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Ch. 4 The Mole: Relating the Microscopic World of Atoms to Laboratory Measurements

Ch. 4 The Mole: Relating the Microscopic World of Atoms to Laboratory Measurements. Brady & Senese, 5th Ed. Index. 3.1 The mole conveniently links mass to number of atoms or molecules 3.2 Chemical formulas relate amounts of substances in a compound

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Ch. 4 The Mole: Relating the Microscopic World of Atoms to Laboratory Measurements

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  1. Ch. 4 The Mole: Relating the Microscopic World of Atoms to Laboratory Measurements Brady & Senese, 5th Ed.

  2. Index 3.1 The mole conveniently links mass to number of atoms or molecules 3.2 Chemical formulas relate amounts of substances in a compound 3.3 Chemical formulas can be determined from experimental mass measurements 3.4 Chemical equations link amounts of substances in a reaction 3.5 The reactant in shortest supply limits the amount of product that can form 3.6 The predicted amount of product is not always obtained experimentally

  3. What Is a Mole? • One mole of any substance contains the same number of units, called Avogadro’s number, N • 1 mole = 6.0223 x 1023 units • 1 mole of an element contains 6.0223 x 1023 atoms • 1 mole of a covalent compound contains 6.0223 x 1023 molecules • 1 mole of an ionic compound contains 6.0223 x 1023 formula units • It is a large quantity of particles because the particles described are so small. 3.1 The mole conveniently links mass to number of atoms or molecules

  4. Why is Molar Mass the Same as Formula Mass? • suppose we start with 12.0107g of C. How many atoms of C are there? • given that the atomic mass of C is 12.0107 u • 12.0107 g C = 6.0223 x 1023 atoms • thus for any substance, the formula/molar mass (in g) corresponds to the same number of atoms, N 3.1 The mole conveniently links mass to number of atoms or molecules

  5. EXAMPLES OF THE MOLE CONCEPT 1 mole H2O = 18.02 g H2O = 6.022x1023 H2O molecules 1 mole CO2 = 44.01 g CO2 = 6.022x1023 CO2 molecules 1 mole NH3 = 17.03 g NH3 = 6.022x1023 NH3 molecules THE MOLE AND CHEMICAL CALCULATIONS The mole concept can be used to obtain factors that are useful in chemical calculations involving both elements and compounds.

  6. Molar Mass • One mole contains the same number of particles as the number of atoms in exactly 12 g of carbon-12 • The molar mass of a substance has the same numeric value as the formula mass • The value is different because the units are different • Thus if the formula mass of Ba3(PO4)2 is 610.332 u, the molar mass of Ba3(PO4)2 is 610.332 g 3.1 The mole conveniently links mass to number of atoms or molecules

  7. Learning Check: Converting Between Mass And Moles Given that the molar mass of CO2 is 44.0098 g/mol • What mass of CO2 is found in 1.55 moles? • How many moles of CO2 are there in 10 g? 68.2 g 0.2 mol 3.1 The mole conveniently links mass to number of atoms or molecules

  8. Using The Chemical Formula • To relate components of a compound to the compound quantity we look at the chemical formula • In Na2CO3 there are 3 relationships: • 2 mol Na: 1 mol Na2CO3 • 1 mol C: 1 mol Na2CO3 • 3 mol O: 1 mol Na2CO3 • We can also use these on the atomic scale ,e.g.: • 1 atom C:1 fu Na2CO3 3.2 Chemical formulas relate amounts of substances in a compound

  9. Learning Check: • Calculate the number of moles of sodium in 2.53 moles of sodium carbonate • Calculate the number of atoms of sodium in 2.53 moles of sodium carbonate 5.06 mol Na 3.05×1024 atoms Na 3.2 Chemical formulas relate amounts of substances in a compound

  10. Percentage Composition Percentage % A =  100 The percentage composition of a compound is the percentage by mass of each element in the compound.

  11. Percentage Composition Example: Determine the percentage composition of table sugar, C12H22O11. Solution: In one mole of C12H22O11: 12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) = 342.30 g C12H22O11 (Solution continued on the next slide)

  12. Percentage Composition 12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) = 342.30 g C12H22O11  100 = 42.10% C  100 = 6.479% H  100 = 51.42% O Check: 42.10% + 6.479% + 51.42% = 100.00%

  13. Percentage Composition In one mole of table sugar: 12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) = 342.30 g C12H22O11 The masses of all elements in a compound and the mass of the compound itself are directly proportional to each other, for example: 12(12.01 g C) PER 342.30 g C12H22O11 12(12.01 g C) PER 22(1.008 g H) Any of these ratios, or their inverses, may be used as a conversion factor from the mass of one species to the mass of the other.

