Chapter 6 – Chemical Composition & The Mole

Writing Chemical Formulas

• Fundamental rules
  • Each atom is written with its element symbol.
  • Quantity of each atom is shown as a subscript to the right of its symbol.
  • If only one atom of a given element is present, the subscript $1$ is omitted.
• Example: $\text{SO}_3$
  • Symbol $S$ represents sulfur.
  • Symbol $O$ represents oxygen.
  • No subscript after $S$ ⇒ one sulfur atom.
  • Subscript $3$ after $O$ ⇒ three oxygen atoms.
• Practice problem (DDT)
  • Contains $14$ C, $9$ H, $5$ Cl atoms.
  • Formula: $\text{C}<em>{14}\text{H}</em>9\text{Cl}_5$.
• Counting atoms in a formula
  • Example: $\text{Mg(NO}<em>3)</em>2$
    • One $Mg^{2+}$ ion ⇒ $1$ Mg atom.
    • Two $\text{NO}_3^-$ groups.
    • $N$: $1 \times 2 = 2$ atoms.
    • $O$: $3 \times 2 = 6$ atoms.

The Mole Concept

• A "collection term" = fixed number of identical items
  • $1$ dozen $= 12$ items
  • $1$ ream $= 500$ sheets
  • $1$ gross $= 144$ items
• Mole (SI amount‐of‐substance unit)
  • $1$ mole $= 6.022 \times 10^{23}$ particles (Avogadro’s number).
  • Provides bridge between microscopic (atoms) & macroscopic (grams) world.
• For elements
  • $1$ mole C $\rightarrow 6.022\times10^{23}$ C atoms.
  • Same numerical value for Na, Au, etc.
    (mass differs → molar mass differences).

Mole Conversions

• Two reciprocal factors:
  • $\dfrac{1\,\text{mol}}{6.022\times10^{23}\,\text{particles}}$
  • $\dfrac{6.022\times10^{23}\,\text{particles}}{1\,\text{mol}}$
• Example: Convert $3.5$ mol He → atoms
  $3.5\,\text{mol He} \times \dfrac{6.022\times10^{23}\,\text{He atoms}}{1\,\text{mol He}} = 2.1\times10^{24}\,\text{atoms}$.
• Quick check questions
  1. Atoms in $2.0$ mol Al ⇒ $1.2\times10^{24}$ (Answer C).
  2. Moles in $1.8\times10^{24}$ atoms S ⇒ $3.0$ mol (Answer B).

Molar Mass (MM)

• Definition: mass of $1$ mole of substance, in $\text{g mol}^{-1}$.
  • For atoms MM equals atomic mass (periodic table) expressed in grams.
  • Examples
    • He: $4.003\,\text{g mol}^{-1}$.
    • Ag: $107.9\,\text{g mol}^{-1}$.
    • C: $12.01\,\text{g mol}^{-1}$.
    • S: $32.07\,\text{g mol}^{-1}$.
• Equal-mole samples contain identical particle count, differing masses.
  E.g., $1$ mol Al (26.98 g) vs $1$ mol Au (196.97 g).

Gram ⇄ Mole ⇄ Particle Conversions

• General pathways
  • Grams ↔ moles via molar mass factor $\dfrac{\text{g}}{\text{mol}}$.
  • Moles ↔ particles via Avogadro factor.
• Worked examples
  1. g → mol (sulfur):
     $57.8\,\text{g S} \times \dfrac{1\,\text{mol S}}{32.07\,\text{g S}} = 1.80\,\text{mol S}$.
  2. mol → g (aluminum):
     $6.73\,\text{mol Al} \times \dfrac{26.98\,\text{g Al}}{1\,\text{mol Al}} = 1.82\times10^{2}\,\text{g}$.
  3. g → atoms (Al can, 16.2 g)
     • Step 1: $16.2\,\text{g} \to \dfrac{16.2}{26.98}=0.600\,\text{mol}$.
     • Step 2: $0.600\,\text{mol} \times 6.022\times10^{23}=3.61\times10^{23}\,\text{atoms}$.
• Concept check (Cu atom mass)
  • Average atomic mass $63.55\,\text{g mol}^{-1}$.
  • Mass per atom $=\dfrac{63.55\,\text{g}}{6.022\times10^{23}}=1.055\times10^{-22}\,\text{g}$ ⇒ option e.

