Stoichiometry, Limiting Reagents, and Chemical Yields in-depth Chemical Yield Notes

Properties and Hazards of Sulfide Compounds

• Color of Copper Sulfate ($CuSO_4$): Described as a "bluish light bluish color," though the instructor personally expresses a lack of fondness for the specific shade.

• Sensory Warning (Smell and Toxicity):

  • Treating copper sulfate with sodium sulfide ($Na_2S$) produces an extremely foul odor.

  • The instructor compares it to a "stink bomb," noting that this reaction is "double" as nasty as standard retail products.

  • Biological Rationale for Disgust: The sensation of wanting to run away from a bad smell is an evolutionary adaptation. Bodies react this way because the substance is toxic.

  • Physiological Impact: Sulfur compounds can enter the bloodstream and interfere with "irons" (referring to heme/iron in hemoglobin), disrupting the entire blood system.

  • Common Sources:

    • Natural: Earth produces these smells in volcanoes and geothermal areas like Yellowstone.

    • Biological: Humans produce small amounts of these substances during flatulence.

• Industrial and Ethical Considerations:

  • Sulfur is often involved in mining. The instructor mentions that illegal mining near volcanoes occurs regardless of legal status.

  • Global Context: The instructor emphasizes that anyone carrying an iPhone or a smartphone with a battery is connected to the mining industry. This reflects how the world works regarding resource demand and environmental/ethical impacts.

Stoichiometry of Copper Sulfate and Sodium Sulfide

• The Chemical Reaction:

  • $CuSO_4 + Na_2S \rightarrow CuS + Na_2SO_4$

  • Bonding: Copper and sulfur form a very strong bond, resulting in the black solid copper sulfide ($CuS$).

  • Real-World Application: This black solid is the same material that forms the "patina" or tarnish often seen on silver.

• Balancing Technique:

  • The instructor recommends a "trick" for balancing equations: if a polyatomic ion like sulfate ($SO_4$) remains intact on both sides of the equation, treat it as a single unit ($1 \times SO_4$) rather than counting individual oxygen and sulfur atoms.

  • Equation Balancing Process:

    • Initially, the instructor displays a potentially incorrect formula to illustrate that if an equation cannot be balanced, either the reaction does not work as written or the formulas provided are incorrect.

    • Corrected balanced equation: $CuSO_4 + Na_2S \rightarrow CuS + Na_2SO_4$.

Theoretical and Percent Yield

• Theoretical Yield Calculation:

  • Scenario: Start with $10\,g$ of copper sulfate ($CuSO_4$) and an unlimited amount of sodium sulfide ($Na_2S$).

  • Step 1 (Convert to Moles): Use the molar mass of $CuSO_4$.

    • Molar mass calculation: $63.546\,g/mol\, (Cu) + 32.06\,g/mol\, (S) + 4 \times 16.00\,g/mol\, (O) \approx 161\,g/mol$.

    • $10\,g\,CuSO_4 \times \frac{1\,mol}{161\,g} = \text{moles of reactant}$.

  • Step 2 (Mole Ratio): Use the balanced equation coefficients (currency of the reaction).

    • $1\,mol\,CuSO_4$ produces $1\,mol\,Na_2SO_4$.

  • Step 3 (Convert back to Mass): Use the molar mass of the product ($Na_2SO_4$).

    • Molar mass of $Na_2SO_4 \approx 142.66\,g/mol$.

    • Total calculated mass: $\approx 8.9\,g\,Na_2SO_4$.

• Percent Yield Definition:

  • Theoretical Yield: The maximum amount of product possible at $100\%$ conversion.

  • Actual/Recovered Yield: The amount actually gathered during an experiment.

  • Scenario: If $7\,g$ of sodium sulfate were actually recovered from the experiment.

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

  • Example Calculation: $\frac{7\,g}{8.9\,g} \times 100 \approx 78\%$.

  • Context: A yield between $70\%$ and $95\%+$ is generally considered "pretty good" in laboratory conditions, as nature rarely allows for a perfect outcome.

