1. © 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation

2. Stoichiometry © 2012 Pearson Education, Inc. 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. Stoichiometry © 2012 Pearson Education, Inc. Chemical Equations Chemical equations are concise representations of chemical reactions.

4. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Reactants appear on the left side of the equation. Products appear on the right side of the equation.

5. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) The states of the reactants and products are written in parentheses to the right of each compound.

6. Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Coefficients are inserted to balance the equation.

7. Stoichiometry © 2012 Pearson Education, Inc. Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule. Coefficients tell the number of molecules.

8. Stoichiometry Sample Exercise 3.1 Interpreting and Balancing Chemical Equations In the following diagram, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH 3 molecules should be shown in the right (products) box? Practice

9. Stoichiometry © 2012 Pearson Education, Inc. Reaction Types

10. Stoichiometry © 2012 Pearson Education, Inc. Combination Reactions (Synthesis) Examples: –2Mg(s) + O 2 (g)  2MgO(s) –N 2 (g) + 3H 2 (g)  2NH 3 (g) –C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l) In combination reactions two or more substances react to form one product.

11. Stoichiometry © 2012 Pearson Education, Inc. In a decomposition reaction one substance breaks down into two or more substances. Decomposition Reactions Examples: –CaCO 3 (s)  CaO(s) + CO 2 (g) –2KClO 3 (s)  2KCl(s) + O 2 (g) –2NaN 3 (s)  2Na(s) + 3N 2 (g)

12. Stoichiometry Sample Exercise 3.3 Writing Balanced Equations for Combination and Decomposition Reactions Write a balanced equation for (a) solid mercury(II) sulfide decomposing into its component elements when heated and (b) aluminum metal combining with oxygen in the air.

13. Stoichiometry © 2012 Pearson Education, Inc. Combustion Reactions Examples: –CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) –2CH 3 OH(l) + 3O 2 (g)  2CO 2 (g) + 4H 2 O(g) Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve hydrocarbons reacting with oxygen in the air.

14. Stoichiometry Sample Exercise 3.4 Writing Balanced Equations for Combustion Reactions Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l), burns in air.

15. Stoichiometry © 2012 Pearson Education, Inc. Formula Weights

16. Stoichiometry © 2012 Pearson Education, Inc. Formula Weight (FW) A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu Formula weights are generally reported for ionic compounds.

17. Stoichiometry © 2012 Pearson Education, Inc. Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.01 amu) 30.08 amu + H: 6(1.01 amu)

18. Stoichiometry © 2012 Pearson Education, Inc. 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: % Element = (number of atoms)(atomic weight) (FW of the compound) x 100

19. Stoichiometry © 2012 Pearson Education, Inc. Percent Composition So the percentage of carbon in ethane is

20. Stoichiometry © 2012 Pearson Education, Inc. Moles

21. Stoichiometry © 2012 Pearson Education, Inc. Avogadro’s Number 6.022 x 10 23 1 mole of 12 C has a mass of 12.000 g.

22. Stoichiometry Sample Exercise 3.7 Estimating Numbers of Atoms Without using a calculator, arrange these samples in order of increasing number of O atoms: 1 mol H 2 O, 1 mol CO 2, 3  10 23 molecules O 3.

23. Stoichiometry Sample Exercise 3.8 Converting Moles to Number of Atoms How many oxygen atoms are in (a) 0.25 mol Ca(NO 3 ) 2 and (b) 1.50 mol of sodium carbonate?

24. Stoichiometry © 2012 Pearson Education, Inc. Molar Mass By definition, a molar mass is 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).

25. Stoichiometry © 2012 Pearson Education, Inc. Using Moles Moles provide a bridge from the molecular scale to the real-world scale.

26. Stoichiometry © 2012 Pearson Education, Inc. 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.

27. Stoichiometry © 2012 Pearson Education, Inc. Calculations Mass, in grams, of 1.50x10 -2 mol CdS? Mass, in grams, of 1.50x10 21 molecules of aspirin, C 9 H 8 O 4 ?

28. Stoichiometry Sample Exercise 3.12 Calculating Numbers of Molecules and Atoms from Mass How many nitric acid molecules are in 4.20 g of HNO 3 ? (b) How many O atoms are in this sample?

29. Stoichiometry © 2012 Pearson Education, Inc. Finding Empirical Formulas

30. Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

31. Stoichiometry Sample Exercise 3.14 Determining a Molecular Formula Ethylene glycol, used in automobile antifreeze, is 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. (a) What is the empirical formula of ethylene glycol? (b) What is its molecular formula?

32. Stoichiometry © 2012 Pearson Education, Inc. Combustion Analysis Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14. –C is determined from the mass of CO 2 produced. –H is determined from the mass of H 2 O produced.

33. Stoichiometry Sample Exercise 3.15 Determining an Empirical Formula by Combustion Analysis (a) Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO 2 and 0.209 g H 2 O.What is the empirical formula of caproic acid? (b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?

34. Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products. –O is determined by difference after the C and H have been determined.

35. Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with 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).

36. Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams. C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

37. Stoichiometry © 2012 Pearson Education, Inc. Calculations 4 KO 2 + 2 CO 2  K 2 CO 3 + 3O 2 How many moles of O 2 are produced when 0.400 mol of KO 2 reacts? How many grams of KO 2 are needed to form 7.50 g of O 2 ? How many grams of CO 2 are used when 7.50 g of O 2 are produced?

38. Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants

39. Stoichiometry © 2012 Pearson Education, Inc. 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 H 2 ).

40. Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants In the example below, the O 2 would be the excess reagent.

41. Stoichiometry © 2012 Pearson Education, Inc. Calculations 2 Al(OH) 3 (s) + 3 H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 6 H 2 O(l) Which is the limiting reactant when 0.500 mol Al(OH) 3 and 0.500 mol H 2 SO 4 react? How many moles of Al 2 (SO 4 ) 3 can form? How many moles of the excess reactant remain after the completion of the reaction?

42. Stoichiometry © 2012 Pearson Education, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. –In other words, it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

43. Stoichiometry © 2012 Pearson Education, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield): Percent yield =x 100 actual yield theoretical yield

44. Stoichiometry © 2012 Pearson Education, Inc. Calculations SA AcAn Asp AcA C 7 H 6 O 3 + C 4 H 6 O 3  C 9 H 8 O 4 + HC 2 H 3 O 2 What is the theoretical yield of Asp if 185 kg of SA reacts 125 kg of AcAn? What is the percentage yield if the the reactions produces 182 kg of Asp?