Chapter 3 stoichiometry


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Chapter 3 stoichiometry

  1. 1. Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten Chapter 3 Stoichiometry:Calculations with Chemical Formulas and Equations Stoichiometry
  2. 2. Law of Conservation of Mass • States that the total mass of a substance present before and after the chemical reaction is the same 2HCl + BaO → H2O + BaCl2 Stoichiometry
  3. 3. Law of Definite Proportions (Joseph Proust, 1799)• States that all samples of a compound have the same composition, i.e. the same proportion of mass by the constituent elements• Elemental composition of a pure compound is always the same regardless of the source Stoichiometry
  4. 4. Law of Multiple Proportions (John Dalton)• States that the mass of one element that combines with a fixed mass of the other element are in ratios of small whole numbers if two elements are combined to form more than one compound. Example: CO and CO2 Stoichiometry
  5. 5. Chemical EquationsConcise representations of chemical reactions Stoichiometry
  6. 6. Anatomy of a Chemical EquationCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Stoichiometry
  7. 7. Anatomy of a Chemical EquationCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the left side of the equation. Stoichiometry
  8. 8. Anatomy of a Chemical EquationCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Products appear on the right side of the equation. Stoichiometry
  9. 9. Anatomy of a Chemical EquationCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)The states of the reactants and products are written in parentheses to the bottom Stoichiometry right of each compound.
  10. 10. Anatomy of a Chemical EquationCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)Coefficients are inserted to balance the equation. Stoichiometry
  11. 11. Subscripts and Coefficients Give Different Information• Subscripts tell the number of atoms of each element in a molecule Stoichiometry
  12. 12. Subscripts and Coefficients Give Different Information• Subscripts tell the number of atoms of each element in a molecule• Coefficients tell the number of Stoichiometry molecules
  13. 13. Types ofChemicalReaction Stoichiometry
  14. 14. Combination Reactions • Two or more substances react to form one product• Examples: N2 (g) + 3 H2 (g) 2 NH3 (g) C3H6 (g) + Br2 (l) C3H6Br2 (l) Stoichiometry 2 Mg (s) + O2 (g) 2 MgO (s)
  15. 15. 2 Mg (s) + O2 (g) 2 MgO (s) Stoichiometry
  16. 16. Decomposition Reactions • One substance breaks down into two or more substances• Examples: CaCO3 (s) CaO (s) + CO2 (g) 2 KClO3 (s) 2 KCl (s) + O2 (g) Stoichiometry 2 NaN3 (s) 2 Na (s) + 3 N2 (g)
  17. 17. Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air• Examples: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Stoichiometry C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
  18. 18. The Atomic andMolecular Mass Stoichiometry
  19. 19. Atomic Mass• The mass of atoms of elements expressed in atomic mass units (amu)• The amu is defined by assigning a mass of exactly 12 amu to the mass of the 12C isotope of carbon. Stoichiometry
  20. 20. Isotopes• Isotopes are two or more atoms (nuclides) having the same atomic number (Z) but different mass numbers. Stoichiometry
  21. 21. Average Atomic Mass/ Atomic Weight• The average of the isotopic masses, weighed according to the naturally occuring abundances of the isotopes of an element.• Unit: amu or u; g/mol Stoichiometry
  22. 22. Sample ProblemsEx1] Calculate the atomic mass of Ne given the percent natural abundances: 90.48%, 0.27%, and 9.26% for the three isotopes respectively (ans : 20.19 amu)Ex2] Two naturally occuring isotopes of boron, boron-10 and boron-11, have the masses of 10.012937 u and 11. 009305 u, respectively. Calculate the percent natural abundances of each isotope given the atomic mass of boron as 10.81 u (ans: B-10:20% and B-11:80%) Stoichiometry
  23. 23. Formula Weight (FW)• Sum of the atomic weights for the atoms in a chemical formula• So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu• These are generally reported for ionic compounds Stoichiometry
  24. 24. Molecular Weight (MW)• Sum of the atomic weights of the atoms in a molecule• For the molecule ethane, C2H6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu Stoichiometry
  25. 25. Percent CompositionThe percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component. (number of A atoms)(atomic weight of A) %A= x 100 (FW of the compound) where A = any element in the periodic table Stoichiometry
  26. 26. Percent CompositionSo the percentage of carbon in ethane, C2H6, is… (2)(12.0 amu) %C = (30.0 amu) 24.0 amu = x 100 30.0 amu = 80.0% Stoichiometry
  27. 27. Sample ProblemsEx3] Calculate the percentage composition of each element in CCl3Br. (MW=198.26) Stoichiometry
  28. 28. Moles Stoichiometry
  29. 29. Avogadro’s Number• defined as the ratio of the number of constituent particles in a sample to the amount of substance.• 6.022 x 1023 particles / 1 mole of entityParticles= atoms/molecules/ions Stoichiometry
