Stoichiometry, Formulas, and Reaction Yields

Empirical, Molecular, and Structural Formulas

• Chemical formulas provide essential quantitative and qualitative information regarding the atomic composition and structure of chemical compounds.

• Types of chemical formulas:

  • Empirical Formula: The simplest chemical formula for a compound that agrees with elemental analysis. It expresses the lowest whole-number ratio of moles and gives the relative number of atoms of each element present in the compound.

  • Example: The empirical formula of hydrogen peroxide is $HO$.

  • Molecular Formula*: Represents the actual number of atoms of each element (intramolecular stoichiometry)present in a single molecule of the compound.

    • need to know the molar mass

    • sufficient for stoichiometric analyses*

  • Example: The molecular formula of hydrogen peroxide is $H_2O_2$.

  • Structural Formula: Illustrates the relative placement, spatial arrangement, and chemical bonds connecting the atoms within a molecule (connectivity).

  • Example: The structural formula of hydrogen peroxide is $H-O-O-H$.

  • Geometric: 4a, 4b, 2-D, 3-D

    • 4b and 3D - non planar

    • 4a and 2D - EX. $H-O-O-H$.

• A series of distinct compounds can share the exact same empirical formula while possessing different molecular formulas, molar masses, and physical or biological functions.

• Compounds sharing the empirical formula $CH_2O$:   

  

  • Formaldehyde:

  • Molecular Formula: $CH_2O$

  • Whole-Number Multiple: $1$

  • Molar Mass ($M$): $30.026\,g/mol$

  • Primary Use or Function: Disinfectant; biological preservative.

  • Acetic Acid:

  • Molecular Formula: $C_2H_4O_2$

  • Whole-Number Multiple: $2$

  • Molar Mass ($M$): $60.052\,g/mol$

  • Primary Use or Function: Production of acetate polymers; main acidic constituent in vinegar ($5\%$ solution).

  • Glucose:

  • Molecular Formula: $C_6H_{12}O_6$

  • Whole-Number Multiple: $6$

  • Molar Mass ($M$): $180.156\,g/mol$

  • Primary Use or Function: Major nutrient for biological energy production in cells.

  • Fructose:

  • Molecular Formula: $C_6H_{12}O_6$

  • Whole-Number Multiple: $6$

  • Molar Mass ($M$): $180.156\,g/mol$

  • Primary Use or Function: Natural sweetener utilized in foods and beverages.

Determining Empirical Formulas from Elemental Mass

• The process of establishing an empirical formula from mass measurement data follows a systematic logical pathway:   

  

  1. Measure or obtain the mass in grams ($g$) of each element present in the compound sample.

  2. Divide the mass of each element by its respective molar mass ($M$ in $g/mol$) to obtain the amount in moles ($mol$) for each element.

  3. Construct a preliminary formula using the calculated mole quantities as subscript values.

  4. Divide each molar value by the smallest molar amount present among all elements to determine relative molar ratios.

  5. Convert any fractional ratios into integer subscripts by multiplying all ratios by the smallest common integer.

• Worked Example (Sample Problem 3.7):

  • Problem: A sample of an unknown chemical compound contains $5.08\,g$ of zinc ($Zn$), $1.60\,g$ of phosphorus ($P$), and $3.32\,g$ of oxygen ($O$). Determine its empirical formula.

