1. Stoichiometry Calculations based on chemical reactions

2. Stoichiometry Stoichiometry is a Greek word that means using chemical reactions to calculate the amount of reactants needed and the amount of products formed. Stoichiometry is a Greek word that means using chemical reactions to calculate the amount of reactants needed and the amount of products formed. Amounts are typically calculated in grams (or kg), but there are other ways to specify the quantities of matter involved in a reaction. Amounts are typically calculated in grams (or kg), but there are other ways to specify the quantities of matter involved in a reaction.

3. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.

4. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l ) 

5. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g)

6. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g) The equation is not yet balanced. Hydrogens come in twos on the left, and three hydrogens are on the right side of the equation. The equation is not yet balanced. Hydrogens come in twos on the left, and three hydrogens are on the right side of the equation.

7. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g) Try a “2” in front of the water. Try a “2” in front of the water.

8. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + 2 H 2 O(l )  NaOH(aq) + H 2 (g) We now have two O atoms on the left, so we need to put a 2 before NaOH. We now have two O atoms on the left, so we need to put a 2 before NaOH.

9. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The two sodium atoms on the right require that we put a 2 in front of Na on the left. The two sodium atoms on the right require that we put a 2 in front of Na on the left.

10. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The two sodium atoms on the right require that we put a 2 in front of Na on the left. The equation is now balanced. The two sodium atoms on the right require that we put a 2 in front of Na on the left. The equation is now balanced.

11. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) Left SideRight Side Na- 2Na- 2 H- 4H- 4 O- 2O- 2

12. Chemical Equations 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 2 atoms of sodium react with 2 molecules of water to produce 2 molecules of sodium hydroxide and a molecule of hydrogen.

13. Chemical Equations 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 200 atoms of sodium react with 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of hydrogen.

14. Chemical Equations 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The balanced chemical equation can be interpreted in a variety of ways. It is usually interpreted as 2 moles of sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen. It is usually interpreted as 2 moles of sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen. The balanced equation tells us nothing about the masses of reactants or products. The balanced equation tells us nothing about the masses of reactants or products.

15. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) How many grams of sodium are needed to produce 50.0g of hydrogen? How many grams of sodium are needed to produce 50.0g of hydrogen? Since the balanced chemical equation gives us information about moles of reactants and products, we cannot answer the question until we convert the mass of hydrogen into moles. Since the balanced chemical equation gives us information about moles of reactants and products, we cannot answer the question until we convert the mass of hydrogen into moles.

16. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) How many grams of sodium are needed to produce 50.0g of hydrogen? How many grams of sodium are needed to produce 50.0g of hydrogen? 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g

17. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g Although the question doesn’t state it, you can assume enough water is present for complete reaction. Although the question doesn’t state it, you can assume enough water is present for complete reaction. We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na

18. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na We use the molar mass of H 2 to go from grams of H 2 to moles of H 2.

19. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 molar mass of H 2

20. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 molar mass of H 2 We use the coefficients from the balanced equation to go from moles of H 2 to moles of Na. We use the coefficients from the balanced equation to go from moles of H 2 to moles of Na.

21. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of H 2 coefficients

22. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of H 2 coefficients We use the molar mass of Na to go from moles of Na to grams of Na. We use the molar mass of Na to go from moles of Na to grams of Na.

23. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of Na molar mass of H 2 coefficients molar mass of Na

24. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of Na molar mass of H 2 coefficients molar mass of Na (50.0 g H 2 ) (1 mol H 2 ) (2 moles Na) ( 22.99 g Na) (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na) (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na)

25. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) (50.0 g H 2 ) (1 mol H 2 ) (2 moles Na) ( 22.99 g Na) = (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na) (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na) = 1,138 grams Na = 1.14 x 10 3 g Na = 1.14 kg Na

26. Limiting Reagent Problems Sometimes you are given quantities of more than one reactant, and asked to calculate the amount of product formed. The quantities of reactants might be such that both react completely, or one might react completely, and the other(s) might be in excess. These are called limiting reagent problems, since the quantity of one of the reacts will limit the amount of product that can be formed. Sometimes you are given quantities of more than one reactant, and asked to calculate the amount of product formed. The quantities of reactants might be such that both react completely, or one might react completely, and the other(s) might be in excess. These are called limiting reagent problems, since the quantity of one of the reacts will limit the amount of product that can be formed.

