Chapter 3: The Mole, Formula Mass, and Composition of Substances and Solutions

Chapter 3 – Composition of Substances and Solutions

Overview

This chapter introduces foundational concepts in General Chemistry, including the mole, Avogadro’s number, formula and molar mass, mass-to-mole conversions, percent composition, empirical and molecular formulas, and molarity. Mastery of these topics is essential for understanding chemical calculations and the quantitative relationships in chemical reactions.

The Mole and Avogadro’s Number

Definition of the Mole

• Mole (mol): The SI base unit for the amount of substance. One mole contains exactly $6.02214076 \times 10^{23}$ entities (Avogadro’s number).

• Avogadro’s Constant ($N_A$): $N_A = 6.02214076 \times 10^{23} \text{ mol}^{-1}$, representing the number of particles (atoms, molecules, ions, etc.) in one mole of a substance.

• Analogies: 1 dozen = 12 items, 1 mole = $6.022 \times 10^{23}$ items.

Example: 1 mole of carbon atoms contains $6.022 \times 10^{23}$ carbon atoms and has a mass of 12.01 g.

Formula Mass vs. Molar Mass

Definitions and Calculations

• Formula Mass (Ionic Substances): The sum of the average atomic masses of all atoms in an ionic compound’s formula unit, expressed in atomic mass units (amu).

• Molecular Mass (Covalent Substances): The sum of the average atomic masses of all atoms in a molecule, also in amu.

• Molar Mass: The mass of one mole of a substance, numerically equal to the formula or molecular mass but expressed in grams per mole (g/mol).

Formulas:

• For a compound $A_xB_y$: $M = xM_A + yM_B$

• General: $\text{Molar mass} = \sum (\text{number of atoms of element}) \times (\text{atomic mass of element})$

Examples:

• NaCl: $M_{\text{NaCl}} = M_{\text{Na}} + M_{\text{Cl}} = 22.99 + 35.45 = 58.44$ amu

• Glucose ($C_6H_{12}O_6$): $M_{C_6H_{12}O_6} = (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.16$ amu

Mass to Mole Conversions

Using Molar Mass and Avogadro’s Number

• Molar Mass ($M$): Used as a conversion factor between mass and moles.

• Avogadro’s Number ($N_A$): Used as a conversion factor between moles and number of particles.

Key Equations:

• $n = \frac{m}{M}$, where $n$ = moles, $m$ = mass (g), $M$ = molar mass (g/mol)

• $N = n \times N_A$, where $N$ = number of particles

Example: To find the number of molecules in 40.0 mg of saccharin ($C_7H_5NO_3S$, $M = 183.18$ g/mol):

• Convert mass to moles: $n = \frac{0.0400\ \text{g}}{183.18\ \text{g/mol}} = 2.18 \times 10^{-4}$ mol

• Convert moles to molecules: $N = 2.18 \times 10^{-4}\ \text{mol} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 1.31 \times 10^{20}$ molecules

Connections Between Mass, Moles, and Particles

Quantity	Conversion Factor	To
Mass (g)	Divide by molar mass (g/mol)	Moles
Moles	Multiply by Avogadro’s number	Particles (atoms, molecules, ions)
Particles	Divide by Avogadro’s number	Moles
Moles	Multiply by molar mass (g/mol)	Mass (g)

Percent Composition

Definition and Calculation

• Percent Composition: The percentage by mass of each element in a compound.

• Formula: $\%\ \text{Element} = \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100\%$

Example: For ammonia ($NH_3$):

• N: $\frac{14.01}{17.03} \times 100\% = 82.3\%$

• H: $\frac{3.03}{17.03} \times 100\% = 17.7\%$

Empirical vs. Molecular Formula

Empirical Formula

• Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.

• Determined from experimental data (masses or percent composition).

Steps to Determine Empirical Formula:

1. Convert mass (or percent) of each element to moles.

2. Divide each by the smallest number of moles to get the simplest ratio.

3. If necessary, multiply by an integer to obtain whole numbers.

Example: A sample contains 1.71 g C and 0.287 g H:

• Moles C: $\frac{1.71}{12.01} = 0.142$ mol

• Moles H: $\frac{0.287}{1.008} = 0.285$ mol

• Ratio: C:H = 1:2 → Empirical formula = $CH_2$

Molecular Formula

• Molecular Formula: The actual number of atoms of each element in a molecule.

• It is a whole-number multiple of the empirical formula.

• Formula: $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

• Molecular formula = (Empirical formula)$_n$

Example: If empirical formula is $CH_2O$ (empirical mass = 30 g/mol) and molar mass is 180 g/mol, $n = 6$, so molecular formula is $C_6H_{12}O_6$.

Solutions and Molarity

Definitions

• Solution: A homogeneous mixture of two or more substances.

• Solvent: The component present in the greatest amount (e.g., water in aqueous solutions).

• Solute: The component present in a lesser amount (e.g., salt, sugar).

Molarity

• Molarity (M): The number of moles of solute per liter of solution.

• Formula: $M = \frac{n}{V}$, where $n$ = moles of solute, $V$ = volume of solution in liters.

• Unit: mol/L (M, molar)

Example: To prepare 1.0 L of 1.0 M NaCl solution, dissolve 1.0 mol (58.44 g) NaCl in enough water to make 1.0 L of solution.

Preparation of Solutions

1. Measure the required mass of solute.

2. Add to a volumetric flask of appropriate volume.

3. Add solvent to about 3/4 full, swirl to dissolve.

4. Add solvent to the calibration mark, mix thoroughly.

Applications and Calculations

• To find moles of solute: $n = M \times V$

• To find mass of solute: $m = n \times M_{\text{molar}}$

• To find volume needed: $V = \frac{n}{M}$

Example: How many moles of HCl are in 25 mL of 0.75 M HCl?

• $n = 0.75\ \text{mol/L} \times 0.025\ \text{L} = 0.019\ \text{mol}$

Concentration of Ions in Solution

• For ionic compounds, multiply the molarity of the compound by the number of each ion per formula unit to get the ion concentration.

Example: What is the concentration of $Cl^-$ in 0.27 M $CaCl_2$?

• Each $CaCl_2$ yields 2 $Cl^-$ ions: $[Cl^-] = 2 \times 0.27\ \text{M} = 0.54\ \text{M}$

Summary Table: Key Quantities and Conversions

Quantity	Symbol/Unit	Conversion
Mass	g, kg	Divide by molar mass to get moles
Moles	mol	Multiply by Avogadro’s number for particles
Particles	atoms, molecules, ions	Divide by Avogadro’s number for moles
Volume (solution)	L, mL	Multiply by molarity for moles of solute

Additional info: For more practice, refer to end-of-chapter exercises and laboratory problems as recommended in your course materials.