Concentration Units & Calculations

Definition of Concentration

  • Concentration = amount of solute dissolved in a solvent.
  • Expressed in multiple standardized units adapted to context.
    • Liquors: volume percent (% v/v); “proof” = 2 × volume percent.
    • Fruit-juice sweetness: degrees Brix (°Bx) = mass percent (g glucose / g solution × 100 %).

Units of Concentration Tested on the MCAT

  • Percent composition by mass
  • Mole fraction
  • Molarity (M)
  • Molality (m)
  • Normality (N)
  • Dilution relationships

Percent Composition by Mass (Mass %)

  • Formula: \text{Mass %}=\frac{\text{mass of solute}}{\text{mass of solution}}\times100\%
  • Applies to aqueous and non-aqueous mixtures (e.g., metal alloys).
  • Example
    • “100 g solution contains 20 g NaCl.”
    • 20 g100 g×100%=20%\frac{20\,g}{100\,g}\times100\%=20\% NaCl by mass.
  • Significance
    • Simple experimental measurement (weigh only).
    • Useful when densities vary or solvents are solids.

Mole Fraction (χ)

  • Formula for component A: χ<em>A=n</em>A∑n<em>i\chi<em>A=\frac{n</em>A}{\sum n<em>i} where ∑χ</em>i=1\sum \chi</em>i =1.
  • Typical Uses
    • Raoult’s law → vapor-pressure depression.
    • Partial pressures of gases (Chapter 8 topics).
  • Example: 184 g glycerol (C₃H₈O₃) + 180 g H₂O
    • Molar masses given: H₂O = 18 g mol⁻¹ ; glycerol = 92 g mol⁻¹.
    • n<em>H</em>2O=180 g18 g mol−1=10 moln<em>{\text{H}</em>2\text{O}}=\frac{180\,g}{18\,g\,mol^{-1}}=10\,mol
    • nglycerol=184 g92 g mol−1=2 moln_{\text{glycerol}}=\frac{184\,g}{92\,g\,mol^{-1}}=2\,mol
    • Total moles = 12.
    • χ<em>H</em>2O=1012=0.83\chi<em>{\text{H}</em>2\text{O}}=\frac{10}{12}=0.83
    • χglycerol=212=0.17\chi_{\text{glycerol}}=\frac{2}{12}=0.17
  • Notes
    • Unitless, temperature-independent.
    • Essential in colligative-property calculations.

Molarity (M)

  • Definition: M=moles of soluteliters of solutionM=\frac{\text{moles of solute}}{\text{liters of solution}}
  • Square-bracket notation: [Na+], [H+], [A][\text{Na}^+],\,[H^+],\,[\text{A}] always implies molarity unless stated otherwise.
  • Denominator = volume of final solution (not solvent alone).
  • Key Applications
    • Rate laws, equilibrium (law of mass action), pH/pOH, osmotic pressure, Nernst equation.
  • Example: 11 g CaCl₂ diluted to 100 mL
    • Molar mass CaCl₂ ≈ 111.1 g mol⁻¹.
    • n=11 g111.1 g mol−1=0.10 moln=\frac{11\,g}{111.1\,g\,mol^{-1}}=0.10\,mol
    • Volume = 0.100 L.
    • M=0.10 mol0.100 L=1.0 MM=\frac{0.10\,mol}{0.100\,L}=1.0\,M
  • Practical Insight
    • Temperature sensitive (volume expands/contracts).

Molality (m)

  • Definition: m=moles of solutekilograms of solventm=\frac{\text{moles of solute}}{\text{kilograms of solvent}}
  • For dilute aqueous solutions at 25 ∘!C25\,^{\circ}!\text{C}, m≈Mm\approx M because ρ<em>H</em>2O≈1 kg L−1\rho<em>{\text{H}</em>2\text{O}}\approx1\,kg\,L^{-1}.
  • Divergence occurs at high solute concentration since solute >> H₂O molar mass → density ↑.
  • Primary Uses
    • Boiling-point elevation.
    • Freezing-point depression.
  • Example: 10 g NaOH in 500 g H₂O
    • Molar mass NaOH = 40 g mol⁻¹.
    • n=10 g40 g mol−1=0.25 moln=\frac{10\,g}{40\,g\,mol^{-1}}=0.25\,mol
    • Solvent mass = 0.500 kg.
    • m=0.25 mol0.500 kg=0.50 mm=\frac{0.25\,mol}{0.500\,kg}=0.50\,m

Normality (N)

  • Definition: N=equivalents of species of interestliter of solutionN=\frac{\text{equivalents of species of interest}}{\text{liter of solution}}
  • “Equivalent” = amount that reacts or supplies one mole of the species of interest (H⁺, e⁻, etc.).
  • Reaction-dependent: same solute can have different N in different contexts.
    • Acid-base: count H⁺ or OH⁻ equivalents.
    • Redox: count e⁻ equivalents.
  • Example scenarios with permanganate MnO4−\text{MnO}_4^-
    • Acidic medium: 1 mol accepts 5 e⁻ → 1 M = 5 N.
    • Alkaline medium: 1 mol accepts 1 e⁻ → 1 M = 1 N.
  • Connections
    • Tied to gram-equivalent weight (Chapter 4).
    • Useful in titration calculations where stoichiometry ≠1:1.

Dilution Calculations

  • When solvent is added, moles remain constant: M<em>iV</em>i=M<em>fV</em>fM<em>iV</em>i = M<em>fV</em>f.
  • Works with any consistent volume units (mL or L).
  • Example: Prepare 300 mL of 1.1 M NaOH from 5.5 M stock
    • V<em>i=M</em>fV<em>fM</em>i=1.1 M×300 mL5.5 M=60 mLV<em>i = \frac{M</em>fV<em>f}{M</em>i} = \frac{1.1\,M \times 300\,mL}{5.5\,M}=60\,mL stock needed.
    • Add water until total volume = 300 mL.

Comparative & Practical Insights

  • Choice of unit depends on property measured, experimental convenience, and sensitivity to temperature or density.
    • Mass % and molality are temperature independent (mass doesn’t change with T).
    • Molarity and normality vary with temperature (dependent on solution volume).
  • Real-world relevance
    • Quality control (proof of alcohol, °Bx for juice sweetness).
    • Clinical labs & pharmaceuticals often use normality for acid/base reagents.
    • Environmental chemistry uses mole fraction for gas mixtures.
  • Ethical/Practical implications
    • Proper labeling of concentration critical for safety (lab reagents, beverages).
    • Misinterpretation of units can lead to dosing errors or experimental failure.