General Chemistry: Electrolytes, Stoichiometry, and Chemical Equations Study Guide

Electrolytes and Dissociation in Water

Strong and Weak Electrolytes

Electrolytes are substances that, when dissolved in water, produce a solution that conducts electricity due to the presence of ions. They are classified as strong or weak based on their degree of dissociation.

• Strong Electrolytes: Completely dissociate into ions in water. Examples include NaCl, HCl, and most soluble salts.

• Weak Electrolytes: Partially dissociate in water, producing fewer ions. Examples include acetic acid (CH3COOH).

Example: The reaction CH3COOH(aq) → CH3COO-(aq) + H+(aq) represents the dissociation of a weak acid.

Solubility of Ionic Compounds

The solubility of ionic compounds in water determines whether they act as strong electrolytes. Some compounds are soluble, while others are insoluble.

• Na3PO4: Soluble in water.

• AgCl: Insoluble in water.

Example: Na3PO4 is soluble, AgCl is insoluble.

Properties of Electrolytes

• Electrolytes increase the concentration of ions in solution and conduct electricity.

• Nonelectrolytes do not produce ions and do not conduct electricity.

Example: An electrolyte conducts electricity when dissolved in water.

Stoichiometry and Chemical Calculations

Mole-to-Mass Conversions

Stoichiometry involves using balanced chemical equations to calculate the amounts of reactants and products. The mole is a central unit in these calculations.

• Mole-to-mass conversion: Use the molar mass to convert between moles and grams.

Formula:

$\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Example: How many grams does 5.71 moles of nickel (Ni) contain?

$5.71\ \text{mol Ni} \times \frac{58.69\ \text{g Ni}}{1\ \text{mol Ni}} = 335\ \text{g Ni}$

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms.

• Empirical formula: Determined from experimental data (e.g., combustion analysis).

• Molecular formula: Determined using the molar mass.

Example: A compound with a molar mass of 78.1 g/mol, made of C and H, produces 2.311 g CO2 and 0.4729 g H2O upon combustion. The empirical formula is CH, and the molecular formula is C6H6.

Stoichiometry in Chemical Reactions

Stoichiometric calculations use balanced equations to determine the amount of reactants or products.

• Steps:

  1. Write the balanced equation.

  2. Convert given quantities to moles.

  3. Use mole ratios from the equation.

  4. Convert moles to desired units (grams, molecules, etc.).

Example: How many grams of NO2 are required to produce 4.30 g of HNO3 in the reaction:

3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

$4.30\ \text{g HNO}_3 \times \frac{1\ \text{mol HNO}_3}{63.02\ \text{g HNO}_3} \times \frac{3\ \text{mol NO}_2}{2\ \text{mol HNO}_3} \times \frac{46.01\ \text{g NO}_2}{1\ \text{mol NO}_2} = 4.7\ \text{g NO}_2$

Limiting Reactant and Yield Calculations

When two or more reactants are present, the limiting reactant is the one that is completely consumed first, determining the maximum amount of product formed.

• Identify limiting reactant: Calculate the amount of product each reactant can produce; the smallest amount is the limiting reactant.

• Theoretical yield: The maximum amount of product possible from the limiting reactant.

Example: Calculating O2 produced from given amounts of KClO3:

$26.2\ \text{g KClO}_3 \times \frac{1\ \text{mol KClO}_3}{122.55\ \text{g KClO}_3} \times \frac{3\ \text{mol O}_2}{2\ \text{mol KClO}_3} \times \frac{32.00\ \text{g O}_2}{1\ \text{mol O}_2} = 8.4\ \text{g O}_2$

Chemical Equations and Ionic Equations

Complete Ionic Equations

Complete ionic equations show all of the ions present in a reaction in aqueous solution.

• Write all strong electrolytes as ions.

• Weak electrolytes, insoluble compounds, and gases are written in molecular form.

Example: The reaction between NaOH(aq) and CuSO4(aq):

$2\text{Na}^+(aq) + 2\text{OH}^-(aq) + \text{Cu}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{Cu(OH)}_2(s) + 2\text{Na}^+(aq) + \text{SO}_4^{2-}(aq)$

Balanced Chemical Equations

Balanced chemical equations have equal numbers of each atom on both sides of the reaction arrow.

• Neutralization reactions: Acid reacts with base to form water and a salt.

Example: Neutralization of phosphoric acid (H3PO4) with potassium hydroxide (KOH):

$\text{H}_3\text{PO}_4(aq) + 3\text{KOH}(aq) \rightarrow 3\text{H}_2\text{O}(l) + \text{K}_3\text{PO}_4(aq)$

Summary Table: Electrolyte Properties

Type	Dissociation in Water	Conducts Electricity?	Example
Strong Electrolyte	Complete	Yes	NaCl, HCl
Weak Electrolyte	Partial	Yes (weakly)	CH3COOH
Nonelectrolyte	None	No	Sugar, ethanol

Additional info: Some context and explanations have been expanded for clarity and completeness, including definitions and stepwise examples for stoichiometry and chemical equations.