  • Step 1: Calculate moles of each constituent element:

  • $\text{Moles of } Zn = 5.08\,g\,Zn \times \frac{1\,mol\,Zn}{65.38\,g\,Zn} = 0.0777\,mol\,Zn$

  • $\text{Moles of } P = 1.60\,g\,P \times \frac{1\,mol\,P}{30.97\,g\,P} = 0.0517\,mol\,P$

  • $\text{Moles of } O = 3.32\,g\,O \times \frac{1\,mol\,O}{16.00\,g\,O} = 0.208\,mol\,O$

  • Step 2: Construct preliminary formula:

  • Preliminary expression: $Zn_{0.0777}P_{0.0517}O_{0.208}$

  • Step 3: Divide subscripts by the smallest molar value ($0.0517$):

  • $Zn: \frac{0.0777}{0.0517} = 1.5$

  • $P: \frac{0.0517}{0.0517} = 1.0$

  • $O: \frac{0.208}{0.0517} = 4.0$

  • Intermediate formula: $Zn_{1.5}P_{1.0}O_{4.0}$

  • Step 4: Convert decimal subscripts into whole integers:

  • Multiply all subscripts by $2$:       $(Zn_{1.5}P_{1.0}O_{4.0}) \times 2 \rightarrow Zn_3P_2O_8$

  • Conclusion: The empirical formula of the compound is $Zn_3P_2O_8$ (zinc phosphate).

Determining Molecular Formulas from Empirical Formula and Molar Mass

• The molecular formula defines the exact molar quantity of each element present in $1\,mol$ of a molecular compound.

• The molecular formula is always an integral whole-number multiple of the empirical formula.

• Equation for calculating the integer multiplier:   $\text{Whole number multiple} = \frac{\text{compound molar mass } (g/mol)}{\text{empirical formula mass } (g/mol)}$

• Worked Example (Sample Problem 3.8):

  • Problem: Lactic acid ($M = 90.08\,g/mol$) forms in muscle tissue during strenuous activity and causes muscle soreness. Elemental analysis reveals that lactic acid consists of $40.0\text{ mass \% } C$, $6.71\text{ mass \% } H$, and 53.3\text{ mass \% } O$.\n - (a) Determine the empirical formula of lactic acid.\n - (b) Determine the molecular formula of lactic acid.\n - **Solution for (a)**:\n - Assume a total sample size of 100.0\,g$, converting mass percentages directly into mass quantities:$40.0\,g\,C$,$6.71\,g\,H$, and$53.3\,g\,O$.

  • Convert masses to moles:

    • $\text{Moles of } C = 40.0\,g\,C \times \frac{1\,mol\,C}{12.01\,g\,C} = 3.33\,mol\,C$

    • $\text{Moles of } H = 6.71\,g\,H \times \frac{1\,mol\,H}{1.008\,g\,H} = 6.66\,mol\,H$

    • $\text{Moles of } O = 53.3\,g\,O \times \frac{1\,mol\,O}{16.00\,g\,O} = 3.33\,mol\,O$

  • Divide each mole quantity by the smallest value ($3.33$):

    • $C: \frac{3.33}{3.33} = 1$

    • $H: \frac{6.66}{3.33} = 2$

    • $O: \frac{3.33}{3.33} = 1$

  • Empirical formula = $CH_2O$

  • Solution for (b):

  • Calculate empirical formula mass for $CH_2O$:       $\text{Empirical mass} = 12.01\,g/mol + 2(1.008\,g/mol) + 16.00\,g/mol = 30.03\,g/mol$

  • Calculate whole-number multiple:       $\text{Multiple} = \frac{\text{molar mass of lactic acid}}{\text{mass of } CH_2O} = \frac{90.08\,g/mol}{30.03\,g/mol} = 3$

  • Multiply empirical subscripts by $3$:       $(C_1H_2O_1) \times 3 = C_3H_6O_3$

  • Molecular formula = $C_3H_6O_3$

Combustion Analysis for Determining Empirical and Molecular Formulas

• Combustion analysis is a standard laboratory method designed to determine the precise amounts of carbon and hydrogen contained within a combustible organic compound sample.

• Combustion System Design:   

  

  • A stream of pure oxygen ($O_2$) gas flows into a high-temperature furnace containing a weighed sample of the compound.

  • Carbon in the sample burns to form gaseous carbon dioxide ($CO_2$), and hydrogen burns to form water vapor ($H_2O$).

