(L10)stoichiometry part1
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    (L10)stoichiometry part1 (L10)stoichiometry part1 Presentation Transcript

    •  The relationship between the number of reactant and product molecules in a chemical equation.  Quantitative
    • 3 Atomic Weights Weighted average of the masses of the constituent isotopes if an element.  Tells us the atomic masses of every known element.  Lower number on periodic chart. How do we know what the values of these numbers are?
    • 4 The Mole A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of things  Just like a dozen = 12 things  One mole = 6.022 x 1023 things Avogadro’s number = 6.022 x 1023  Symbol for Avogadro’s number is NA.
    • 5 The Mole How do we know when we have a mole?  count it out  weigh it out Molar mass - mass in grams numerically equal to the atomic weight of the element in grams. H has an atomic weight of 1.00794 g  1.00794 g of H atoms = 6.022 x 1023 H atoms Mg has an atomic weight of 24.3050 g  24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
    • 6 The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. Mgg?
    • 7 The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. atomMg1Mgg?
    • 8 The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. atomsMg106.022 atomsMgmol1 atomMg1Mgg? 23
    • 9 The Mole Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures. Mgg104.04 atomsMgmol1 Mgg24.30 atomsMg106.022 atomsMgmol1 atomMg1Mgg? 23 23
    • 10 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? atomsMg?
    • 11 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? Mgmol1.67atomsMg?
    • 12 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? Mgmol1 atomsMg106.022 Mgmol1.67atomsMg? 23
    • 13 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? atomsMg101.00 Mgmol1 atomsMg106.022 Mgmol1.67atomsMg? 24 23
    • 14 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? atomsMg101.00 Mgmol1 atomsMg106.022 Mgmol1.67atomsMg? 24 23
    • 15 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? You do it!
    • 16 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? Mgg4.73Mgmol?
    • 17 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? Mgg24.30 atomsMgmol1 Mgg4.73Mgmol?
    • 18 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? Mgmol02.3 Mgg24.30 atomsMgmol1 Mgg4.73Mgmol? IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS
    • 19 Formula Weights, Molecular Weights, and Moles How do we calculate the molar mass of a compound?  add atomic weights of each atom The molar mass of propane, C3H8, is: amu44.11massMolar amu8.08amu1.018H8 amu36.03amu12.013C3
    • 20 Formula Weights, Molecular Weights, and Moles The molar mass of calcium nitrate, Ca(NO3)2 , is: You do it!
    • 21 Formula Weights, Molecular Weights, and Moles amu164.10massMolar amu96.00amu16.006O6 amu28.02amu14.012N2 amu40.08amu40.081Ca1
    • 22
    • 23 Formula Weights, Molecular Weights, and Moles One Mole of Contains  Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms  C3H8 You do it!
    • 24 Formula Weights, Molecular Weights, and Moles One Mole of Contains  Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms  C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms
    • 25 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 8383 HCg74.6moleculesHC?
    • 26 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 83 83 8383 HCg44.11 HCmole1 HCg74.6moleculesHC?
    • 27 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 83 83 23 83 83 8383 HCg44.11 moleculesHC106.022 HCg44.11 HCmole1 HCg74.6moleculesHC?
    • 28 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. molecules1002.1 HCg44.11 moleculesHC106.022 HCg44.11 HCmole1 HCg74.6moleculesHC? 24 83 83 23 83 83 8383
    • 29 Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. You do it!
    • 30 Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. atomsO106.49 COLiunitformula1 atomsO3 COLimol1 COLiunitsform.106.022 COLig73.8 COLimol1 COLig26.5atomsO? 23 3232 32 23 32 32 32
    • 31 Formula Weights, Molecular Weights, and Moles Occasionally, we will use millimoles.  Symbol - mmol  1000 mmol = 1 mol For example: oxalic acid (COOH)2  1 mol = 90.04 g  1 mmol = 0.09004 g or 90.04 mg
    • 32 Formula Weights, Molecular Weights, and Moles Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. You do it!
    • 33 Formula Weights, Molecular Weights, and Moles Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. 2 2 2 22 (COOH)mmol2.60 (COOH)g0.09004 (COOH)mmol1 (COOH)g234.0(COOH)mmol?
    • 34 Percent Composition and Formulas of Compounds % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C3H8. 81.68% 100% g44.11 g12.013 100% HCmass Cmass C% 83
    • 35 Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? You do it!
    • 36 Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? 81.68%100%18.32% or %18.32100% g44.11 g1.018 100% HC H8 100% HCmass Hmass H% 83 83
    • 37 Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures. You do it!
