CHAPTER 6 Chemical Composition & The Mole – Detailed Study Notes

Counting by Weighing – Everyday Analogy

  • Hardware stores often sell nails by the pound rather than by individual count.
    • Rationale: customers frequently need hundreds of nails; physically counting them is impractical.
  • Example problem presented in lecture:
    • Customer buys 2.60 lb of medium-sized nails.
    • Given: 1 dozen nails weigh 0.150 lb.
    • Desired: number of nails purchased (solve by dimensional analysis: 2.60\,\text{lb}\times\frac{12\,\text{nails}}{0.150\,\text{lb}} = 208\,\text{nails}).
  • Key pedagogical point: mass can stand in for number when individual items are too numerous to count.

The Mole – Chemist’s "Dozen"

  • A simple "dozen" (12 objects) is far too small for atoms because atoms are extremely tiny and numerous.
  • Chemists use the mole (mol) as the counting unit.
    • Definition: 1 mol = 6.022 \times 10^{23} entities (Avogadro’s number).
    • Named after Amedeo Avogadro (1776–1856).
  • Scale illustrations:
    • 1\,\text{mol} of marbles = 6.022\times10^{23} marbles – would form a sphere larger than Earth.
    • 1\,\text{mol} of eggs would blanket Earth in a layer several miles thick.
    • 22 copper pennies ≈ 1 mol of Cu atoms (tangible analogy).
    • 2 large helium balloons ≈ 1 mol of He atoms.

Avogadro’s Number & Official Mole Definition

  • The numerical value of the mole is fixed by stating that exactly 12 g of ^{12}C contains 6.022 140 76 \times 10^{23} atoms.
  • Significance:
    • Links mass (g) to count (atoms or molecules).
    • Enables chemists to count atoms by weighing samples.

Converting Between Moles and Number of Atoms

  • Core relationships:
    • \text{atoms} = \text{moles} \times 6.022\times10^{23}
    • \text{moles} = \dfrac{\text{atoms}}{6.022\times10^{23}}
  • Example 1 (lecture):
    • Convert 3.5 mol He → atoms.
    • 3.5\,\text{mol}\times6.022\times10^{23}\,\frac{\text{atoms}}{\text{mol}} = 2.1\times10^{24}\,\text{He atoms}.
  • Example 2:
    • Convert 1.1\times10^{22} Ag atoms → moles.
    • \frac{1.1\times10^{22}}{6.022\times10^{23}}=1.8\times10^{-2}\,\text{mol Ag}.
  • Visual reminder: all sample pictures in slides have equal atom count, but total mass differs with element identity (heavier atoms → heavier 1-mol sample).

Atomic Mass, Molar Mass & the amu

  • Atomic mass unit (amu): defined as \tfrac{1}{12} the mass of one ^{12}C atom.
  • Molar mass of an element = atomic mass (in amu) expressed as g/mol.
    • Ex: Cu: single atom 63.55 amu → 1\,\text{mol} Cu = 63.55 g.
  • Trend: lighter atoms → smaller molar mass; heavier atoms → larger molar mass.

Comparative Atom Counts at Fixed Mass

  • Question posed: "One gram of which element contains the largest number of atoms?"
    • Options: N (14.01 g/mol), H (1.01), P (30.97), O (16.00).
    • Answer: Hydrogen (smallest molar mass → most moles → most atoms).

Mass ↔ Moles Conversions (Elements)

  • Generic equations:
    • \text{moles} = \dfrac{\text{mass (g)}}{\text{molar mass (g/mol)}}
    • \text{mass (g)} = \text{moles}\times\text{molar mass}
  • Example problems (lecture):
    • Moles of S in 57.8 g S: \frac{57.8}{32.07}=1.80\,\text{mol}.
    • Mass of Al in 6.73 mol Al: 6.73\times26.98=182\,\text{g}.
    • Atoms in 16.2 g Al can: first moles =\frac{16.2}{26.98}=0.601\,\text{mol}; atoms =0.601\times6.022\times10^{23}=3.62\times10^{23}\,\text{atoms}.
    • Mass of S that contains 3.0\times10^{23} atoms: moles =0.498\,\text{mol}; mass =0.498\times32.07=16.0\,\text{g}.
  • Conceptual True/False question:
    • Statement (a): Equal atoms → equal moles (TRUE, inherent definition).
    • Statement (b): Equal mass → equal moles (FALSE, depends on identity).
    • Statement (c): Equal mass → equal atoms (FALSE).

Counting Molecules by the Gram (Compounds)

  • For compounds, molar mass = mass of one mole of molecules/formula units.
  • Requires formula mass (sum of atomic masses in the formula).

Formula Mass & Molar Mass Calculations

  • Example: CO_2.
    • Atomic masses: C = 12.01, O = 16.00.
    • Formula mass =12.01+2(16.00)=44.01\,\text{amu}.
    • Therefore molar mass =44.01\,\text{g/mol}.
  • Task from lecture: Calculate molar mass of ammonium carbonate (NH4)2CO_3.
    • (2\times14.01)+(8\times1.01)+12.01+3(16.00)=96.09\,\text{g/mol}.
  • Conversions illustrated:
    • Mass of 1.75 mol H$_2$O: 1.75\times18.02=31.5\,\text{g}.
    • Moles in 35.0 g H$_2$O: \frac{35.0}{18.02}=1.94\,\text{mol}.
    • Mass of 4.78\times10^{24} molecules NO$_2: moles =7.94\,\text{mol}; mass =7.94\times46.01=365\,\text{g}.

