Chapter 9 - Introduction to Stoichiometry and Yield Analysis

Introduction to Stoichiometry

• Definition of Stoichiometry: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. Much of chemical knowledge is based on careful quantitative analysis of substances.

• Composition Stoichiometry: This branch deals specifically with the mass relationships of elements within compounds.

• Reaction Stoichiometry: This branch involves the mass relationships between reactants and products in a chemical reaction. It is the primary focus of this study material.

• Theoretical Basis: Reaction stoichiometry is based on chemical equations and the Law of Conservation of Mass.

• Starting Point: All reaction stoichiometry calculations must begin with a balanced chemical equation. This equation provides the relative numbers of moles of reactants and products.

• Key Units:

  • Amount of a substance is expressed in units of moles ($mol$).

  • Mass of a substance is typically expressed in grams ($g$), though large-scale or micro-scale problems may use kilograms ($kg$) or milligrams ($mg$).

Problem Classification in Stoichiometry

Stoichiometric problems are classified by the information given and the information sought (the unknown). They are categorized into four types:

• Problem Type 1: Given and unknown are amounts in moles:

  • General Plan: $\text{amount of given substance (mol)} \rightarrow \text{amount of unknown substance (mol)}$

• Problem Type 2: Given is moles, unknown is mass:

  • General Plan: $\text{amount of given substance (mol)} \rightarrow \text{amount of unknown substance (mol)} \rightarrow \text{mass of unknown substance (g)}$

• Problem Type 3: Given is mass, unknown is moles:

  • General Plan: $\text{mass of given substance (g)} \rightarrow \text{amount of given substance (mol)} \rightarrow \text{amount of unknown substance (mol)}$

• Problem Type 4: Given is mass, unknown is mass:

  • General Plan: $\text{mass of given substance (g)} \rightarrow \text{amount of given substance (mol)} \rightarrow \text{amount of unknown substance (mol)} \rightarrow \text{mass of unknown substance (g)}$

The Mole Ratio

• Definition: A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. It is obtained directly from the coefficients in a balanced chemical equation.

• Significance: Coefficients represent the relative amounts in moles of reactants and products, satisfying the Law of Conservation of Matter.

• Example: Electrolysis of Aluminum Oxide:

  • Equation: $2Al_2O_3(l) \rightarrow 4Al(s) + 3O_2(g)$

  • Possible Mole Ratios for $Al_2O_3$ to $Al$: $\frac{2\,mol\,Al_2O_3}{4\,mol\,Al}$ or $\frac{4\,mol\,Al}{2\,mol\,Al_2O_3}$

  • Possible Mole Ratios for $Al_2O_3$ to $O_2$: $\frac{2\,mol\,Al_2O_3}{3\,mol\,O_2}$ or $\frac{3\,mol\,O_2}{2\,mol\,Al_2O_3}$

  • Possible Mole Ratios for $Al$ to $O_2$: $\frac{4\,mol\,Al}{3\,mol\,O_2}$ or $\frac{3\,mol\,O_2}{4\,mol\,Al}$

• Significant Figures: Mole ratios are considered exact numbers and do not limit the number of significant figures in a calculation. The final answer's precision is determined by the measured quantities provided in the problem.

Molar Mass as a Conversion Factor

• Definition: Molar mass is the mass in grams of one mole of a substance ($g/mol$). It is found using the periodic table.

• Function: It relates the mass of a substance to the amount in moles.

• Conversion Factors from Aluminum Oxide Example:

  • $1\,mol\,Al_2O_3 = 101.96\,g$

  • $1\,mol\,Al = 26.98\,g$

  • $1\,mol\,O_2 = 32.00\,g$

  • Factors for $Al$: $\frac{26.98\,g\,Al}{1\,mol\,Al}$ or $\frac{1\,mol\,Al}{26.98\,g\,Al}$

Case Study: The History of Combustion and Lavoisier

• Early Perspectives: In 1772, Daniel Rutherford explained a mouse dying in a closed container using the "Phlogiston Theory." He believed the mouse emitted phlogiston until the air became "phlogisticated" and could hold no more, causing death.

• Joseph Priestley: In the 1770s, Priestley heated mercury(II) oxide to produce metallic mercury and a gas he called "dephlogisticated air," which supported intense burning.

