Moles, Molar Mass, Empirical Formulas, and Solution Concentrations in General Chemistry

The Mole and Its Importance

Definition and Purpose

• The mole is the SI unit for the amount of substance, providing a bridge between the atomic/molecular scale and the laboratory scale.

• It allows chemists to count particles (atoms, molecules, ions) by weighing measurable amounts of material.

• Avogadro's number (NA) defines one mole as containing $6.022 \times 10^{23}$ representative particles.

Formula Mass vs. Molar Mass

Definitions and Differences

• Formula Mass: The sum of the atomic masses (in amu) of all atoms in a chemical formula. Used for both ionic and molecular compounds.

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

Example: For water (H2O): $\text{Formula Mass} = 2 \times 1.008\ \text{amu} + 16.00\ \text{amu} = 18.02\ \text{amu}$ $\text{Molar Mass} = 18.02\ \text{g/mol}$

Calculating Molar Mass and Formula Mass

Step-by-Step Example

• Given: Sucrose (C12H22O11)

• Calculate formula mass in amu:

$\text{Formula Mass} = (12 \times 12.01) + (22 \times 1.008) + (11 \times 16.00) = 342.30\ \text{amu}$

• Convert to molar mass (g/mol):

$342.30\ \text{amu} \times \left(1.66054 \times 10^{-24}\ \text{g/amu}\right) \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 342.3\ \text{g/mol}$

Mass, Moles, and Number of Particles Calculations

Relationships and Conversions

• Key relationships:

  • Mass (g) ↔ Moles (mol) ↔ Number of particles

  • Use molar mass to convert between mass and moles.

  • Use Avogadro's number to convert between moles and number of particles.

Example 1: Calculate the mass in grams of 4.50 moles of Al. $4.50\ \text{mol Al} \times 26.98\ \text{g/mol} = 121\ \text{g Al}$ Example 2: Number of NH3 molecules in 45 g NH3: $45\ \text{g NH}_3 \times \frac{1\ \text{mol}}{17.03\ \text{g}} \times \frac{6.022 \times 10^{23}\ \text{molecules}}{1\ \text{mol}} = 1.6 \times 10^{24}\ \text{molecules}$ Example 3: Number of oxygen atoms in 14.3 g of Cr2(SO4)3: $14.3\ \text{g} \times \frac{1\ \text{mol}}{392.17\ \text{g}} \times \frac{12\ \text{mol O}}{1\ \text{mol}} \times \frac{6.022 \times 10^{23}}{1\ \text{mol}} = 2.64 \times 10^{23}\ \text{atoms O}$

Mass Percent Composition

Definition and Calculation

• Mass percent composition is the percentage by mass of each element in a compound.

• Calculated as:

$\%\text{A in A}_x\text{B}_y = \frac{x \times \text{molar mass of A}}{\text{molar mass of A}_x\text{B}_y} \times 100\%$ Example: For H2O: $\%\text{H} = \frac{2 \times 1.008}{18.02} \times 100\% = 11.19\%$ $\%\text{O} = 100\% - 11.19\% = 88.81\%$

Mass Percent Composition from Combustion Data

Combustion Analysis

• For organic compounds containing C, H, and O, combustion produces CO2 and H2O.

• Mass percent of C is determined from the mass of CO2 produced.

• Mass percent of H is determined from the mass of H2O produced.

• Mass percent of O is found by subtracting the C and H percentages from 100%.

Example: Butyric acid (4.30 g sample) produces 8.59 g CO2 and 3.52 g H2O.

• $\%\text{C} = \frac{2.34\ \text{g}}{4.30\ \text{g}} \times 100\% = 54.4\%$

• $\%\text{H} = \frac{0.394\ \text{g}}{4.30\ \text{g}} \times 100\% = 9.16\%$

• $\%\text{O} = 100\% - 54.4\% - 9.16\% = 36.4\%$

Empirical and Molecular Formulas

Definitions

• Empirical Formula: The simplest whole-number ratio of elements in a compound.

• Molecular Formula: The actual number of each type of atom in a molecule of the compound.

Example:

Compound	Molecular Formula	Empirical Formula
Benzene	C6H6	CH
Acetylene	C2H2	CH

Determining the Empirical Formula

1. Use mass data to calculate moles of each element.

2. Divide all mole amounts by the smallest number of moles.

3. If necessary, multiply ratios by a small whole number to obtain whole-number subscripts.

4. Write the empirical formula using these subscripts.

Example: Adipic acid (49.3% C, 6.9% H, 43.8% O):

• Assume 100 g sample: 49.3 g C, 6.9 g H, 43.8 g O

• Moles: 4.10 mol C, 6.8 mol H, 2.74 mol O

• Divide by smallest: C 1.5, H 2.5, O 1

• Multiply by 2: C3H5O2

Determining the Molecular Formula

• Calculate the empirical formula mass.

• Divide the given molecular mass by the empirical formula mass to find the multiplier (n).

• Multiply the subscripts in the empirical formula by n to get the molecular formula.

$n = \frac{\text{molecular mass}}{\text{empirical formula mass}}$ Example: Adipic acid, molecular mass = 146 amu, empirical formula mass = 73.07 amu: $n = 146 / 73.07 = 2$ Molecular formula: (C3H5O2)2 = C6H10O4

Solutions and Concentration

Parts of a Solution

• Solute: The substance being dissolved.

• Solvent: The substance doing the dissolving (often water in aqueous solutions).

• Solutions are homogeneous mixtures.

Molarity (Molar Concentration)

• Molarity (M): Moles of solute per liter of solution.

$\text{Molarity of A} = [A] = \frac{\text{moles of A}}{\text{liters of solution}}$

Calculating Molarity

• Need: moles of solute (from mass) and volume of solution in liters.

Example: 5.55 g CaCl2 in 125 mL solution: $5.55\ \text{g} \times \frac{1\ \text{mol}}{110.98\ \text{g}} \times \frac{1}{0.125\ \text{L}} = 0.400\ \text{M}$

Using Molarity in Calculations

• Relate volume, moles, and mass using molarity and molar mass.

Example: What volume of 0.120 M CuSO4 contains 1.50 g CuSO4? $1.50\ \text{g} \times \frac{1\ \text{mol}}{159.62\ \text{g}} \times \frac{1}{0.120\ \text{mol/L}} \times 1000\ \text{mL} = 78.3\ \text{mL}$

Dilution Calculations

• Dilution: Decreasing the concentration of a solution by adding more solvent.

• The number of moles of solute remains constant during dilution.

• Use the formula:

$M_1V_1 = M_2V_2$ Example: What volume of 15.9 M HNO3 is needed to prepare 500 mL of 3.75 M HNO3? $V_1 = \frac{M_2V_2}{M_1} = \frac{3.75\ \text{M} \times 500\ \text{mL}}{15.9\ \text{M}} = 118\ \text{mL}$

Summary Table: Key Relationships

Quantity	Conversion	Equation
Mass ↔ Moles	Use molar mass	$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
Moles ↔ Particles	Use Avogadro's number	$\text{particles} = \text{moles} \times 6.022 \times 10^{23}$
Moles ↔ Volume (solution)	Use molarity	$\text{moles} = \text{M} \times \text{L}$
Dilution	Concentration and volume	$M_1V_1 = M_2V_2$

Additional info: These notes cover foundational concepts in stoichiometry, solution chemistry, and chemical composition, which are essential for success in General Chemistry.