Comprehensive Study Guide for General Chemistry: Stoichiometry, Chemical Formulas, and Reaction Analysis

Core Foundations and Chemical Conversions

  • Essential Problem-Solving Tools:

    • Periodic table of the elements for atomic masses and molar mass calculations.

    • Scientific calculator for scientific notation and dimensional analysis.

  • Mass-to-Mole and Mole-to-Mole Conversions:

    • Conversion of Sodium Nitrate (NaNO3\text{NaNO}_3) Mass to Moles:

    • Given quantity: 3.2104 kg NaNO33.2104\,\text{kg NaNO}_3.

    • Conversion factor from kilograms to grams: 1000 g=1 kg1000\,\text{g} = 1\,\text{kg}.

    • Molar mass of NaNO3\text{NaNO}_3: 85 g/mol85\,\text{g/mol}.

    • Dimensional analysis setup:       3.2104 kg NaNO3×1000 g1 kg×1 mol NaNO385 g NaNO3=37.77 mol NaNO33.2104\,\text{kg NaNO}_3 \times \frac{1000\,\text{g}}{1\,\text{kg}} \times \frac{1\,\text{mol NaNO}_3}{85\,\text{g NaNO}_3} = 37.77\,\text{mol NaNO}_3

    • Result: 37.77 mol NaNO337.77\,\text{mol NaNO}_3.

    • Conversion of Potassium Carbonate (K2CO3\text{K}_2\text{CO}_3) to Moles of Carbonate Ion (CO32−\text{CO}_3^{2-}):

    • Given mass: 10.3 g K2CO310.3\,\text{g K}_2\text{CO}_3.

    • Molar mass of K2CO3\text{K}_2\text{CO}_3: 138 g/mol138\,\text{g/mol}.

    • Stoichiometric ratio within the formula unit: 1 mol K2CO31\,\text{mol K}_2\text{CO}_3 contains 1 mol CO31\,\text{mol CO}_3.

    • Dimensional analysis setup:       10.3 g K2CO3×1 mol K2CO3138 g K2CO3×1 mol CO31 mol K2CO3=0.0746 mol CO310.3\,\text{g K}_2\text{CO}_3 \times \frac{1\,\text{mol K}_2\text{CO}_3}{138\,\text{g K}_2\text{CO}_3} \times \frac{1\,\text{mol CO}_3}{1\,\text{mol K}_2\text{CO}_3} = 0.0746\,\text{mol CO}_3

    • Result: 0.0746 mol CO30.0746\,\text{mol CO}_3.

  • Atom and Molecule Counting with Avogadro's Number:

    • Avogadro's Constant: 6.022×1023 particles/mol6.022 \times 10^{23}\,\text{particles/mol}.

    • Calculating Oxygen Atoms in Ammonium Nitrate (NH4NO3\text{NH}_4\text{NO}_3):

    • Given quantity: 21.002 kg NH4NO321.002\,\text{kg NH}_4\text{NO}_3.

    • Molar mass of NH4NO3\text{NH}_4\text{NO}_3: 80.06 g/mol80.06\,\text{g/mol}.

    • Formula unit stoichiometry: 1 mol NH4NO31\,\text{mol NH}_4\text{NO}_3 contains 3 mol O3\,\text{mol O}.

    • Step 1: Convert mass to moles of compound and moles of oxygen atoms:       21.002 kg NH4NO3×1000 g1 kg×1 mol NH4NO380.06 g NH4NO3×3 mol O1 mol NH4NO3=787 mol O21.002\,\text{kg NH}_4\text{NO}_3 \times \frac{1000\,\text{g}}{1\,\text{kg}} \times \frac{1\,\text{mol NH}_4\text{NO}_3}{80.06\,\text{g NH}_4\text{NO}_3} \times \frac{3\,\text{mol O}}{1\,\text{mol NH}_4\text{NO}_3} = 787\,\text{mol O}

    • Step 2: Convert moles of oxygen to total oxygen atoms:       787 mol O×6.022×1023 atoms O1 mol O=4.74×1026 atoms O787\,\text{mol O} \times \frac{6.022 \times 10^{23}\,\text{atoms O}}{1\,\text{mol O}} = 4.74 \times 10^{26}\,\text{atoms O}

    • Result: 4.74×1026 atoms O4.74 \times 10^{26}\,\text{atoms O}.

