Lesson 12: Stoichiometry - Mole-Mass, Mass-Mass, Limiting Reactants, and Percent Yield

Mole-Mass and Mass-Mass Calculations in Stoichiometry

Learning Objective: The primary goal is to convert from the mass or moles of one substance to the mass or moles of another substance involved in a chemical reaction.
General Principles: - A balanced chemical equation is balanced in terms of moles as well as atoms or molecules. - Balanced equations allow for the creation of molar ratios used as conversion factors for stoichiometric questions (e.g., determining how many moles of substance A react with a specific amount of reactant B). - Molar amount is related to mass amount using the molar mass ( $g/mol$ ).
Mole-Mass Calculation Sequence: - This sequence is used to answer stoichiometry questions in terms of the masses of a particular substance based on the moles of another. - Process: $ext{mol first substance} \rightarrow \text{mol second substance} \rightarrow \text{mass second substance}$ .
Step-by-Step Example: Iron(III) Oxide and Sulfur Trioxide: - Equation: $Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3$ - Question: Given $3.59\,mol$ of $Fe_2O_3$ , calculate the grams of $SO_3$ required to react. - Step 1: Construct a molar ratio from the balanced equation. In this case, $\frac{3\,mol\,SO_3}{1\,mol\,Fe_2O_3}$ . - Step 2: Calculate moles of $SO_3$ : $3.59\,mol\,Fe_2O_3 \times \frac{3\,mol\,SO_3}{1\,mol\,Fe_2O_3} = 10.77\,mol\,SO_3$ . - Step 3: Use the molar mass of $SO_3$ ( $80.06\,g/mol$ ) to find mass: $10.77\,mol\,SO_3 \times \frac{80.06\,g\,SO_3}{1\,mol\,SO_3} = 862.2462\,g\,SO_3$ . - Significant Figures: The final answer expressed to three significant figures is $862\,g\,SO_3$ . - Single-Line Calculation: $3.59\,mol\,Fe_2O_3 \times \frac{3\,mol\,SO_3}{1\,mol\,Fe_2O_3} \times \frac{80.06\,g\,SO_3}{1\,mol\,SO_3} = 862\,g\,SO_3$ .
EXAMPLE 1: Calcium Carbonate and Hydrochloric Acid: - Equation: $CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$ - Problem: Calculate the grams of $CO_2$ produced from $2.09\,mol$ of $HCl$ . - Strategy: Convert moles of $HCl$ to moles of $CO_2$ , then to grams of $CO_2$ using molar mass ( $44.01\,g/mol$ ). - Calculation: $2.09\,mol\,HCl \times \frac{1\,mol\,CO_2}{2\,mol\,HCl} \times \frac{44.01\,g\,CO_2}{1\,mol\,CO_2} = 46.0\,g\,CO_2$ .
SKILL-BUILDING EXERCISE 1: - Equation: $6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$ - Question: Calculate grams of glucose ( $C_6H_{12}O_6$ ) produced from $17.3\,mol$ of $H_2O$ .

Mass-Mass Calculation Procedures

General Sequence: If starting with a known mass of one substance, follow this three-part process: - Sequence: $\text{mass first substance} \rightarrow \text{mol first substance} \rightarrow \text{mol second substance} \rightarrow \text{mass second substance}$ . - Step 1: Convert known mass to moles using the first substance's molar mass. - Step 2: Use the balanced chemical equation to construct a conversion factor (molar ratio) to find moles of the second substance. - Step 3: Convert the moles of the second substance to its corresponding mass using its molar mass.
EXAMPLE 2: Methane and Chlorine Reaction: - Equation: $CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl$ - Goal: Find grams of $HCl$ produced from $100.0\,g$ of $CH_4$ . - Stepwise Breakdown: - Moles of $CH_4$ : $100.0\,g\,CH_4 \times \frac{1\,mol\,CH_4}{16.05\,g\,CH_4} = 6.231\,mol\,CH_4$ . - Moles of $HCl$ : $6.231\,mol\,CH_4 \times \frac{4\,mol\,HCl}{1\,mol\,CH_4} = 24.92\,mol\,HCl$ . - Mass of $HCl$ : $24.92\,mol\,HCl \times \frac{36.46\,g\,HCl}{1\,mol\,HCl} = 908.6\,g\,HCl$ . - Single-Line Calculation: $100.0\,g\,CH_4 \times \frac{1\,mol\,CH_4}{16.05\,g\,CH_4} \times \frac{4\,mol\,HCl}{1\,mol\,CH_4} \times \frac{36.46\,g\,HCl}{1\,mol\,HCl} = 909.1\,g\,HCl$ . - Note on Discrepancy: The small difference ( $908.6\,g$ vs $909.1\,g$ ) arises because the stepwise method rounds intermediate significant figures, whereas the single-line method only restricts the final answer.
SKILL-BUILDING EXERCISE 2: - Equation: $CH_3CH_2CHO + 2K_2Cr_2O_7 \rightarrow CH_3CH_2COOH + \text{other products}$ - Question: Calculate the grams of propionic acid ( $CH_3CH_2COOH$ ) produced from $135.8\,g$ of $K_2Cr_2O_7$ .