  • The effluent gas stream passes through a selectively absorbed $H_2O$ absorber trap, followed immediately by a selectively absorbed $CO_2$ absorber trap. Unabsorbed gases exit the system.

  • The increase in mass of the $H_2O$ absorber trap reflects the mass of water formed; the increase in mass of the $CO_2$ absorber trap reflects the mass of carbon dioxide formed.

  • If the original compound contains oxygen or other elements, the mass of oxygen is calculated by subtracting the determined masses of carbon and hydrogen from the initial sample mass.

• Worked Example (Sample Problem 3.9):

  • Problem: A $1.000\,g$ sample of vitamin C ($M = 176.12\,g/mol$) is burned in a combustion apparatus. Experimental mass measurements:

  • Mass of $CO_2$ absorber after combustion = $85.35\,g$

  • Mass of $CO_2$ absorber before combustion = $83.85\,g$

  • Mass of $H_2O$ absorber after combustion = $37.96\,g$

  • Mass of $H_2O$ absorber before combustion = $37.55\,g$

  • Determine the molecular formula of vitamin C.

  • Step 1: Calculate masses of $CO_2$ and $H_2O$ formed:

  • $\text{Mass of } CO_2 = 85.35\,g - 83.85\,g = 1.50\,g\,CO_2$

  • $\text{Mass of } H_2O = 37.96\,g - 37.55\,g = 0.41\,g\,H_2O$

  • Step 2: Calculate mass of C and H present in sample:

  • $\text{Mass of } C = 1.50\,g\,CO_2 \times \frac{12.01\,g\,C}{44.01\,g\,CO_2} = 0.409\,g\,C$

  • $\text{Mass of } H = 0.41\,g\,H_2O \times \frac{2.016\,g\,H}{18.02\,g\,H_2O} = 0.046\,g\,H$

  • Step 3: Calculate mass of O in sample by difference:

  • $\text{Mass of } O = \text{total sample mass} - (\text{mass of } C + \text{mass of } H)$

  • $\text{Mass of } O = 1.000\,g - (0.409\,g + 0.046\,g) = 0.545\,g\,O$

  • Step 4: Convert element masses to moles:

  • $\text{Moles of } C = \frac{0.409\,g\,C}{12.01\,g/mol\,C} = 0.0341\,mol\,C$

  • $\text{Moles of } H = \frac{0.046\,g\,H}{1.008\,g/mol\,H} = 0.046\,mol\,H$

  • $\text{Moles of } O = \frac{0.545\,g\,O}{16.00\,g/mol\,O} = 0.0341\,mol\,O$

  • Step 5: Determine relative mole ratios and empirical formula:

  • $C: \frac{0.0341}{0.0341} = 1$

  • $H: \frac{0.046}{0.0341} = 1.3$

  • $O: \frac{0.0341}{0.0341} = 1$

  • Preliminary formula: $C_1H_{1.3}O_1$

  • Multiply subscripts by $3$ to clear decimals: $C_3H_{3.9}O_3 \rightarrow C_3H_4O_3$

  • Empirical formula = $C_3H_4O_3$

  • Step 6: Determine molecular formula from empirical formula mass and molar mass:

  • Empirical formula mass ($C_3H_4O_3$) = $3(12.01) + 4(1.008) + 3(16.00) = 88.06\,g/mol$

  • Multiplier = $\frac{176.12\,g/mol}{88.06\,g/mol} = 2.000$

  • Molecular formula = $C_{(3 \times 2)}H_{(4 \times 2)}O_{(3 \times 2)} = C_6H_8O_6$

Fundamentals of Chemical Equations and Balancing

• A chemical equation uses chemical formulas to represent the qualitative identities and quantitative proportions of substances undergoing physical or chemical transformations.   

  

• Key Principles of Chemical Equations:

  • Reactants are listed on the left side of the equation; products are listed on the right side.