  • Practice Problems for Further Study:

    • Calculate the grams of oxygen present in 30.6×102 moles30.6 \times 10^2\,\text{moles} of potassium dichromate (K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7).

Percent Composition, Empirical Formulas, and Molecular Formulas

  • Foundational Definitions:

    • Empirical Formula: The lowest whole-number molar ratio of elements in a chemical compound.

    • Molecular Formula: The actual, unreduced integer ratio of each element present in a single discrete molecule of a compound.

  • Distinguishing Empirical Formulas from Molecular Formulas:

    • CH2O3\text{CH}_2\text{O}_3: Empirical formula (the subscripts 1:2:31:2:3 share no common integer divisor greater than 1).

    • H2O2\text{H}_2\text{O}_2: Molecular formula (subscripts 2:22:2 reduce to an empirical formula of HO\text{HO}).

    • C6H12O6\text{C}_6\text{H}_{12}\text{O}_6: Molecular formula (subscripts 6:12:66:12:6 reduce to an empirical formula of CH2O\text{CH}_2\text{O}).

    • N2O3\text{N}_2\text{O}_3: Empirical formula (the subscripts 2:32:3 share no common integer divisor greater than 1).

  • Mass Percent Composition:

    • General Formula:     Mass % of Element=Total Mass of Element in 1 mol of CompoundMolar Mass of Compound×100%\text{Mass \% of Element} = \frac{\text{Total Mass of Element in 1 mol of Compound}}{\text{Molar Mass of Compound}} \times 100\%

    • Sodium in Sodium Chloride (NaCl\text{NaCl}):

    • Atomic mass of Na\text{Na}: 22.99 g/mol22.99\,\text{g/mol}.

    • Atomic mass of Cl\text{Cl}: 35.45 g/mol35.45\,\text{g/mol}.

    • Total molar mass of NaCl\text{NaCl}: 22.99+35.45=58.44 g/mol22.99 + 35.45 = 58.44\,\text{g/mol}.

    • Percent sodium calculation:       22.99 g/mol58.44 g/mol×100%=39.34%\frac{22.99\,\text{g/mol}}{58.44\,\text{g/mol}} \times 100\% = 39.34\%

    • Component Breakdown in Magnesium Carbonate (MgCO3\text{MgCO}_3):

    • Atomic mass of Mg\text{Mg}: 24.3 g/mol24.3\,\text{g/mol}.

    • Atomic mass of C\text{C}: 12.01 g/mol12.01\,\text{g/mol}.

    • Mass of 3 mol O3\,\text{mol O}: 3×16=48 g/mol3 \times 16 = 48\,\text{g/mol}.

    • Molar mass of MgCO3\text{MgCO}_3: 24.3+12.01+48=84.3 g/mol24.3 + 12.01 + 48 = 84.3\,\text{g/mol}.

    • Percent magnesium:       24.384.3×100%=28.8% Mg\frac{24.3}{84.3} \times 100\% = 28.8\%\,\text{Mg}

    • Percent carbon:       12.0184.3×100%=14.25% C\frac{12.01}{84.3} \times 100\% = 14.25\%\,\text{C}

    • Percent oxygen:       4884.3×100%=57% O\frac{48}{84.3} \times 100\% = 57\%\,\text{O}

  • Determining Empirical and Molecular Formulas from Percent Composition:

    • Problem 1 (ThoughtCo Reference: https://www.thoughtco.com/molecular-formula-practice-test-questions-604125):

    • A compound has a molecular mass of 74.1 g/mol74.1\,\text{g/mol} and contains 64.8% C64.8\%\,\text{C}, 13.5% H13.5\%\,\text{H}, and 21.7% O21.7\%\,\text{O}.

