General Chemistry Study Guide: Mole Calculations, Chemical Reactions, and Solution Chemistry

Chapter 3: Mole Calculations and Solution Chemistry

Difference Between Molecular and Molar Mass

The molecular mass (or molecular weight) is the sum of the atomic masses of all atoms in a molecule, measured in atomic mass units (amu). The molar mass is the mass of one mole of a substance (atoms, molecules, or formula units), measured in grams per mole (g/mol).

• Molecular mass is used for individual molecules.

• Molar mass is used for bulk quantities (1 mole = 6.022 × 1023 particles).

• Example: The molecular mass of H2O is 18.02 amu; its molar mass is 18.02 g/mol.

Using Molar Mass to Convert Between Grams and Moles

The molar mass allows conversion between the mass of a substance and the number of moles.

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

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

• Example: 36.04 g of H2O is 2.00 moles ($36.04 \div 18.02 = 2.00$).

Using Avogadro’s Number to Convert Particles to Moles and Vice Versa

Avogadro’s number ($6.022 \times 10^{23}$) is the number of particles in one mole of a substance.

• Formula: $\text{moles} = \frac{\text{number of particles}}{6.022 \times 10^{23}}$

• To find number of particles: $\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

• Example: 1.5 moles of NaCl contains $1.5 \times 6.022 \times 10^{23} = 9.03 \times 10^{23}$ formula units.

Solution Concentration Formulas

Concentration is the amount of solute dissolved in a given quantity of solvent or solution. The most common unit is molarity (M).

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

• Example: Dissolving 0.5 moles of NaCl in 1.0 L of water gives a 0.5 M solution.

Dilution Formula

Dilution involves adding solvent to decrease the concentration of a solution.

• Formula: $M_1V_1 = M_2V_2$

• Where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume.

• Example: To dilute 100 mL of 2.0 M HCl to 1.0 M, use $2.0 \times 100 = 1.0 \times V_2$, so $V_2 = 200$ mL.

Chapter 4: Chemical Reactions and Stoichiometry

Law of Conservation of Mass

The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.

• Application: All chemical equations must be balanced to reflect this law.

Balancing Chemical Equations

Balancing ensures the same number of each type of atom on both sides of the equation.

• Adjust coefficients (not subscripts) to balance atoms.

• Example: $2H_2 + O_2 \rightarrow 2H_2O$

Types of Chemical Reactions

Chemical reactions can be classified by how reactants are transformed into products.

• Synthesis (Combination): Two or more substances combine to form one product. $A + B \rightarrow AB$

• Decomposition: One substance breaks down into two or more products. $AB \rightarrow A + B$

• Single Displacement: One element replaces another in a compound. $A + BC \rightarrow AC + B$

• Double Displacement: Exchange of ions between two compounds. $AB + CD \rightarrow AD + CB$

Classes of Chemical Reactions

Reactions can also be grouped by their chemical behavior:

• Acid-Base Neutralization: Acid reacts with base to form salt and water.

• Precipitation: Two solutions form an insoluble solid (precipitate).

• Redox (Oxidation-Reduction): Transfer of electrons between species.

Predicting Precipitates Using Solubility Charts

Solubility rules help predict if a precipitate will form in a double displacement reaction.

• Insoluble products form precipitates.

• Example: Mixing AgNO3 and NaCl forms AgCl (a precipitate).

Calculating Oxidation Numbers

Oxidation numbers indicate the charge an atom would have if electrons were transferred completely.

• Rules assign oxidation numbers based on element type and position in a compound.

• Example: In H2O, H is +1, O is -2.

Identifying Oxidized and Reduced Elements in Redox Reactions

In a redox reaction:

• Oxidation: Loss of electrons (increase in oxidation number).

• Reduction: Gain of electrons (decrease in oxidation number).

• Example: In $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$, Zn is oxidized, Cu2+ is reduced.

Stoichiometry: Mole and Mass Conversions Using Balanced Equations

Stoichiometry uses balanced equations to relate amounts of reactants and products.

• Mole-to-mole conversion: Use coefficients from the balanced equation.

• Mass-to-mass conversion: Convert mass to moles, use mole ratio, then convert back to mass.

• Example: $2H_2 + O_2 \rightarrow 2H_2O$; 4 moles H2 produce 4 moles H2O.

Percent Yield Formula

Percent yield compares actual yield to theoretical yield.

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

• Example: If theoretical yield is 10 g and actual yield is 8 g, percent yield is $\frac{8}{10} \times 100\% = 80\%$.