  • The equation must strictly observe the Law of Conservation of Mass: the same total number and type of each atom must appear on both sides of the yield arrow ($\rightarrow$).

• Sequential Steps for Balancing Chemical Equations:

  1. Translate the Statement: Convert the qualitative chemical description into an unbalanced skeleton equation containing correct chemical formulas.

  2. Balance Atoms: Place stoichiometric coefficients before chemical formulas to match atom totals for every element across reactant and product sides.

  3. Adjust Coefficients: Convert fractional coefficients into the lowest possible set of whole-number integers by multiplying through by the denominator.

  4. Check Atom Balance: Systematically re-verify that atom counts for every single element are identical on both sides.

  5. Specify Physical States: Include state designators following each chemical formula: solid ($s$), liquid ($l$), gas ($g$), or aqueous solution ($aq$).

• Multi-Level Representation of Chemical Reactions:   

  

  • Macroscopic Level: Observable physical phenomena (e.g., bright flare and smoke formation during magnesium ribbon combustion in air).

  • Atomic-Scale (Microscopic) Level: Interaction and rearrangement of discrete structural units (e.g., $2$ elemental $Mg$ atoms reacting with $1$ diatomic $O_2$ molecule to yield $2$ formula units of ionic $MgO$ lattice containing $Mg^{2+}$ and $O^{2-}$ ions).

  • Symbolic Level (Balanced Equation):     $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

• Worked Example: Balancing Hydrocarbon Combustion (Sample Problem 3.10):

  • Problem: Octane ($C_8H_{18}$), a component of gasoline, reacts with oxygen in a vehicle cylinder to form carbon dioxide and water vapor. Write a balanced chemical equation.

  • Step 1 (Skeleton Equation):     $\text{\_\_\_}C_8H_{18} + \text{\_\_\_}O_2 \rightarrow \text{\_\_\_}CO_2 + \text{\_\_\_}H_2O$

  • Step 2 (Balance C and H):     $1C_8H_{18} + \text{\_\_\_}O_2 \rightarrow 8CO_2 + 9H_2O$

  • Step 3 (Balance O using fraction):     $1C_8H_{18} + \frac{25}{2}O_2 \rightarrow 8CO_2 + 9H_2O$

  • Step 4 (Clear fraction by multiplying by 2):     $2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O$

  • Step 5 (Include physical states):     $2C_8H_{18}(l) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$

  • Molecular visual model for octane combustion:     

    

• Worked Example: Balancing from a Molecular Scene (Sample Problem 3.11):   

  

  • Problem: Analyze molecular scenes where blue spheres represent nitrogen ($N$) and red spheres represent oxygen ($O$).

  • Reactant Analysis: The initial mixture contains $4$ identical molecules composed of $2$ nitrogen atoms and $5$ oxygen atoms ($N_2O_5$).

  • Product Analysis: The completed reaction mixture contains $8$ molecules composed of $1$ nitrogen and $2$ oxygen atoms ($NO_2$) and $2$ diatomic oxygen molecules ($O_2$).

  • Unadjusted Reaction: $4N_2O_5 \rightarrow 8NO_2 + 2O_2$

  • Balanced Equation (Smallest Integers and States):     $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$

Stoichiometric Calculations and Molar Relationships

• In a balanced chemical equation, molar quantities of all participating species are stoichiometrically equivalent to each other.

• Quantitative relationships derived from stoichiometric coefficients serve as molar ratio conversion factors in chemical calculations.

