Empirical and Molecular Formula Determination & Exam Prep

Determining Empirical and Molecular Formulas

Elemental Analysis (EA) and Empirical Formula

• Elemental Analysis (EA) provides data in the form of weight percentages of elements in a substance.
• To find the composition or empirical formula (the base ratio of atoms), weight percentages must be converted to moles.
• Process to convert weight percentages to an empirical formula:
  1. Assume a $100 ext{ g}$ sample: This allows direct conversion of weight percentages to grams (e.g., $50\% ext{ by weight} = 50 ext{ g}$ in a $100 ext{ g}$ sample).
  2. Convert grams to moles: Divide the mass of each element by its respective molar mass (from the periodic table).
     • Example: If you have $50 ext{ g}$ of sulfur, divide by $32 ext{ g/mol}$ (molar mass of sulfur) to get moles of sulfur.
     • Example: If you have $50 ext{ g}$ of oxygen, divide by $16 ext{ g/mol}$ (molar mass of oxygen) to get moles of oxygen.
  3. Find the simplest whole-number ratio: Divide the number of moles of each element by the smallest number of moles calculated.
     • If the resulting ratios are not whole numbers (e.g., $1.67, 1.34$), multiply all ratios by a common small integer (e.g., $2, 3, 4$) until whole numbers are obtained. This ensures the empirical formula represents the simplest whole-number ratio of atoms.
• The resulting ratio defines the empirical formula, which shows the simplest atomic ratio in the compound.

Determining the Molecular Formula Using Mass Spectrometry (MS)

• The empirical formula represents the base ratio of elements.
• The molecular formula is a multiple of the empirical formula. For example, if the empirical formula is $ext{XY}<em>2$, possible molecular formulas could be $ext{XY}</em>2$, $ext{X}<em>2 ext{Y}</em>4$, $ext{X}<em>3 ext{Y}</em>6$, etc., all maintaining the same $1:2$ ratio.
• Mass Spectrometry (MS) is used to determine the molecular mass of the compound.
  • The heaviest peak in the mass spectrum typically corresponds to the molecular ion, which gives the compound's overall molecular mass.
• Process to find the molecular formula:
  1. Calculate the molar mass of the empirical formula.
  2. Identify potential molecular formula options by scaling up the empirical formula (e.g., $ext{n} imes ( ext{empirical formula})$, where $ext{n}$ is an integer like $1, 2, 3$…). Calculate the molar mass for each option.
  3. Compare these calculated molar masses with the molecular mass observed in the mass spectrum.
  4. The molecular formula whose calculated molar mass matches the highest peak in the mass spectrum is the correct molecular formula.

Examples and Walk-Throughs

• Example 1: Sulfur and Oxygen (50% S, 50% O by weight)
  • Assume $100 ext{ g}$ sample: $50 ext{ g S}$, $50 ext{ g O}$.
  • Moles:
    • $ext{Mol S} = \frac{50 ext{ g}}{32 ext{ g/mol}} = 1.56 ext{ mol}$
    • $ext{Mol O} = \frac{50 ext{ g}}{16 ext{ g/mol}} = 3.13 ext{ mol}$
  • Ratio: Divide by $1.56$
    • $ext{S}: \frac{1.56}{1.56} = 1$
    • $ext{O}: \frac{3.13}{1.56} hickapprox 2$
  • Empirical Formula: $ext{SO}_2$.
  • Molecular Options: $ext{SO}<em>2$, $ext{S}</em>2 ext{O}<em>4$, $ext{S}</em>3 ext{O}_6$.
  • Molar Masses:
    • $ext{SO}_2 = (32 + 2 imes 16) = 64 ext{ g/mol}$
    • $ext{S}<em>2 ext{O}</em>4 = (2 imes 32 + 4 imes 16) = 128 ext{ g/mol}$
  • If Mass Spec peak is at $64 ext{ g/mol}$, then the Molecular Formula is $ext{SO}_2$.
• Example 2: Carbon, Hydrogen, Chlorine (Hypothetical %s leading to non-whole numbers)
  • Given moles (after conversion from percentages): $ext{C} = 2.4 ext{ mol}$, $ext{H} = 1.464 ext{ mol}$, $ext{Cl} = 1.9526 ext{ mol}$.
  • Divide by lowest ( $1.464$):
    • $ext{C}: \frac{2.4}{1.464} hickapprox 1.67$
    • $ext{H}: \frac{1.464}{1.464} = 1$
    • $ext{Cl}: \frac{1.9526}{1.464} hickapprox 1.34$
  • Since these are not whole numbers, multiply by $3$ (the smallest integer to get approximate whole numbers):
    • $ext{C}: 1.67 imes 3 hickapprox 5$
    • $ext{H}: 1 imes 3 = 3$
    • $ext{Cl}: 1.34 imes 3 hickapprox 4$
  • Empirical Formula: $ext{C}<em>5 ext{H}</em>3 ext{Cl}_4$. Its molar mass is e.g. $205 ext{ g/mol}$.
  • If Mass Spec peak is at $410 ext{ g/mol}$, which is twice the empirical formula mass ($2 imes 205 = 410$), then scale the empirical formula by $2$.
  • Molecular Formula: $ext{C}<em>{10} ext{H}</em>6 ext{Cl}_8$.
• Example 3: Nitrogen and Oxygen
  • Given moles (after conversion from percentages): $ext{O} = 4.34 ext{ mol}$, $ext{N} = 2.17 ext{ mol}$.
  • Ratio: Divide by $2.17$
    • $ext{N}: \frac{2.17}{2.17} = 1$
    • $ext{O}: \frac{4.34}{2.17} hickapprox 2$
  • Empirical Formula: $ext{NO}_2$.
  • Molecular Options: $ext{NO}<em>2$, $ext{N}</em>2 ext{O}<em>4$, $ext{N}</em>3 ext{O}_6$.
  • Molar Mass of $ext{NO}_2$: $(14 + 2 imes 16) = 46 ext{ g/mol}$.
  • If Mass Spec peak is at $46 ext{ g/mol}$, then the Molecular Formula is $ext{NO}_2$.

Exam Logistics (Thursday)

• Location: Check ICON for your assigned room based on your last name.
• Time:
  • Be in your seat well before $06:30$ (AM/PM). Aim to arrive at least $15 ext{ minutes}$ prior.
  • Exam starts precisely at $06:30 ext{ (AM/PM)}$.
• What to bring:
  • Pencil
  • Pen
  • Calculator
  • Student ID (required to collect the exam)
• Resources for Study:
  • Information page on ICON.
  • Sample exam with detailed video walkthroughs on ICON:
    • Q1-11 walkthrough
    • Q11-22 walkthrough (Note: The instructor mentioned 11-22 but it could be 12-22 or a typo for the range)

Unit 1 Review and Integration

• Homework: Due tomorrow.
• **