Chapter 6 – Chemical Composition: Avogadro’s Number, the Mole, and Molar Mass

Avogadro’s Number & the Concept of the Mole

• Avogadro’s number: $6.022 \times 10^{23}$ objects.

  • Symbolically: $1\ \text{mol} = 6.022 \times 10^{23}\ \text{units (atoms, molecules, ions, etc.)}$

  • Named after Amedeo Avogadro (1776–1856).

• Analogy

  • Just as “1 dozen = 12,” “1 mole = $6.022\times10^{23}$.”

  • If you own a mole of marbles, you possess $6.022\times10^{23}$ marbles.

Defining the Mole via Carbon-12

• Official SI definition (simplified): the amount of substance containing as many elementary entities as there are atoms in exactly 12 g of $^{12}\text{C}$.

• Reasoning chain:

  1. One $^{12}\text{C}$ atom has a mass of $12\ \text{u}$ (atomic-mass units).

  2. Determine how many such atoms weigh 12 g → result is $6.022\times10^{23}$.

  3. Therefore, any substance with $6.022\times10^{23}$ entities is 1 mol of that substance.

Mass Depends on the Identity of the Atoms

• Same number of particles ≠ same mass.

  • 1 mol C atoms → $12.01\ \text{g}$.

  • 1 mol S atoms → $32.07\ \text{g}$.

• Visual slide analogy: different “dozens” of nails weigh differently; likewise 1 mol of different atoms/compounds weighs differently.

Molar Mass (M)

• Definition: mass of 1 mol of a substance.

  • For elements: numerically equal to the atomic mass on the periodic table, but expressed in $\text{g mol}^{-1}$.

  • For compounds: sum of the atomic masses of each atom contained in the formula.

• Mathematical form: $M = \sum<em>{i} n</em>i M_i$

  • $n_i$ = number of atoms of element $i$ in the formula.

  • $M_i$ = atomic mass of element $i$ (u → g mol$^{-1}$).

• Examples

  • Copper: one Cu atom = $63.55\ \text{u}$, so $1\ \text{mol Cu} = 63.55\ \text{g}$.

  • \ce{CO2}: $M = 1(12.01) + 2(16.00) = 44.01\ \text{g mol}^{-1}$.

  • \ce{H2O}: $M = 2(1.008) + 1(15.999) \approx 18.02\ \text{g mol}^{-1}$.

Reading Molar Masses from the Periodic Table

• Hydrogen: atomic mass $=1.008$ ⇒ M_{\ce H}=1.008\ \text{g mol}^{-1}.

• Oxygen: atomic mass $=16.00$ ⇒ M_{\ce O}=16.00\ \text{g mol}^{-1}.

• Always keep at least two decimal places in calculated molar masses for accuracy.

Mole Ratios Embedded in Chemical Formulas

• Chemical formula provides a conversion factor between moles of compound and moles of constituent atoms/ions.

  • \ce{H2O}: $2\ \text{mol H} : 1\ \text{mol O} : 1\ \text{mol H2O}$.

  • \ce{CCl4}: $4\ \text{mol Cl} : 1\ \text{mol CCl4}$.

  • \ce{CO2}: $2\ \text{mol O} : 1\ \text{mol CO2}$.

• Analogy: “1 spider → 8 legs” ; “1 molecule \ce{H2O} → 2 H atoms.”

Dimensional-Analysis Roadmap

Number of entities  ⇄  Moles  ⇄  Mass (g)
        |                 |
  Avogadro's #         Molar mass

• Avogadro’s number converts between number ↔ moles.

• Molar mass (g mol$^{-1}$) converts between moles ↔ grams.

• Problems may chain multiple conversions (e.g., grams → moles → number of atoms in a sub-element).

Worked Examples

1. Molecules → Moles (HCl)

• Given: $4.3\times10^{23}$ molecules \ce{HCl}.

• Set-up: $4.3\times10^{23}\ \text{molecules} \times \frac{1\ \text{mol}}{6.022\times10^{23}\ \text{molecules}}$.

• Result (2 sig figs): $7.1\ \text{mol HCl}$.

2. Grams → Molecules (Water)

• Given: $35\ \text{g H2O}$.

• Steps:

  1. $35\ \text{g} \times \frac{1\ \text{mol}}{18.02\ \text{g}}$ = $1.94\ \text{mol}$.

  2. $1.94\ \text{mol} \times 6.022\times10^{23}$ = $1.17\times10^{24}$ molecules.

• Rounded to 2 sig figs: $1.2\times10^{24}$ water molecules.

3. Grams (Water) → Hydrogen Atoms

• Given: $46\ \text{g H2O}$.

• Chain:

  1. $\frac{46}{18.02}=2.55\ \text{mol H2O}$.

  2. Mole ratio: $2.55\ \text{mol H2O} \times \frac{2\ \text{mol H}}{1\ \text{mol H2O}} = 5.10\ \text{mol H}$.

  3. Convert to atoms: $5.10\ \text{mol} \times 6.022\times10^{23} = 3.07\times10^{24}$ H atoms.

• Rounded (2 sig figs): $3.1\times10^{24}$ H atoms.

4. Molecules (CO₂) → Grams of Carbon

• Given: $1.2\times10^{23}$ molecules \ce{CO2}.

• Path: molecules → mol CO₂ → mol C → g C.

  1. $\frac{1.2\times10^{23}}{6.022\times10^{23}}=0.199\ \text{mol CO2}$.

  2. Mole ratio: $0.199\ \text{mol CO2} \times \frac{1\ \text{mol C}}{1\ \text{mol CO2}}=0.199\ \text{mol C}$.

  3. Mass: $0.199\ \text{mol} \times 12.01\ \text{g mol}^{-1}=2.39\ \text{g}$.

• 2 sig figs → $2.4\ \text{g C}$.

5. Grams (Carbonic Acid) → Hydrogen Atoms

• Compound: \ce{H2CO3}; $M = 63.02\ \text{g mol}^{-1}$.

• Given: $23\ \text{g}$.

  1. $23\ \text{g} \times \frac{1\ \text{mol}}{63.02\ \text{g}} = 0.365\ \text{mol H2CO3}$.

  2. Mole ratio: 2 mol H per 1 mol compound → $0.729\ \text{mol H}$.

  3. Convert: $0.729\ \text{mol} \times 6.022\times10^{23}=4.39\times10^{23}$ H atoms.

• Rounded (2 sig figs): $4.4\times10^{23}$ H atoms.

Conceptual & Real-World Connections

• Ore analysis: compute % Fe in iron ore to gauge economic value.

• Environmental chemistry: determine Cl mass in chlorofluorocarbons for ozone-depletion studies.

• Industrial batching: recipes scaled by moles ensure stoichiometric precision (e.g., pharmaceuticals).

• Counting by mass is essential because individual atoms/molecules are far too small to tally directly.

• Ethical dimension: precise mole-based dosing prevents under/over-medication and minimizes waste.

Summary Cheat-Sheet

• Always identify given and wanted quantities with units & substance labels.

• Choose the correct bridge:

  • Avogadro’s number for number ↔ moles.

  • Molar mass for moles ↔ grams.

  • Formula subscripts for compound moles ↔ element moles.

• Keep significant figures consistent with the least precise given data.

• Track both units and chemical identity throughout calculations to avoid cancelation errors.