Chapter 14 lecture 2

Concentration Units

A. Molarity

Molarity is a fundamental unit of concentration in chemistry, defined as the number of moles of solute per liter of solution. It is widely used for solutions in laboratory and industrial settings.

• Definition: Molarity (M) is given by the formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Millimoles: Sometimes calculations use millimoles (mmol), where 1 mmol = 1/1000 mole. Molarity can also be expressed as millimoles per milliliter: $M = \frac{\text{mmol solute}}{\text{mL solution}}$

• Example: To find the molarity of a solution made by dissolving 6.00 g ethanol (EtOH, molar mass = 46.07 g/mol) in water and diluting to 250.0 mL: $\text{M}_{\text{EtOH}} = \frac{6.00\,\text{g} \times \frac{1\,\text{mol}}{46.07\,\text{g}}}{0.2500\,\text{L}} = 0.521\,\text{M}$

• Preparation Example: To prepare 250.0 mL of 0.290 M sucrose solution (molar mass = 342.3 g/mol): $\text{mass}_{\text{sucrose}} = 0.290\,\text{mol/L} \times 0.2500\,\text{L} \times 342.3\,\text{g/mol} = 24.8\,\text{g}$ Weigh 24.8 g sucrose, dissolve in water, transfer to a volumetric flask, dilute to 250 mL, mix, and label.

B. Molality

Molality is another concentration unit, defined as the number of moles of solute per kilogram of solvent. It is especially useful when temperature changes, as it is independent of volume.

• Definition: Molality (m) is given by: $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

• Millimoles: Molality can also be expressed as millimoles per gram of solvent: $m = \frac{\text{mmol solute}}{\text{g solvent}}$

• Example: Molality of a solution with 9.60 g glucose (molar mass = 180.02 g/mol) in 250.0 g water: $m = \frac{9.60\,\text{g} \times \frac{1\,\text{mol}}{180.02\,\text{g}}}{0.2500\,\text{kg}} = 0.213\,m$

• Preparation Example: To prepare 0.350 m sucrose solution in 300.0 g water: $\text{mass}_{\text{sucrose}} = 0.350\,\text{mol/kg} \times 0.300\,\text{kg} \times 342.3\,\text{g/mol} = 35.9\,\text{g}$ Weigh 35.9 g sucrose, add to 300.0 g water, stir, transfer, and label.

C. Mole Fraction

Mole fraction is the ratio of the number of moles of a component to the total number of moles in the solution. It is a dimensionless quantity used in thermodynamics and colligative properties.

• Definition: $X_A = \frac{n_A}{n_{\text{total}}}$ Where $n_A$ is moles of component A, and $n_{\text{total}}$ is total moles.

• Example: For 8.60 g glucose (molar mass = 180.02 g/mol) in 26.5 g water (molar mass = 18.02 g/mol): $\text{mol glucose} = 0.0477$ $\text{mol water} = 1.4705$ $X_{\text{glucose}} = \frac{0.0477}{1.4705 + 0.0477} = 0.0314$

D. Mass Percent (% m/m)

Mass percent expresses the mass of solute as a percentage of the total mass of the solution. It is commonly used in commercial and laboratory settings.

• Definition: $\text{mass}\% = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

• Example: 23.40 g naphthalene in 99.60 g benzene: $\text{mass}\% = \frac{23.40}{23.40 + 99.60} \times 100 = 19.02\%$

• Conversion Example: For a 30.0% by mass H2O2 solution (density = 1.135 g/mL, molar mass = 34.02 g/mol):

  • Molality: Assume 1.00 kg solution, 30.0% H2O2 = 300 g H2O2, 700 g water. $\text{mol H}_2\text{O}_2 = \frac{300}{34.02} = 8.82$ $m = \frac{8.82}{0.700} = 12.6\,m$

  • Molarity: Volume of 1.00 kg solution: $V = \frac{1000\,\text{g}}{1.135\,\text{g/mL}} = 881\,\text{mL} = 0.881\,\text{L}$ $M = \frac{8.82}{0.881} = 10.0\,M$

  • Mole Fraction: Moles of water: $\text{mol H}_2\text{O} = \frac{700}{18.02} = 38.8$ $X_{\text{H}_2\text{O}_2} = \frac{8.82}{8.82 + 38.8} = 0.185$

E. Mass/Volume Percent (% m/v)

Mass/volume percent is used for dilute aqueous solutions, especially of solids. It expresses grams of solute per 100 mL of solution.

• Definition: $\text{mass/volume}\% = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$

• Example: 10% (m/v) NaCl solution: 10 g NaCl in enough water to make 100 mL solution.

F. Volume/Volume Percent (% v/v)

Volume/volume percent is used for solutions of liquids, expressing milliliters of solute per 100 mL of solution.

