Chapter 7: Chemical Quantities - Moles, Avogadro's Number, and Formulas

Introduction to Chemical Quantities and the Mole

  • The study of chemical quantities involves understanding how to count atoms, molecules, and ions using units that bridge the gap between the microscopic and macroscopic worlds.
  • Examples of common counting units include:
    • 1 dozen=121\text{ dozen} = 12
    • 1 baker’s dozen=131\text{ baker's dozen} = 13
    • 1 case=24 cans1\text{ case} = 24\text{ cans}
    • 1 gross=1441\text{ gross} = 144
  • Atoms are so small that it requires an extremely large number of them to weigh and use in a laboratory setting.
  • Avogadro’s Number is defined as 6.022×10236.022 \times 10^{23}.
  • The Mole (mol): This is a unit representing the same number of particles as there are carbon atoms in 12.01 g12.01\,g of carbon.
  • One mole of any element contains 6.022×10236.022 \times 10^{23} atoms of that element.
    • 1 mol C=6.022×1023 C atoms1\text{ mol C} = 6.022 \times 10^{23}\text{ C atoms}
    • 1 mol Au=6.022×1023 Au atoms1\text{ mol Au} = 6.022 \times 10^{23}\text{ Au atoms}

Avogadro’s Number as a Conversion Factor

  • One mole represents a specific quantity of different particles depending on the substance:
    • For an element, it contains 6.022×1023 atoms6.022 \times 10^{23}\text{ atoms}.
    • For a molecule, it contains 6.022×1023 molecules6.022 \times 10^{23}\text{ molecules}.
    • For an ion, it contains 6.022×1023 ions6.022 \times 10^{23}\text{ ions}.
    • For an ionic compound (formula unit), it contains 6.022×1023 formula units6.022 \times 10^{23}\text{ formula units}.
  • Avogadro’s number can be written as an equality for use in conversion factors:
    • Equality: 1 mole=6.022×1023 particles1\text{ mole} = 6.022 \times 10^{23}\text{ particles}
    • Conversion Factors:
    • 6.022×1023 particles1 mole\frac{6.022 \times 10^{23}\text{ particles}}{1\text{ mole}}
    • 1 mole6.022×1023 particles\frac{1\text{ mole}}{6.022 \times 10^{23}\text{ particles}}

Using Avogadro’s Number in Calculations

  • Converting Moles to Atoms/Molecules:
    • To determine the number of atoms in 0.50 moles0.50\text{ moles} of Cu\text{Cu}, multiply the moles by Avogadro's number:
    • 0.50 moles Cu×(6.022×1023 molecules1 mole)=3.0×1023 molecules0.50\text{ moles Cu} \times \left( \frac{6.022 \times 10^{23}\text{ molecules}}{1\text{ mole}} \right) = 3.0 \times 10^{23}\text{ molecules}
  • Converting Particles to Moles:
    • To determine the number of moles of CO2\text{CO}_2 in 2.50×1024 molecules2.50 \times 10^{24}\text{ molecules}, divide by Avogadro's number:
    • 2.50×1024 molecules CO2×(1 mole6.022×1023 molecules)=4.15 mole CO22.50 \times 10^{24}\text{ molecules CO}_2 \times \left( \frac{1\text{ mole}}{6.022 \times 10^{23}\text{ molecules}} \right) = 4.15\text{ mole CO}_2
  • Aluminum Example:
    • The number of atoms in 2.0 moles2.0\text{ moles} of aluminum Al\text{Al} is calculated as:
    • 2.0 moles×(6.02×1023 atoms1 moles)=1.2×1024 Al atoms2.0\text{ moles} \times \left( \frac{6.02 \times 10^{23}\text{ atoms}}{1\text{ moles}} \right) = 1.2 \times 10^{24}\text{ Al atoms}

