General Chemistry Study Notes: Stoichiometry, Solutions, Electrolytes, and Chemical Equations

Stoichiometry and Chemical Equations

Balancing Chemical Equations

Balancing chemical equations is essential to ensure the conservation of mass and atoms in a chemical reaction. The process involves adjusting coefficients to have equal numbers of each atom on both sides of the equation.

• Step 1: Balance carbon (C) and hydrogen (H) atoms first.

• Step 2: Balance oxygen (O) atoms last.

Example:

• Unbalanced: $\mathrm{C_8H_{10}(l) + O_2(g) \rightarrow 8CO_2(g) + 4H_2O}$

• Balanced: $\mathrm{C_8H_{10}(l) + 10O_2(g) \rightarrow 8CO_2(g) + 4H_2O}$

Molar Mass and Percent Composition

Calculating Molar Mass

The molar mass of a compound is the sum of the atomic masses of all atoms in its formula, expressed in grams per mole (g/mol).

• Example: For $\mathrm{C_2H_5NO_2}$:

  • C: $2 \times 12.0 = 24.0$ g/mol

  • H: $5 \times 1.0 = 5.0$ g/mol

  • N: $1 \times 14.0 = 14.0$ g/mol

  • O: $2 \times 16.0 = 32.0$ g/mol

  • Total: $75.0$ g/mol

Calculating Moles

Moles are calculated by dividing the mass of a substance by its molar mass.

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

• Example: $\frac{100.0\ \text{g}}{75.0\ \text{g/mol}} = 1.333\ \text{mol}$

Percent Composition

Percent composition by mass of an element in a compound is calculated as:

• $\%\ N = \frac{14.0\ \text{g}}{75.0\ \text{g/mol}} \times 100 = 18.7\%$

Molecules and Atoms Calculations

Avogadro's Number

Avogadro's number ($6.022 \times 10^{23}$) is used to convert between moles and number of molecules or atoms.

• Number of molecules: $\text{moles} \times 6.022 \times 10^{23}$

• Example: $8.14 \times 10^{-2}\ \text{mol} \times 6.022 \times 10^{23}\ \text{molecules/mol} = 5.087 \times 10^{22}\ \text{molecules}$

Atoms in a Compound

• To find the number of atoms of a specific element in a compound, multiply the number of molecules by the number of that atom per molecule.

• Example: $3.76 \times 10^{21}\ \text{molecules}\ \mathrm{Al(NO_3)_3} \times 9\ \text{atoms O/molecule} = 3.39 \times 10^{22}\ \text{atoms O}$

Limiting Reactant and Percent Yield

Limiting Reactant

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Calculate moles of each reactant.

• Compare mole ratios using the balanced equation.

• Example:

  • $30.0\ \text{g}\ \mathrm{C_6H_6}$ $\rightarrow 0.385\ \text{mol}$

  • $65.0\ \text{g}\ \mathrm{Br_2}$ $\rightarrow 0.407\ \text{mol}$

  • Limiting reactant: $\mathrm{C_6H_6}$

Theoretical and Percent Yield

• Theoretical yield: Maximum amount of product possible from limiting reactant.

• Percent yield: $\frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

• Example: $\frac{42.3\ \text{g}}{60.3\ \text{g}} \times 100 = 70.1\%$

Electrolytes and Acids

Classification of Electrolytes

Electrolytes are substances that conduct electricity when dissolved in water. They are classified as strong or weak based on their degree of ionization.

• Strong electrolytes: Completely ionize in solution (e.g., HBr).

• Weak electrolytes: Partially ionize in solution (e.g., acetic acid).

Classification of Acids

• Strong acids: Ionize completely (e.g., HBr).

• Weak acids: Ionize partially (e.g., acetic acid, $\mathrm{CH_3COOH}$).

Precipitation and Net Ionic Equations

Precipitation Reactions

Precipitation occurs when two aqueous solutions react to form an insoluble solid (precipitate).

• Example: $\mathrm{K_2CO_3(aq) + MgSO_4(aq) \rightarrow K_2SO_4(aq) + MgCO_3(s)}$

• Net ionic equation: $\mathrm{Mg^{2+}(aq) + CO_3^{2-}(aq) \rightarrow MgCO_3(s)}$

Neutralization Reactions

Neutralization is the reaction of an acid and a base to form water and a salt.

• Example: $\mathrm{2CH_3COOH(aq) + Ba(OH)_2(aq) \rightarrow 2H_2O(l) + Ba(CH_3COO)_2(aq)}$

• Net ionic equation: $\mathrm{2H^+(aq) + 2OH^-(aq) \rightarrow 2H_2O(l)}$

Oxidation Numbers and Redox Reactions

Assigning Oxidation Numbers

Oxidation numbers are assigned to atoms to track electron transfer in reactions.

• H in acids: +1

• O: -2 (except in peroxides)

• Sum of oxidation numbers in a neutral compound is zero.

Redox Reactions

Redox (oxidation-reduction) reactions involve the transfer of electrons between species.

• Example: $\mathrm{Fe(s) + Cu(NO_3)_2(aq) \rightarrow Cu(s) + Fe(NO_3)_2(aq)}$

• Net ionic equation: $\mathrm{Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)}$

Solutions and Dilutions

Calculating Molarity

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

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

• Example: $\frac{175\ \text{g}}{60.05\ \text{g/mol}} \div 2.00\ \text{L} = 1.46\ \text{M}$

Dilution Calculations

To dilute a solution, use the formula $M_1V_1 = M_2V_2$.

• Example: To dilute $0.30\ \text{M}$ NaCl to three times its volume:

• $0.30\ \text{M} \times V_1 = M_2 \times 3V_1$

• $M_2 = \frac{0.30\ \text{M}}{3} = 0.10\ \text{M}$

Summary Table: Electrolyte and Acid Classification

Substance	Acid/Base Strength	Electrolyte Strength
Hydrobromic acid (HBr)	Strong acid	Strong electrolyte
Acetic acid (CH3COOH)	Weak acid	Weak electrolyte
Carbonic acid (H2CO3)	Weak acid	Weak electrolyte

Additional info: These notes cover foundational topics in General Chemistry, including stoichiometry, solution chemistry, acid-base and redox reactions, and classification of electrolytes. All calculations and equations are based on standard chemical principles and conventions.