Empirical vs Molecular Formulas

Chapter 1: Introduction

• Comparison of Empirical and Molecular Formulas

  • Introduction to empirical formulas as the first step in solving molecular formulas.

  • Upcoming lab assignment to calculate an empirical formula from lab data.

• Definitions

  • Empirical Formula: A formula that gives the simplest whole number ratio of atoms in a molecule.

  • Molecular Formula: A formula that gives the actual whole number ratio of atoms in a molecule.

• Examples

  • For the molecular formula $C<em>2H</em>6$:

  • Explanation of bonding between 2 carbons and 6 hydrogens.

  • Simplification: Given that 2 carbons (C) and 6 hydrogens (H) can be simplified to a ratio of 1:3, the empirical formula is $CH_3$.

  • For the molecular formula $C<em>6H</em>{12}O_6$:

  • Identification of the molecular formula.

  • Simplification: Dividing by 6 results in the empirical formula $CH_2O$.

Chapter 2: Atoms of Carbon

• Finding Empirical Formulas with Data

  • Example problem: A compound is 80% carbon and 20% hydrogen. Task is to find the empirical formula.

  • Mass percent overview: 80% is the mass percent of carbon and 20% is the mass percent of hydrogen.

• Understanding Ratios

  • Discussion on how quantities relate to the ratio of parts in a whole using a metaphor (the cheese sandwich metaphor).

  • Example: For $CO_2$, one molecule contains 1 atom of carbon and 2 atoms of oxygen, illustrating the need for ratios in molecular representation.

• Scaling Ratios

  • If scaling from 1 sandwich to 12 sandwiches, the ratio of bread to sandwiches remains the same at 2 pieces.

  • Mole concept introduced: One mole equals $6.02 imes 10^{23}$ as a scaling factor for large quantities.

Chapter 3: Need a Dozen

• Ratios in Molecules

  • For individual molecules like $CH<em>3$, scaling to a dozen or a mole is established: 1 mole of $CH</em>3$ needs 1 mole of carbon and 3 moles of hydrogen.

• Mole Abbreviation

  • Discussion on the abbreviation for mole, which is often written without the 'e'.

Chapter 4: Grams of Carbon

• Converting Percentages to Mass

  • Importance of knowing how to translate percentages into grams to compute empirical formulas.

  • If 80% is carbon, it means that in any sample, the mass of carbon can be calculated directly.

• Mass/Mole Conversion

  • Mass to Moles Conversion:

  • Carbon Molar Mass: $ext{Molar Mass of Carbon} = 12 ext{ g/mol}$

  • Divide mass by molar mass to find number of moles.

Chapter 5: Grams of Carbon (Continued)

• Example Calculation

  • For 80 grams of carbon: $80 ext{ g} imes rac{1 ext{ mol}}{12 g} = 6.666… ext{ mol}$ (repeating 6).

  • For hydrogen: $20 ext{ g} imes rac{1 ext{ mol}}{1 g} = 20 ext{ mol}$.

Chapter 6: Grams of Carbon (Reiteration)

• Establishing Ratios

  • Ratio from previous calculations: $C<em>{6.666…}H</em>{20}$ lacks whole numbers.

  • To simplify ratios:

  • Divide by the smallest mole value.

  • For example: $C<em>{6.666…}$ divided by itself results in 1, and $20$ divided by $6.666…$ gives 3 to simplify to $C</em>1H_3$.

Chapter 7: Conclusion

• Summary of Steps for Empirical Formulas

  • Repeat process for empirical formulas—always: convert mass percent to mass, mass to moles, and divide by the smallest whole number.

• Practice Assignments

  • Suggestions for practice with simpler problems before advanced ones. Check answers provided in textbook for validation.