Chemical Formulas and Subscripts

  • Subscripts in a chemical formula represent two distinct relationships:
    • The relationship of individual atoms within a single formula unit or molecule.
    • The ratio of moles of each element per one mole of the total compound.
  • Ammonia Example (N1H3\text{N}_1\text{H}_3):
    • Per 1 molecule: 1 atom N1\text{ atom N} and 3 atoms H3\text{ atoms H}.
    • Per 1 mole of NH3\text{NH}_3: 1 mole N1\text{ mole N} and 3 moles H3\text{ moles H}.
  • Aspirin Example (C9H8O4\text{C}_9\text{H}_8\text{O}_4):
    • Aspirin is used to reduce pain and inflammation.
    • The subscripts express the ratio of atoms or moles of Carbon, Hydrogen, and Oxygen.
    • Conversion factors can be derived directly from subscripts: 1 mole aspirin=4 moles oxygen1\text{ mole aspirin} = 4\text{ moles oxygen}.
  • Calculating Oxygen Moles in Aspirin:
    • For 2.50 moles2.50\text{ moles} of aspirin:
    • 2.50 moles aspirin×(4 mole oxygen1 mole aspirin)=10.0 moles Oxygen2.50\text{ moles aspirin} \times \left( \frac{4\text{ mole oxygen}}{1\text{ mole aspirin}} \right) = 10.0\text{ moles Oxygen}
  • Molar Ratios in Sodium Sulfate (Na2SO4\text{Na}_2\text{SO}_4):
    • To find moles of Na\text{Na} in 3.5 moles3.5\text{ moles} of Na2SO4\text{Na}_2\text{SO}_4:
    • 3.5 mol Na2SO4×(2 mol Na1 mol Na2SO4)=7.0 moles Na3.5\text{ mol Na}_2\text{SO}_4 \times \left( \frac{2\text{ mol Na}}{1\text{ mol Na}_2\text{SO}_4} \right) = 7.0\text{ moles Na}

Formula Weight (Molar Mass)

  • Formula Weight is defined as:
    • The quantity in grams equal to the atomic mass of an element.
    • The mass of one mole of an element or compound.
    • The mass of 6.022×10236.022 \times 10^{23} atoms, molecules, or formula units.
  • This mass can be located on the periodic table for elements.
  • Calculating Formula Weight for Compounds:
    • Multiply the molar mass of every element by its subscript.
    • Sum the results and round to 0.01 g0.01\,g.
  • Formula Weight of Ammonia (NH3\text{NH}_3):
    • 1 mole N=1×14.01 g=14.01 g1\text{ mole N} = 1 \times 14.01\,g = 14.01\,g
    • 3 mole H=3×1.01 g=3.03 g3\text{ mole H} = 3 \times 1.01\,g = 3.03\,g
    • Total FW=17.04 g mol−1\text{Total FW} = 17.04\,g\,mol^{-1}
  • Formula Weight of Lithium Carbonate (Li2CO3\text{Li}_2\text{CO}_3):
    • 2 mol Li×6.94=13.88 g2\text{ mol Li} \times 6.94 = 13.88\,g
    • 1 mol C×12.01=12.01 g1\text{ mol C} \times 12.01 = 12.01\,g
    • 3 mol O×16.00=48.00 g3\text{ mol O} \times 16.00 = 48.00\,g
    • Total FW=73.89 g mol−1\text{Total FW} = 73.89\,g\,mol^{-1}
  • Formula Weight of Potassium Chloride (KCl\text{KCl}):
    • 1 mol K×39.10=39.10 g1\text{ mol K} \times 39.10 = 39.10\,g
    • 1 mol Cl×35.45=35.45 g1\text{ mol Cl} \times 35.45 = 35.45\,g
    • Total FW=74.55 g mol−1\text{Total FW} = 74.55\,g\,mol^{-1}
  • Formula Weight of Calcium Nitrate (Ca(NO3)2\text{Ca(NO}_3)_2):
    • 1 Ca×40=401\text{ Ca} \times 40 = 40
    • 2 N×14=282\text{ N} \times 14 = 28
    • 6 O×16=966\text{ O} \times 16 = 96
    • Total FW=164 g mol−1\text{Total FW} = 164\,g\,mol^{-1}
  • Representative One-Mole Quantities:
    • Sulfur (S\text{S}): 32.07 g32.07\,g
    • Iron (Fe\text{Fe}): 55.85 g55.85\,g
    • Salt (NaCl\text{NaCl}): 58.44 g58.44\,g
    • Potassium dichromate (K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7): 294.2 g294.2\,g
    • Sucrose (C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11}): 342.3 g342.3\,g

