mole of any element.

  • Avogadro's Number: 6.02 x 10²³, a constant that relates the number of particles in a mole.

  • Empirical Formula: The simplest whole number ratio of the atoms of each element in a compound.

  • Molecular Formula: The actual number of atoms of each element in a molecule of a compound, which is a whole number multiple of the empirical formula.

Finding Molar Mass

  • To determine the molar mass of a compound such as Potassium Nitrate (KNO₃):
    • Potassium (K): 39.0983 g/mol
    • Nitrogen (N): 14.01 g/mol
    • Oxygen (O): 15.999 g/mol (multiply this by 3 for KNO₃)
    • Calculation: 39.0983 + 14.01 + (15.999 x 3) = 101.103 g/mol.

Relative Mass

  • Relative Mass: Compared to the smallest atom in the sample, typically using hydrogen as a base reference.
  • Example: Hydrogen (0.042 g) sets the baseline (relative mass = 1).

Examples of Mole Calculations

  • Example Problem 1: Convert 25 g of Cl₂ gas to moles.

    • Molar Mass Cl₂ = 70.9 g/mol.
    • Calculation: 25 g x (1 mol/70.9 g) = 0.353 moles of Cl₂.

  • Example Problem 2: Convert 0.15 moles of NaNO₃ to grams.

    • Molar Mass NaNO₃ = 84.994 g/mol.
    • Calculation: 0.15 mol x 84.994 g/mol = 12.75 g of NaNO₃.

Empirical and Molecular Formula Determination

  1. Start with mass of elements (in grams).

    • Convert to moles by dividing by the respective molar mass of each element.

  2. Divide by the smallest mole value to obtain ratio.

    • If ratio yields non-whole numbers (e.g., x/2.5), multiply to achieve whole numbers.
  • Example with Nitrogen and Oxygen:
    • If 4.2g N and 12g O:
      • N: 4.2g / 14.007g/mol = 0.300 mol, O: 12g / 15.999g/mol = 0.750 mol.
      • Ratio: N=1 (0.300/0.300), O=2.5 (0.750/0.300) → multiply by 2→ N₂O₅ (empirical).

Molecular Formula Calculations

  • To find the molecular formula from the empirical formula:
    1. Determine the molar mass of the empirical formula.
    2. Divide the molecular mass by the empirical mass to find the multiplier.
    3. Apply this multiplier to the empirical formula.
      • Example: C₁H₁ (empirical) and 78 g/mol (molecular).
      • Empirical mass = 13 g/mol, Factor = 78 g/mol / 13 g/mol = 6 → C₆H₆.

Percent Composition

  • To find percent composition of a compound:

    1. Calculate molar mass of the entire compound (H₂O).
    2. Divide the mass of each element by total molar mass and multiply by 100%.

  • Example for H₂O:

    • H: 2.016 g (2 * 1.008 g), O: 15.999 g, Total = 18.015 g.
    • Percent H = (2.016 / 18.015) * 100 = 11.2%, Percent O = (15.999 / 18.015) * 100 = 88.81%.

Final Problem Example

  • Problem involves percent by mass calculation for Iron and Oxygen:
    • Example: 20g Fe and 8g O = Total 36g.
    • Fe% = (20/36)100 = 55.56%, O% = (8/36)100 = 22.22%.