Chemical Reactions, Stoichiometry, and Aqueous Solutions: Study Notes

Stoichiometry and Chemical Reactions

Avogadro’s Number and Mole Calculations

Avogadro’s number is a fundamental constant used to relate the number of particles (atoms, molecules, ions) to the amount of substance in moles.

• Avogadro’s Number: $6.022 \times 10^{23}$ particles per mole.

• Converting Mass to Number of Molecules:

  • Step 1: Convert mass to moles using molar mass.

  • Step 2: Multiply moles by Avogadro’s number to get number of molecules.

• Formula: $\text{Number of molecules} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \times 6.022 \times 10^{23}$

• Example: How many molecules are in 18 g of H2O?

  • Moles: $\frac{18\ \text{g}}{18.02\ \text{g/mol}} = 1.00\ \text{mol}$

  • Molecules: $1.00\ \text{mol} \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ molecules

Empirical Formula from Experimental Data

The empirical formula represents the simplest whole-number ratio of elements in a compound.

• Steps to Determine Empirical Formula:

  1. Convert mass of each element to moles.

  2. Divide each by the smallest number of moles to get the simplest ratio.

  3. Multiply to get whole numbers if necessary.

• Example: A compound contains 40.0 g C and 6.7 g H. Find the empirical formula.

  • Moles C: $\frac{40.0}{12.01} = 3.33$ mol

  • Moles H: $\frac{6.7}{1.008} = 6.65$ mol

  • Ratio: C : H = 3.33 : 6.65 ≈ 1 : 2

  • Empirical formula: CH2

Chemical Equations and Stoichiometry

Balancing Chemical Equations

Balancing chemical equations ensures the law of conservation of mass is obeyed.

• Steps:

  1. Write the unbalanced equation.

  2. Balance atoms one element at a time using coefficients.

  3. Check that all atoms are balanced.

• Example: $\ce{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$ Balanced: $\ce{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$

Stoichiometry of Reactions

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction.

• General Steps:

  1. Balance the chemical equation.

  2. Convert given quantities to moles.

  3. Use mole ratios from the balanced equation to find moles of desired substance.

  4. Convert moles to desired units (mass, volume, etc.).

• Formula: $\text{Moles of B} = \text{Moles of A} \times \frac{\text{Coefficient of B}}{\text{Coefficient of A}}$

• Example: How many grams of CO2 are produced from 44 g of C3H8?

  • Moles C3H8: $\frac{44}{44.1} = 1.00$ mol

  • Mole ratio: 1 mol C3H8 : 3 mol CO2

  • Moles CO2: $1.00 \times 3 = 3.00$ mol

  • Mass CO2: $3.00 \times 44.0 = 132$ g

Limiting Reactant, Theoretical Yield, and Percent Yield

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

• Limiting Reactant: Compare the mole ratio of reactants to determine which is limiting.

• Theoretical Yield: The maximum amount of product that can be formed from the limiting reactant.

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

• Example: If 10.0 g of A reacts with 15.0 g of B to produce 12.0 g of C, and the theoretical yield is 14.0 g, percent yield is:

  • $\frac{12.0}{14.0} \times 100\% = 85.7\%$

Combustion, Alkali Metal, and Halogen Reactions

Certain types of reactions have characteristic patterns.

• Combustion Reactions: Hydrocarbon + O2 → CO2 + H2O

• Alkali Metal Reactions: Alkali metal + Halogen → Alkali metal halide (e.g., $\ce{2Na + Cl_2 \rightarrow 2NaCl}$)

Solutions and Aqueous Reactions

Molarity and Dilution

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

• Formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Dilution Equation: $M_1V_1 = M_2V_2$

• Example: To prepare 250 mL of 0.10 M NaCl from 1.0 M NaCl:

  • $V_1 = \frac{M_2V_2}{M_1} = \frac{0.10 \times 0.250}{1.0} = 0.025$ L = 25 mL

Stoichiometry in Aqueous Reactions

Stoichiometric calculations in solution use molarity and volume to determine moles of reactants and products.

• Formula: $\text{Moles} = M \times V$ (V in liters)

• Example: How many moles of AgNO3 are in 50.0 mL of 0.200 M solution?

  • $0.200 \times 0.0500 = 0.0100$ mol

Solubility and Electrolytes

Compounds in water can be classified by their solubility and ability to conduct electricity.

• Soluble Compounds: Dissolve in water (e.g., NaCl, KNO3).

• Insoluble Compounds: Do not dissolve significantly (e.g., AgCl, BaSO4).

• Electrolytes: Substances that conduct electricity when dissolved (strong: NaCl; weak: CH3COOH).

• Nonelectrolytes: Do not conduct electricity (e.g., sugar, ethanol).

Precipitation Reactions

When two aqueous solutions are mixed, an insoluble product (precipitate) may form.

• Example: $\ce{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}$

Molecular, Complete Ionic, and Net Ionic Equations

Chemical equations for aqueous reactions can be written in different forms to show the species involved.

• Molecular Equation: Shows all compounds as neutral molecules.

• Complete Ionic Equation: Shows all strong electrolytes as ions.

• Net Ionic Equation: Shows only the species that change during the reaction.

• Example:

  • Molecular: $\ce{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}$

  • Complete Ionic: $\ce{Ag^+(aq) + NO_3^-(aq) + Na^+(aq) + Cl^-(aq) \rightarrow AgCl(s) + Na^+(aq) + NO_3^-(aq)}$

  • Net Ionic: $\ce{Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)}$

Acid-Base and Titration Reactions

Neutralization reactions occur between acids and bases, producing water and a salt.

• Strong Acid + Strong Base: $\ce{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)}$

• Weak Acid Example: $\ce{CH_3COOH(aq) + NaOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)}$

• Titration: A technique to determine concentration by reacting a known volume with a standard solution.

• Calculation: Use $M_1V_1 = M_2V_2$ for monoprotic acids and bases.

Gas-Evolution Reactions

Certain reactions in solution produce a gas as a product.

• Example: $\ce{Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)}$

Redox Reactions and Oxidation States

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

• Oxidation State: A number assigned to an atom to indicate its degree of oxidation.

• Rules:

  • Elemental form: 0

  • Monatomic ion: charge of ion

  • Oxygen: usually -2

  • Hydrogen: +1 with nonmetals, -1 with metals

  • Sum in compound: 0; in ion: equals charge

• Identifying Redox Reactions: Look for changes in oxidation states.

• Oxidizing Agent: Causes oxidation, is reduced.

• Reducing Agent: Causes reduction, is oxidized.

• Spontaneity: A redox reaction is spontaneous if the overall cell potential is positive (see electrochemistry for details).

Type of Compound	Solubility in Water	Electrolyte Type
NaCl	Soluble	Strong Electrolyte
AgCl	Insoluble	Nonelectrolyte
CH3COOH	Soluble	Weak Electrolyte
C6H12O6 (glucose)	Soluble	Nonelectrolyte

Additional info: For redox spontaneity, the cell potential (Ecell) is calculated using standard reduction potentials, but this is typically covered in more detail in electrochemistry chapters.