Mass Relationships in Chemical Reactions

Atomic Mass & Isotopes

• Atomic mass (amu): mass of a single atom on the $^{12}C$-scale
  • By definition $1\;^{12}C$ atom "weighs" $12\,\text{amu}$
  • Typical values: $^{1}H = 1.008\,\text{amu},\;^{16}O = 16.00\,\text{amu}$
• Average (weighted) atomic mass = Σ(isotopic mass × fractional abundance)
  • Copper example (Example 3.1):
  • $^{63}_{\ 29}Cu$: 62.93 amu, 69.09 % (→0.6909)
  • $^{65}_{\ 29}Cu$: 64.9278 amu, 30.91 % (→0.3091)
  • $\bar M_{Cu}=0.6909(62.93)+0.3091(64.9278)=63.55\,\text{amu}$
  • Check: result lies between the two isotopic masses and is closer to the more abundant isotope.
• Mass spectrometer separates ions by $m/q$; Ne spectrum shows three peaks (amu ≈19, 20, 22) with relative intensities 0.26 %, 90.92 %, 8.82 %.

Atomic Mass on the Periodic Table

• Each element square lists atomic number (Z) and average atomic mass (weighted).
• Atomic masses (amu) numerically equal molar masses (g mol⁻¹).

The Mole & Avogadro’s Number

• Mole (mol): quantity containing $N_A = 6.022\,141\,5\times10^{23}$ elementary entities.
• By definition: $12.00\,\text{g}$ $^{12}C$ contains $N_A$ atoms.
• Analogous to “dozen” = 12, “pair” = 2.
• Molar mass (M): mass of 1 mol of substance in grams.
  • For elements: numerical value equals atomic mass (amu).
  • Illustrations: 1 mol of Li atoms = 6.941 g; 1 mol $^{12}C$ atoms = 12.00 g.
• Unit relations
  • $1\,\text{amu}=1.66\times10^{-24}\,\text{g}$
  • $1\,\text{g}=6.022\times10^{23}\,\text{amu}$

Mass↔Moles↔Particles Conversions (Key Formulae)

• $\text{moles}=\dfrac{\text{mass (g)}}{M}$
• $\text{particles}=\text{moles}\times N_A$

Worked Examples

• Example 3.2 (He): $6.46\,\text{g}\;He \Rightarrow 1.61\,\text{mol}$ (using $M_{He}=4.003$ g mol⁻¹).
• Example 3.3 (Zn): $0.356\,\text{mol}\,Zn \Rightarrow 23.3\,\text{g}$ (using $M_{Zn}=65.39$ g mol⁻¹).
• Example 3.4 (S): $16.3\,\text{g}\,S \Rightarrow 3.06\times10^{23}\,\text{atoms}$ (two-step: g→mol→atoms).
• Example 3.6 (CH₄): $6.07\,\text{g}\,CH_4 \Rightarrow 0.378\,\text{mol}$.

Molecular, Formula & Formula Unit Masses

• Molecular mass (aka molecular weight): Σ atomic masses of all atoms in a molecule.
  • $SO_2$: $32.07 + 2(16.00)=64.07\,\text{amu}$
  • Caffeine $C<em>8H</em>{10}N<em>4O</em>2$: $8(12.01)+10(1.008)+4(14.01)+2(16.00)=194.20\,\text{amu}$.
• Formula mass: analogous sum for ionic compounds.
  • $NaCl$: $22.99+35.45=58.44\,\text{amu}$.
• Identity rule: “amu-value” = “g per mole”.

Percent Composition by Mass

• General formula:
  $\%\,\text{element}=\dfrac{n\,M<em>{\text{element}}}{M</em>{\text{compound}}}\times100\%$
  where $n$ = number of that element’s atoms per formula unit.
• Ethanol example $C<em>2H</em>6O$:
  $\%C=52.14,\;\%H=13.13,\;\%O=34.73$ (%s sum to 100).
• Example 3.8 ($H<em>3PO</em>4$):
  $\%H=3.09,\;\%P=31.61,\;\%O=65.31$ (round-off ↔100.01 %).