    • Step 1: Assume a 100 g100\,\text{g} sample to convert percentages directly to grams:

      • Mass of C=64.8 g\text{C} = 64.8\,\text{g}

      • Mass of H=13.5 g\text{H} = 13.5\,\text{g}

      • Mass of O=21.7 g\text{O} = 21.7\,\text{g}

    • Step 2: Convert grams to moles of each element:       64.8 g C×1 mol C12.01 g C=5.4 mol C64.8\,\text{g C} \times \frac{1\,\text{mol C}}{12.01\,\text{g C}} = 5.4\,\text{mol C}       13.5 g H×1 mol H1.01 g H=13.4 mol H13.5\,\text{g H} \times \frac{1\,\text{mol H}}{1.01\,\text{g H}} = 13.4\,\text{mol H}       21.7 g O×1 mol O16 g O=1.36 mol O21.7\,\text{g O} \times \frac{1\,\text{mol O}}{16\,\text{g O}} = 1.36\,\text{mol O}

    • Step 3: Divide each molar value by the smallest molar value (1.36 mol1.36\,\text{mol}):       C:5.41.36=3.9≈4\text{C}: \frac{5.4}{1.36} = 3.9 \approx 4       H:13.41.36=9.8≈10\text{H}: \frac{13.4}{1.36} = 9.8 \approx 10       O:1.361.36=1\text{O}: \frac{1.36}{1.36} = 1

    • Empirical formula: C4H10O\text{C}_4\text{H}_{10}\text{O}.

    • Step 4: Determine the empirical formula mass:       (12.0×4)+(1.01×10)+16=74.12 g/mol(12.0 \times 4) + (1.01 \times 10) + 16 = 74.12\,\text{g/mol}

    • Step 5: Calculate the ratio of molecular mass to empirical formula mass:       74.1 g/mol74.12 g/mol≈1\frac{74.1\,\text{g/mol}}{74.12\,\text{g/mol}} \approx 1

    • Molecular formula: C4H10O\text{C}_4\text{H}_{10}\text{O} (identical to the empirical formula).

    • Problem 2 (Nitrogen-Oxygen Compound):

    • A compound contains 46.7% N46.7\%\,\text{N} and 53.3% O53.3\%\,\text{O}, with an experimentally determined molecular mass of 60.0 g/mol60.0\,\text{g/mol}.

    • Step 1: Convert mass in 100 g100\,\text{g} sample to moles:       46.7 g N×1 mol N14.01 g N=3.33 mol N46.7\,\text{g N} \times \frac{1\,\text{mol N}}{14.01\,\text{g N}} = 3.33\,\text{mol N}       53.3 g O×1 mol O16.0 g O=3.33 mol O53.3\,\text{g O} \times \frac{1\,\text{mol O}}{16.0\,\text{g O}} = 3.33\,\text{mol O}

    • Step 2: Determine empirical mole ratio by dividing by the smallest mole value (3.333.33):       N:3.333.33=1\text{N}: \frac{3.33}{3.33} = 1       O:3.333.33=1\text{O}: \frac{3.33}{3.33} = 1

    • Empirical formula: NO\text{NO}.

    • Step 3: Compute empirical formula mass:       14.01+16.0=30.01 g/mol14.01 + 16.0 = 30.01\,\text{g/mol}

    • Step 4: Compute the scaling factor (nn) for the molecular formula:       n=60.0 g/mol30.01 g/mol≈2n = \frac{60.0\,\text{g/mol}}{30.01\,\text{g/mol}} \approx 2

    • Molecular formula: (NO)2=N2O2(\text{NO})_2 = \text{N}_2\text{O}_2.

    • Problem 3 (Unworked Prompt):

    • Find the empirical formula of a compound composed of 30% oxygen30\%\,\text{oxygen}, 20% hydrogen20\%\,\text{hydrogen}, and 50% nitrogen50\%\,\text{nitrogen}.