• Phlogiston Theory (pre-1700s): Scientists believed combustion involved the emission of a substance called "phlogiston" (Greek for "burned"). It was thought that air had a saturation point for phlogiston.

• Antoine Laurent Lavoisier: Known as the "Father of Chemistry," Lavoisier shifted the field toward quantitative analysis using balances.

  • Tin Experiment: Lavoisier burned tin in a closed vessel. He observed that the mass of the burned metal increased, and the weight gain was equal to the mass of the air that rushed in when the vessel was opened.

  • Conclusion: Lavoisier concluded that air is a mixture of gases. He renamed Priestley’s "dephlogisticated air" to Oxygen (meaning "acid former") and Rutherford’s "phlogisticated air" to Azote (mostly nitrogen).

  • Scientific Contribution: His work supported the Law of Conservation of Mass and established a common naming system for elements and compounds.

Ideal Stoichiometric Calculations

• Ideal Conditions: Calculations assume that all reactants are completely converted into products without the formation of side products or losses.

• Type 1: Mole-to-Mole Calculations:

  • Conversion factor: Stoichiometric mole ratio from the balanced equation.

  • Formula: $\text{given quantity (mol)} \times \frac{\text{unknown (mol)}}{\text{given (mol)}} = \text{unknown quantity (mol)}$

  • Sample Problem A: Spacecraft occupants exhale $20\,mol\,CO_2$ daily.

    • Reaction: $CO_2(g) + 2LiOH(s) \rightarrow Li_2CO_3(s) + H_2O(l)$

    • Solution: $20\,mol\,CO_2 \times \frac{2\,mol\,LiOH}{1\,mol\,CO_2} = 40\,mol\,LiOH$

• Type 2: Mole-to-Gram Calculations:

  • Requires two factors: (1) Mole ratio and (2) Molar mass of the unknown.

  • Formula: $\text{mol given} \times \frac{\text{mol unknown}}{\text{mol given}} \times \frac{\text{g unknown}}{1\,mol\,unknown} = \text{g unknown}$

  • Sample Problem B: Photosynthesis producing glucose ($C_6H_12O_6$) from $3.00\,mol\,H_2O$.

    • Reaction: $6CO_2(g) + 6H_2O(l) \rightarrow C_6H_{12}O_6(s) + 6O_2(g)$

    • Solution: $3.00\,mol\,H_2O \times \frac{1\,mol\,C_6H_{12}O_6}{6\,mol\,H_2O} \times \frac{180.18\,g\,C_6H_{12}O_6}{1\,mol\,C_6H_{12}O_6} = 90.1\,g\,C_6H_{12}O_6$

• Type 3: Gram-to-Mole Calculations:

  • Requires two factors: (1) Molar mass of the given and (2) Mole ratio.

  • Formula: $\text{g given} \times \frac{1\,mol\,given}{\text{molar mass given}} \times \frac{\text{mol unknown}}{\text{mol given}} = \text{mol unknown}$

  • Sample Problem D: Catalytic oxidation of ammonia ($NH_3$).

    • Balanced Reaction: $4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$

    • Calculating $NO$ from $824\,g\,NH_3$:

    • Solution: $824\,g\,NH_3 \times \frac{1\,mol\,NH_3}{17.04\,g\,NH_3} \times \frac{4\,mol\,NO}{4\,mol\,NH_3} = 48.4\,mol\,NO$

• Type 4: Mass-to-Mass Calculations:

  • Requires three factors: (1) Molar mass of given, (2) Mole ratio, (3) Molar mass of unknown.

  • Formula: $\text{g given} \times \frac{1\,mol\,given}{\text{molar mass given}} \times \frac{\text{mol unknown}}{\text{mol given}} \times \frac{\text{g unknown}}{1\,mol\,unknown} = \text{g unknown}$

  • Sample Problem E: Producing Tin(II) fluoride ($SnF_2$) from $30.00\,g\,HF$.

    • Reaction: $Sn(s) + 2HF(g) \rightarrow SnF_2(s) + H_2(g)$

    • Solution: $30.00\,g\,HF \times \frac{1\,mol\,HF}{20.01\,g\,HF} \times \frac{1\,mol\,SnF_2}{2\,mol\,HF} \times \frac{156.71\,g\,SnF_2}{1\,mol\,SnF_2} = 117.5\,g\,SnF_2$

Limiting Reactants

• Limiting Reactant (Limiting Reagent): The substance that is completely consumed first in a reaction. It limits the amount of product that can form.