Limiting Reactants

Learning Objective: Identify a limiting reactant and calculate product yield based on that reactant.
Definitions: - Stoichiometric Amounts: The relative amounts of reactants and products as represented exactly by the coefficients in a balanced equation. - Limiting Reactant: The substance that is entirely consumed in a reaction, thereby limiting the amount of product that can form. - Excess Reactant: The substance that remains after the limiting reactant is completely consumed.
Sandwich Analogy: - Recipe: $1\,slice\,cheese + 2\,slices\,bread \rightarrow 1\,sandwich$ . - Inventory: $28\,slices\,bread$ and $11\,slices\,cheese$ . - Result: Only $11\,sandwiches$ can be made (limited by cheese). $6\,slices\,bread$ remain (excess).
Chemical Application: Hydrogen and Chlorine Reaction: - Equation: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$ . - Analysis: Stoichiometric ratio is $1:1$ . If combining $3\,moles$ of $H_2$ and $2\,moles$ of $Cl_2$ , the ratio is $1.5:1$ . - Result: Chlorine is the limiting reactant. Reaction of all $2\,mol\,Cl_2$ consumes $2\,mol\,H_2$ , leaving $1\,mol\,H_2$ unreacted.
Technique 1: Comparison of Molar Ratios: - Compute provided molar amounts and compare to the coefficients in the balanced equation.
Technique 2: Comparison of Product Yields: - Calculate the amount of product expected from the complete reaction of each reactant individually. - The reactant yielding the lesser amount of product is the limiting reactant. - Example with $H_2$ and $Cl_2$ : - Product from $3\,mol\,H_2$ : $3\,mol\,H_2 \times \frac{2\,mol\,HCl}{1\,mol\,H_2} = 6\,mol\,HCl$ . - Product from $2\,mol\,Cl_2$ : $2\,mol\,Cl_2 \times \frac{2\,mol\,HCl}{1\,mol\,Cl_2} = 4\,mol\,HCl$ . - Since Chlorine yields fewer moles of $HCl$ , it is the limiting reactant.
EXAMPLE 3: Silicon Nitride Preparation: - Equation: $3Si(s) + 2N_2(g) \rightarrow Si_3N_4(s)$ . - Inventory: $2.00\,g\,Si$ and $1.50\,g\,N_2$ . - Product from $Si$ : $2.00\,g\,Si \times \frac{1\,mol\,Si}{28.09\,g\,Si} \times \frac{1\,mol\,Si_3N_4}{3\,mol\,Si} \times \frac{140.3\,g\,Si_3N_4}{1\,mol\,Si_3N_4} = 3.33\,g\,Si_3N_4$ . - Product from $N_2$ : $1.50\,g\,N_2 \times \frac{1\,mol\,N_2}{28.02\,g\,N_2} \times \frac{1\,mol\,Si_3N_4}{2\,mol\,N_2} \times \frac{140.3\,g\,Si_3N_4}{1\,mol\,Si_3N_4} = 3.76\,g\,Si_3N_4$ . - Conclusion: Silicon ( $Si$ ) is the limiting reactant as it produces the lesser amount of product ( $3.33\,g$ ).

Percent Yield

Learning Objective: Determine the percent yield given actual product quantity.
Definitions: - Theoretical Yield: The maximum amount of product that can be generated as calculated by stoichiometry. - Actual Yield: The amount of product actually obtained in practice (lab environment). - Percent Yield: The ratio of actual yield to theoretical yield expressed as a percentage.
Formula: - $\text{Percent yield} = \frac{ ext{actual yield}}{\text{theoretical yield}} \times 100\%$
Reasons for Reduced Yield: - Inherent inefficiency of reactions. - Side reactions generating unintended products. - Incomplete reactions (e.g., partial reactions of weak acids/bases). - Physical loss of product during collection/recovery.
Note on Units: Yields can be in mass, moles, or volume. Units must be consistent to cancel during calculation.
EXAMPLE 4: Copper Sulfate and Zinc Reaction: - Equation: $CuSO_4(aq) + Zn(s) \rightarrow Cu(s) + ZnSO_4(aq)$ . - Data: $1.274\,g$ of $CuSO_4$ (limiting reactant) produces $0.392\,g$ of $Cu$ metal (actual yield). - Step 1: Calculate Theoretical Yield of $Cu$ : - $1.274\,g\,CuSO_4 \times \frac{1\,mol\,CuSO_4}{159.62\,g\,CuSO_4} \times \frac{1\,mol\,Cu}{1\,mol\,CuSO_4} \times \frac{63.55\,g\,Cu}{1\,mol\,Cu} = 0.5072\,g\,Cu$ . - Step 2: Calculate Percent Yield: - $\frac{0.392\,g}{0.5072\,g} \times 100\% = 77.3\%$ .
SKILL-BUILDING EXERCISE 4: - Equation: $CCl_4 + 2HF \rightarrow CF_2Cl_2 + 2HCl$ . - Question: Find percent yield if $32.9\,g$ of $CCl_4$ produces $12.5\,g$ of Freon ( $CF_2Cl_2$ ).

