Chapter 6—Mass Percent Composition, Empirical Formulas & Molecular Formulas

Mass Percent Composition of Compounds

• Concept & Definition
  • Mass-percent composition ("mass %" or "percent by mass") of an element X in a compound measures what fraction of the compound’s total mass is contributed by X.
  • Generic mathematical form:
    \text{mass \% of element }X = \frac{\text{mass of }X\text{ in sample}}{\text{total mass of compound sample}} \times 100\%
• Key requirements
  • A chemical formula or experimentally measured masses of each element.
  • Atomic (molar) masses from the periodic table.

Worked Formula-Based Example — Sodium Carbonate (\mathrm{Na2CO3})

• Molar-mass determination
  • \text{Na: }23.00\,\mathrm{g\,mol^{-1}} \times 2 = 46.00
  • \text{C: }12.01\,\mathrm{g\,mol^{-1}} \times 1 = 12.01
  • \text{O: }16.00\,\mathrm{g\,mol^{-1}} \times 3 = 48.00
  • M{\mathrm{Na2CO_3}} = 46.00 + 12.01 + 48.00 = 106.01\,\mathrm{g\,mol^{-1}}
• Percentages
  • Sodium: \frac{46.00}{106.01}\times100 = 43.39\%
  • Carbon: \frac{12.01}{106.01}\times100 = 11.33\%
  • Oxygen: \frac{48.00}{106.01}\times100 = 45.28\%
  • Check: 43.39+11.33+45.28 \approx 100\% (round-off agreement).
• Significance
  • Enables quality-control, formulation, nutritional labeling, and stoichiometric planning.

Worked Experimental Example — Fe/Cl Compound

1. Measured data: m{\text{Fe}} = 2.74\,\mathrm{g},\; m{\text{Cl}} = 5.24\,\mathrm{g}.
2. Total mass of compound: m_{\text{compound}} = 2.74 + 5.24 = 7.98\,\mathrm{g}.
3. Percent iron: \frac{2.74}{7.98}\times100 = 34.8\% (3 sig figs).
4. Percent chlorine: \frac{5.24}{7.98}\times100 = 65.7\%.
5. Demonstrates that percent composition can be found without knowing the formula, useful for unknown samples.

Using Mass Percent as a Conversion Factor

• Any percentage can be rewritten as "g of part per 100 g of whole" to link grams of element ↔ grams of compound.
• Example: \mathrm{NaCl} is 39\% Na by mass ⟹ \frac{39\,\mathrm{g\,Na}}{100\,\mathrm{g\,NaCl}} or its reciprocal can serve as a conversion ratio.
• Dietary application: FDA limit <2.4\,\mathrm{g\,Na\,day^{-1}}. 2.4\,\mathrm{g\,Na}\times \frac{100\,\mathrm{g\,NaCl}}{39\,\mathrm{g\,Na}} = 6.2\,\mathrm{g\,NaCl} permitted per day.
  • Illustrates health relevance and real-world importance of stoichiometry.

Empirical Formula

• Definition: The simplest whole-number ratio of each type of atom in a compound, determined experimentally (often from percent composition).
• Relationship to molecular formula:
  \text{(molecular formula)} = n \times \text{(empirical formula)},\; n \in \mathbb{N}
• Examples
  • \mathrm{C2H4} \rightarrow CH_2 (empirical).
  • \mathrm{C3H6} \rightarrow CH_2 as well.

Five-Step Procedure for Determining an Empirical Formula

1. Obtain masses of each element.
   • If given percent composition, assume a 100\,\mathrm{g} sample so that percent values numerically equal grams.
2. Convert grams → moles using each element’s molar mass.
3. Write pseudo-formula using the non-integer mole values as subscripts.
4. Normalize: Divide every subscript by the smallest subscript to aim for simplest ratio.
5. Clear fractions: If any subscript remains non-integral, multiply all subscripts by the smallest whole number that converts them to integers.
   • Guideline table: endings 0.1 or 0.9 ×10; 0.2 or 0.8 ×5; 0.25 ×4; 0.33 or 0.67 ×3; 0.5 ×2; etc.

