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Slide 1 ___________________________________ ___________________________________ ___________________________________ Stoichiometry ___________________________________ Chapter 3 ___________________________________ ___________________________________ ___________________________________ Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slide 2 ___________________________________ Micro World atoms & molecules Macro World grams ___________________________________ ___________________________________ Atomic mass is the mass of an atom in atomic mass units (amu) ___________________________________ By definition: 1 atom 12C “weighs” 12 amu ___________________________________ ___________________________________ On this scale 1H = 1.008 amu 16O = 16.00 amu ___________________________________ 2 Slide 3 ___________________________________ The average atomic mass is the weighted average of all of the naturally occurring ___________________________________ isotopes of the element. ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ 3 Slide ___________________________________ 4 ___________________________________ Naturally occurring lithium is: 7.42% 6Li (6.015 amu) ___________________________________ 92.58% 7Li (7.016 amu) ___________________________________ ___________________________________ Average atomic mass of lithium: ___________________________________ (7.42 x 6.015) + (92.58 x 7.016) = 6.941 amu 100 ___________________________________ 4 Slide ___________________________________ 5 ___________________________________ Average atomic mass (6.941) ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ 5 Slide 6 ___________________________________ The Mole (mol): A unit to count numbers of particles ___________________________________ Dozen = 12 ___________________________________ ___________________________________ Pair = 2 ___________________________________ The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C ___________________________________ 1 mol = NA = 6.0221367 x 1023 ___________________________________ Avogadro’s number (NA) 6 Slide 7 eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu ___________________________________ ___________________________________ ___________________________________ ___________________________________ 1 mole 12C atoms = 12.00 g 12C ___________________________________ 1 mole lithium atoms = 6.941 g of Li ___________________________________ For any element ___________________________________ atomic mass (amu) = molar mass (grams) 7 Slide ___________________________________ One Mole of: 8 ___________________________________ S C ___________________________________ ___________________________________ ___________________________________ Hg ___________________________________ ___________________________________ Cu Fe 8 Slide 9 ___________________________________ 1 12C atom 12.00 g 1.66 x 10-24 g x = 12.00 amu 6.022 x 1023 12C atoms 1 amu ___________________________________ ___________________________________ 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu ___________________________________ ___________________________________ ___________________________________ M = molar mass in g/mol ___________________________________ NA = Avogadro’s number 9 Slide ___________________________________ 10 How many atoms are in 0.551 g of potassium (K) ? ___________________________________ ___________________________________ 1 mol K = 39.10 g K ___________________________________ 1 mol K = 6.022 x 1023 atoms K 1 mol K 6.022 x 1023 atoms K 0.551 g K x x = 1 mol K 39.10 g K 8.49 x 1021 atoms K ___________________________________ ___________________________________ ___________________________________ 10 Slide 11 ___________________________________ Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1S SO2 2O SO2 ___________________________________ 32.07 amu ___________________________________ + 2 x 16.00 amu 64.07 amu ___________________________________ ___________________________________ For any molecule molecular mass (amu) = molar mass (grams) ___________________________________ 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 ___________________________________ 11 Slide ___________________________________ 12 ___________________________________ How many H atoms are in 72.5 g of C3H8O ? ___________________________________ 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O ___________________________________ 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 72.5 g C3H8O x ___________________________________ 1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms x x = 1 mol C3H8O 1 mol H atoms 60 g C3H8O 5.82 x 1024 atoms H ___________________________________ ___________________________________ 12 Slide 13 ___________________________________ Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na NaCl ___________________________________ 22.99 amu ___________________________________ 1Cl + 35.45 amu NaCl 58.44 amu ___________________________________ ___________________________________ For any ionic compound formula mass (amu) = molar mass (grams) ___________________________________ 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl ___________________________________ 13 Slide 14 ___________________________________ What is the formula mass of Ca3(PO4)2 ? ___________________________________ ___________________________________ 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2P 8O 2 x 30.97 ___________________________________ ___________________________________ + 8 x 16.00 310.18 amu ___________________________________ ___________________________________ Mass Spectrometer 15 ___________________________________ Heavy Slide Light 14 ___________________________________ ___________________________________ Light ___________________________________ Heavy ___________________________________ ___________________________________ Mass Spectrum of Ne ___________________________________ 15 Slide 16 ___________________________________ Percent composition of an element in a compound = ___________________________________ n x molar mass of element x 100% molar mass of compound ___________________________________ n is the number of moles of the element in 1 mole of the compound 2 x (12.01 g) x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g %C = C2H6O 52.14% + 13.13% + 34.73% = 100.00% ___________________________________ ___________________________________ ___________________________________ ___________________________________ 16 Slide 17 ___________________________________ Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. nK = 24.75 g K x nMn = 34.77 g Mn x 1 mol K = 0.6330 mol K 39.10 g K 1 mol Mn = 0.6329 mol Mn 54.94 g Mn nO = 40.51 g O x 1 mol O = 2.532 mol O 16.00 g O ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ 17 Slide ___________________________________ Percent Composition and Empirical Formulas 18 nK = 0.6330, nMn = 0.6329, nO = 2.532 ___________________________________ ___________________________________ 0.6330 ~ K: ~ 1.0 0.6329 ___________________________________ 0.6329 = 1.0 Mn : 0.6329 ___________________________________ 2.532 ~ O: ~ 4.0 0.6329 ___________________________________ KMnO4 ___________________________________ 18 Slide ___________________________________ 19 ___________________________________ ___________________________________ Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O ___________________________________ g CO2 mol CO2 mol C gC 6.0 g C = 0.5 mol C g H2O mol H2O mol H gH 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Slide 20 ___________________________________ ___________________________________ Divide by smallest subscript (0.25) Empirical formula C2H6O ___________________________________ 19 ___________________________________ How to “Read” Chemical Equations ___________________________________ 2 Mg + O2 2 MgO ___________________________________ 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO ___________________________________ 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO NOT ___________________________________ ___________________________________ 2 grams Mg + 1 gram O2 makes 2 g MgO ___________________________________ 20 Slide 21 ___________________________________ Amounts of Reactants and Products ___________________________________ ___________________________________ ___________________________________ ___________________________________ 1. Write balanced chemical equation ___________________________________ 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units 21 ___________________________________ Slide 22 ___________________________________ Methanol burns in air according to the equation ___________________________________ 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH molar mass CH3OH 209 g CH3OH x moles H2O ___________________________________ grams H2O molar mass coefficients H2O chemical equation ___________________________________ ___________________________________ 4 mol H2O 18.0 g H2O 1 mol CH3OH = x x 32.0 g CH3OH 2 mol CH3OH 1 mol H2O ___________________________________ ___________________________________ 235 g H2O 22 Slide 23 ___________________________________ Limiting Reagent: ___________________________________ Reactant used up first in the reaction. ___________________________________ 2NO + O2 2NO2 ___________________________________ NO is the limiting reagent ___________________________________ O2 is the excess reagent ___________________________________ ___________________________________ 23 Slide 24 ___________________________________ In one process, 124 g of Al are reacted with 601 g of Fe 2O3 2Al + Fe2O3 ___________________________________ Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al g Fe2O3 124 g Al x mol Al mol Fe2O3 needed x g Fe2O3 needed OR mol Al needed mol Fe2O3 1 mol Al 27.0 g Al ___________________________________ 1 mol Fe2O3 2 mol Al Start with 124 g Al x 160. g Fe2O3 = 1 mol Fe2O3 g Al needed 367 g Fe2O3 ___________________________________ ___________________________________ ___________________________________ need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent ___________________________________ 24 Slide 25 ___________________________________ Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 2Al + Fe2O3 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al g Al2O3 ___________________________________ ___________________________________ Al2O3 + 2Fe x 102. g Al2O3 = 1 mol Al2O3 ___________________________________ 234 g Al2O3 ___________________________________ At this point, all the Al is consumed and Fe2O3 remains in excess. ___________________________________ ___________________________________ 25 Slide 26 ___________________________________ Reaction Yield ___________________________________ Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. ___________________________________ ___________________________________ Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield ___________________________________ ___________________________________ x 100% Theoretical Yield ___________________________________ 26 Slide 27 ___________________________________ CH-3 (part) ___________________________________ ___________________________________ Questions and Problems Page 88 - 93 ___________________________________ 3.6, 3.8, 3.14, 3.16, 3.18, 3.20, 3.22, 3.24, 3.26, 3.28, 3.33, 3.40, 3.42, 3.44, 3.46, 3.48, 3.50, 3.52, 3.54, 3.68, 3.70, 3.72, ___________________________________ ___________________________________ ___________________________________ 27