Chemical Yields and Aqueous Reactions

Theoretical and Actual Yield

• Theoretical Yield:

  • Represents the maximum amount of product that can be made given the amounts of reactants available.

  • It is calculated based on stoichiometry and is determined by the limiting reagent.

  • This is typically the maximum amount of product that can be formed under ideal conditions.

• Actual Yield:

  • The amount of product actually obtained from a chemical reaction in a laboratory or industrial setting.

  • The actual yield is always less than or equal to the theoretical yield.

  • Reasons for Actual Yield being Less than Theoretical Yield:

    • Incomplete Reactions: Not all reactants may completely convert to products.

    • Side Reactions: Unwanted reactions may occur simultaneously, consuming reactants and forming undesired byproducts.

    • Difficulty in Recovery/Purification: During purification steps (e.g., filtration, distillation, chromatography like column chromatography), some product may be lost.

    • Example: Porphyrins are notoriously difficult to synthesize and purify in organic chemistry, with a $20\%$ to $25\%$ yield considered excellent. A $25\%$ conversion rate is generally undesirable, comparable to a car only utilizing $25\%$ of the fuel put into it, which would be inefficient.

Percent Yield

• Definition: A quantitative measure of the efficiency of a chemical reaction, expressing the actual yield as a percentage of the theoretical yield.

• Equation: This equation must be memorized for exams. $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

  • The theoretical yield is the value you calculate from the stoichiometry of the reaction.

  • The actual yield is the value given in the problem, representing what was experimentally obtained.

• Units: It is crucial that both the actual and theoretical yields are expressed in the same units (e.g., grams) for the calculation to be valid.

Percent Yield Example Problem

• Problem: Upon reaction of a $1.274$ gram sample of copper (II) sulfate ($\text{CuSO}<em>4$) with excess zinc metal, $0.392$ grams of copper metal ($\text{Cu}$) was obtained according to the equation: $\text{CuSO}</em>{4(aq)} + \text{Zn}<em>{(s)} \rightarrow \text{Cu}</em>{(s)} + \text{ZnSO}_{4(aq)}$
  What is the percent yield?

• Given Information:

  • Mass of $\text{CuSO}_4$ (limiting reactant) = $1.274 \text{ g}$

  • Zinc metal ($\text{Zn}$) is in excess.

  • Actual Yield of $\text{Cu}$ (product obtained) = $0.392 \text{ g}$

• Step 1: Write down knowns and balanced equation. (Provided, and is balanced)

• Step 2: Calculate the Theoretical Yield of Copper.

  • We use the mass of the limiting reactant, $\text{CuSO}_4$.

  • **Convert grams of $\text{CuSO}<em>4$ to moles of $\text{CuSO}</em>4$:}

    • Molar mass of $\text{CuSO}<em>4$ = $159.602 \text{ g/mol}$. $1.274 \text{ g } \text{CuSO}</em>4 \times \frac{1 \text{ mol } \text{CuSO}<em>4}{159.602 \text{ g } \text{CuSO}</em>4}$

  • **Use the mole-to-mole ratio from the balanced equation ($1:1$ for $\text{CuSO}<em>4$ to $\text{Cu}$):} $\ldots \times \frac{1 \text{ mol } \text{Cu}}{1 \text{ mol } \text{CuSO}</em>4}$

  • **Convert moles of $\text{Cu}$ to grams of $\text{Cu}$ (Theoretical Yield):}

    • Molar mass of $\text{Cu}$ = $63.546 \text{ g/mol}$.
      $\ldots \times \frac{63.546 \text{ g } \text{Cu}}{1 \text{ mol } \text{Cu}} = 0.5072468 \text{ g } \text{Cu}$

    • Therefore, the Theoretical Yield of $\text{Cu}$ is approximately $0.5072 \text{ g}$.

• Step 3: Calculate the Percent Yield using the formula.
  $\text{Percent Yield} = \frac{0.392 \text{ g } \text{Cu}}{0.5072468 \text{ g } \text{Cu}} \times 100\%$
  $\text{Percent Yield} = 77.279\%$

• Step 4: Round to the correct number of significant figures.

  • The actual yield ($0.392 \text{ g}$) has $3$ significant figures.

  • Final Answer: The percent yield is $77.3\%$

Precipitation Reactions

• Definition: These are chemical reactions that occur when two aqueous solutions are mixed, leading to the formation of an insoluble ionic solid, known as a precipitate. This precipitate typically separates from the solution.