• Excess Reactant: The substance that is not used up completely; some remains at the end of the reaction.

• Concept Illustration: If you have $5\,mol\,C$ and $10\,mol\,O_2$ for the reaction $C + O_2 \rightarrow CO_2$, carbon is the limiting reactant because it will run out after forming $5\,mol\,CO_2$. Oxygen is in excess by $5\,mol$.

• Sample Problem F: Neutralizing Acid Spills:

  • Reaction: $Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$

  • Given: $6.3\,mol\,HCl$ and $2.8\,mol\,Ca(OH)_2$.

  • Calculation: $2.8\,mol\,Ca(OH)_2 \times \frac{2\,mol\,HCl}{1\,mol\,Ca(OH)_2} = 5.6\,mol\,HCl\,needed$.

  • Identification: Since $6.3\,mol$ is available but only $5.6\,mol$ is needed, $HCl$ is the excess reactant and $Ca(OH)_2$ is the limiting reactant.

• Complex Scenario (Magnetite Synthesis):

  • Reaction: $3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$

  • Given: $36.0\,g\,H_2O$ and $67.0\,g\,Fe$.

  • Step 1: Calculate output for each reactant.

    • $67.0\,g\,Fe \rightarrow 0.400\,mol\,Fe_3O_4$

    • $36.0\,g\,H_2O \rightarrow 0.499\,mol\,Fe_3O_4$

  • Step 2: Compare. $Fe$ yields less product, so it is the limiting reactant.

  • Step 3: Mass of product: $0.400\,mol\,Fe_3O_4 \times 231.55\,g/mol = 92.6\,g\,Fe_3O_4$.

  • Step 4: Excess remaining: Calculate $H_2O$ consumed based on $0.400\,mol\,Fe_3O_4$. $H_2O\,consumed = 28.8\,g$. Remaining: $36.0\,g - 28.8\,g = 7.2\,g\,H_2O$.

Percentage Yield

• Theoretical Yield: The maximum amount of product that can be produced from a given amount of reactant, as predicted by ideal stoichiometry.

• Actual Yield: The measured amount of product obtained from a reaction performed in a laboratory.

• Reasons for Discrepancy:

  • Reactant impurities.

  • Formation of byproducts through side reactions.

  • Incomplete reaction (not all reactants convert to products).

• Formula for Percentage Yield:

  • $\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

• Sample Problem H:

  • Reaction: $HgO + Cl_2 \rightarrow HgCl_2 + Cl_2O$

  • Given: Theoretical yield = $0.86\,g\,Cl_2O$, Actual yield = $0.71\,g\,Cl_2O$.

  • Solution: $\frac{0.71\,g}{0.86\,g} \times 100 = 83\%$

Questions & Discussion

• Chemical Technician Career Profile:

  • Role: These professionals help develop products, process materials, manage hazardous waste, and ensure regulatory compliance.

  • Example: Working in water treatment plants testing pH and lead levels.

  • Preparation: Associate’s degree in applied science or chemical technology; many also hold bachelor’s degrees in chemistry or related fields.

• History Q&A:

  • Question: Why does the mass of tin increase when heated in air?

  • Answer: Because it chemically combines with oxygen in the air to form an oxide.

  • Question: What was the composition of Priestley’s "dephlogisticated air" and Rutherford’s "phlogisticated air"?

  • Answer: Priestley's was oxygen; Rutherford's was primarily nitrogen.

• Critical Thinking: Determining Limiting Reactant in a Recipe:

  • A cookie recipe uses $1\,egg$ for $24\,cookies$. If you have one dozen ($12$) eggs, you could theoretically make $288\,cookies$. However, the limiting reactant is determined by whichever ingredient (sugar, flour, chocolate chips, etc.) produces the fewest total cookies.

Table of Simple Substances (Lavoisier, 1789)

Summary Key Terms

• Mole Ratio: Conversion factor relating amounts in moles of two substances in a reaction.

• Composition Stoichiometry: Mass relationships of elements in compounds.

• Reaction Stoichiometry: Mass relationships between reactants and products.

• Limiting Reactant: Limits product formation.

• Excess Reactant: Leftover reactant.

• Theoretical Yield: Predicted maximum product.

• Actual Yield: Realized product in laboratory.

• Percentage Yield: Efficiency metric comparing actual to theoretical yield.