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Zumdahl’s Chapter 3

Zumdahl’s Chapter 3. Stoichiometry. Aston’s Atomic Mass The Mole Molar Mass % Composition Molecular Formulae Chemical Equations. Balancing Chemical Equations Stoichiometric Calculations Limiting Reactant Calculations. Chapter Contents. Aston’s Atomic Masses (works for molecules too!).

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Zumdahl’s Chapter 3

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  1. Zumdahl’s Chapter 3 Stoichiometry

  2. Aston’s Atomic Mass The Mole Molar Mass % Composition Molecular Formulae Chemical Equations Balancing Chemical Equations Stoichiometric Calculations Limiting Reactant Calculations Chapter Contents

  3. Aston’s Atomic Masses(works for molecules too!) • Mass Spectrometer • Fast atomic ions (current) bend near magnet • Deflection varies inversely with inertia! • Multiple isotopes differ in mass (inertia) and so give multiple beam deflections. • 98.892 % 12C at 12 amu by definition. • 1.108 % 13C at 13.00338 amu • averageC  12.011 atomic mass units

  4. Avogadro’s Mole • Definition: One mole of 12C atoms weighs exactly 0.012 kg (12 g) • Thus, 1 amu  1 g / NAv • Since atoms combine by numbers, NAv has the advantage of showing combinations by weights. • NAv = 6.02213671023, the SI count

  5. Molar Masses & % Composition • Trivial; weigh them in mass spectrometer? • Since atomic masses of elements never vary, MW = sum of atomic weights times number of atoms in molecule (subscripts). • MW(P4O10) = 4(AWP) + 10(AWO) = 4(30.97) + 10(16.00) = 283.9 g mol–1 • % O is 100%  160.0/283.9 = 56.36%

  6. Formula Weight Analysis • Atomic Absorption Spectrometry (AA) • Intensity of atomic glows in controlled flame gives proportion of atom in molecule. • Organic combustion analysis for CxHyOz • Burn in excess oxygen, O2 • Trap and weigh resultant H2O and CO2 • Convert weights to moles H and C, resp. • Get mass O by difference with original mass • Scale moles to find simplest integer x, y, & z.

  7. Empirical formula from elemental composition must be scaled by ratio of MW:FW to obtain molecular formula. Standard formulae show combinations: Hg6(PO4)2 (why OK?) Structural formula give geometric information as well: ClH2CCH2COOH, 3-chloropropanoic acid, & digests to C3H5O2Cl Mineral formulae show co•crystalites like Y2(CO3)3•3H2O Molecular Formulae

  8. Chemical Equations • How many reactants  how many products? • Chemical equations not only codify the perfect proportions but also note conditions: • CaCO3(s) + 2 HCl(aq)  Ca2+(aq) + 2 Cl–(aq) . + CO2(g) + H2O(l) • Catalysts, photons (h), or heat () may stand above the reaction arrow (). • The key is molecular consumption/production.

  9. Balancing Chemical Reactions • First know all the reactants and products! • Balance first atoms appearing in only one molecule on each side. Let’s burn TNT: • C7H5(NO2)3 + O2 7CO2 + H2O + N2 • C7H5(NO2)3 + O2 7CO2 + H2O + 1.5N2 • C7H5(NO2)3 + O2 7CO2 + 2.5H2O + 1.5N2 • C7H5(NO2)3 + 5.25O2 7CO2 + 2.5H2O + 1.5N2 • 4C7H5(NO2)3 + 21O2 28CO2 + 10H2O + 6N2

  10. Stoichiometric Calculations • While TNT is solid, the O2 and the products are gases. 1 mol TNT is worth ? moles gas? • 4C7H5(NO2)3 + 21O2 28CO2 + 10H2O + 6N2 •  Moles gas = ¼(28 + 10 + 6 – 21) = 5.75 • Since each gas mole is ~1000 the volume of a solid mole, TNT is destructive by rapidly increasing its volume by a factor of 5,750!

  11. More Calculations • A “megaton” of TNT (measure of nuclear bomb destructivity) would produce what weight of CO2? • 106 tons [ 2000 lbs / ton ] [ 0.4536 kg / lbs ] = 9.072108 kg • MW[C7H5(NO2)3] = 7(12)+5(1)+6(16)+3(14) = 227 g = 0.227 kg • 4C7H6(NO2)3 + 21O2 28CO2 + 10H2O + 6N2 • 9.072  108 kg TNT  [1 mol TNT/0.227 kg TNT]  [28 mol CO2/4 mol TNT]  [0.044 kg CO2/1 mol CO2] • 1.23  109 kg CO2 = 1.23 megatons CO2

  12. Limiting Reactant Calculations • Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l) • What weight of molten iron is produced by 1 kg each of the reactants? • Needed weights are MW(Fe2O3) and 2AW(Al) or 159.7 g Fe2O3 and 53.96 g Al, respectively. • Smallest ratio Available:Needed is limiting! • 1000 g/159.7 g = 6.26 Fe2O3< 1000 g/53.96 g = 18.52 Al • 6.26 mol Fe2O3 [2 mol Fe/1 mol Fe2O3]  [55.85 g Fe/1 mol Fe] = 699 g Fe

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