The Limiting Reagent Concept

• Definition: The limiting reagent is the reactant that is entirely consumed first, thereby causing the production/reaction to stop.

• Hamburger Metaphor:

  • Recipe: 1 Patty + 2 Tomatoes + 8 Pickles + 2 Buns = 1 Hamburger.

  • Scenario A: If you have unlimited materials but only 1 bun, you can make zero burgers (since 2 buns are required per burger).

  • Scenario B: If you have 10 pickles, you can only make a specific number of burgers before the pickles run out, regardless of how many patties or buns remain.

  • Core Principle: Production is dictated by the component that runs out first. Once it is gone, everything else is left over.

Advanced Stoichiometry Examples

Example 1: Water Formation

• Reaction: $2H_2 + O_2 \rightarrow 2H_2O$

• Initial Amounts: $4\,g$ of Hydrogen ($H_2$) and $10\,g$ of Oxygen ($O_2$).

• Identify Limiting Reagent:

  • Hydrogen Path:

    • $4\,g\,H_2 \times \frac{1\,mol\,H_2}{2\,g} \times \frac{2\,mol\,H_2O}{2\,mol\,H_2} \times \frac{18\,g\,H_2O}{1\,mol\,H_2O} = 36\,g\,H_2O$.

  • Oxygen Path:

    • $10\,g\,O_2 \times \frac{1\,mol\,O_2}{32\,g} \times \frac{2\,mol\,H_2O}{1\,mol\,O_2} \times \frac{18\,g\,H_2O}{1\,mol\,H_2O} \approx 11.25\,g\,H_2O$.

• Result: Oxygen is the limiting reagent because it produces the least amount of product ($11.25\,g$). The theoretical yield is $11.25\,g$.

Example 2: Aluminum Fluoride

• Reaction: Aluminum ($Al$) + Fluorine ($F_2$) $\rightarrow$ Aluminum Fluoride ($AlF_3$).

• Balanced Equation: $2Al + 3F_2 \rightarrow 2AlF_3$

• Initial Amounts: $10\,g\,Al$ and $10\,g\,F_2$. Actual recovery: $5\,g\,AlF_3$.

• Step 1 (Aluminum Calculation):

  • Using molar mass ($27\,g/mol$ for $Al$), $10\,g$ produces $\approx 31.1\,g\,AlF_3$.

• Step 2 (Fluorine Calculation):

  • Using molar mass ($38\,g/mol$ for $F_2$), $10\,g\,F_2 \times \frac{1\,mol}{38\,g} \times \frac{2\,mol\,AlF_3}{3\,mol\,F_2} \times Molar\,Mass\,AlF_3 \approx 15\,g\,AlF_3$.

• Conclusion: Fluorine is the limiting reagent. Theoretical yield = $15\,g$.

• Percent Yield: $\frac{5\,g}{15\,g} \times 100 \approx 33\%$.

Questions & Discussion

• Q: Why are there two aluminums in the balanced recipe?

• A: The instructor explains that balancing is essential to discover the correct mole ratio. The ratio is not necessarily 1:1; it is determined by the specific reaction chemistry to ensure that the mass and charge are conserved.

• Q: If a reaction has two products, do you have to calculate for both?

• A: No. You only need to choose one product to perform the comparison between reactants. Once a reagent runs out, it stops the production of all products simultaneously.

• Q: Where can students find practice problems?

• A: There was confusion regarding the PDF version of the book. The instructor clarifies that the course uses the OER (Open Educational Resource) version, which should contain questions at the end of chapters. Some older versions or alternate links may lack these questions. There are multiple copies of older versions available that contain similar problems for practice.

• Q: What is the most important part of these calculations?

• A: The mole-to-mole conversion (the mole ratio from the balanced equation). This is the "truth" for every calculation in the chapter, allowing one to equate products to products, products to reactants, or reactants to reactants.

• Advice for Exam Anxiety: The instructor jokingly suggests waking up at 4:00 AM to solve chemistry problems while highly anxious to simulate exam conditions, noting a former student who used to solve equations in the passenger seat of a car to manage his high intelligence and high anxiety level.