  14. Percentage Composition Example: What mass of table sugar is needed to have a sample that contains 10.0 grams of carbon atoms? Solution: 10.0 g C  = 23.8 g C12H22O11

  15. Empirical Formula Empirical Formula The lowest whole-number ratio of atoms of the elements in a compound. The empirical formula of C2H4 is CH2. Likewise, the empirical formula of C6H12O6 is CH2O All compounds with the general formula CnH2n have the same empirical formula and therefore the same percentage composition.

  16. Compounds with Empirical Formula CH2O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) Molecular Formula M (g/mol) Whole-Number Multiple Name Use or Function 30.03 1 CH2O formaldehyde disinfectant; biological preservative 60.05 2 C2H4O2 acetic acid acetate polymers; vinegar(5% soln) 3 90.09 C3H6O3 lactic acid sour milk; forms in exercising muscle 4 120.10 C4H8O4 erythrose part of sugar metabolism 5 150.13 C5H10O5 ribose component of nucleic acids and B2 6 180.16 C6H12O6 glucose major energy source of the cell CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6

  17. Empirical Formula How to Find an Empirical Formula • Find the masses of different elements in a sample of the compound. • Convert the masses into moles of atoms of the different elements. • Determine the ratio of moles of atoms by dividing through by the smallest number of moles. • Express the moles of the atoms as the smallest possible ratio of integers. • Write the empirical formula, using the number for each atom in the integer ratio as the subscript in the formula.

  18. Empirical Formula Example: What is the empirical formula of a compound that analyzes as 79.95% carbon, 9.40% hydrogen, and 10.65% oxygen? Solution: It is usually helpful in an empirical formula problem to organize the calculations in a table with the following headings: Mole Formula Empirical Element Grams Moles Ratio Ratio Formula

  19. Empirical Formula Mole Formula Empirical Element Grams Moles Ratio Ratio Formula C 79.95 10 6.657 10.00 H 9.40 14 C10H14O 9.33 14.0 O 10.65 1 0.6656 1.000

  20. mol Na mol O mol Cl 16.00 g O 35.45 g Cl 22.99 g Na Determining the Empirical Formula PROBLEM: Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? PLAN: Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). SOLUTION: 2.82 g Na = 0.123 mol Na mass(g) of each element divide by M(g/mol) 4.35 g Cl = 0.123 mol Cl amount(mol) of each element 7.83 g O = 0.489 mol O use # of moles as subscripts preliminary formula Na1 Cl1 O3.98 Na1 Cl1 O3.98 NaClO4 NaClO4 change to integer subscripts empirical formula NaClO4 is sodium perchlorate.

  21. Molecular Formula The molecular formula of a compound can be found by determination of the number of empirical formula units in the molecule. For example, if the empirical formula of a compound is CH2, its molecular formula must be: CH2, C2H4, C2H6, C4H8, etc. Empirical formula units in 1 molecule =

  22. Molecular Formula How to Find the Molecular Formula of a Compound • Determine the empirical formula of the compound. • Calculate the molar mass of the empirical formula unit. • Determine the molar mass of the compound (which will be given in problems in the textbook). • Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n, the number of empirical formula units per molecule. • Write the molecular formula.

  23. Molecular Formula Example: What is the molecular formula of a compound with the empirical formula C5H10O and a molar mass of 258 g/mol? Solution: The molar mass of the empirical formula unit, C5H10O, is 5(12.01 g/mol C) + 10(1.008 g/mol H) + 1(16.00 g/mol O) = 86.13 g/mol C5H10O The number of empirical formula units per molecule is (C5H10O)3 = C15H30O3

  24. Molecular Formula PROBLEM: During physical activity. lactic acid (M=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. PLAN: assume 100g lactic acid and find the mass of each element divide each mass by mol mass(M) amount(mol) of each element molecular formula use # mols as subscripts divide mol mass by mass of empirical formula to get a multiplier preliminary formula convert to integer subscripts empirical formula

  25. mol O mol H mol C 1.008 g H 16.00 g O 12.01g C 3.33 3.33 3.33 90.08 g molar mass of lactate 30.03 g mass of CH2O Molecular Formula continued SOLUTION: Assuming there are 100. g of lactic acid, the constituents are 40.0 g C 6.71 g H 53.3 g O 3.33 mol C 6.66 mol H 3.33 mol O C3.33 H6.66 O3.33 CH2O empirical formula C3H6O3 is the molecular formula 3

  26. THE MOLE AND CHEMICAL EQUATIONS The mole concept can be applied to balanced chemical equations and used to calculate mass relationships in chemical reactions. Balanced equations can be interpreted in terms of the mole concept and the results used to provide factors for use in factor-unit solutions to numerical problems.