Converisons Between Grams, Moles, and Particles

  • The Formula Weight (Molar Mass) acts as a conversion factor between mass in grams and amount in moles.
  • Converting Moles to Grams:
    • For silver (Ag\text{Ag}), 1 mole=107.9 g1\text{ mole} = 107.9\,g.
    • Jewelry requiring 0.750 moles Ag0.750\text{ moles Ag}
    • 0.750 moles Ag×(107.9 grams Ag1 mole Ag)=80.9 grams Ag0.750\text{ moles Ag} \times \left( \frac{107.9\text{ grams Ag}}{1\text{ mole Ag}} \right) = 80.9\text{ grams Ag}
  • Converting Grams to Moles:
    • Salt (NaCl\text{NaCl}) formula weight: 1 Na (22.99)+1 Cl (35.45)=58.44 g1\text{ Na }(22.99) + 1\text{ Cl }(35.45) = 58.44\,g.
    • A box containing 737 g NaCl737\,g\text{ NaCl}
    • 737 g NaCl×(1 mole NaCl58.44 g NaCl)=12.6 moles NaCl737\,g\text{ NaCl} \times \left( \frac{1\text{ mole NaCl}}{58.44\,g\text{ NaCl}} \right) = 12.6\text{ moles NaCl}
  • Three-Step Conversion (Mass to Particles):
    • Path: Grams→Moles→Atoms/Molecules\text{Grams} \rightarrow \text{Moles} \rightarrow \text{Atoms/Molecules}.
    • Example: A 10.00-lb10.00\text{-lb} bag of table sugar containing 4536 g4536\,g of sucrose (C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11}, FW=342 g/molFW = 342\,g/mol).
    • 4536 g sucr.×(1 mole sucr.342 g sucr.)×(6.02×1023 molec.1 mole sucr.)=7.980×1024 molecules4536\,g\text{ sucr.} \times \left( \frac{1\text{ mole sucr.}}{342\,g\text{ sucr.}} \right) \times \left( \frac{6.02 \times 10^{23}\text{ molec.}}{1\text{ mole sucr.}} \right) = 7.980 \times 10^{24}\text{ molecules}
    • Reverse Example: Find grams in 12.04×1023 molecules12.04 \times 10^{23}\text{ molecules} of sucrose.
    • 12.04×1023 molec.×(1 mole sucr.6.02×1023 molec.)×(342 g sucr.1 mole sucr.)=684.0 grams sucrose12.04 \times 10^{23}\text{ molec.} \times \left( \frac{1\text{ mole sucr.}}{6.02 \times 10^{23}\text{ molec.}} \right) \times \left( \frac{342\,g\text{ sucr.}}{1\text{ mole sucr.}} \right) = 684.0\text{ grams sucrose}

Mass Percent Composition

  • This value represents the mass of a specific element divided by the total mass of the compound, multiplied by 100%100\%.
  • Propyl Acetate (C5H10O2\text{C}_5\text{H}_{10}\text{O}_2):
    • Molar mass = 102.13 g/mol102.13\,g/mol.
    • Percent Composition: 58.80% C+9.870% H+31.33% O=100.00%58.80\%\text{ C} + 9.870\%\text{ H} + 31.33\%\text{ O} = 100.00\%