Combustion Analysis

  • Problem Statement (FSU Chemistry Department Practice Problems: https://www.chem.fsu.edu/chemlab/chm1045/Combustion%20Analysis%20Practice%20Problems.pdf):

    • A 0.658 g0.658\,\text{g} sample of a compound containing only carbon, hydrogen, and oxygen is completely combusted in excess O2\text{O}_2.

    • Products generated: 1.285 g CO21.285\,\text{g CO}_2 and 0.658 g H2O0.658\,\text{g H}_2\text{O}.

    • Mass spectrometry determines the compound's molar mass is 90 g/mol90\,\text{g/mol}.

  • Step-by-Step Analytical Derivation:

    • Step 1: Determine the mass and moles of carbon from produced CO2\text{CO}_2:     1.285 g CO2×1 mol CO244.01 g CO2×1 mol C1 mol CO2=0.0292 mol C1.285\,\text{g CO}_2 \times \frac{1\,\text{mol CO}_2}{44.01\,\text{g CO}_2} \times \frac{1\,\text{mol C}}{1\,\text{mol CO}_2} = 0.0292\,\text{mol C}     0.0292 mol C×12.01 g C1 mol C=0.351 g C0.0292\,\text{mol C} \times \frac{12.01\,\text{g C}}{1\,\text{mol C}} = 0.351\,\text{g C}

    • Step 2: Determine the mass and moles of hydrogen from produced H2O\text{H}_2\text{O}:     0.658 g H2O×1 mol H2O18.02 g H2O×2 mol H1 mol H2O=0.073 mol H0.658\,\text{g H}_2\text{O} \times \frac{1\,\text{mol H}_2\text{O}}{18.02\,\text{g H}_2\text{O}} \times \frac{2\,\text{mol H}}{1\,\text{mol H}_2\text{O}} = 0.073\,\text{mol H}     0.073 mol H×1.01 g H1 mol H=0.0738 g H0.073\,\text{mol H} \times \frac{1.01\,\text{g H}}{1\,\text{mol H}} = 0.0738\,\text{g H}

    • Step 3: Determine the mass and moles of oxygen by mass difference:     Mass of O=0.658 g total−0.0738 g H−0.351 g C=0.233 g O\text{Mass of O} = 0.658\,\text{g total} - 0.0738\,\text{g H} - 0.351\,\text{g C} = 0.233\,\text{g O}     0.233 g O×1 mol O16.0 g O=0.0146 mol O0.233\,\text{g O} \times \frac{1\,\text{mol O}}{16.0\,\text{g O}} = 0.0146\,\text{mol O}

    • Step 4: Compute the lowest whole-number mole ratios (divide by 0.01460.0146):     O:0.0146 mol0.0146=1\text{O}: \frac{0.0146\,\text{mol}}{0.0146} = 1     C:0.0292 mol0.0146=2\text{C}: \frac{0.0292\,\text{mol}}{0.0146} = 2     H:0.073 mol0.0146=5\text{H}: \frac{0.073\,\text{mol}}{0.0146} = 5

    • Empirical formula: C2H5O\text{C}_2\text{H}_5\text{O}.

    • Step 5: Determine the molecular formula:

    • Empirical formula weight of C2H5O\text{C}_2\text{H}_5\text{O}: (2×12.01)+(5×1.01)+16.00=45.07 g/mol(2 \times 12.01) + (5 \times 1.01) + 16.00 = 45.07\,\text{g/mol}.

    • Ratio of molar mass to empirical mass:       90 g/mol45.07 g/mol≈2\frac{90\,\text{g/mol}}{45.07\,\text{g/mol}} \approx 2

    • Molecular formula: (C2H5O)2=C4H10O2(\text{C}_2\text{H}_5\text{O})_2 = \text{C}_4\text{H}_{10}\text{O}_2.

