NWTC General Chemistry Ch 07

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NWTC General Chemistry Ch 07

  1. 1. Chapter 7 Quantitative Composition ofThese black Compoundspearls are madeof layers ofcalciumcarbonate.They can bemeasured bycounting orweighing. Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena
  2. 2. Chapter Outline7.1 The Mole 7.5 Calculating Empirical7.2 Molar Mass of Compounds Formulas7.3 Percent Composition of 7.6 Calculating the Molecular Compounds Formula from the Empirical Formula7.4 Empirical Formula versus Molecular Formula Copyright 2012 John Wiley & Sons, Inc
  3. 3. Convenient Ways of Counting Items • 1 dozen eggs = 12 eggs • 1 ream paper = 500 sheets • 1 gross pencils = 144 pencils 1×10 -10 m How do chemists count really small things, like atoms? 1 mole atoms =Review Question 1 6.022×1023 atoms This number is known as Avogadro’s number. 2012 John Wiley & Sons, Inc Copyright
  4. 4. The MoleReview Question 7 & 8 1 mole of anything contains Avogadro’s number (6.022×1023 ) of particles. • 1 mol atoms = 6.022×1023 atoms • 1 mol molecules = 6.022×1023 molecules • 1 mol ions = 6.022×1023 ions Avogadro’s number can be used as a conversion factor: 6.022 1023 particles 1 mol 1 mol 6.022 1023 particles Copyright 2012 John Wiley & Sons, Inc
  5. 5. The MoleCalculate the number of atoms in 2.4 mol Na. Plan 2.5 mol Na  Na atoms 6.022 x 1023 Na atoms 1 mol Na 1 mol Na 6.022 x 1023 Na atoms Calculate 6.022 x 1023 Na atoms 2.4 mol Na = 1.4 1024 Na atoms 1 mol Na Copyright 2012 John Wiley & Sons, Inc
  6. 6. Your Turn!How many moles of HCl are present in 4.3×1023 molecules?a. 2.6×1047 molb. 0.71 molc. 2.6 mold. 7.1×1045 mol 4.3 x10 23 molecules 1 mol = 0.71 mol 6.022 x10 23 molecules Copyright 2012 John Wiley & Sons, Inc
  7. 7. Convenient Ways of CountingIt is often convenient to count using mass.For example, a canning recipe calls for 150 apples to be peeled and cored.The average mass of an individual apple is 235 g. How many kg are needed to complete this recipe? 235 g 1 kg 150 apples × = 35 kg 1 apple 1000 gThe canner should buy 35 kg of apple. Copyright 2012 John Wiley & Sons, Inc
  8. 8. Counting Atoms with Mass • Chemists count atoms by using mass since individual atoms are too small to count. • Average mass of an atom = atomic mass in amu • Average mass of 1 mole of atoms = atomic mass expressed in grams (molar mass).Review Question 5 1 mol atoms = atomic mass in grams 1 mol atoms = 6.022×1023 atoms atomic mass in grams = 6.022×1023 atoms Copyright 2012 John Wiley & Sons, Inc
  9. 9. The MoleCopyright 2012 John Wiley & Sons, Inc
  10. 10. Your Turn!Which of these is not correct? 6a. The mass of 1 atom of C is 12.01 amu. Cb. The mass of 1 mole of C atoms is 12.01 g. 12.01c. Avogadro’s number of atoms has a mass of 12.01 amu.d. Avogadro’s number of atoms has a mass of 12.01 g. Copyright 2012 John Wiley & Sons, Inc
  11. 11. The MoleCalculate the mass of 2.4 mol C. atomic mass Plan 2.4 mol C  g C 12.01 Molar mass is a conversion factor: 12.01g C 1 mol C 1 mol C 12.01g C Calculate 12.01g C 2.4 mol C = 29 g C 1 mol C Copyright 2012 John Wiley & Sons, Inc
