5/7/25 CCP Chemistry Notes (Final Exam Review)

Solubility and Precipitation Reactions

• To identify precipitates, determine the solubility of reactants and products using solubility rules.
• Solubility trends:
  • Start with trend number one and work downwards until a trend applies to one of the ions in the compound.
  • It's not necessary to match both ions, just one.
• Example: Iron(II) Sulfate ($FeSO_4$)
  • Trend 5 states: Sulfates are soluble except for a few exceptions, and iron is not one of them. Therefore Iron(II) Sulfate ($FeSO_4$) is soluble (aqueous).
• Example: Iron(II) Hydroxide ($Fe(OH)_2$)
  • Trend 7 states: Hydroxides are generally insoluble, with exceptions. Iron is not one of the exceptions. Therefore, Iron(II) Hydroxide ($Fe(OH)_2$) is insoluble (solid).
• Aqueous solutions of ionic compounds and soluble acids dissociate into ions.
• Insoluble compounds do not dissociate and remain in their solid, liquid, or gaseous phase.

Spectator Ions and Net Ionic Equations

• Spectator ions: Ions that remain unchanged from reactant to product.
• To determine spectator ions, represent all soluble compounds in their dissociated form (ions).
• Identify ions that appear on both sides of the equation without change; these are spectator ions.
• Net ionic equation: Rewrite the reaction without the spectator ions.

Stoichiometry

• Stoichiometry: Converting between amounts of different substances in a chemical reaction.
• General steps:
  1. Start with the given amount of a substance.
  2. Convert the given amount to moles using molar mass: $moles = \frac{mass}{molar\,mass}$. For Lithium Hydroxide ($LiOH$) molar mass is approximately $23.95 \frac{g}{mol}$.
  3. Use the mole ratio from the balanced chemical equation to convert moles of the given substance to moles of the desired substance.
  4. Convert moles of the desired substance to the desired units (e.g., grams) using molar mass.
• Example: Convert 45 grams of lithium hydroxide ($LiOH$) to grams of iron(II) hydroxide ($Fe(OH)_2$)
  • $45 \, g \, LiOH * \frac{1 \, mol \, LiOH}{23.95 \, g \, LiOH} * \frac{1 \, mol \, Fe(OH)<em>2}{2 \, mol \, LiOH} * \frac{84.8 \, g \, Fe(OH)</em>2}{1 \, mol \, Fe(OH)<em>2} = 79.7 \, g \, Fe(OH)</em>2$

Gas-Forming Reactions

• Most gas-forming reactions are double displacement reactions with an ion swap, except for metal-acid reactions (single displacement).
• Examples:
  • Sulfuric acid ($H<em>2SO</em>4$) reacting with potassium sulfide ($K<em>2S$) produces potassium sulfate ($K</em>2SO<em>4$) and hydrogen sulfide ($H</em>2S$) gas.
    • $H<em>2SO</em>4(aq) + K<em>2S(aq) \rightarrow K</em>2SO<em>4(aq) + H</em>2S(g)$
  • Aluminum ($Al$) reacting with sulfuric acid ($H<em>2SO</em>4$) produces aluminum sulfate ($Al<em>2(SO</em>4)<em>3$) and hydrogen gas ($H</em>2$).
    • $2Al(s) + 3H<em>2SO</em>4(aq) \rightarrow Al<em>2(SO</em>4)<em>3(aq) + 3H</em>2(g)$
• Exceptions:
  • Carbonic acid ($H<em>2CO</em>3$) decomposes into water ($H<em>2O$) and carbon dioxide ($CO</em>2$).
    • $H<em>2CO</em>3(aq) \rightarrow H<em>2O(l) + CO</em>2(g)$
  • Sulfurous acid ($H<em>2SO</em>3$) decomposes into water ($H<em>2O$) and sulfur dioxide ($SO</em>2$).
    • $H<em>2SO</em>3(aq) \rightarrow H<em>2O(l) + SO</em>2(g)$

Oxidation Numbers

• Rules for assigning oxidation numbers:
  1. Elements in their standard state have an oxidation number of 0.
  2. Monatomic ions have an oxidation number equal to their charge.
  3. Specific elements in compounds:
     • Fluorine is always -1.
     • Oxygen is usually -2 (except in peroxides).
     • Hydrogen is usually +1 (except in metal hydrides).
  4. The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the charge of the ion.
• When multiple atoms are covered by rule 3, apply priority order: F > O > H.
• Example 1: sulfur dioxide ($SO_2$)
  • Oxygen is -2. The sum of the oxidation numbers is 0 ($(-2)*2 + x = 0$). Therefore, sulfur is +4.
• Example 2: the sulfate ion ($SO_4^{-2}$)
  • Oxygen is -2. The sum of the oxidation numbers is -2 ($(-2)*4 + x = -2$). Therefore, sulfur is +6
• The loss of electrons is oxidation and the gain of electrons is reduction.

Molarity Calculations

• Molarity (M) is defined as moles of solute per liter of solution: $M = \frac{moles}{liter}$.
• To convert grams to moles, use molar mass: $moles = \frac{grams}{molar\,mass}$.
• Example: To find the liters for 25 grams of sodium chloride (NaCl) 0.17 M solution, first convert grams to moles and then solve for volume using the molarity equation if molar mass of $NaCl$ is $58.44 \frac{g}{mol}$:
  • $25 \, g \, NaCl * \frac{1 \, mol \, NaCl}{58.44 \, g \, NaCl}$. Number of moles is 0.428.
  • $Volume = \frac{0.428 \, mol}{1.7 \frac{mol}{L}} = 0.25 \,L$

Stoichiometry with Molarity

• Use molarity to convert between volume and moles of reactants or products in a chemical reaction.
• Steps:
  1. Convert the given volume of a solution to moles using molarity: $moles = Molarity * Volume$.
  2. Use the mole ratio from the balanced equation to convert moles of the given substance to moles of the desired substance.
  3. Convert moles of the desired substance to volume using molarity: $Volume = \frac{moles}{Molarity}$.

Gas Laws: Charles's Law

• Charles's Law: Relates volume and temperature of a gas at constant pressure: $\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2}$.
• Temperature must be in Kelvin: $K = \, ^\circ C +273.15$
• Keep sets of variables together for initial and final conditions.
• Example:
  • $V<em>1 = 3.5 L$, $T</em>1 = 26 ^\circ C = 299 K$, $V<em>2 = 5.2 L$, find $T</em>2$
• $\frac{3.5 L}{299 K} = \frac{5.2 L}{T_2}$
• $T_2 = 444 K = 171 ^\circ C$

Ideal Gas law: Molar Mass Determination

• Ideal Gas Law: $PV = nRT$
• To determine the molar mass:
  1. Solve for $n$ using Ideal Gas Law, where:
     • P is pressure (in atmospheres).
     • V is volume (in liters).
     • n is the number of moles.
     • R is the ideal gas constant (0.0821 L atm / (mol K)).
     • T is temperature (in Kelvin).
  2. Calculate molar mass using: $Molar \, Mass = \frac{grams}{moles}$.
• Ensure all units match the units in the gas constant.
• Example: pressure is 6.3 atm, Volume is 2L, and T is 300K. Mass is 65 grams.
  • First solve for $n$. $n = \frac{PV}{RT} = \frac{6.3 * 2}{0.0821 * 300} = 0.512$. 65 grams contains 0.512 moles.
  • $Molar \, Mass = \frac{65}{0.512} = 127 \frac{grams}{mole}$.