  27. How many moles of Fe2O3 can be produced from 6.0 moles O2? 4Fe(s) + 3O2(g) 2Fe2O3(s) Relationship: 3 mole O2 = 2 mole Fe2O3 Write a mole-mole factor to determine the moles of Fe2O3. 6.0 mole O2 x 2 mole Fe2O3 = 4.0 moles Fe2O3 3 mole O2 Calculations with Mole Factors

  28. Moles to Grams Suppose we want to determine the mass (g) of NH3 that can form from 2.50 moles N2. N2(g) + 3H2(g) 2NH3(g) The plan needed would be moles N2 moles NH3 grams NH3 The factors needed would be: mole factor NH3/N2and themolar mass NH3

  29. Moles to Grams The setup for the solution would be: 2.50 mole N2 x 2 moles NH3 x 17.0 g NH3 1 mole N2 1 mole NH3 given mole-mole factor molar mass = 85.0 g NH3

  30. The reaction between H2 and O2 produces 13.1 g water. How many grams of O2 reacted? 2H2(g) + O2(g) 2H2O(g) ? g 13.1 g The plan and factors would be g H2O mole H2O mole O2 g O2 molar mole-mole molar mass H2O factor mass O2 Calculating the Mass of a Reactant

  31. The setup would be: 13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2 18.0 g H2O 2 moles H2O 1 mole O2 molar mole-mole molar mass H2O factor mass O2 = 11.6 g O2 Calculating the Mass of a Reactant

  32. Calculating the Mass of Product When 18.6 g ethane gas C2H6 burns in oxygen, how many grams of CO2 are produced? 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 18.6 g ? g The plan and factors would be g C2H6 mole C2H6 mole CO2 g CO2 molar mole-mole molar mass C2H6 factor mass CO2

  33. Calculating the Mass of Product The setup would be 18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2 30.1 g C2H6 2 moles C2H6 1 mole CO2 molar mole-mole molar mass C2H6 factor mass CO2 = 54.4 g CO2

  34. Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product, which is calculated using the balanced equation. Actual yield The amount of product obtained when the reaction takes place. Percent yield The ratio of actual yield to theoretical yield. percent yield = actual yield (g) x 100 theoretical yield (g)

  35. You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burn and you throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies Calculating Percent Yield

  36. Limiting Reactant A limiting reactant in a chemical reaction is the substance that is used up first. Limits the amount of product that can form and stops the reaction.

  37. Example of An Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

  38. Example of An Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.

  39. Limiting Reactant When 4.00 moles H2 is mixed with 2.00 moles Cl2,how many moles of HCl can form? H2(g) + Cl2(g)  2HCl (g) 4.00 moles 2.00 moles ??? moles Calculate the moles of product that each reactant, H2 and Cl2, could produce. The limiting reactant is the one that produces the smallest amount of product.

  40. Limiting Reactant HCl from H2 4.00 moles H2 x 2 moles HCl = 8.00 moles HCl 1 moles H2(not possible) HCl from Cl2 2.00 moles Cl2 x 2 moles HCl = 4.00 moles HCl 1 mole Cl2 (smaller number of moles, Cl2will be used up first) The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl. 40

  41. Check Calculations

  42. Limiting Reactants Using Mass If 4.80 moles Ca are mixed with 2.00 moles N2, which is The limiting reactant? 3Ca(s) + N2(g)  Ca3N2(s) moles of Ca3N2 from Ca 4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N2 3 moles Ca (Ca is used up) moles of Ca3N2 from N2 2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles Ca3N2 1 mole N2 (not possible) Ca is used up when 1.60 mole Ca3N2 forms. Thus, Ca is the limiting reactant.

  43. Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H2 and 24.0 g O2 react? 2H2(g) + O2(g) 2H2O(l)

  44. Limiting Reactants Using Mass Moles H2O from H2: 8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O 2.02 g H2 2 moles H2 (not possible) Moles H2O from O2: 24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O 32.0 g O2 1 mole O2 O2 is limiting The maximum amount of product is 1.50 moles H2O, which is converted to grams. 1.50 moles H2O x 18.0 g H2O = 27.0 g H2O 1 mole H2O

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