Lesson 12 Homework Problems and Selected Solutions

Problem 1: Reaction $H_3PO_4 + NaOH \rightarrow H_2O + Na_3PO_4$ (unbalanced). If $2.35\,mol$ of $H_3PO_4$ react, what mass of $H_2O$ is produced? (Answer: $127\,g$ ).
Problem 2: Reaction $C_2H_6 + Br_2 \rightarrow C_2H_4Br_2 + HBr$ (unbalanced). What mass of $HBr$ is produced from $0.884\,mol$ of $C_2H_6$ ?
Problem 3: Fat reaction to make soap. How many moles of glycerol from $1,000.0\,g$ of fat? (Answer: $1.236\,mol$ ).
Problem 4: Photosynthesis $6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$ . Moles of glucose from $544\,g$ of $CO_2$ ?
Problem 5: Precipitation: $Ba(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaNO_3(aq)$ . Grams of $Na_2SO_4$ needed for $43.9\,g$ of $Ba(NO_3)_2$ ? (Answer: $23.9\,g$ ).
Problem 6: Nitroglycerin formation: $C_3H_5(OH)_3 + 3HNO_3 \rightarrow C_3H_5(ONO_2)_3 + 3H_2O$ . Mass made from $87.4\,g$ of $HNO_3$ with excess glycerol?
Problem 7: Antacid neutralization: $Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O$ . Grams of $HCl$ neutralized by $200\,mg$ of $Mg(OH)_2$ ? (Answer: $0.251\,g$ ).
Problem 8: Acid rain: $3NO_2 + H_2O \rightarrow 2HNO_3 + NO$ . Yearly yield of $HNO_3$ from $1.82 \times 10^{13}\,g$ of $NO_2$ .
Problem 9: Iron ore processing: $2Fe_2O_3 + 3C \rightarrow 4Fe + 3CO_2$ . Grams of Carbon for $1.00 \times 10^9\,g$ of $Fe$ ? (Answer: $1.61 \times 10^8\,g$ ).
Problem 10: SS Hindenburg fire ( $H_2$ gas burned in $1937$ ): $2H_2 + O_2 \rightarrow 2H_2O$ . Grams of $H_2O$ from $5.33 \times 10^5\,g$ of $H_2$ .
Problem 11 (Limiting Reactants): - a. $2Al + 3Cl_2 \rightarrow 2AlCl_3$ . $20.0\,g$ each reactant. (Answer: $25.1\,g\,AlCl_3$ ). - b. $4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$ . $20.0\,g$ each. (Answer: $13.5\,g\,H_2O$ ). - c. $CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$ . $20.0\,g$ each. (Answer: $26.7\,g\,SO_2$ ).
Problem 12 (Limiting Reactants): - a. $2SO_2 + O_2 \rightarrow 2SO_3$ ( $25.0\,g\,SO_2$ , $40.0\,g\,O_2$ ). (Answer: $31.2\,g\,SO_3$ ). - b. $3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2$ ( $25.0\,g\,Fe$ , $40.0\,g\,H_2O$ ). (Answer: $34.6\,g\,Fe_3O_4$ ). - c. $C_7H_{16} + 11O_2 \rightarrow 7CO_2 + 8H_2O$ ( $25.0\,g$ , $40.0\,g$ ). (Answer: $35.0\,g\,CO_2$ ).
Problem 13 (Percent Yield): - a. $40.0\,g\,C$ produces $36.0\,g\,CS_2$ . (Answer: $71.0\%$ ). - b. $32.0\,g\,SO_2$ produces $12.0\,g\,CS_2$ . (Answer: $63.2\%$ ).
Problem 14: $4Al(s) + 3O_2(g) \rightarrow 2Al_2O_3(s)$ . Mass of $Al_2O_3$ with $50.0\,g\,Al$ and $75.0\%$ yield. (Answer: $70.9\,g$ ).
Problem 15: $SiO_2(s) + 3C(s) \rightarrow SiC(s) + 2CO(g)$ . $30.0\,g\,C$ and $28.2\,g\,CO$ produced. (Answer: $60.5\%$ ).