1. Mass Relationships By Doba Jackson, Ph.D.

2. Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2NaCl(s)2Na(s) + Cl 2 (g) right side: 2 Na 2 Cl left side: 2 Na 2 Cl

3. Chemical Symbols on a Different Level 2H 2 O(l)2H 2 (g) + O 2 (g) 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. microscopic:

4. Chemical Symbols on a Different Level 2H 2 O(l)2H 2 (g) + O 2 (g) 0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas to yield 5.00 kg of liquid water. macroscopic: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. microscopic:

5. Chemical Arithmetic: Stoichiometry How many grams of each reactant is needed and how many grams of the product are expected?

6. 2(12.0 amu) + 4(1.0 amu) = 28.0 amuC2H4:C2H4: Chemical Arithmetic: Stoichiometry HCl:1.0 amu + 35.5 amu = 36.5 amu Molecular Mass: Sum of atomic masses of all atoms in a molecule. Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.

7. 1 mole = 28.0 gC2H4:C2H4: 6.022 x 10 23 molecules = 28.0 g Chemical Arithmetic: Stoichiometry One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 36.5 g 6.022 x 10 23 molecules = 36.5 g HCl:

8. Chemical Arithmetic: Stoichiometry How many moles of chlorine gas, Cl 2, are in 25.0 g? 25.0 g Cl 2 70.9 g Cl 2 1 mol Cl 2 x= 0.353 mol Cl 2

9. Chemical Arithmetic: Stoichiometry How many grams of sodium hypochlorite, NaOCl, are in 0.705 mol? 0.705 mol NaOCl 1 mol NaOCl 74.5 g NaOCl x= 52.5 g NaOCl Na = 22.99 amu O = 15.99 amu Cl = 35.45 amu

10. Chemical Arithmetic: Stoichiometry Stoichiometry: The relative proportions in which elements form compounds or in which substances react. aA + bBcC + dD Moles of A Grams of A Moles of B Grams of B Mole Ratio Between A and B (Coefficients) Molar Mass of B Molar Mass of A

11. Chemical Arithmetic: Stoichiometry How many grams of NaOH are needed to react with 25.0 g Cl 2 ? 2NaOH(aq) + Cl 2 (g)NaOCl(aq) + NaCl(aq) + H 2 O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: Moles of Cl 2 Grams of Cl 2 Moles of NaOH Grams of NaOH Mole RatioMolar Mass

12. Chemical Arithmetic: Stoichiometry How many grams of NaOH are needed to react with 25.0 g Cl 2 ? 2NaOH(aq) + Cl 2 (g)NaOCl(aq) + NaCl(aq) + H 2 O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: 1 mol NaOH 40.0 g NaOH25.0 g Cl 2 70.9 g Cl 2 1 mol Cl 2 2 mol NaOH = 28.2 g NaOH xxx

13. Yields of Chemical Reactions The amount actually formed in a reaction. The amount predicted by calculations. Actual Yield: Theoretical Yield: Actual yield of product Theoretical yield of product x 100%Percent Yield =

14. Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant. C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l)

15. Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol, which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l)

16. Problem: Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: If 80.0 g of water are to be removed and 65.0 g of Li 2 O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced?

17. Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) Which reactant is limiting? 65.0 g Li 2 O 1 mol Li 2 O 1 mol H 2 O 29.9 g Li 2 O 1 mol Li 2 O = 4.44 moles H 2 O 80.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O = 2.17 moles H 2 O Amount of H 2 O given: Amount of H 2 O that will react with 65.0 g Li 2 O: Li 2 O is limiting x x x

18. Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 80.0 g H 2 O - 39.1 g H 2 O = 40.9 g H 2 O 2.17 mol H 2 O 1 mol H 2 O 18.0 g H 2 O = 39.1 g H 2 O (consumed) How many grams of excess H 2 O remain? remaininginitialconsumed x

19. Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 2.17 mol H 2 O 1 mol LiOH 23.9 g LiOH = 104 g LiOH How many grams of LiOH are produced? 1 mol H 2 O 2 mol LiOH xx

20. Concentrations of Reactants in Solution: Molarity Molarity (M): The number of moles of a substance dissolved in each liter of solution. In practice, a solution of known molarity is prepared by weighing an appropriate amount of solute, placing it in a container called a volumetric flask, and adding enough solvent until an accurately calibrated final volume is reached. Solution: A homogeneous mixture. Solute: The dissolved substance in a solution. Solvent: The major component in a solution.