Chemical Equations and Balancing Methodology

  • Anatomy of a Chemical Equation:

    • Reactants: The starting chemical substances undergoing change, positioned to the left of the reaction arrow.

    • Products: The new chemical substances created by the transformation, positioned to the right of the reaction arrow.

    • Coefficients: Multiplicative whole numbers placed in front of chemical formulas specifying the relative molar, molecular, or atomic quantities involved.

    • Subscripts: Fixed integers inside formulas indicating the exact stoichiometry of elements per chemical unit. Subscripts must never be modified when balancing an equation.

    • Reaction Arrow (→\rightarrow): Signifies the directional transition from reactants to products.

    • States of Matter: Solid (ss), liquid (ll), gas (gg), and aqueous solution (aqaq).

Anatomy of a chemical equation diagram showing coefficients, subscripts, states of matter, and reactants converting to products

  • Algorithmic Balancing Process:

    • Step 1: Write out the skeletal equation with correct formulas and physical states.

    • Step 2: Construct an atom balance sheet listing total counts of each element on the reactant side versus product side. Conserve intact polyatomic ions as single distinct units where applicable.

    • Step 3: Adjust stoichiometric coefficients one element at a time, leaving elements in elemental form (such as O2\text{O}_2, Al\text{Al}, Fe\text{Fe}) for the final adjustment step.

    • Step 4: Re-tally all atoms on both sides to verify absolute balance according to the law of conservation of mass.

  • Worked Example 1: Thermite Displacement Reaction:

    • Unbalanced equation:     Al(s)+Fe2O3(s)→Al2O3(s)+Fe(l)\text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{Al}_2\text{O}_3(s) + \text{Fe}(l)

    • Initial atom inventory:

    • Reactants: Al=1\text{Al} = 1, Fe=2\text{Fe} = 2, O=3\text{O} = 3

    • Products: Al=2\text{Al} = 2, Fe=1\text{Fe} = 1, O=3\text{O} = 3

    • Balancing sequence:

    • Balance Al\text{Al}: Adjust reactant coefficient of Al\text{Al} from 1 to 2.

    • Balance Fe\text{Fe}: Adjust product coefficient of Fe\text{Fe} from 1 to 2.

    • Balance O\text{O}: Oxygen is already balanced with 3 atoms on each side.

    • Balanced equation:     2Al(s)+Fe2O3(s)→Al2O3(s)+2Fe(l)2\text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2\text{Fe}(l)

  • Worked Example 2: Double Displacement Phosphate Precipitation:

    • Unbalanced equation:     Na3PO4+MgCl2→NaCl+Mg3(PO4)2\text{Na}_3\text{PO}_4 + \text{MgCl}_2 \rightarrow \text{NaCl} + \text{Mg}_3(\text{PO}_4)_2

    • Initial atom and ion inventory:

    • Reactants: Na=3\text{Na} = 3, PO4=1\text{PO}_4 = 1, Mg=1\text{Mg} = 1, Cl=2\text{Cl} = 2

    • Products: Na=1\text{Na} = 1, PO4=2\text{PO}_4 = 2, Mg=3\text{Mg} = 3, Cl=1\text{Cl} = 1

    • Balancing sequence:

    • Balance phosphate groups (PO4\text{PO}_4): Place a coefficient of 2 in front of Na3PO4\text{Na}_3\text{PO}_4, bringing PO4\text{PO}_4 to 2 and Na\text{Na} to 3×2=63 \times 2 = 6.

    • Balance magnesium (Mg\text{Mg}): Place a coefficient of 3 in front of MgCl2\text{MgCl}_2, bringing Mg\text{Mg} to 3 and Cl\text{Cl} to 2×3=62 \times 3 = 6.

    • Balance sodium (Na\text{Na}) and chloride (Cl\text{Cl}): Place a coefficient of 6 in front of NaCl\text{NaCl}, bringing both Na\text{Na} and Cl\text{Cl} to 6.

