Mass and Mole Relationships, Energy, and Solution Calculations in GOB Chemistry

Mass and Mole Relationships

Mole Concept and Molar Mass

The mole is a fundamental unit in chemistry used to count particles such as atoms, molecules, or ions. One mole of any substance contains Avogadro’s number of particles, which is $6.022 \times 10^{23}$.

• Molar mass is the mass of one mole of a substance, expressed in grams.

• It is numerically equal to the molecular (formula) weight of the substance in grams.

• Avogadro’s number ($6.022 \times 10^{23}$) links the number of particles to the amount in moles.

Calculating Molar Mass

Molar mass is determined by summing the atomic masses of all atoms in a chemical formula.

• For example, diethyl ether (C4H10O):

  • C: $4 \times 12.01 = 48.04$

  • H: $10 \times 1.008 = 10.08$

  • O: $1 \times 16.00 = 16.00$

  • Molar mass = 74.12 g/mol

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

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

Stoichiometry and Limiting Reactant

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction using balanced equations.

• Limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed.

• Example: If you have 14 slices of bread, 15 slices of ham, and 20 slices of cheese, and each sandwich requires 2 bread, 3 ham, and 1 cheese, ham is the limiting reactant.

Percent Yield

Percent yield measures the efficiency of a reaction by comparing the actual yield to the theoretical yield.

• Formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

• Example: If 15.8 g of water is produced but the theoretical yield is 23.6 g, percent yield is $\frac{15.8}{23.6} \times 100 = 66.9\%$

Chemical Reactions: Energy, Rate, and Equilibrium

Energy Changes in Reactions

Chemical reactions involve energy changes, often in the form of heat.

• Bond dissociation energy is the energy required to break a bond in a molecule.

• Endothermic reactions absorb heat (bond breaking).

• Exothermic reactions release heat (bond formation).

• Law of conservation of energy: Energy cannot be created or destroyed.

• Heat of reaction is the difference between energy absorbed and released.

Entropy and Free Energy

• Entropy change (ΔS): Positive if disorder increases; negative if disorder decreases.

• Exergonic processes release free energy and are spontaneous.

• Endergonic processes require free energy and are nonspontaneous.

Le Châtelier’s Principle

When a stress is applied to a system at equilibrium, the system shifts to relieve the stress.

Reaction Rate and Rate Law

• Rate law example: $\text{rate} = k[A]^3[B]^2$

• If concentrations of A and B are doubled, the rate increases by a factor of 32.

Solutions and Concentration Calculations

Solution Concentration

Concentration expresses the amount of solute in a given volume of solution.

• Molarity (M): $\text{M} = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Percent (m/v): $\frac{\text{grams of solute}}{100 \text{ mL solution}}$

• Percent (v/v): $\frac{\text{volume of solute}}{100 \text{ mL solution}}$

Solution Preparation and Dilution

• To dilute a solution: $M_1V_1 = M_2V_2$

• Example: To prepare 75.00 mL of 1.25 M solution from 4.00 M stock, $V_1 = \frac{1.25 \times 75.00}{4.00} = 23.4$ mL

Mass and Volume Calculations in Solutions

• To find mass of solute: $\text{mass} = \text{moles} \times \text{molar mass}$

• To find volume of solute in a percent (v/v) solution: $\text{volume} = \text{percent} \times \text{total volume}$

Additional info:

• All equations are provided in LaTeX format for clarity.

• Images included are directly relevant to the explanation of molar mass, Avogadro’s number, and solution concentration.