• Information content contained within a balanced chemical equation ($C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$):

  • Molecules: $1\text{ molecule } C_3H_8 + 5\text{ molecules } O_2 \rightarrow 3\text{ molecules } CO_2 + 4\text{ molecules } H_2O$

  • Amount (mol): $1\,mol\,C_3H_8 + 5\,mol\,O_2 \rightarrow 3\,mol\,CO_2 + 4\,mol\,H_2O$

  • Mass (amu): $44.09\,amu\,C_3H_8 + 160.00\,amu\,O_2 \rightarrow 132.03\,amu\,CO_2 + 72.06\,amu\,H_2O$

  • Mass (g): $44.09\,g\,C_3H_8 + 160.00\,g\,O_2 \rightarrow 132.03\,g\,CO_2 + 72.06\,g\,H_2O$

  • Total Mass (g): $204.09\,g\text{ total reactants} \rightarrow 204.09\,g\text{ total products}$

• General Interconversion Schema:   

  

  • Interconverting mass and moles: $\text{Amount (mol)} = \frac{\text{Mass (g)}}{\text{Molar Mass } M\,(g/mol)}$

  • Interconverting moles and entities: $\text{Entities} = \text{Amount (mol)} \times N_A$ (where Avogadro's number $N_A = 6.022 \times 10^{23}\,\text{entities/mol}$).

  • Interconverting amounts between Substance A and Substance B: Uses the molar ratio $\frac{\text{moles of B}}{\text{moles of A}}$ obtained directly from equation coefficients.

• Worked Example: Stoichiometry of Copper Roasting (Sample Problem 3.12):

  • Reaction: $2Cu_2S(s) + 3O_2(g) \rightarrow 2Cu_2O(s) + 2SO_2(g)$

  • Problem (a): How many moles of $O_2$ are required to roast $10.00\,mol$ of copper(I) sulfide ($Cu_2S$)?

  • Calculation:       $10.00\,mol\,Cu_2S \times \frac{3\,mol\,O_2}{2\,mol\,Cu_2S} = 15.0\,mol\,O_2$

  • Problem (b): How many grams of sulfur dioxide ($SO_2$) form when $10.00\,mol$ of $Cu_2S$ reacts?

  • Calculation:       $10.00\,mol\,Cu_2S \times \frac{2\,mol\,SO_2}{2\,mol\,Cu_2S} \times \frac{64.06\,g\,SO_2}{1\,mol\,SO_2} = 641\,g\,SO_2$

  • Road map for this calculation:       

    

• Worked Example: Mass-to-Mass Calculation (Sample Problem 3.13):

  • Problem: How many kilograms of oxygen ($O_2$) are required to produce $2.86\,kg$ of copper(I) oxide ($Cu_2O$) during roasting?

  • Road Map:     

    

  • Step 1: Convert mass of $Cu_2O$ from $kg$ to moles:     $2.86\,kg\,Cu_2O \times \frac{10^3\,g}{1\,kg} \times \frac{1\,mol\,Cu_2O}{143.10\,g\,Cu_2O} = 20.0\,mol\,Cu_2O$

  • Step 2: Calculate mass of $O_2$ required in $kg$:     $20.0\,mol\,Cu_2O \times \frac{3\,mol\,O_2}{2\,mol\,Cu_2O} \times \frac{32.00\,g\,O_2}{1\,mol\,O_2} \times \frac{1\,kg}{10^3\,g} = 0.960\,kg\,O_2$

Limiting Reactants and Reaction Tables

• In actual chemical processes, reactants are rarely present in exact stoichiometric proportions.

• Definitions:

  • Limiting Reactant: The reactant that is completely consumed first in a reaction. It governs and limits the maximum theoretical quantity of product formed.

  • Excess Reactant: The reactant present in a quantity greater than required to react completely with the limiting reactant. Unreacted excess reactant remains after the reaction terminates.

• Sundae Analogy for Limiting Reactants:   

  

  • Recipe for $1$ sundae: $12\text{ oz (2 scoops)}$ ice cream + $1$ cherry + $50\,mL$ syrup.

  • Initial ingredients available: $8$ scoops ice cream ($4$ sundaes worth), $6$ cherries ($6$ sundaes worth), and $100\,mL$ syrup ($2$ sundaes worth).

  • Outcome: Syrup limits production to exactly $2$ sundaes. Ice cream ($4$ scoops left over) and cherries ($4$ cherries left over) are in excess.

