The Mole Concept, Chemical Formulas, and Solution Concentration

The Mole Concept and Molecular Mass

Definition and Importance

The mole concept is fundamental in chemistry for relating the mass of substances to the number of particles they contain. It allows chemists to count atoms, molecules, or ions by weighing them.

• Mole (mol): The amount of substance containing as many entities (atoms, molecules, ions) as there are atoms in exactly 12 grams of carbon-12.

• Avogadro's Number (NA): $6.022 \times 10^{23}$ entities per mole.

• Molecular mass: The sum of the average atomic masses of all atoms in a molecule's formula.

• Formula mass: For ionic compounds, the sum of the average atomic masses of the ions as they appear in the formula unit.

Examples and Applications

• Covalent substances exist as discrete molecules; their mass is called molecular mass.

• Ionic compounds do not exist as molecules; their mass is called formula mass.

Example: Calculating Molecular Mass

The average mass of an aspirin molecule (C9H8O4) is 180.15 amu.

Element	Quantity	Average Atomic Mass (amu)	Subtotal (amu)
C	9	12.01	108.09
H	8	1.008	8.064
O	4	16.00	64.00
Molecular mass (amu)			180.15

Avogadro's Number and the Mole

Definition and Use

The mole is a unit for counting particles at the atomic scale. Avogadro's number ($N_A$) bridges the gap between the atomic scale and the macroscopic scale.

• 1 mole = $6.022 \times 10^{23}$ entities (atoms, molecules, ions, etc.).

• Used to relate mass, number of particles, and volume in chemical calculations.

Example: Table of Atomic Masses and Moles

Element	Average Atomic Mass (AMU)	Mole Mass (g/mol)	Atoms/Mole
C	12.01	12.01	6.022×1023
H	1.008	1.008	6.022×1023
O	16.00	16.00	6.022×1023
Na	22.99	22.99	6.022×1023
Cl	35.45	35.45	6.022×1023

Example: Converting Grams to Moles

• Given: 4.7 g of potassium (K).

• Molar mass of K = 39.10 g/mol.

• Calculation: $\frac{4.7\,\text{g K}}{39.10\,\text{g/mol K}} = 0.12\,\text{mol K}$

Calculations Involving Moles

Converting Between Mass, Moles, and Number of Particles

• To convert grams to moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• To convert moles to number of particles: $\text{number of particles} = \text{moles} \times N_A$

Example: Mass of Argon in a Liter of Air

• Given: 9.2 × 10-4 mol Ar in 1 L air.

• Molar mass of Ar = 39.95 g/mol.

• Calculation: $9.2 \times 10^{-4}\,\text{mol} \times 39.95\,\text{g/mol} = 0.037\,\text{g Ar}$

Example: Number of Atoms in a Sample

• Given: 0.0400 g of saccharin (C7H5NO3S), molar mass = 183.18 g/mol.

• Calculation steps:

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

  2. Convert moles to molecules: $2.18 \times 10^{-4}\,\text{mol} \times 6.022 \times 10^{23}\,\text{molecules/mol} = 1.31 \times 10^{20}\,\text{molecules}$

  3. Calculate number of C atoms: $1.31 \times 10^{20}\,\text{molecules} \times 7 = 9.17 \times 10^{20}\,\text{C atoms}$

Percent Composition

Definition and Calculation

Percent composition is the percentage by mass of each element in a compound.

• Formula: $\%\,\text{element} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100$

• Example: In 10.0 g of a compound, 2.5 g is hydrogen and 7.5 g is carbon. Percent H = $\frac{2.5}{10.0} \times 100 = 25\%$; Percent C = $\frac{7.5}{10.0} \times 100 = 75\%$

Empirical and Molecular Formulas

Definitions

• Empirical formula: Simplest whole-number ratio of atoms in a compound (e.g., CH2O).

• Molecular formula: Actual number of atoms of each element in a molecule (e.g., C6H12O6).

Determining Empirical Formulas

1. Convert element masses to moles using molar masses.

2. Divide each number of moles by the smallest number of moles.

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

• Example: A compound contains 1.71 g C and 0.287 g H.

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

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

  • Ratio: C:H = 0.142:0.285 ≈ 1:2

  • Empirical formula: CH2

Solutions and Molarity

Solutions and Solvents

A solution is a homogeneous mixture of two or more substances. The solvent is the component present in greater amount, and the solute is present in lesser amount.

• When water is the solvent, the solution is called aqueous.

Molarity (M)

Molarity is a common unit of concentration, defined as the number of moles of solute per liter of solution.

• Formula: $M = \frac{\text{mol solute}}{\text{L solution}}$

• Example: A 355 mL soft drink contains 0.133 mol of sucrose. Molarity = $\frac{0.133\,\text{mol}}{0.355\,\text{L}} = 0.375\,\text{M}$

Summary Table: Formula Mass of NaCl

Element	Quantity	Average Atomic Mass (AMU)	Subtotal (amu)
Na	1	22.99	22.99
Cl	1	35.45	35.45
Formula mass (amu)			58.44

Additional info: The notes above include all major concepts and calculations relevant to the mole concept, chemical formulas, percent composition, and solution concentration, as presented in the provided materials.