Introduction to the Mole in Chemistry

Counting by Weighing

• Analogy: Factory counting small hardware pieces (nuts, bolts, screws, washers) by weighing rather than individually counting due to large orders.
  • Common Issue: Slight variations in individual weights.
  • Solution: Average weight using a random sample of 100 pieces for better accuracy.
• Application to Chemistry: Similar issues arise when counting atoms and molecules due to their small size and isotopic variations.

The Mole (mol)

• Definition: A counting unit in chemistry similar to a dozen but much larger.
  • 1 mole = $6.02 \times 10^{23}$ (Avogadro's number).
• Purpose: To count atoms and molecules by weighing.
• Molar Mass: The mass (in grams) of one mole of a substance. Periodic table average atomic masses (amu) converted to grams.

Counting and Weighing with Moles

• Periodic Table as Conversion Tool:
  • Example: Carbon (C): 12.01 amu $ ightarrow$ 12.01 g/mol.
  • Example: Phosphorus (P): 30.97 amu $ ightarrow$ 30.97 g/mol.

Conversion Relationships

• Moles and Counting Numbers: Use Avogadro’s number ($1 \text{ mole} = 6.02 \times 10^{23}$).
• Moles and Grams: Use the periodic table for molar masses.
• No Direct Conversion: Between counting numbers and grams; always go through moles to avoid overly complex or inaccurate conversions.

Practical Examples

Example 1: Atoms in a Diamond

• Problem: How many atoms in a 315 mg diamond (pure C)?
• Steps:
  1. Convert mg to g: $315 \text{ mg} \rightarrow 0.315 \text{ g}$.
  2. Convert g to moles using molar mass of C: $0.315 \text{ g} \rightarrow 0.0262 \text{ mol}$ (Step: $\frac{0.315}{12.01}$).
  3. Convert moles to atoms: $0.0262 \text{ mol} \times 6.02 \times 10^{23} = 1.58 \times 10^{22}$ atoms.

Example 2: Mass from Number of Atoms

• Problem: Mass of $5.78 \times 10^{22}$ atoms of iron (Fe)?
• Steps:
  1. Convert atoms to moles: $5.78 \times 10^{22} \rightarrow 0.0961 \text{ mol}$ (Step: $\frac{5.78 \times 10^{22}}{6.02 \times 10^{23}}$).
  2. Convert moles to grams using molar mass of Fe: $0.0961 \times 55.85 = 5.36 \text{ g}$.

Complex Compounds

Example 3: Molar Mass of a Compound

• N$_2$Br$_4$ Calculation:
  • 2 N atoms: $2 \times 14.01 = 28.02$ g/mol.
  • 4 Br atoms: $4 \times 79.91 = 319.64$ g/mol.
  • Total molar mass: $28.02 + 319.64 = 347.66$ g/mol.

Example 4: Converting Mass to Molecules (CO$_2$ Example)

• Problem: Molecules in 1.5 lb (0.68 kg) dry ice.
• Steps:
  1. Convert lbs to grams: $1.5 \text{ lb} \rightarrow 680 \text{ g}$.
  2. Convert grams to moles (using 44.01 g/mol for CO$_2$): $680 \text{ g} \rightarrow 15.45 \text{ mol}$ (Step: $\frac{680}{44.01}$).
  3. Convert moles to molecules: $15.45 \text{ mol} \times 6.02 \times 10^{23} = 9.3 \times 10^{24}$ molecules.

Example 5: Number of Atoms in a Compound

• Problem: Atoms of oxygen in 436 mg of calcium nitrate (Ca(NO$_3$)$_2$)?
• Steps:
  1. Convert mg to g: $436 \text{ mg} \rightarrow 0.436 \text{ g}$.
  2. Calculate molar mass of Ca(NO$_3$)$_2$: 164.10 g/mol (using atomic masses).
  3. Convert grams to moles: $0.436 \text{ g} \rightarrow 0.00266 \text{ mol}$ (Step: $\frac{0.436}{164.10}$).
  4. Convert moles to molecules using Avogadro's number: $0.00266 \text{ mol} \rightarrow 1.60 \times 10^{21}$ molecules.
  5. Calculate atoms of O (6 atoms/O2 molecule): $1.60 \times 10^{21} \times 6 = 9.60 \times 10^{21}$ atoms of O.

Summary

• Two Definitions of the Mole Used:
  1. Counting Number: $6.02 \times 10^{23}$.
  2. Counting by Weighing: Molar Mass (g/mol).
• Technique: Dimensional analysis for unit cancellation and conversions.
• Periodic Table: Integral for conversion factors and calculations.
• Relevance: Fundamental starting point for advanced chemistry calculations.

Key Takeaway: Mastering the basic mole calculations is essential as the concept of the mole is pervasive in all areas of chemistry.