Empirical & Molecular Formulas

• Workflow (Fig 3.7):
  mass % → g → divide by molar mass → moles → divide by smallest → ratio → integer subscripts.
• Vitamin C Example 3.9: 40.92 %C, 4.58 %H, 54.50 %O.
  • From 100 g sample → moles → $C<em>{3.407}H</em>{4.54}O_{3.406}$.
  • Divide by 3.406 → $CH_{1.33}O$.
  • Multiply subscripts by 3 → empirical formula $C<em>3H</em>4O_3$.
• NOₓ Example 3.11: 30.46 % N, 69.54 % O.
  • Empirical formula $NO_2$ (46.01 g mol⁻¹).
  • Experimental molar mass 90–95 g → ratio ≈2 → molecular formula $N<em>2O</em>4$ with exact $M=92.02\,\text{g mol}^{-1}$.

Chemical Equations & Balancing

• Chemical reaction: Reactants → Products.
• Reading equations: coefficients express numbers of atoms, moles, or masses in molar masses—not plain grams (e.g.
  $2\,Mg + O_2 → 2\,MgO$ corresponds to 48.6 g Mg + 32.0 g O₂ → 80.6 g MgO).
• Balancing procedure (ethane example):
  1. Write correct formulas.
  2. Adjust coefficients; do not alter subscripts.
  3. Balance elements appearing in single reactant/product first (C, H).
  4. Balance remaining (O); fractions allowed, then clear denominators.
  5. Verify atom counts.
• Aluminum oxidation Example 3.12: final balanced $4Al + 3O<em>2 → 2Al</em>2O_3$.

Stoichiometric Calculations (Quantities of Reactants & Products)

• General roadmap:
  1. Balanced equation.
  2. Convert known masses → moles.
  3. Apply mole ratio to find unknown moles.
  4. Convert moles → requested units (mass, volume, particles).
• Glucose degradation Example 3.13: 856 g $C<em>6H</em>{12}O<em>6$ produces $1.25\times10^{3}\,\text{g}$ $CO</em>2$.
• Lithium & water Example 3.14: 9.89 g $H_2$ needs 68.1 g Li.

Limiting Reactants & Excess

• Limiting reagent: reactant consumed first → limits product.
  • Illustration: $CO + 2H<em>2 → CH</em>3OH$ (if $H_2$ < stoichiometric, it limits).
• Urea synthesis Example 3.15:
  • Reaction $2NH<em>3 + CO</em>2 → (NH<em>2)</em>2CO + H<em>2O$ with 637.2 g $NH</em>3$ & 1142 g $CO_2$.
  • $NH_3$ limits (produces 18.71 mol product).
  • Mass of urea formed: $1124\,\text{g}$.
  • Excess $CO_2$ left: 319 g.
• Complex organic example 3.16: for 10 g $CH<em>3OH$ need 29.6 g $CH</em>3Br$ (stoich) and 50.0 g $LiC<em>4H</em>9$ (2.5 equiv).

Reaction Yield

• Theoretical yield: product amount calculated from limiting reagent.
• Actual yield: amount isolated.
• Percent yield: $\%\text{yield}=\dfrac{\text{actual}}{\text{theoretical}}\times100\%$.
• Titanium extraction Example 3.17:
  • Limiting $TiCl_4$ gives theoretical $8.95\times10^{6}\,\text{g}$ Ti.
  • Actual $7.91\times10^{6}\,\text{g}$ → $88.4\%$ yield.

Experimental Determination of Empirical Formula (Combustion-type)

• Example path:
  • Measure g $CO_2$ → mol C → g C.
  • Measure g $H_2O$ → mol H → g H.
  • g O = sample mass − (g C + g H).
  • Convert g → mol → divide → ratio → empirical formula (e.g.
    $C<em>2H</em>6O$ from C 0.5, H 1.5, O 0.25 → divide by 0.25).

Practical Chemistry Context: Fertilizers

• Essential plant macronutrients: N, P, K, Ca, S, Mg.
• Ammonia synthesis: $3H<em>2 + N</em>2 → 2NH_3$.
• Nitric acid neutralization: $NH<em>3 + HNO</em>3 → NH<em>4NO</em>3$ (fertilizer).
• Phosphate rock (fluorapatite) with sulfuric acid:
  $2Ca<em>5(PO</em>4)<em>3F + 7H</em>2SO<em>4 → 3Ca(H</em>2PO<em>4)</em>2 + 7CaSO_4 + 2HF$
  yields soluble phosphate fertilizers.