• Definition: $\text{volume/volume}\% = \frac{\text{volume of solute (mL)}}{\text{volume of solution (mL)}} \times 100$

• Example: 10% (v/v) methanol solution: 10 mL methanol diluted to 100 mL with water.

G. Parts per Million (ppm)

Parts per million is used for trace concentrations, especially in environmental chemistry.

• Definition: $\text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6$

• Example: 20.60 μg Cu in 10.40 mL water (density = 1.00 g/mL): $\text{ppm Cu} = \frac{20.60 \times 10^{-6}\,\text{g}}{10.40\,\text{g}} \times 10^6 = 1.981$

• Special Note: In dilute aqueous solution, 1 ppm = 1 mg solute per 1 L solution.

H. Parts per Billion (ppb)

Parts per billion is used for even lower concentrations.

• Definition: $\text{ppb} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^9$

• Example: 0.01800 μg Pb in 15.68 mL water: $\text{ppb Pb} = \frac{0.01800 \times 10^{-6}\,\text{g}}{15.68\,\text{g}} \times 10^9 = 1.148$

• Special Note: In dilute aqueous solution, 1 ppb = 1 μg solute per 1 L solution.

I. Parts per Trillion (ppt)

Parts per trillion is used for extremely dilute solutions.

• Definition: $\text{ppt} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^{12}$

• Example: 0.000400 μg Pb in 11.76 mL water: $\text{ppt Pb} = \frac{0.000400 \times 10^{-6}\,\text{g}}{11.76\,\text{g}} \times 10^{12} = 34.0$

• Special Note: In dilute aqueous solution, 1 ppt = 1 ng solute per 1 L solution.

Practice: Converting Between Units

Conversion between concentration units is often required in laboratory and environmental chemistry.

• Example: Convert 8.90 × 10–9 M C21H44 (molar mass = 296.5 g/mol) in water to ppb:

  • Calculate mass in 1 L: $8.90 \times 10^{-9}\,\text{mol} \times 296.5\,\text{g/mol} = 2.64 \times 10^{-6}\,\text{g}$

  • ppb: $\frac{2.64 \times 10^{-6}\,\text{g}}{1\,\text{g}} \times 10^9 = 2.64$

Preparing a Solution by Dilution

Solutions of desired concentration can be prepared by diluting a more concentrated stock solution. The number of moles of solute remains constant before and after dilution.

• Dilution Equation: $M_{\text{conc}} V_{\text{conc}} = M_{\text{dilute}} V_{\text{dilute}}$

• Example: To prepare 2.00 L of 0.968 M HCl from 12.1 M HCl: $V_{\text{conc}} = \frac{M_{\text{dilute}} V_{\text{dilute}}}{M_{\text{conc}}} = \frac{0.968 \times 2.00}{12.1} = 0.160\,\text{L} = 160\,\text{mL}$ Procedure: Add 160 mL concentrated HCl to water in a volumetric flask, dilute to 2.00 L, mix, and label.

Summary Table: Concentration Units

Unit	Definition	Formula	Typical Use
Molarity (M)	Moles solute per liter solution	$M = \frac{\text{mol solute}}{\text{L solution}}$	General lab solutions
Molality (m)	Moles solute per kg solvent	$m = \frac{\text{mol solute}}{\text{kg solvent}}$	Colligative properties
Mole Fraction (X)	Ratio of moles of component to total moles	$X_A = \frac{n_A}{n_{\text{total}}}$	Thermodynamics
Mass Percent (%)	Mass solute per mass solution × 100	$\text{mass}\% = \frac{\text{mass solute}}{\text{mass solution}} \times 100$	Commercial products
Mass/Volume Percent (% m/v)	Mass solute per volume solution × 100	$\text{mass/volume}\% = \frac{\text{g solute}}{\text{mL solution}} \times 100$	Dilute aqueous solutions
Volume/Volume Percent (% v/v)	Volume solute per volume solution × 100	$\text{volume/volume}\% = \frac{\text{mL solute}}{\text{mL solution}} \times 100$	Liquid mixtures
ppm	Mass solute per mass solution × 106	$\text{ppm} = \frac{\text{g solute}}{\text{g solution}} \times 10^6$	Trace analysis
ppb	Mass solute per mass solution × 109	$\text{ppb} = \frac{\text{g solute}}{\text{g solution}} \times 10^9$	Ultra-trace analysis
ppt	Mass solute per mass solution × 1012	$\text{ppt} = \frac{\text{g solute}}{\text{g solution}} \times 10^{12}$	Extremely dilute solutions

Additional info: Normality and weight molarity are mentioned but not discussed in detail; these are less commonly used in general chemistry and are typically reserved for specific titration or analytical contexts.