    • Balanced equation:     2Na3PO4+3MgCl2→6NaCl+Mg3(PO4)22\text{Na}_3\text{PO}_4 + 3\text{MgCl}_2 \rightarrow 6\text{NaCl} + \text{Mg}_3(\text{PO}_4)_2

  • Worked Example 3: Propane Combustion:

    • Unbalanced statement: Hydrocarbon propane (C3H8\text{C}_3\text{H}_8) reacts with oxygen gas to yield carbon dioxide and water.

    • Balanced chemical equation:     C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)

Reaction Stoichiometry

  • Stoichiometry of Propane Combustion:

    • Given: Production of 36.1 g CO236.1\,\text{g CO}_2.

    • Question: Determine the corresponding moles and molecules of O2\text{O}_2 required.

    • Molar mass of CO2\text{CO}_2: 44.01 g/mol44.01\,\text{g/mol}.

    • Stoichiometric ratio from balanced equation: 5 mol O2:3 mol CO25\,\text{mol O}_2 : 3\,\text{mol CO}_2.

    • Moles of O2\text{O}_2 calculation:     36.1 g CO2×1 mol CO244.01 g CO2×5 mol O23 mol CO2=1.37 mol O236.1\,\text{g CO}_2 \times \frac{1\,\text{mol CO}_2}{44.01\,\text{g CO}_2} \times \frac{5\,\text{mol O}_2}{3\,\text{mol CO}_2} = 1.37\,\text{mol O}_2

    • Molecules of O2\text{O}_2 calculation:     1.37 mol O2×6.022×1023 molecules O21 mol O2=8.25×1023 molecules O21.37\,\text{mol O}_2 \times \frac{6.022 \times 10^{23}\,\text{molecules O}_2}{1\,\text{mol O}_2} = 8.25 \times 10^{23}\,\text{molecules O}_2

  • Iodine Formation via Redox Precipitation:

    • Balanced equation:     2CuCl2+4KI→2CuI+4KCl+I22\text{CuCl}_2 + 4\text{KI} \rightarrow 2\text{CuI} + 4\text{KCl} + \text{I}_2

    • Given: Complete reaction of 0.4235 mol CuCl20.4235\,\text{mol CuCl}_2.

    • Part A: Molecules of I2\text{I}_2 produced:     0.4235 mol CuCl2×1 mol I22 mol CuCl2×6.022×1023 molecules I21 mol I2=1.28×1023 molecules I20.4235\,\text{mol CuCl}_2 \times \frac{1\,\text{mol I}_2}{2\,\text{mol CuCl}_2} \times \frac{6.022 \times 10^{23}\,\text{molecules I}_2}{1\,\text{mol I}_2} = 1.28 \times 10^{23}\,\text{molecules I}_2

    • Part B: Mass in grams of I2\text{I}_2 produced (molar mass of I2=254 g/mol\text{I}_2 = 254\,\text{g/mol}):     1.28×1023 molecules I2×1 mol I26.022×1023 molecules I2×254 g I21 mol I2=53.8 g I21.28 \times 10^{23}\,\text{molecules I}_2 \times \frac{1\,\text{mol I}_2}{6.022 \times 10^{23}\,\text{molecules I}_2} \times \frac{254\,\text{g I}_2}{1\,\text{mol I}_2} = 53.8\,\text{g I}_2

  • Glucose Fermentation Stoichiometry (LibreTexts Reference: https://chem.libretexts.org/):

    • Balanced equation:     C6H12O6→2C2H5OH+2CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2

    • Reaction classification: Fermentation / decomposition.