Empirical and Molecular Formulas

  • Empirical Formula: The simplest whole-number ratio of atoms in a compound.
    • Example: Molecular formula C5H10O5\text{C}_5\text{H}_{10}\text{O}_5 reduces to empirical formula CH2O\text{CH}_2\text{O} by dividing subscripts by 55.
  • Molecular Formula: Represents the actual number of atoms in a compound; it is always a whole-number multiple of the empirical formula.
  • Determining Empirical Formula from Percent Composition:
    1. Assume a 100 g100\,g sample to convert percent to grams.
    2. Convert grams to moles by dividing by atomic mass.
    3. Normalize the values by dividing all mole results by the smallest calculated mole value.
    4. If results are not whole numbers, multiply all values by a small integer to eliminate fractions.
  • Normalization Caveat: Decimals greater than 0.10.1 or less than 0.90.9 should not be rounded off; they must be multiplied by a factor.
  • Example: C-H Compound (74.9% C, 25.1% H):
    • Carbon: 74.9 g÷12=6.24 mol74.9\,g \div 12 = 6.24\,mol. Divide by smaller: 6.24÷6.24=16.24 \div 6.24 = 1.
    • Hydrogen: 25.1 g÷1=25.1 mol25.1\,g \div 1 = 25.1\,mol. Divide by smaller: 25.1÷6.24=4.04≈425.1 \div 6.24 = 4.04 \approx 4.
    • Empirical Formula: C1H4\text{C}_1\text{H}_4.
  • Example: Aspirin (60.0% C, 4.5% H, 35.5% O):
    • Carbon: 60.0 g÷12.01=5.0060.0\,g \div 12.01 = 5.00. Divide: 5.00÷2.22=2.255.00 \div 2.22 = 2.25. Multiply by 4: 99.
    • Hydrogen: 4.5 g÷1.008=4.504.5\,g \div 1.008 = 4.50. Divide: 4.50÷2.22=2.034.50 \div 2.22 = 2.03. Multiply by 4: 88.
    • Oxygen: 35.5 g÷16.00=2.2235.5\,g \div 16.00 = 2.22. Divide: 2.22÷2.22=12.22 \div 2.22 = 1. Multiply by 4: 44.
    • Empirical Formula: C9H8O4\text{C}_9\text{H}_8\text{O}_4.

Molecular Formula from Empirical Formula and Weight

  • To find the molecular formula:
    1. Calculate the Factor using the actual molecular weight and the empirical formula weight:
    • Small Integer Factor=Actual Formula Weight (Molar Mass)Empirical Formula Weight\text{Small Integer Factor} = \frac{\text{Actual Formula Weight (Molar Mass)}}{\text{Empirical Formula Weight}}
    1. Multiply empirical subscripts by this factor.
  • Ascorbic Acid (Vitamin C):
    • Empirical formula C3H4O3\text{C}_3\text{H}_4\text{O}_3. Actual weight 176.12 g/mol176.12\,g/mol.
    • Multiplying empirical subscripts by 22 yields molecular formula C6H8O6\text{C}_6\text{H}_8\text{O}_6.
  • Nitrogen-Sulfur-Chlorine Compound:
    • Composition: 27.4% S27.4\%\text{ S}, 12.0% N12.0\%\text{ N}, 60.6% Cl60.6\%\text{ Cl}. Actual weight 351 g351\,g.
    • Empirical moles: S=0.854S = 0.854, N=0.857N = 0.857, Cl=1.71Cl = 1.71.
    • Normalized: S=1S=1, N=1N=1, Cl=2Cl=2. Empirical Formula: SNCl2\text{SNCl}_2.
    • Empirical weight: 117117.
    • Factor: 351÷117=3351 \div 117 = 3.
    • Molecular Formula: S3N3Cl6\text{S}_3\text{N}_3\text{Cl}_6.
  • CHCl Compound:
    • Composition: 37.2% C37.2\%\text{ C}, 7.8% H7.8\%\text{ H}, 55.0% Cl55.0\%\text{ Cl}. Actual weight 516 g516\,g.
    • Empirical calculations yield C2H5Cl\text{C}_2\text{H}_5\text{Cl}.
    • Empirical weight: C(24)+H(5)+Cl(35.5)=64.5\text{C}(24) + \text{H}(5) + \text{Cl}(35.5) = 64.5.
    • Factor: 516÷64.5=8516 \div 64.5 = 8.
    • Molecular Formula: C16H40Cl8\text{C}_{16}\text{H}_{40}\text{Cl}_8.