22. Concentrations of Reactants in Solution: Molarity Molarity converts between mole of solute and liters of solution: Molarity = Moles of solute Liters of solution L mol or 1.00 M 1.00 L 1.00 mol = 1.00 1.00 mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to:

23. Concentrations of Reactants in Solution: Molarity Molar mass C 6 H 12 O 6 = 180.0 g/mol How many grams of solute would you use to prepare 1.50 L of 0.250 M glucose, C 6 H 12 O 6 ? 1 mol 0.275 mol180.0 g = 49.5 g 1 L 1.50 L0.250 mol = 0.275 mol x x

25. Diluting Concentrated Solutions dilute solutionconcentrated solution + solvent Since the number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent. M i x V i = M f x V f finalinitial

26. Diluting Concentrated Solutions Add 6.94 mL 18.0 M sulfuric acid to enough water to make 250.0 mL of 0.500 M solution. M i = 18.0 MM f = 0.500 M V i = ? mLV f = 250.0 mL = 6.94 mL 18.0 M 250.0 mL V i = VfVf 0.500 M = Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare 250.0 mL of 0.500 M aqueous H 2 SO 4 ? x MiMi MfMf x

27. Solution Stoichiometry aA + bBcC + dD Moles of A Volume of Solution of A Moles of B Volume of Solution of B Mole Ratio Between A and B (Coefficients) Molar Mass of B Molarity of A

28. Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) What volume of 0.250 M H 2 SO 4 is needed to react with 50.0 mL of 0.100 M NaOH? Moles of H 2 SO 4 Volume of Solution of H 2 SO 4 Moles of NaOH Volume of Solution of NaOH Mole Ratio Between H 2 SO 4 and NaOH Molarity of NaOH Molarity of H 2 SO 4

29. Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) 2 mol NaOH 1 mol H 2 SO 4 0.250 mol H 2 SO 4 1 L solution 1 L 0.100 mol 1 L 1000 mL = 0.00500 mol NaOH Volume of H 2 SO 4 needed: 1000 mL 1 L 10.0 mL solution (0.250 M H 2 SO 4 ) 0.00500 mol NaOH 50.0 mL NaOH Moles of NaOH available: x x x xx

30. Titration How can you tell when the reaction is complete? HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. Once the reaction is complete you can calculate the concentration of the unknown solution.

31. Titration unknown concentration solution Erlenmeyer flask buret standard solution (known concentration) An indicator is added which changes color once the reaction is complete

33. Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 48.6 mL of a 0.100 M NaOH solution is needed to react with 20.0 mL of an unknown HCl concentration. What is the concentration of the HCl solution? Moles of NaOH Volume of Solution of NaOH Moles of HCl Volume of Solution of HCl Mole Ratio Between NaOH and HCl Molarity of HCl Molarity of NaOH

34. Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 20.0 mL solution 0.00486 mol HCl = 0.243 M HCl Concentration of HCl solution: Moles of NaOH available: 1 L 0.100 mol = 0.00486 mol NaOH 48.6 mL NaOH 1000 mL 1 L Moles of HCl reacted: 1 mol NaOH 1 mol HCl = 0.00486 mol HCl 0.00486 mol NaOH 1 L 1000 mL x x x x

35. Problem: What is the molecular mass of table sugar (sucrose, C 12 H 22 O 11 ), and what is its molar mass in g/mol?

36. Problem: How many moles of sucrose are in a tablespoon of sugar containing 2.85 g?

37. Problem: How many grams are in 0.0626 mol of NaHCO 3, the main ingredient of Alka-Seltzer tablets?

38. Problem: Aqueous solutions of sodium hypochlorite (NaOCl, household bleach) are prepared by reaction of sodium hydroxide with chlorine. How many grams of NaOH are needed to react with 25.0 g of Cl 2 ?

39. Problem: Cisplatin, an anticancer agent used for the treatment of solid tumors, is prepared by the reaction of ammonia with potassium tetrachloroplatinate. Assume that 10.0 g of K 2 PtCl 4 and 10.0 g of NH 3 are allowed to react. (a) Which reactant is limiting, and which is in excess? (b) How many grams of the excess reactant are consumed, and how many grams remain? (c) How many grams of cisplatin are formed?