• General Form: These reactions commonly involve the exchange of ions (metathesis or double displacement), often represented as: $\text{AB}<em>{(aq)} + \text{YX}</em>{(aq)} \rightarrow \text{AY}<em>{(s)} + \text{BX}</em>{(aq)}$

  • One of the products ($\text{AY}$ or $ext{BX}$) must be an insoluble solid.

  • The key to predicting these reactions and identifying the precipitate is knowledge of solubility rules.

• Key Concept: In most cases, only one new solid will form. If the potential products are both soluble (stay in the aqueous phase), then no precipitation reaction occurs.

Solubility Rules

• Importance: Understanding these rules is essential for predicting whether a compound will dissolve (be soluble and form an aqueous solution) or form a precipitate (be insoluble and form a solid). These rules are not provided on exams and must be learned.

• Soluble Substances (usually aqueous, $(aq)$):

  • All compounds containing Group 1A cations ($\text{Li}^+, \text{Na}^+, \text{K}^+, \text{Rb}^+, \text{Cs}^+$) are soluble.

  • All compounds containing the Ammonium ion ($\text{NH}_4^+$) are soluble.

  • All compounds containing Nitrate ($\text{NO}3^-), Bicarbonate ($\text{HCO}3^-), Acetate ($\text{CH}3\text{COO}^-), and Chlorate ($\text{ClO}3^-) anions are always soluble. There are no exceptions for these anions.

  • Most Chlorides ($\text{Cl}^-), Bromides ($\text{Br}^-), and Iodides ($\text{I}^-) are soluble.

    • Exceptions (Insoluble): Compounds formed with $\text{Ag}^+, \text{Pb}^{2+}$, and $\text{Hg}<em>2^{2+}$ (e.g., $\text{AgCl}, \text{PbBr}</em>2, \text{Hg}<em>2\text{I}</em>2$) are insoluble.

  • Most Sulfates ($\text{SO}_4^{2-}) are soluble.

    • Exceptions (Insoluble): Compounds formed with $\text{Ag}^+, \text{Pb}^{2+}, \text{Hg}<em>2^{2+}$, $\text{Ca}^{2+}, \text{Sr}^{2+}$, and $\text{Ba}^{2+}$ (e.g., $\text{BaSO}</em>4, \text{PbSO}_4$) are insoluble.

• Insoluble Substances (usually precipitates, $(s)$):

  • Most Carbonates ($\text{CO}3^{2-}), Chromates ($\text{CrO}4^{2-}), Phosphates ($\text{PO}_4^{3-}), and Sulfides ($\text{S}^{2-}) are insoluble.

    • Exceptions (Soluble): Compounds formed with Group 1A cations and Ammonium ($\text{NH}_4^+$) are soluble.

  • Most Hydroxides ($\text{OH}^-) are insoluble.

    • Exceptions (Soluble): Compounds formed with Group 1A cations, Ammonium ($\text{NH}_4^+$), and heavier Group 2A cations such as Barium ($\text{Ba}^{2+}$) and Strontium ($\text{Sr}^{2+}$) are soluble. Calcium hydroxide is sparingly soluble.

Types of Chemical Equations

When writing equations for reactions in aqueous solutions, particularly precipitation reactions, there are three common forms:

1. Molecular Equation:

   • This is the standard balanced chemical equation where all reactants and products are written as neutral compounds, even if they exist as ions in solution.

   • It explicitly indicates the physical states of all substances using ($(s), (l), (g), (aq)$).

   • Example: When calcium chloride solution reacts with silver nitrate solution: $\text{CaCl}<em>{2(aq)} + 2\text{AgNO}</em>{3(aq)} \rightarrow 2\text{AgCl}<em>{(s)} + \text{Ca(NO}</em>3)_{2(aq)}$

     • Here, $ext{AgCl}$ is shown as a solid precipitate, while other compounds remain dissolved.

2. Complete Ionic Equation:

   • This equation shows all dissolved ionic compounds (those indicated with $(aq)$) separated into their individual ions.

   • Solids ($(s)$), liquids ($(l)$), and gases ($(g)$) are written as intact molecules or compounds because they do not dissociate into ions in solution.

   • Derivation from Molecular: For the example above, dissociate the aqueous compounds:
     $\text{Ca}^{2+}<em>{(aq)} + 2\text{Cl}^-</em>{(aq)} + 2\text{Ag}^+<em>{(aq)} + 2\text{NO}</em>3^-<em>{(aq)} \rightarrow 2\text{AgCl}</em>{(s)} + \text{Ca}^{2+}<em>{(aq)} + 2\text{NO}</em>3^-_{(aq)}$

   • Analogy: This is like your