Mole Relationships in Chemical Formulas

  • Chemical formula acts as a conversion factor between moles of compound and moles of constituent atoms.
    • CO2 contains 2 O atoms per molecule → 2\,\text{mol O} : 1\,\text{mol CO}2.
  • Examples from slides:
    • Moles of O in 1.7 mol CaCO$_3: 1.7\times3=5.1\,\text{mol O}.
    • Moles of Cl$^-$ in 36.0 g CaCl$_2: mass→moles compound =0.324\,\text{mol}; moles Cl$^-$ =0.324\times2=0.648\,\text{mol}.

Elemental Mass Percent Composition

  • Mass percent (w/w %) of element E in compound: \frac{\text{mass of }E\text{ in 1 mol}}{\text{molar mass of compound}}\times100\%.
  • Lecture practice:
    • Mass % of each element in Na2CO3.
    • Na: 2(22.99)=45.98\,\text{g}
    • C: 12.01\,\text{g}
    • O: 3(16.00)=48.00\,\text{g}
    • Molar mass: 105.99\,\text{g/mol}
    • \%\,Na=43.4\%\,,\;\%C=11.3\%\,,\;\%O=45.3\%.
    • Percent H$2$O in CoCl3·6H_2O (hydrate): compare mass of 6 waters (108.12 g) to full molar mass (237.86 g) → 45.5 % water.
  • Quick-choice question: Which chromium oxide has highest mass % O? Answer: CrO$_2$ (highest O:Cr ratio by mass).

Empirical vs. Molecular Formulas

  • Empirical formula: simplest whole-number ratio of atoms (e.g., glucose C6H{12}O6 → empirical CH2O).
  • Molecular formula: actual counts; always an integer multiple of empirical: \text{Molecular} = (\text{Empirical})\times n where n\in\mathbb{Z^+}.
  • Many different molecules can share the same empirical formula.

Procedure to Derive an Empirical Formula from % Composition

  1. Assume 100 g of compound → numeric grams equal to percentages.
  2. Convert grams → moles for each element.
  3. Divide all mole values by the smallest mole to obtain ratios.
  4. If a ratio is not near whole number, multiply all ratios by the smallest integer that yields whole numbers.
  • Additional note: if, after division, subscripts are not integers (e.g., 1.5, 2.33), scale by 2, 3, etc., until whole numbers emerge.

Worked Empirical-Formula Examples (Lecture)

  • Example 1: 71.65 % Cl, 24.27 % C, 4.07 % H.
    • Assume 100 g → 71.65 g Cl, 24.27 g C, 4.07 g H.
    • Moles: Cl = 2.02, C = 2.02, H = 4.04.
    • Divide by 2.02 → Cl = 1, C = 1, H = 2 → empirical formula CClH$2$ (often written CH$2$Cl).
  • Example 2: 72.2 % Mg, 27.8 % N.
    • Masses: 72.2 g Mg, 27.8 g N.
    • Moles: Mg = 2.97, N = 1.98.
    • Divide by 1.98 → Mg = 1.5, N = 1 → multiply by 2 → Mg$3$N$2$.
  • Example 3: Ti + O reaction: 3.24 g Ti → 5.40 g oxide.
    • Mass O = 5.40 − 3.24 = 2.16 g.
    • Moles: Ti = 0.0676, O = 0.135.
    • Ratio: Ti:O = 1:2 → TiO$_2$.

From Empirical to Molecular Formula

  • Relation: \text{Molar mass} = n\times(\text{Empirical-formula molar mass}).
    • Solve n = \dfrac{\text{given molar mass}}{\text{empirical-formula MM}} (must be a whole number).
  • Example (lecture): Fructose.
    • Empirical formula = CH$_2$O; empirical MM = 30.03 g/mol.
    • Molar mass given = 180.2 g/mol.
    • n=\frac{180.2}{30.03}=6.
    • Molecular formula = (CH2O)6 = C6H{12}O_6.
  • Example: Unknown compound 65.45 % C, 5.492 % H, 29.06 % O; molar mass ≈ 110 g/mol.
    • Empirical procedure → C$3$H$3$O$_2$ (MM = 71.06).
    • n=\frac{110}{71.06}\approx1.55$$ → multiply empirical subscripts by 2 → C$6$H$6$O$_4$ (matches ~110 g/mol).

Conceptual Connections & Take-Away Principles

  • Counting by weighing unites macroscopic lab practice with atomic-scale theory.
  • The mole concept provides the crucial bridge linking laboratory mass measurements (grams) to microscopic counts (atoms, molecules).
  • Molar mass acts as the universal conversion factor; its numerical value equals the atomic/formula mass but in units of g/mol.
  • Percent composition data → empirical formula → (with molar mass) molecular formula; this chain underpins many analytical techniques (e.g., combustion analysis, mass spectrometry).
  • Always watch units: atoms ↔ molecules ↔ moles ↔ grams; clear factor-label/dimensional analysis prevents errors.
  • Conceptual pitfalls addressed in lecture:
    • Equal mass does not automatically imply equal moles or atoms across different elements.
    • Hydrates include water as part of their formula mass and percent composition.
  • Ethical & practical note: Accurate molar mass knowledge is essential in pharmacology, environmental testing, and materials science; dosage or pollutant limits hinge on mole-based calculations.