7.2 Concentration Units: Molarity, Molality, Mole Fraction & Mass Percent

7.2 Concentration Units: Molarity, Molality, Mole Fraction & Mass Percent

Quick Summary: Quantitative chemistry requires rigorous expressions of solution composition. Molarity ($M$) expresses moles of solute per liter of solution and varies with temperature due to thermal expansion. Molality ($m$), mole fraction ($X$), and mass percent ($\%\text{ m/m}$) depend strictly on mass and molar ratios, rendering them completely independent of temperature. Solution density serves as the critical mathematical bridge for converting between volumetric and mass-based concentration frameworks. Dilutions follow the conservation of solute moles ($M_1 V_1 = M_2 V_2$).

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1. Quantitative Definitions of Concentration Units

Concentration quantifies the amount of solute present in a specified volume of solution or mass of solvent.

Molarity ($M$)

Molarity is the moles of solute per liter of total solution: $M = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{n_{\text{solute}}}{V_{\text{soln (L)}}} \quad \left[\frac{\text{mol}}{\text{L}} \text{ or M}\right]$ Because liquid volume expands with increasing temperature, molarity is temperature-dependent; heating a solution decreases its molarity even though solute moles remain invariant.

Molality ($m$)

Molality is the moles of solute per kilogram of pure solvent: $m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}} \quad \left[\frac{\text{mol}}{\text{kg}} \text{ or } m\right]$ Because mass is unaffected by thermal expansion, molality is temperature-independent. Molality is therefore the concentration unit required for colligative property analyses over variable temperatures. In dilute aqueous solutions at 20 °C, $1\text{ L } \text{H}_2\text{O} \approx 1\text{ kg}$, so $M \approx m$.

Mole Fraction ($X_i$)

The mole fraction of component $i$ is the dimensionless ratio of moles of $i$ to total moles in the mixture: $X_i = \frac{n_i}{n_{\text{total}}} = \frac{n_i}{n_A + n_B + \dots}$ The sum of mole fractions in any solution equals unity ($\sum X_i = 1$). Mole fraction is independent of temperature.

Mass Percent ($\%\text{ m/m}$)

Mass percent expresses solute mass as a percentage of total solution mass: $\%\text{ by mass} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 100\%$ where $\text{total mass} = \text{mass of solute} + \text{mass of solvent}$.

Trace Units: Parts per Million (ppm) & Parts per Billion (ppb)

For very dilute solutions (e.g., environmental contaminants, water purity assays): $\text{ppm} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 10^6$ $\text{ppb} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 10^9$ In dilute aqueous media ($\rho \approx 1.00\text{ g/mL}$): $1\text{ ppm} \approx 1\text{ mg/L}$, and $1\text{ ppb} \approx 1\text{ }\mu\text{g/L}$.

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2. Concentration Units Master Summary

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3. Converting Between Concentration Units: The Density Bridge

Converting between volumetric units ($M$) and mass-based units ($m, X, \%$) requires the solution density ($\rho = m_{\text{soln}} / V_{\text{soln}}$).

Systematic Conversion Strategy

1. Assume a convenient basis: For mass percent, assume $100.0\text{ g}$ of solution; for molarity, assume $1.000\text{ L}$ of solution.
2. Find component masses: $\text{Mass of solvent} = \text{total solution mass} - \text{mass of solute}$.
3. Convert masses to moles: $n = m / M_w$.
4. Use density to find volume: $V_{\text{soln}} = m_{\text{soln}} / \rho$.

Worked Conversion: Concentrated Nitric Acid

Problem: Concentrated nitric acid ($\text{HNO}_3$, $63.01\text{ g/mol}$) is $68.0\%$ by mass with density $\rho = 1.41\text{ g/mL}$. Calculate its molarity, molality, and mole fraction.

Step 1: Basis of $100.0\text{ g}$ solution

• $\text{Mass of HNO}_3 = 68.0\text{ g}$
• $\text{Mass of H}_2\text{O} = 100.0\text{ g} - 68.0\text{ g} = 32.0\text{ g} = 0.0320\text{ kg}$

Step 2: Convert masses to moles $n_{\text{HNO}_3} = \frac{68.0\text{ g}}{63.01\text{ g/mol}} = 1.079\text{ mol}$ $n_{\text{H}_2\text{O}} = \frac{32.0\text{ g}}{18.02\text{ g/mol}} = 1.776\text{ mol}$ $n_{\text{total}} = 1.079 + 1.776 = 2.855\text{ mol}$

Step 3: Solution volume from density $V_{\text{soln}} = \frac{100.0\text{ g}}{1.41\text{ g/mL}} = 70.92\text{ mL} = 0.07092\text{ L}$

Step 4: Compute concentrations $M = \frac{1.079\text{ mol}}{0.07092\text{ L}} = 15.2\text{ M}$ $m = \frac{1.079\text{ mol}}{0.0320\text{ kg}} = 33.7\text{ m}$ $X_{\text{HNO}_3} = \frac{1.079\text{ mol}}{2.855\text{ mol}} = 0.378$

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4. Dilution Principles & Laboratory Practice

During dilution, solvent is added to concentrated stock without altering solute moles ($n_{\text{initial}} = n_{\text{final}}$): $M_1 V_1 = M_2 V_2$

Standard Solution Preparation Protocol

1. Weigh Solute: Measure solid solute on an analytical balance to $\pm 0.0001\text{ g}$.
2. Dissolve: Dissolve solute in a beaker with a fraction of total deionized water.
3. Quantitative Transfer: Pour into a volumetric flask using a funnel; rinse beaker and funnel repeatedly into the flask.
4. Dilute to Calibration Mark: Add water until the bottom of the curved meniscus rests tangent to the etched calibration ring at eye level.
5. Invert to Homogenize: Stopper tightly and invert $15 - 20$ times to ensure uniform mixing.