1 / 12

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!).

tuvya
Download Presentation

Zumdahl’s Chapter 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related