• Worked Example: Molecular Scene Limiting Reactant Problem (Sample Problem 3.14):

  • Reaction: $Cl_2(g) + 3F_2(g) \rightarrow 2ClF_3(g)$

  • Initial Scene: Contains $3$ molecules of $Cl_2$ (green) and $6$ molecules of $F_2$ (yellow).

  • Step 1: Determine limiting reactant by calculating yield from each:

  • From $Cl_2$: $3\text{ molecules } Cl_2 \times \frac{2\text{ molecules } ClF_3}{1\text{ molecule } Cl_2} = 6\text{ molecules } ClF_3$

  • From $F_2$: $6\text{ molecules } F_2 \times \frac{2\text{ molecules } ClF_3}{3\text{ molecules } F_2} = 4\text{ molecules } ClF_3$

  • Diatomic fluorine ($F_2$) is the limiting reactant because it yields fewer product molecules.

  • Step 2: Construct Reaction Table:     | Quantity (Molecules) | $Cl_2(g)$ | $+$ | $3F_2(g)$ | $\rightarrow$ | $2ClF_3(g)$ |     |---|---|---|---|---|---|     | Initial | $3$ | | $6$ | | $0$ |     | Change | $-2$ | | $-6$ | | $+4$ |     | Final | $1$ | | $0$ | | $4$ |

  • Step 3: Visual Description of Final Scene: Consists of $4$ synthesized $ClF_3$ molecules and $1$ unreacted excess $Cl_2$ molecule.

• Worked Example: Molar Limiting Reactant Problem (Sample Problem 3.15):

  • Problem: $0.750\,mol$ of $Cl_2$ reacts with $3.00\,mol$ of $F_2$ to form ClF_3$.\n - **Determine Yields**:\n - From Cl_2$:$0.750\,mol\,Cl_2 \times \frac{2\,mol\,ClF_3}{1\,mol\,Cl_2} = 1.50\,mol\,ClF_3\n - From F_2$:$3.00\,mol\,F_2 \times \frac{2\,mol\,ClF_3}{3\,mol\,F_2} = 2.00\,mol\,ClF_3\n - Cl_2 is the limiting reactant.\n - **Reaction Table**:\n    | Amount (mol) | Cl_2(g)$|$+$|$3F_2(g)$|$\rightarrow$|$2ClF_3(g) |\n    |---|---|---|---|---|---|\n    | **Initial** | 0.750$| |$3.00$| |$0 |\n    | **Change** | -0.750$| |$-2.25$| |$+1.50 |\n    | **Final** | 0$| |$0.75$| |$1.50 |\n- **Worked Example: Mass-Based Limiting Reactant Problem (Sample Problem 3.16)**:\n - **Reaction**: Rocket fuel reaction of hydrazine and dinitrogen tetroxide:\n    2N_2H_4(l) + N_2O_4(l) \rightarrow 3N_2(g) + 4H_2O(g)\n - **Given**: 100.\,g$of$N_2H_4$mixed with$125\,g$of$N_2O_4$.

  • Road Map:     

    

  • Part (a): Determine grams of $N_2$ gas formed:

  • Moles of $N_2H_4 = 100.\,g\,N_2H_4 \times \frac{1\,mol\,N_2H_4}{32.05\,g\,N_2H_4} = 3.12\,mol\,N_2H_4$

  • Moles of $N_2$ from $N_2H_4 = 3.12\,mol\,N_2H_4 \times \frac{3\,mol\,N_2}{2\,mol\,N_2H_4} = 4.68\,mol\,N_2$

  • Moles of $N_2O_4 = 125\,g\,N_2O_4 \times \frac{1\,mol\,N_2O_4}{92.02\,g\,N_2O_4} = 1.36\,mol\,N_2O_4$