    • Problem 1: Moles of CO2\text{CO}_2 produced from 2.5 mol C6H12O62.5\,\text{mol C}_6\text{H}_{12}\text{O}_6:     2.5 mol C6H12O6×2 mol CO21 mol C6H12O6=5 mol CO22.5\,\text{mol C}_6\text{H}_{12}\text{O}_6 \times \frac{2\,\text{mol CO}_2}{1\,\text{mol C}_6\text{H}_{12}\text{O}_6} = 5\,\text{mol CO}_2

    • Problem 2: Mass of glucose (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6) needed to generate 13.5 g C2H5OH13.5\,\text{g C}_2\text{H}_5\text{OH}:

    • Molar mass of ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}): 46.07 g/mol46.07\,\text{g/mol}.

    • Molar mass of glucose (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6): 180 g/mol180\,\text{g/mol}.

    • Stoichiometric ratio: 1 mol C6H12O6:2 mol C2H5OH1\,\text{mol C}_6\text{H}_{12}\text{O}_6 : 2\,\text{mol C}_2\text{H}_5\text{OH}.

    • Calculation sequence:       13.5 g C2H5OH×1 mol C2H5OH46.07 g C2H5OH×1 mol C6H12O62 mol C2H5OH×180 g C6H12O61 mol C6H12O6=26.37 g C6H12O613.5\,\text{g C}_2\text{H}_5\text{OH} \times \frac{1\,\text{mol C}_2\text{H}_5\text{OH}}{46.07\,\text{g C}_2\text{H}_5\text{OH}} \times \frac{1\,\text{mol C}_6\text{H}_{12}\text{O}_6}{2\,\text{mol C}_2\text{H}_5\text{OH}} \times \frac{180\,\text{g C}_6\text{H}_{12}\text{O}_6}{1\,\text{mol C}_6\text{H}_{12}\text{O}_6} = 26.37\,\text{g C}_6\text{H}_{12}\text{O}_6

  • Practice Problem for Further Study:

    • If 824 g824\,\text{g} of NH3\text{NH}_3 and excess oxygen gas react according to the equation:     NH3+O2→H2O+NO\text{NH}_3 + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{NO}     Determine the total moles of nitrogen oxide formed upon complete reaction.

Limiting Reactants, Excess Reactants, and Percent Yield

  • Core Conceptual Principles:

    • Limiting Reactant (LR): The reagent completely consumed first in a reaction, which determines and limits the maximum theoretical quantity of product that can form.

    • Excess Reactant: The reagent present in a quantity greater than necessary to react completely with the limiting reactant; a portion remains unreacted upon reaction completion.

    • Baking Metaphor: If an unlimited quantity of flour is available but only 2 cups of sugar are present, the total number of cakes produced is strictly governed and constrained by the sugar.

  • Percent Yield:

    • Theoretical Yield: The calculated maximum mass of product obtainable assuming 100% stoichiometric conversion of the limiting reactant.

    • Actual Yield: The physical mass of product recovered from an experiment in a laboratory setting.

    • Percent Yield Equation:     Percent Yield=Actual YieldTheoretical Yield×100%\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

    • Sample Percent Yield Calculation:

    • Given theoretical yield: 97 g97\,\text{g}.

    • Actual laboratory yield produced: 80.2 g80.2\,\text{g}.

    • Calculation:       80.2 g97 g×100%=82.68%\frac{80.2\,\text{g}}{97\,\text{g}} \times 100\% = 82.68\%

  • Limiting and Excess Reactant Determination: Magnesium Oxidation:

    • Balanced equation:     2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}

    • Initial conditions: 3.0 g Mg3.0\,\text{g Mg} reacted with 11.6 g O211.6\,\text{g O}_2.

    • Molar masses: Mg=24.3 g/mol\text{Mg} = 24.3\,\text{g/mol}, O2=32 g/mol\text{O}_2 = 32\,\text{g/mol}, MgO=40.31 g/mol\text{MgO} = 40.31\,\text{g/mol}.