  12. 12. Your Turn!What is the correct set up to calculate number of moles of atoms contained in 3.52 g Al? atomic massa. 1 mol Al Al 26.98 3.52g Al 26.98 g Alb. 26.98g Al 3.52g Al 1 mol Alc. 6.022 x 10 23 Al atoms 3.52g Al 1 g Al Copyright 2012 John Wiley & Sons, Inc
  13. 13. The MoleCalculate the number of atoms in 36.0 g C. atomic massPlan 36.0 g C  moles C  atoms C C 12.01 1 mol C 6.022 1023 atoms C 12.01 g C 1 mol C 1 mol C 6.022 1023 atoms CCalculate 36.0 g C 12.01 g C 1 mol C = 1.8 1024 atoms C Copyright 2012 John Wiley & Sons, Inc
  14. 14. Your Turn!What is the mass of 3.01 ×1023 atoms of lead?a. 104 g atomic massb. 414 g Pb 207.2c. 0.500 gd. 1.04×1048 g3.01 x10 23 atoms 1 mol 207.2 gram = 104 gram 6.022 x10 23 atoms 1 mol Copyright 2012 John Wiley & Sons, Inc
  15. 15. Your Turn! 1 gram of which of the following elements would contain the largest number of atoms? atomic mass a. nitrogenReview Question 3 N 14.01 H 1.01 b. hydrogen P 30.97 c. phosphorus O 16.00 d. oxygen 1 gram H 1 mol 6.022 x10 23 atom = 5.96 x10 23 atoms 1.01 g 1 mol 1 gram P 1 mol 6.022 x10 23 atom = 1.94 x10 22 atoms 30.97 g 1 mol Copyright 2012 John Wiley & Sons, Inc
  16. 16. Your Turn! Which has a higher mass, mol of K or mol of Au? a. potassiumReview Question 2 atomic mass b. gold K 39.0983 Au 196.96654 1 Mol of K 39.0983 gram = 39.0983 gram 1 mol 1 Mol of Au 196.9665 gram = 196.9665 gram 1 mol Copyright 2012 John Wiley & Sons, Inc
  17. 17. Your Turn! Which has more electrons, mol of K or mol of Au? a. potassiumReview Question 4 atomic number b. gold K 19 Au 79 1 Mol K 6.022 x10 23 atom 19 e- = 1.144 x10 25 E- 1 mol 1 atom 1 Mole Au 6.022 x10 23 atom 79 e- = 4.757 x10 25 E- 1 mol atom Copyright 2012 John Wiley & Sons, Inc
  18. 18. Review Question 9 a. Mole of O atoms = ? Atoms 6.022 x10 23 b. Mole of O2 molecules = ? Molecules 6.022 x10 23 c. Mole of O2 molecules = ? atoms 1.204 x10 24 d. Mole of O atoms= ? grams 15.99 g e. Mole of O2 molecules = ? grams 31.98 g 1 Mole of O 15.99 gram = 15.99 gram 1 mole 1 Mole O 2 2 Mole of O 15.99 gram = 31.98 gram 1 Mole O 2 1 moleReview Question 9 Copyright 2012 John Wiley & Sons, Inc
  19. 19. Molar Mass of Compounds1 mol compound = 6.022×1023 formula units compoundMolar mass of compound = mass of 1 mol compoundThe molar mass of a compound is the sum of the atomic masses of each atom in the compound.What is the molar mass of CO2? atomic mass 1C 1(12.01g) C 12.01 2O 2(16.00g) O 16.00 CO2 44.01g/mol Copyright 2012 John Wiley & Sons, Inc
  20. 20. Molar Mass of CompoundsWhat is the molar mass of Al2(CO3)3? 2Al 2(26.98g) atomic mass 3C 3(12.01g) Al 26.98 C 12.01 9O 9(16.00g) O 16.00 Al2(CO3)3 233.99g/mol Copyright 2012 John Wiley & Sons, Inc
  21. 21. Your Turn!What is the molar mass of (NH4)3PO4?a. 141.04g/mol atomic massb. 144.07g/mol N 14.01 H 1.01c. 146.09g/mol P 30.97d. 149.12g/mol O 16.00 3*1 N 3(14.01g) 3*4 H 12(1.01g) 1P 1(30.97g) 4O 4(16.00g) Copyright 2012 John Wiley & Sons, Inc
  22. 22. Your Turn!What is the molar mass of Mg(ClO4)2?a. 301.01g/mol atomic massb. 191.21g/mol Mg 24.31 Cl 35.45c. 223.21g/mol O 16.00d. 123.76g/mol __ Mg __(____g) __ Cl __(____g) __ O __(____g) Copyright 2012 John Wiley & Sons, Inc
