Chapter 3: Mole Concept, Molar Mass, Percent Composition, and Solution Concentration in General Chemistry

Mole Concept and Molar Mass

Avogadro's Number and the Mole

The mole is a fundamental unit in chemistry used to count numbers of atoms, ions, or molecules in a sample. It allows chemists to relate macroscopic quantities of substances to the number of particles they contain.

• Avogadro's Number (NA): The number of entities in one mole is $6.022 \times 10^{23}$.

• Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol).

• Atomic Mass Unit (amu): 1 amu is defined as $1~\text{g/mol}$ for practical calculations.

Example: Calculating the molar mass of acetaminophen (C8H9NO2):

• Sum the atomic masses of all atoms in the formula:

$\text{Molar mass} = (8 \times 12.01) + (9 \times 1.008) + (1 \times 14.01) + (2 \times 16.00) = 180.159~\text{g/mol}$

Conversions Between Grams, Moles, Molecules, and Atoms

Dimensional Analysis in Chemical Calculations

Chemists often convert between mass, moles, molecules, and atoms using conversion factors derived from molar mass and Avogadro's number.

• Grams to Moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Moles to Molecules: $\text{number of molecules} = \text{moles} \times 6.022 \times 10^{23}$

• Molecules to Atoms: Multiply by the number of atoms per molecule.

Example: How many sulfur atoms are in 0.45 mol BaSO4?

• 0.45 mol BaSO4 $\times$ $6.022 \times 10^{23}$ molecules/mol $\times$ 1 S atom/molecule = $2.7 \times 10^{23}$ S atoms

Percent Composition by Mass

Calculating Mass Percent of Elements in Compounds

Percent composition expresses the mass percentage of each element in a compound. It is useful for determining empirical formulas and analyzing chemical samples.

• Formula:

$\%~\text{composition of element} = \frac{\text{number of atoms} \times \text{atomic weight}}{\text{formula weight of substance}} \times 100\%$

Example: ATP (C10H16N5O13P3):

• Calculate the molar mass of ATP.

• Find the mass percent for each element using the formula above.

Empirical and Molecular Formulas

Determining Formulas from Mass Data

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms in a molecule.

• Steps to Determine Empirical Formula:

  1. Convert mass of each element to moles.

  2. Divide by the smallest number of moles to get ratios.

  3. Multiply to get whole numbers if necessary.

• Molecular Formula: Determined by dividing the molar mass by the empirical formula mass.

Example: Hematite sample with 34.97 g Fe and 15.03 g O:

• Fe: $\frac{34.97~\text{g}}{55.85~\text{g/mol}} = 0.626~\text{mol}$

• O: $\frac{15.03~\text{g}}{16.00~\text{g/mol}} = 0.939~\text{mol}$

• Ratio: Fe:O = 1:1.5, multiply by 2: Fe2O3

Solution Concentration: Molarity and Dilution

Molarity and Solution Preparation

Molarity (M) is the concentration of a solution, defined as moles of solute per liter of solution.

• Formula:

$M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$

• Dilution Equation:

$M_i V_i = M_f V_f$

Example: If 15.0 mL of acetic acid (d = 1.048 g/mL) is dissolved to make 500.0 mL solution:

• Calculate mass, moles, and molarity using the formulas above.

Example: If 275 mL of 0.105 M NaCl is evaporated to 237 mL, find new concentration:

• Use $M_i V_i = M_f V_f$ to solve for $M_f$.

Applications and Practice Problems

Sample Calculations and Real-World Examples

• Calculating the number of molecules in a given volume at a specified concentration (e.g., chloroform in water).

• Determining the mass of a sample from the number of atoms (e.g., Si atoms in aquamarine).

Example: How many molecules of CHCl3 are present in 350 mL of water at 0.8 ppb?

• Convert ppb to mass, then to moles, then to molecules using Avogadro's number.

Example: What is the mass in mg of a sample containing $3.0 \times 10^{18}$ atoms of Si?

• Convert atoms to moles, then to mass using molar mass.

Summary Table: Key Formulas and Relationships

Quantity	Formula	Units
Moles	$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$	mol
Number of Particles	$\text{number} = \text{moles} \times 6.022 \times 10^{23}$	atoms, molecules
Percent Composition	$\% = \frac{\text{mass of element}}{\text{total mass}} \times 100$	%
Molarity	$M = \frac{\text{moles}}{\text{volume (L)}}$	mol/L
Dilution	$M_i V_i = M_f V_f$	mol/L, L

Additional info: These notes expand on the handwritten calculations and diagrams, providing full academic context and explanations for each concept. All formulas and relationships are standard for introductory college-level General Chemistry.