Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

1. Chapter 3Stoichiometry:Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community College St. Peters, MO ? 2006, Prentice-Hall

2. Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789

3. Chemical Equations Concise representations of chemical reactions

4. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

5. Anatomy of a Chemical Equation Reactants appear on the left side of the equation. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

6. Anatomy of a Chemical Equation Products appear on the right side of the equation. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

7. Anatomy of a Chemical Equation The states of the reactants and products are written in parentheses to the right of each compound. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

8. Anatomy of a Chemical Equation Coefficients are inserted to balance the equation. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

9. Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule

10. Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules

11. Reaction Types

12. Why Categorize Reaction Types? There are three basic reaction types considered in this chapter It is worthwhile recognizing basic patterns of reactivity so that products of chemical reactions can be predicted before reactions are actually carried out

13. Combination Reactions Examples: N2 (g) + 3 H2 (g) ??? 2 NH3 (g) C3H6 (g) + Br2 (l) ??? C3H6Br2 (l) 2 Mg (s) + O2 (g) ??? 2 MgO (s) Two or more substances react to form one product

14. 2 Mg (s) + O2 (g) ??? 2 MgO (s)

15. Decomposition Reactions Examples: CaCO3 (s) ??? CaO (s) + CO2 (g) 2 KClO3 (s) ??? 2 KCl (s) + 3O2 (g) 2 NaN3 (s) ??? 2 Na (s) + 3 N2 (g) One substance breaks down into two or more substances. Usually brought on by heating.

16. Combustion Reactions Examples: CH4 (g) + 2 O2 (g) ??? CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g) ??? 3 CO2 (g) + 4 H2O (g) Rapid reactions that produce a flame Most often involve hydrocarbons reacting with oxygen in the air

17. Atomic Weights, Formula Weights, and Molecular Weights

18. Atomic Weight and Formula Weight The weight of an atom of some element (atomic weight) is equivalent to its periodic table mass, expressed in atomic mass units (amu) Ionic compounds involve huge numbers of ions held together by electrostatic attraction – the formula of an ionic compound is represented by the formula unit (example, NaCl represents the formula unit for sodium chloride) So, the “formula weight” (FW) of calcium chloride, NaCl, would be

19. Molecular Weight (MW) Sum of the atomic weights of the atoms in a molecule For the molecule ethane, C2H6, the molecular weight would be

20. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

21. Percent Composition So the percentage of carbon in ethane (C2H6) is…

22. Example – Percentage Composition Calculate the percentage carbon by mass in sodium acetate

23. Moles

24. Moles and Dozens The term “dozen” is used to simplify descriptions of larger numbers of items (for example, two dozen apples instead of twenty four apples) Chemists use the term “mole” to describe large numbers of atoms/ions/molecules/particles, etc. A dozen means twelve objects, a mole means 6.022 x 1023 objects (Avogadro’s number) There are 6.022 x 1023 atoms of 12C in a 12g sample of isotopically pure 12C

25. Avogadro’s Number How many atoms of 12C are there in one mole of 12C? How many H2O molecules are there in one mole of H2O? How many frosh are there in one mole of frosh?...

26. Using Avogadro’s Number How many oxygen atoms are there in 1 mol of H2O? How many hydrogen atoms are there in 1 mol of H2O? How many H atoms are there in 0.400 mol H2O?

27. Avogadro’s Number 1 atom of 12C has a mass of 12 amu 1 mole of 12C has a mass of 12 g

28. Using the molar mass and dimensional analysis, it is an easy exercise to convert masses to moles, and to numbers of atoms/molecules How many moles of glucose (C6H12O6) are there in a teaspoon (5.0g)? How many C-atoms are there in this mass of glucose?

29. Finding Empirical Formulas

30. Empirical Formula of Hydrogen Peroxide In an analysis of hydrogen peroxide, it was found that the sample contained 5.93% hydrogen, and 94.07% oxygen, by mass. What is the empirical formula of hydrogen peroxide?

31. Empirical Formula of Hydrogen Peroxide In an analysis of hydrogen peroxide, it was found that the sample contained 5.93% hydrogen, and 94.07% oxygen, by mass. What is the empirical formula of hydrogen peroxide?