  • Moles of $N_2$ from $N_2O_4 = 1.36\,mol\,N_2O_4 \times \frac{3\,mol\,N_2}{1\,mol\,N_2O_4} = 4.08\,mol\,N_2$

  • $N_2O_4$ is the limiting reactant; maximum theoretical N_2 = 4.08\,mol$.\n - \text{Mass of } N_2 = 4.08\,mol\,N_2 \times \frac{28.02\,g\,N_2}{1\,mol\,N_2} = 114\,g\,N_2\n - **Part (b): Determine grams of excess reactant (N_2H_4) remaining**:\n - \text{Mass } N_2H_4\text{ reacted} = 125\,g\,N_2O_4 \times \frac{1\,mol\,N_2O_4}{92.02\,g\,N_2O_4} \times \frac{2\,mol\,N_2H_4}{1\,mol\,N_2O_4} \times \frac{32.05\,g\,N_2H_4}{1\,mol\,N_2H_4} = 87.1\,g\,N_2H_4\n - \text{Mass } N_2H_4\text{ in excess} = 100.\,g - 87.1\,g = 13\,g\,N_2H_4\text{ unreacted}\n - **Part (c): Reaction Table**:\n    | Amount (mol) | 2N_2H_4(l)$|$+$|$N_2O_4(l)$|$\rightarrow$|$3N_2(g)$|$+$|$4H_2O(g) |\n    |---|---|---|---|---|---|---|---|\n    | **Initial** | 3.12$| |$1.36$| |$0$| |$0 |\n    | **Change** | -2.72$| |$-1.36$| |$+4.08$| |$+5.43 |\n    | **Final** | 0.40$| |$0$| |$4.08$| |$5.43 |\n- **Comprehensive Overview of Stoichiometric Relationships**:\n  \n\n# Theoretical Yield, Actual Yield, and Percent Yield\n\n- **Key Yield Concepts**:\n - **Theoretical Yield**: The maximum calculated amount of product that can form from given quantities of reactants according to the balanced chemical equation.\n - **Actual Yield**: The mass or amount of product actually isolated and measured experimentally in a laboratory setting. Actual yield is almost universally less than theoretical yield due to incomplete reactions, side reactions, or product losses during isolation.\n - **Percent Yield Formula**:\n    \% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\n- **Effect of Side Reactions**:\n  \n - When reactants A + B$react, competing side reactions can divert material to form undesirable side products ($D$) rather than the intended main product ($C$), thereby significantly reducing the actual yield of$C$.

• Worked Example: Calculating Percent Yield (Sample Problem 3.17):

  • Reaction: Production of silicon carbide ($SiC$) from sand ($SiO_2$) and carbon ($C$):     $SiO_2(s) + 3C(s) \rightarrow SiC(s) + 2CO(g)$

  • Problem: What is the percent yield if $51.4\,kg$ of $SiC$ is recovered from processing $100.0\,kg$ of sand ($SiO_2$)?

  • Road Map:     

    

  • Step 1: Calculate moles of $SiO_2$ processed:     $\text{Moles of } SiO_2 = 100.0\,kg\,SiO_2 \times \frac{10^3\,g}{1\,kg} \times \frac{1\,mol\,SiO_2}{60.08\,g\,SiO_2} = 1664\,mol\,SiO_2$

  • Step 2: Calculate theoretical mass of $SiC$ formed:     $\text{Moles of } SiC = 1664\,mol\,SiO_2 \times \frac{1\,mol\,SiC}{1\,mol\,SiO_2} = 1664\,mol\,SiC$     $\text{Theoretical mass of } SiC = 1664\,mol\,SiC \times \frac{40.10\,g\,SiC}{1\,mol\,SiC} \times \frac{1\,kg}{1000\,g} = 66.71\,kg\,SiC$

  • Step 3: Calculate percent yield:     $\% \text{ yield} = \frac{51.4\,kg\,SiC}{66.71\,kg\,SiC} \times 100 = 77.0\%$