Example 1 — 52.14\%\,\mathrm{C},\;13.12\%\,\mathrm{H},\;34.73\%\,\mathrm{O}

• Assume 100\,\mathrm{g}: masses are 52.14\,\mathrm{g},\;13.12\,\mathrm{g},\;34.73\,\mathrm{g}.
• Convert to moles
  n{\text{C}} = \frac{52.14}{12.01}=4.341\,\mathrm{mol} n{\text{H}} = \frac{13.12}{1.008}=13.02\,\mathrm{mol}
  n_{\text{O}} = \frac{34.73}{16.00}=2.171\,\mathrm{mol}
• Pseudo-formula: \mathrm{C{4.341}H{13.02}O_{2.171}}.
• Divide by smallest (2.171): \mathrm{C2H6O1} (all integers) → Empirical formula \boxed{\mathrm{C2H_6O}}.

Example 2 — 43.64\%\,\mathrm{P},\;56.36\%\,\mathrm{O}

• Assume 100\,\mathrm{g}: 43.64\,\mathrm{g\,P},\;56.36\,\mathrm{g\,O}.
• Moles:
  n{\text{P}}=\frac{43.64}{30.97}=1.409 n{\text{O}}=\frac{56.36}{16.00}=3.523
• Pseudo-formula: \mathrm{P{1.409}O{3.523}}.
• Divide by 1.409: \mathrm{P1O{2.5}} (fractional subscript).
  • Ends in 0.5 ⇒ multiply all by 2.
  • Empirical formula \boxed{\mathrm{P2O5}}.

Molecular Formula

• Definition: The actual formula of a molecule; always an integral multiple of its empirical formula.
• Finding the multiplier n n = \frac{\text{molar mass of compound (given)}}{\text{molar mass of empirical formula}}
  • n must be an integer; non-integral result signals an error in preceding work.

Example — Fructose

• Given empirical formula \mathrm{CH_2O}, molar mass M = 180.2\,\mathrm{g\,mol^{-1}}.
• Empirical-unit mass:
  M_{\text{emp}} = 12.01 + 2(1.008) + 16.00 = 30.03\,\mathrm{g\,mol^{-1}}.
• Multiplier:
  n = \frac{180.2}{30.03} \approx 6.
• Molecular formula: n \times\text{empirical} = (CH2O)6 \rightarrow \boxed{\mathrm{C6H{12}O_6}}.

Conceptual & Practical Connections

• Stoichiometry: Mass-percent data seamlessly integrate with mole-based stoichiometric calculations (reactant needs, yield predictions).
• Analytical Chemistry: Percent composition is foundational in gravimetric analysis, combustion analysis, and quality-assurance testing.
• Nutrition & Medicine: Mass-percent calculations underpin dietary recommendations (e.g., sodium limits) and pharmaceutical formulation.
• Materials Science: Empirical formulas assist in characterizing unknown inorganic compounds, alloys, and minerals.
• Ethical/Regulatory Impact: Accurate composition reporting is essential for consumer safety, environmental compliance, and fair trade.

Quick Reference Equations & Relationships

• Mass percent: \%X = \frac{mX}{m{\text{compound}}}\times100
• Conversion factor from percent: \frac{\%X\,\text{(in g)}}{100\,\text{ g compound}}
• Empirical formula algorithm (grams → moles → divide → multiply): see five-step list above.
• Molecular formula: \text{MF} = n(\text{EF}),\; n = \frac{M{\text{compound}}}{M{\text{EF}}}

Study Tips

• Always include units in dimensional-analysis set-ups; they guide placement of numerators/denominators.
• After computing mass %, sum check ⇒ approximately 100\% to catch arithmetic errors.
• Keep a table of common fractional endings (0.25, 0.33, 0.5, 0.67, 0.75) and corresponding multipliers handy during empirical-formula work.
• Remember that empirical formulas must have whole numbers; never leave subscripts as decimals or fractions.
• When uncertain, redo molar-mass calculations directly from the periodic table to avoid propagation of rounding mistakes.