32. Empirical Formula of Hydrogen Peroxide

33. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition

34. Calculating Empirical Formulas

35. Calculating Empirical Formulas

36. Calculating Empirical Formulas

37. Calculating Empirical Formulas

38. Combustion Analysis 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 determined

39. Combustion Analysis Problem (CxHy) Combustion of toluene, an organic compound that contains only carbon and hydrogen, produces 5.86 mg of CO2 and 1.37 mg H2O. What is the empirical formula for toluene?

40. Combustion Analysis Problem (CxHy) massCO2 and massH2O ? mol C and mol H: Divide by smallest number of moles to get subscript for empirical formula

41. Combustion Analysis Problem (CxHy) Multiply each of these numbers by whole numbers until both are whole numbers

42. A Combustion Analysis Problem (CxHyOz) A 0.1005 g sample of menthol, a compound consisting of carbon, hydrogen, and oxygen, is combusted, producing 0.2829 g CO2 and 0.1159 g H2O. What is the empirical formula of menthol?

44. I’ve got the empirical formula…now what? We are usually more interested in a compound’s molecular formula than its empirical formula – how do we get the molecular formula once we know the empirical formula? If the molar mass of the substance is known, then the molecular formula can be determined as follows:

45. Stoichiometric Calculations Let O mean “is stoichiometrically equivalent to” 2 mol H2 O 1 mol O2 O 2 mol H2O

46. Stoichiometric Relationships In the combustion problem we looked at earlier, we converted a certain number of grams of CO2 to moles of carbon (and g H2O to mol H). In doing this, we used a stoichiometric relationship: 1 atom of C per 1 molecule of CO2 1 mol CO2 O 1 mol C The same relationships allow us to indirectly relate masses of reactants and products together in some balanced equation (or reactants and reactants, products and products) – example, what mass of water can be made from 1.00 g of C6H12O6 and enough O2 to ensure complete reaction?

47. Stoichiometric Calculations – What mass of H2O can be made from 1.00 g C6H12O6? Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams

48. Stoichiometric Calculations From the mass of substance A you can use the ratio of the coefficients of A and B to calculate the mass of substance B formed (if it’s a product) or used (if it’s a reactant)

49. Or, do it all in one calculation: Stoichiometric Calculations

50. We can also use the stoichiometric relationships in the balanced equation to calculate other things What mass of O2 is required to react with 1.00g C6H12O6? What mass of CO2 will be produced? What mass of C6H12O6 is required to produce 1.00g H2O? Stoichiometric Calculations

51. Limiting Reactants

52. How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

53. How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

54. 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, H2)

55. Limiting Reactants In the example below, the O2 would be the excess reagent (i.e. there would be some O2 left when the reaction has finished). H2 is the limiting reagent.

56. Theoretical Yield The theoretical yield is the amount of product that can be made upon complete reaction of the limiting reactant In other words it’s the amount of product possible as calculated through the stoichiometry problem The theoretical yield is calculated from the number of moles of limiting reactant This is different from the actual yield, the amount one actually produces and measures

57. Q: The most important commercial process for the production of ammonia (NH3) is the Haber process: 3H2(g) + N2(g) ? 2NH3(g) How many moles of NH3 can be formed from reaction of 3.0 mol N2 with 6.0 mol H2? Theoretical Yield

58. Which reactant is limiting? Three common ways to check this: Solve for the amount of NH3 produced by each amount of reactant. The reactant that produces a smaller amount of NH3 must be limiting. In this case, we would convert the number of moles of each reactant to moles of NH3: Thus, since the H2 initially present produces a smaller amount of NH3, it must be limiting (if we wanted to produce 6.0 mol NH3 from the N2 present, we’d have to add more H2) 3H2(g) + N2(g) ? 2NH3(g)

59. Find how many moles of H2 are needed to react with this much N2: Thus, it would take 9.0 mol H2 to react completely with 3.0 mol N2. There is not that much H2 present initially (rem: there are only 6.0 mol H2), so H2 must be limiting. 3H2(g) + N2(g) ? 2NH3(g)

60. Take the number of moles of each reactant present and divide it by the coefficient that corresponds to it in the balanced chemical equation. The smallest result indicates the limiting reactant: Thus H2 is limiting (since 2.0 is smaller than 3.0). 3H2(g) + N2(g) ? 2NH3(g)

61. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make