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}{14}\text{H}9\text{Cl}_5.
• Counting atoms in a formula
  • Example: \text{Mg(NO}3)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}3\text{PO}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}{17}\text{H}{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}3\text{H}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}6\text{H}{12}\text{O}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}6\text{H}{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.