  23. 23. Using Molar Masses of Compounds Molar mass = 1 mol = 6.022×1023 formula unitsCalculate the mass of 0.150 mol Mg(ClO4)2 Plan 0.150 mol Mg(ClO4)2  g Mg(ClO4)2 1 mol Mg(ClO 4 ) 2 223.21 g Mg(ClO 4 ) 2 223.21 g Mg(ClO 4 ) 2 1 mol Mg(ClO 4 ) 2 Calculate 223.21 g Mg(ClO 4 ) 20.150 mol Mg(ClO4 ) 2 = 33.5 g Mg(ClO 4 ) 2 1 mol Mg(ClO 4 ) 2 Copyright 2012 John Wiley & Sons, Inc
  24. 24. Using Molar Masses of Compounds Molar mass = 1 mol = 6.022×1023 formula unitsCalculate the number of moles in 35 g H2O. Plan 35 g H2O  moles H2O atomic mass 1 mol H 2 O H 1.01 18.02 g H 2 O O 16.00 18.02 g H 2 O 1 mol H 2 O 1 mol H 2 O Calculate 35 g H 2 O = 1.9 g H 2 O 18.02 g H 2O Copyright 2012 John Wiley & Sons, Inc
  25. 25. Using Molar Masses of Compounds Molar mass = 1 mol = 6.022×1023 formula unitsCalculate the number of molecules in 35 g H2O. Plan 35 g H2O  mol H2O molecules H2O 18.02 g H 2 O 6.022 1023 molecules H2O 1 mol H 2 O 1 mol H2O Calculate 1 mol H 2O 6.022 1023 molecules H 2O 35.0 g H 2O 18.02 g H 2O 1 mol H 2O = 1.2 10 24 molecules H 2 O Copyright 2012 John Wiley & Sons, Inc
  26. 26. Preview Question 10How many molecules are present in 1 molar mass of sulfuric acid (H2SO4)? atomic mass1. What is the molar mass of H2SO4? S 32.06 2 H 2(1.01g) H 1.01 1 S 1(32.06g) O 16.00 4 O 4(16.00g) = 98.08g/mol2. How many atoms in one molecule? 2+1+4=7 How many atoms in 1 molar mass? ___98.08 g 1 mol 6.022 x10 23 molecules = 6.022 x10 23 molecules 98.08 g 1 mol Copyright 2012 John Wiley & Sons, Inc
  27. 27. Your Turn!How many moles of molecules are present in 146 g of glucose (C6H12O6)? atomic massa. 180. mol C 12.01 H 1.01b. 0.810 mol __ C __(____g) O 16.00c. 26300 mol __ H __(____g)d. 4.88×1023 mol __ O __(____g) = _____ g/mol g = molecules Copyright 2012 John Wiley & Sons, Inc
  28. 28. Your Turn!What is the mass of 1.20 ×1023 molecules of CH3OH?a. 161g atomic massb. 38.5g C 12.01 H 1.01c. 32.1g __ C __(____g) O 16.00d. 6.39g __ H __(____g) __ O __(____g) = _____ g/mol molecules = g Copyright 2012 John Wiley & Sons, Inc
  29. 29. Your Turn!How many molecules are present in 4.21 moles of HBr?a. 2.53×1023 molecules atomic massb. 2.53×1024 molecules H 1.01 -24 molecules Br 79.90c. 6.99×10d. 3.97×102 moleculese. 6.99×1024 molecules Copyright 2012 John Wiley & Sons, Inc
  30. 30. Percent Composition of CompoundsPercent composition is a list of the mass percent of each element in a compound.Na2CO3 is 43.38% Na 11.33% C 45.29% OHow do we calculate the mass percent of Na2CO3? Copyright 2012 John Wiley & Sons, Inc
  31. 31. Calculating Percent CompositionFirst determine the molar mass of Na2CO32(22.99g Na) + 1(12.01g C) + 3(16.00 g O) = 105.99g/mol Na2CO3Then find ratio of the mass of each element to the mass ofthe compound. 2 22.99 g Na x100 43.38% Na 105.99 g Na2CO3 1 12.01 g C x100 11.33% C 105.99 g Na2 CO3 3 16.00 g O x100 45.29% O 105.99 g N a2 CO3
  32. 32. Calculating Percent CompositionA compound is found to consist of 2.74g of iron and 5.24g of chlorine. What is the percent composition of the compound?1. Calculate the mass of the product formed: 2.74g Fe + 5.24g Cl = 7.98g product2. Calculate the percent for each element. 2.74 g Fe 5.24 g Cl x100 = 34.3% Fe x100 = 65.7% Cl 7.98 g 7.98 g Copyright 2012 John Wiley & Sons, Inc