    • 38 Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 sig. fig. 100%Total O48.0%100% g399.9 g16.012 100% )(SOFe O12 O% S24.1%100% g399.9 g32.13 100% )(SOFe S3 S% Fe27.9%100% g399.9 g55.82 100% )(SOFe Fe2 Fe% 342 342 342
    • 39 Derivation of Formulas from Elemental Composition Empirical Formula - smallest whole-number ratio of atoms present in a compound  CH2 is the empirical formula for alkenes  No alkene exists that has 1 C and 2 H’s Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound  Ethene – C2H4  Pentene – C5H10 We determine the empirical and molecular formulas of a compound from the percent composition of the compound.  percent composition is determined experimentally
    • 40 Derivation of Formulas from Elemental Composition Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are:  24.74 g of K  34.76 g of Mn  40.50 g of O
    • 41 Derivation of Formulas from Elemental Composition Kmol0.6327 Kg39.10 Kmol1 Kg24.74Kmol?
    • 42 Derivation of Formulas from Elemental Composition Mnmol0.6327 Mng54.94 Mnmol1 Mng34.76Mnmol? Kmol0.6327 Kg39.10 Kmol1 Kg24.74Kmol?
    • 43 Derivation of Formulas from Elemental Composition rationumberwholesmallestobtain Omol2.531 Og16.00 Omol1 Og40.50Omol? Mnmol0.6327 Mng54.94 Mnmol1 Mng34.76Mnmol? Kmol0.6327 Kg39.10 Kmol1 Kg24.74Kmol?
    • 44 Derivation of Formulas from Elemental Composition Mn1 0.6327 0.6327 MnforK1 0.6327 0.6327 Kfor rationumberwholesmallestobtain Omol2.531 Og16.00 Omol1 Og40.50Omol? Mnmol0.6327 Mng54.94 Mnmol1 Mng34.76Mnmol? Kmol0.6327 Kg39.10 Kmol1 Kg24.74Kmol?
    • 45 Derivation of Formulas from Elemental Composition 4KMnOisformulachemicalthethus O4 0.6327 2.531 Ofor Mn1 0.6327 0.6327 MnforK1 0.6327 0.6327 Kfor rationumberwholesmallestobtain Omol2.531 Og16.00 Omol1 Og40.50Omol? Mnmol0.6327 Mng54.94 Mnmol1 Mng34.76Mnmol? Kmol0.6327 Kg39.10 Kmol1 Kg24.74Kmol?
    • 46 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? You do it!
    • 47 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? rationumberwholesmallestfind Omol0.1480 Og16.00 Omol1 Og2.368Omol? Comol0.1110 Cog58.93 Comol1 Cog6.541Comol?
    • 48 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? 43OCo :isformulascompound'theThus O43O1.333Co33Co1 numberwholetofractionturnto3bybothmultipy O1.333 0.1110 0.1480 OforCo1 0.1110 0.1110 Cofor
    • 49 Determination of Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula?  short cut method Hofg8.100.1437g56.1 Cofg48.00.8563g56.1 His14.37%andCis85.63% g56.1containsmol1
    • 50 Determination of Molecular Formulas 84HC :isformulatheThus Hmol8 Hg1.01 Hmol1 Hofg8.10 Cmol4 Cg12.0 Cmol1 Cofg48.0 molestomassesconvert
    • 51 Some Other Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? Nmol1.07 Ng14.0 Nmol1 Nofg15.0Nmol? g/mol149.0PO)(NHofmassmolar 434
    • 52 Some Other Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? 434 434 434 PO)(NHmol0.357 Nmol3 PO)(NHmol1 Nmol1.07 Nmol1.07 Ng14.0 Nmol1 Nofg15.0Nmol? g/mol149.0PO)(NHofmassmolar
    • 53 Some Other Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? 434 434 434 434 434 434 434 PO)(NHg53.2 PO)(NHmol1 PO)(NHg149.0 PO)(NHmol0.357 PO)(NHmol0.357 Nmol3 PO)(NHmol1 Nmol1.07 Nmol1.07 Ng14.0 Nmol1 Nofg15.0Nmol? g/mol149.0PO)(NHofmassmolar
    • 54 Purity of Samples The percent purity of a sample of a substance is always represented as impuritiesincludessampleofmass %100 sampleofmass substancepureofmass =purity%
    • 55 Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? sampleg100.0 PONag98.3 factorunit 43
    • 56 Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? 43 43 43 43 PONag246= sampleg100.0 PONag98.3 sampleg0.250PONag? sampleg100.0 PONag98.3 factorunit
    • 57 Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? impuritiesg4.00= PONag246-sampleg250.0=impuritiesg? PONag246= sampleg100.0 PONag98.3 sampleg0.250PONag? sampleg100.0 PONag98.3 factorunit 43 43 43 43 43