  30. 30. Avogadro’s Number• 6.022 x 1023• 1 mole of 12C has a mass of 12 g Stoichiometry
  31. 31. Molar Mass• By definition, these are the mass of 1 mol of a substance (i.e., g/mol) – The molar mass of an element is the mass number for the element that we find on the periodic table – The formula weight (in amu’s) will be the same number as the molar mass (in g/mol) Stoichiometry
  32. 32. Using MolesMoles provide a bridge from the molecular scale to the real-world scale Stoichiometry
  33. 33. Mole Relationships• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound Stoichiometry
  34. 34. Sample ProblemsEx4] a) How many molecules are there in 3.10 moles of H2O? (1.87x1024) b) How many grams are there in 3.8 moles in H2O? (68 g) Stoichiometry
  35. 35. FindingEmpirical & Molecular Formulas Stoichiometry
  36. 36. • Empirical Formula – Simplest whole-number ratio of atoms.• Molecular Formula – Represents the molecule of a substance and shows the actual number of atoms in a molecule. Stoichiometry
  37. 37. EMPIRICAL MOLECULAR n CH2 C2H4 2 CH C6H6 6 CO2 CO2 1 CH2O C5H10O5 5 Stoichiometry
  38. 38. Calculating Empirical Formulas Stoichiometry
  39. 39. Guidelines in Determining Mole Ratio• If the mole ratio has the following value: Mole Ratio Multiply by Factor a.50 2 a.33 or a.66 3 a.25 or a.75 4 a.20, a.40, a.60, a.80 5 Where a = 0, 1, 2, 3... Stoichiometry
  40. 40. Calculating Molecular Formula from Empirical Formula • Molecular formula (MF) can be determined from empirical formula (EF) using the following guidelines: MF = (EF) n Molecular Weight, MW where n = Empirical Formula Weight, EFW Stoichiometry
  41. 41. Sample ProblemDibutyl succinate is an insect repellent used against roaches. Its composition is 62.58% C, 9.63%H and 27.79%O.Its experimentally determined molecular mass is 230 amu. Calculate the empirical formula and the molecular formula. Stoichiometry
  42. 42. Sample ProblemsEx6] Sorbitol, used as a sweetener in some sugar-free foods, has a molecular mass of 182 u and a mass percent composition: 39.56% C, 7.74% H, and 52.7% O. What are the empirical formula and the molecular formula of sorbitol? Stoichiometry
  43. 43. Calculating Empirical FormulasEx 5]Ascorbic acid, commonly known asVitamin C, is essential for supportingdifferent metabollic reactions in ourbody. It contains 40.92 %C, 4.58%H, 54.50 %O by mass. What is theempirical formula of vitamin C? Stoichiometry
  44. 44. Combustion Analysis CxHyOz + O2 → CO2 + H2O• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO2 produced – H is determined from the mass of H2O produced – O is determined by difference after the C and H have been Stoichiometry determined
  45. 45. Sample ProblemsEx7] Caproic acid, the substance responsible for the aroma of dirty gym socks and running shoes, contain C, H, and O only. On combustion, 0.450g of sample of caproic acid gives 0.418g of H2O and 1.023g CO2. What is the EF of caproic acid? If the molecular weight of caproic acid is 116.2 g/mol, what is its MF? Stoichiometry
  46. 46. Stoichiometric CalculationsThe coefficients in the balanced equation give the ratio of moles of reactants and products Stoichiometry
  47. 47. Stoichiometric CalculationsFrom the mass ofSubstance A you canuse the ratio of thecoefficients of A andB to calculate themass of Substance Bformed (if it’s aproduct) or used (ifit’s a reactant) Stoichiometry
  48. 48. Sample ProblemsEx8] Given the unbalanced equation: Al(s) + O2(g)  Al2O3a.How many mole of O2 is needed to produce 10 mole of Al2O3 (15 mol)b.How many moles of Al are needed to completely react with 52.00 g O2? (2.167 mol)c.How many grams of O2 are needed to produce 158.00 g Al2O3 (74.382 g) Stoichiometry
  49. 49. LimitingReactants Stoichiometry
  50. 50. • Limiting Reactant – The reactant in a chemical reaction that limits the amount of product that can be formed.• Excess Reactant – The reactant in a chemical reaction that remains when the limiting reactant is completely consumed. Stoichiometry
  51. 51. Analogy: Cheese SandwichSupposing:10 slices loaf bread + 10 slices cheese = X sandwiches Questions: 1. How many sandwich can we make? 2. What is the limiting ingredient? 3. What is the excess reactant? Stoichiometry
  52. 52. Limiting Reactants• The limiting reactant is the reactant present in the smallest stoichiometric amount – In other words, it’s the reactant you’ll run out of first (in this case, the H2) Stoichiometry
  53. 53. Theoretical Yield• The theoretical yield is the amount of product that can be produced based on calculation. – In other words it’s the amount of product possible as calculated through the stoichiometry problem.• This is different from the actual yield, the amount one actually produces and measures. Stoichiometry
  54. 54. Percent Yield A comparison of the amount actually obtained to the amount predicted. Actual Yield, (AY)Percent Yield = x 100 Theoretical Yield, (TY) Stoichiometry
  55. 55. Sample ProblemsEx9] How many grams of NO(g) can be produced in the reaction of 1.00 mol of NH3 and 1.00 mol O2? NH3 (g) + O2 (g)  NO(g) + H2O(l) Stoichiometry