Calculating Molar Mass of Compounds

• Sum individual atom masses multiplied by subscripts.
• Example: $\text{K}<em>3\text{PO}</em>4$
  • $3(39.1)+31.0+4(16.0)=212.3\,\text{g mol}^{-1}$.
• Exercise: NiCO$_3$
  • $58.69+12.01+3(16.00)=118.7\,\text{g mol}^{-1}$ ⇒ option a.
• NO$_2$ conversion prompt
  (Method: particles → mol via Avogadro, mol → g via MM $46.01\,\text{g mol}^{-1}$).

Percent Composition

• Mass percent formula
  $\%\,\text{element}=\dfrac{\text{mass of element in sample}}{\text{total sample mass}}\times100$.
• Chromium oxide example
  $\%\,\text{Cr}=\dfrac{0.358\,\text{g}}{0.523\,\text{g}}\times100=68.5\%$.
• Morphine $\text{C}<em>{17}\text{H}</em>{19}\text{NO}_3$
  • Total molar mass $=17(12.01)+19(1.008)+14.01+3(16.00)=285.34\,\text{g}$.
  • Carbon mass $=17(12.01)=204.17\,\text{g}$.
  • $\%\,\text{C}=\dfrac{204.17}{285.34}\times100=71.6\%$ (choice d).

Empirical Formula Determination

• Procedure
  1. Assume $100$ g sample, convert percentages to grams.
  2. Convert g → mol for each element.
  3. Divide by smallest mole value to get simplest ratios.
  4. If any ratio is non-integer (0.5, 0.33, etc.), multiply all by appropriate factor (2, 3, 4…) to obtain whole numbers.
• Hydrocarbon example (83.65% C, remainder H)
  1. g C $=83.65$; g H $=16.35$.
  2. mol C $=\dfrac{83.65}{12.01}=6.965$; mol H $=\dfrac{16.35}{1.008}=16.22$.
  3. Divide: $\text{C}:1$, $\text{H}:2.33$.
  4. Multiply by $3$ ⇒ $\text{C}<em>3\text{H}</em>7$.
• Exercise: adipic acid (49.3% C, 6.9% H, 43.8% O)
  (Students apply same 4-step method).

Molecular Formula

• Relationship:
  $\text{Molecular formula}=n(\text{empirical formula})$ where $n$ is an integer.
• Find $n$ via
  $n=\dfrac{\text{molar mass (given)}}{\text{empirical molar mass}}$.
• Fructose
  • Empirical formula CH$2$O, empirical MM $30.03\,\text{g mol}^{-1}$. • Given molar mass $180.2\,\text{g mol}^{-1}$. • $n=\dfrac{180.2}{30.03}=6$. • Molecular formula $\text{C}</em>6\text{H}<em>{12}\text{O}</em>6$.
• Hydrocarbon continuation
  • Empirical formula C$3$H$7$, empirical MM $43.086\,\text{g mol}^{-1}$.
  • Molecular MM $86.2\,\text{g mol}^{-1}$.
  • $n=2$ ⇒ molecular formula $\text{C}<em>6\text{H}</em>{14}$.

Summary of All Possible Conversions

• Diagrammatically:
  • Grams (element/compound) $\xleftrightarrow{\text{MM}}$ Moles $\xleftrightarrow{N_A}$ Atoms / Molecules / Formula units.
• Key constants
  • Avogadro’s number $N_A=6.022\times10^{23}$.
  • Molar mass from periodic table or compound sum.
• Always verify units cancel, retain correct significant figures, and include scientific notation for very large/small values.