  33. 33. First determine the molar mass of Na2CO3 Your Turn!What is the percent carbon in acetic acid, HC2H3O2?a. 41.01% C __ C __(____g) atomic massb. 20.00% C __ H __(____g) C 12.01 __ O __(____g) H 1.01c. 6.73% C = _____ g/mol O 16.00d. 39.99% C Copyright 2012 John Wiley & Sons, Inc
  34. 34. Your Turn!A 6.00g sample of calcium sulfide is found to contain 3.33g of calcium. What is the percent by mass of sulfur in the compound?a. 80.2% Sb. 55.5% Sc. 44.5% Sd. 28.6% S Copyright 2012 John Wiley & Sons, Inc
  35. 35. Empirical Formula versus Molecular FormulaReview Question 11 The molecular formula for a substance is the C6H12O6 actual number of atoms of each element. The empirical formula is the lowest whole CH2O number ratio of atoms in a compound. Note that the molecular formula is a whole (CH2O)6 number multiple of the empirical formula. Copyright 2012 John Wiley & Sons, Inc
  36. 36. Empirical Formula versus Molecular FormulaIt is possible for several different molecules to have the same empirical formula. Copyright 2012 John Wiley & Sons, Inc
  37. 37. Your Turn!What is the empirical formula for the compound P4O10?a. P4O10b. P2O5c. PO2.5d. PO3 Copyright 2012 John Wiley & Sons, Inc
  38. 38. Calculating Empirical Formulas Steps for calculating an empirical formula:Review Question 12 100g of compound and express the mass of 1. Assume each element in grams. 2. Convert the grams of each element to moles. 3. Find the mole ratio of each element. Round to nearest whole number if it is close to the whole number. 4. If necessary, multiply the ratios by the smallest whole number that will convert them to a whole number. Copyright 2012 John Wiley & Sons, Inc
  39. 39. Calculating Empirical FormulasCalculate the empirical formula of a compound that is 63.19% Mn and 36.81% O. atomic mass1. Assume 100 g of material. Mn 54.94 0 16.00 63.19 g Mn 36.81 g O2. Convert grams of each element to moles: 1 mol Mn 63.19 g Mn× =1.1502 mol Mn 54.94 g Mn 1 mol O 36.81 g O× = 2.3006 mol O 16.00 g O Copyright 2012 John Wiley & Sons, Inc
  40. 40. Calculating Empirical Formulas3. Change the numbers of atoms to whole numbers by dividing by the smallest number. 1.1502 mol Mn 2.3006 mol OMn = = 1.000 O= = 2.000 1.1502 mol Mn 1.1502 mol Mn The simplest ratio of Mn:O is 1:2. Empirical formula = MnO2 Copyright 2012 John Wiley & Sons, Inc
  41. 41. Calculating Empirical FormulasCalculate the empirical formula of a compound that is 72.2% Mg and 27.8% N. atomic mass1. Assume 100 g of material. Mg 24.31 N 14.01 72.2 g Mg 27.8 g O2. Convert grams of each element to moles: 1mol Mg 72.2g Mg× =2.970 mol Mg 24.31g Mg 1mol N 27.8g N× = 1.984 mol N 14.01g N Copyright 2012 John Wiley & Sons, Inc
  42. 42. Calculating Empirical Formulas3. Change the numbers of atoms to whole numbers by dividing by the smallest number. 2.970 mol Mg 1.984 mol N Mg = = 1.500 N= = 1.000 1.984 mol N 1.984 mol N4. Multiply by a number that will give whole numbers. Mg: (1.500)2 = 3.00 N: (1.000)2 = 2.00 Empirical formula = Mg3N2 Copyright 2012 John Wiley & Sons, Inc