    • Step 1: Calculate theoretical yield of MgO\text{MgO} from each starting reagent:

    • Yield from Mg\text{Mg}:       3.0 g Mg×1 mol Mg24.3 g Mg×2 mol MgO2 mol Mg×40.31 g MgO1 mol MgO=4.98 g MgO3.0\,\text{g Mg} \times \frac{1\,\text{mol Mg}}{24.3\,\text{g Mg}} \times \frac{2\,\text{mol MgO}}{2\,\text{mol Mg}} \times \frac{40.31\,\text{g MgO}}{1\,\text{mol MgO}} = 4.98\,\text{g MgO}

    • Yield from O2\text{O}_2:       11.6 g O2×1 mol O232 g O2×2 mol MgO1 mol O2×40.31 g MgO1 mol MgO=29.2 g MgO11.6\,\text{g O}_2 \times \frac{1\,\text{mol O}_2}{32\,\text{g O}_2} \times \frac{2\,\text{mol MgO}}{1\,\text{mol O}_2} \times \frac{40.31\,\text{g MgO}}{1\,\text{mol MgO}} = 29.2\,\text{g MgO}

    • Step 2: Identify limiting reactant and theoretical product yield:

    • Magnesium yields the lesser amount of product (4.98 g4.98\,\text{g} vs. 29.2 g29.2\,\text{g}).

    • Limiting reactant: Mg\text{Mg}.

    • Theoretical yield: 4.98 g MgO4.98\,\text{g MgO}.

    • Step 3: Determine mass of excess reactant (O2\text{O}_2) consumed:     4.98 g MgO×1 mol MgO40.31 g MgO×1 mol O22 mol MgO×32 g O21 mol O2=1.97 g O2 consumed4.98\,\text{g MgO} \times \frac{1\,\text{mol MgO}}{40.31\,\text{g MgO}} \times \frac{1\,\text{mol O}_2}{2\,\text{mol MgO}} \times \frac{32\,\text{g O}_2}{1\,\text{mol O}_2} = 1.97\,\text{g O}_2\,\text{consumed}

    • Step 4: Calculate mass of unreacted excess reactant left over:     11.6 g O2 initial−1.97 g O2 consumed=9.63 g O2 remaining11.6\,\text{g O}_2\,\text{initial} - 1.97\,\text{g O}_2\,\text{consumed} = 9.63\,\text{g O}_2\,\text{remaining}

  • Limiting Reactant Precipitation: Calcium Chloride and Silver Nitrate:

    • Balanced precipitation equation:     CaCl2(aq)+2AgNO3(aq)→Ca(NO3)2(aq)+2AgCl(s)\text{CaCl}_2(aq) + 2\text{AgNO}_3(aq) \rightarrow \text{Ca}(\text{NO}_3)_2(aq) + 2\text{AgCl}(s)

    • Problem parameters: 2.00 mol CaCl2(aq)2.00\,\text{mol CaCl}_2(aq) reacts in solution with excess AgNO3(aq)\text{AgNO}_3(aq). CaCl2\text{CaCl}_2 is designated as the limiting reactant. Yield is assumed to be 100%100\%. Molar mass of AgCl=143.32 g/mol\text{AgCl} = 143.32\,\text{g/mol}.

    • Step 1: Calculate theoretical moles of precipitate AgCl\text{AgCl}:     2.00 mol CaCl2(aq)×2 mol AgCl(s)1 mol CaCl2(aq)=4 mol AgCl(s)2.00\,\text{mol CaCl}_2(aq) \times \frac{2\,\text{mol AgCl}(s)}{1\,\text{mol CaCl}_2(aq)} = 4\,\text{mol AgCl}(s)

    • Step 2: Convert moles of AgCl\text{AgCl} to theoretical mass in grams:     4 mol AgCl(s)×143.32 g AgCl(s)1 mol AgCl(s)=573.28 g AgCl(s)4\,\text{mol AgCl}(s) \times \frac{143.32\,\text{g AgCl}(s)}{1\,\text{mol AgCl}(s)} = 573.28\,\text{g AgCl}(s)

    • Theoretical yield: 573.28 g AgCl(s)573.28\,\text{g AgCl}(s) (reported to three significant figures as 573 g AgCl573\,\text{g AgCl}).