JP

Empirical and Molecular Formulas

Overview of Mass Percent and Its Applications

  • Introduces mass percent as a useful tool in chemistry.
  • Emphasizes the relationship between mass percent, empirical formula, molecular formula, and balanced equations.
  • The end goal is to create a balanced equation from the empirical and molecular formulas.

Empirical Formula

  • Definition: Empirical formulas show the lowest whole number ratio of different atoms in a compound.
  • Example: For a molecular formula like C2H4, the empirical formula is CH2.
  • Purpose: It indicates how many of each atom (e.g., hydrogens per carbon) are present but does not convey the exact number of atoms like the molecular formula.
  • Value: While the empirical formula can represent multiple molecular formulas, it is the molecular formula that provides detailed information needed for chemical reactions.

Percent Composition

  • Percent composition is addressed to connect mass percent with empirical and molecular formulas.
  • The process of determining empirical and molecular formulas involves several steps that utilize percent composition.

Six Steps to Deriving Empirical Formula

  1. Convert Percent to Grams: Assume a 100-gram sample for ease of calculation.
    • Example: If it’s 10% carbon, it becomes 10 g of carbon in the sample.
  2. Convert Grams to Moles: Use molar mass for conversion using the formula:
    • ext{Moles} = rac{ ext{grams}}{ ext{molar mass}}
  3. Write a Pseudo Formula: Using the mole quantities calculated, write a not-yet-finalized formula.
  4. Divide by the Smallest Subscript: Normalize the ratios of the atoms by dividing all subscripts in the pseudo formula by the smallest subscript.
  5. Ensure Whole Numbers: Multiply to achieve whole numbers for all subscripts if necessary.
  6. Determine the Empirical Formula: Finalize the empirical formula based on calculated whole numbers.

Example Calculation Steps

  • Example: Consider a sample with the following percentages: 59% selenium, 12% oxygen, 28% fluorine.
    1. Convert to grams: 59 g Se, 12 g O, 28 g F.
    2. Convert to moles using molar masses:
      • Selenium: 78.96 g/mol
      • Oxygen: 16 g/mol
      • Fluorine: 19 g/mol
    3. Calculate moles:
      • Se: rac{59}{78.96} = 0.7478 moles;
      • O: rac{12}{16} = 0.75 moles;
      • F: rac{28}{19} = 1.4737 moles.
    4. Pseudo formula becomes Se0.7478O0.75F1.4737.
    5. Divide by the smallest subscript (0.7478) to normalize to:
      • Se: 1, O: 1, F: 1.968
    6. Convert to whole numbers if necessary, here resulting in SeO1F2, or SeOF2.

Converting Grams Directly to Empirical Formula

  • If the lab results are given in grams instead of percentages, skip the conversion step.
  • Follow the subsequent steps to normalize the values and derive the empirical formula.

Example with Molecular Connection

  • Hypothetical Lab Results: 60% carbon, 4.48% hydrogen, 35.5% oxygen. (Aspiring to find empirical formula.)
  • Steps executed similarly as previous examples ultimately leading to C9H8O4 as both the empirical and molecular formula.

Molecular vs. Empirical Formulas

  • Definition: The molecular formula represents the actual number of atoms in a molecule.
  • An empirical formula may simplify to a smaller ratio.
  • Example: For glucose, if the empirical formula (CH2O) is known, finding the molecular formula involves knowing the molar mass and scaling the empirical formula.
  • The relationship is that the molecular formula can be derived by multiplying the empirical formula by a whole number to match the molar mass.

The Balanced Equation

  • Definition: A balanced equation ensures that the same amount of each element is on both sides of the equation per the law of conservation of mass.
  • Balancing methods are discussed, emphasizing that coefficients can be adjusted but subscripts must not be altered to maintain chemical identities.
  • Examples cover how to balance both combustion reactions and others, using coefficients derived from elemental counts.

Key Concepts in Balancing Chemical Equations

  • When balancing equations, consider the number of each type of atom on both sides, adjusting coefficients as necessary without changing subscripts.
  • Example: The combustion of methane (CH4) results in CO2 and H2O, requiring adjustments to balance carbon and hydrogen before oxygen.

Practical Applications of Balanced Equations

  • Understanding balanced equations allows for quantitative analysis in chemical reactions.
  • Stoichiometric ratios can be derived for calculations relating to yields, reactant consumption, and product formation based on the balanced equation.
  • Emphasizes that moles are crucial for conversions between substances in reactions; grams cannot be directly compared without this conversion.

Conclusion

  • Reiterates the significance of mastering both empirical and molecular formulas, along with the ability to balance equations, for successful application in chemistry, especially in stoichiometric calculations.