  43. 43. Your Turn!What is the empirical formula of an alcohol that is 52.13% C, 13.15% H and 34.72% O. atomic massa. CH2O 1. Assume 100 g so 52.13g C, 13.15g H and 34.72g O C 12.01b. C4HO3 2. Moles H 1.01 52.13g C/12.01= 4.34 mol Cc. C2H6O 13.15g H/1.01= 13.02 mol H O 16.00d. C2H3O2 34.72g O/16.00= 2.17mol O 3. Fractional part 4.34 mol C /2.17= 2 13.02 mol H /2.17= 6 2.17mol O /2.17= 1 4. Whole numubers Already done 5. Empirical formula: 2 C & 6 H & 1 O, so … Copyright 2012 John Wiley & Sons, Inc Give it a try
  44. 44. Calculating the Molecular Formula from the Empirical FormulaThe molecular formula will be either equal to the empirical formula or some integer multiple of it.The ratio of the molecular mass to the mass predicted by the empirical formula tells us how many times larger the molecular formula is. molecular formula massn= = number of empirical formula units empirical formula mass Copyright 2012 John Wiley & Sons, Inc
  45. 45. Calculating the Molecular Formula from the Empirical FormulaDetermine the molecular formula for glyceraldehyde which has a molar mass of 90.08 g/mol and an empirical formula of CH2O. molecular formula mass n= empirical formula mass 90.09 g n= = 3 (CH2O)3 = C3H6O3 30.03 g CH 2O Copyright 2012 John Wiley & Sons, Inc
  46. 46. Calculating the Molecular Formula from the Empirical FormulaDetermine the molecular formula of a nitrogen oxide compound (NxOy) with a molar mass of 92.011 g/mol and a empirical formula of NO2. molecular formula mass n= empirical formula mass 92.011 g N x O y (NO2)2 = N2O4n 2 46.0055 g NO2 Copyright 2012 John Wiley & Sons, Inc
  47. 47. Your Turn!What is the molecular formula of a compound with the empirical formula CH2Cl and molar mass of 197.92 g/mol?a. CH2Cl PLAN THE STEPSb. C2H4Cl2 atomic mass A. Empirical formula mass C 12.01c. C3H6Cl3 B. Divide molar by empirical H 1.01d. C4H8Cl4 Cl 35.45 C. Multiple result by formula A. 1(12.01g C) + 2(1.01g H) + 1(35.45 g Cl) = 49.48g/mol CH2Cl B. 197.92 g/mol / 49.48 g/mol = 4Plan – Set up - Calcualte C. (CH Cl) *4= 2 Copyright 2012 John Wiley & Sons, Inc
  48. 48. Calculating the Molecular FormulaA disinfectant is known to be 76.57% C , 6.43% H, and 17.00% O. It has a molar mass of 188.24 g/mol Determine its molecular formula.Determine the mass and moles of C, H and O.C: 76.57%(188.24g) = 144.1 g C/(12.01g/mol) = 12H: 6.43% (188.24g) = 12.1 g H/(1.01g/mol) = 12O: 17.00%(188.24g) = 32.00 g O/(16.00 g/mol) = 2Molecular formula: C12H12O2 Copyright 2012 John Wiley & Sons, Inc
  49. 49. Your Turn!What is the molecular formula of a substance that consists of 85.60% C and 14.40% H and has a molar mass of 28.08 g/mol? atomic massa. CH2 C 12.01b. C2H2 H 1.01c. CH3d. C2H4e. C2H6 Copyright 2012 John Wiley & Sons, Inc
  50. 50. QuestionsReview Questions – Did in classPaired Questions (pg 139) – Do 1, 3, 7, 9, 11, 15, 21, 27, 29, 31, 35 , 39, 43 – Practice later 2, 6, 12, 16, 20, 24, 28, 32, 36, 40, 44 Copyright 2012 John Wiley & Sons, Inc 1-50

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