27. Limiting Reagent - Problem Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 1. First write the formulas for reactants and products. Al + Br 2  AlBr 3

28. Limiting Reagent - Problem Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 2. Now balance the equation by adding coefficients. 2 Al +3 Br 2  2 AlBr 3

29. Limiting Reagent - Problem Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 2 Al +3 Br 2  2 AlBr 3 The theoretical yield is the maximum amount of product that can be formed, given the amount of reactants. It is usually expressed in grams.

30. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g500.g ? grams There are several ways to solve this problem. One method is to solve the problem twice. Once, assuming that all of the aluminum reacts, the other assuming that all of the bromine reacts. The correct answer is whichever assumption provides the smallest amount of product. There are several ways to solve this problem. One method is to solve the problem twice. Once, assuming that all of the aluminum reacts, the other assuming that all of the bromine reacts. The correct answer is whichever assumption provides the smallest amount of product.

31. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g500.g ? grams The problem can be mapped: The problem can be mapped: Al: grams Al  moles Al  moles AlBr 3  g AlBr 3 molar mass Al coefficients molar mass AlBr 3 molar mass Al coefficients molar mass AlBr 3

32. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given:50.0g 500.g ? grams The problem can be mapped: The problem can be mapped: Al: grams Al  moles Al  moles AlBr 3  g AlBr 3 molar mass Al coefficients molar mass AlBr 3 (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr 3 ) (266.7 g AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 )

33. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g ? grams Al: (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr 3 ) (266.7 g AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 ) = 494. g AlBr 3 (if all of the Al reacts)

34. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g ? grams The calculation is repeated for Br 2. g Br 2  moles Br 2  moles AlBr 3  g AlBr 3 molar mass Br 2 coefficients molar mass AlBr 3

35. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g ? grams The calculation is repeated for Br 2. g Br 2  moles Br 2  moles AlBr 3  g AlBr 3 molar mass Br 2 coefficients molar mass AlBr 3 (500. g Br 2 ) (1 mol Br 2 ) (2 moles AlBr 3 ) (266.7 g AlBr 3 ) (159.8 g Br 2 ) (3 moles Br 2 ) (1 mol AlBr 3 ) (159.8 g Br 2 ) (3 moles Br 2 ) (1 mol AlBr 3 ) = 556. grams of AlBr 3

36. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g500.g ? grams Summary: Summary: We have enough Al to produce 494. g AlBr 3 We have enough Br 2 to produce 556. grams of AlBr 3 The theoretical yield is 494. grams of AlBr 3.

37. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g 494. g Summary: All of the Al reacts, so Al is limiting. All of the Al reacts, so Al is limiting. Bromine is in excess. Bromine is in excess. Additional questions: Additional questions: 1. How much bromine is left over? 2. If 418 grams of AlBr 3 is obtained, what is the % yield?

38. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g 494. g 1. How much bromine is left over? Since all 50.0 g of the Al reacts, the product must contain 494.g -50.g = 444. g of bromine. Therefore, 500.g- 444.g = 56. g of Br 2 are left over.

39. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g 494. g 2.If 418 grams of AlBr 3 is obtained, what is the % yield? The percent yield is (actual yield) (100%) (theoretical yield) % yield = (418 g) (100%) = 84.6 % (494g) (494g)

40. Concentration of Solutions There are many ways to express the concentration of a solution. The one most commonly used for solution stoichiometry is molarity (M). Molarity = M = moles of solute liter of solution

41. Concentration of Solutions Solutions are often used to determine how much of a certain substance is present in a sample. As a result, it is important to know how to perform calculations using solution stoichiometry.

42. Concentration of Solutions 40.0 grams of sodium hydroxide are dissolved in enough water to make 250. mL of solution. Calculate the molarity of the solution. 40.0 grams of sodium hydroxide are dissolved in enough water to make 250. mL of solution. Calculate the molarity of the solution. grams NaOH  moles NaOH  M NaOH

43. Concentration of Solutions 40.0 grams of sodium hydroxide are dissolved in enough water to make 250. mL of solution. Calculate the molarity of the solution. 40.0 grams of sodium hydroxide are dissolved in enough water to make 250. mL of solution. Calculate the molarity of the solution. grams NaOH  moles NaOH  M NaOH molar mass of NaOHM=moles/volume

44. Concentration of Solutions What volume of the previous solution contains 0.125 moles of NaOH? What volume of the previous solution contains 0.125 moles of NaOH?

45. Concentration of Solutions How would you make 100. mL of a 0.075M solution of NaCl? How would you make 100. mL of a 0.075M solution of NaCl? molarity = moles solute _________________ (liter of solution)

46. Concentration of Solutions How would you make 100. mL of a 0.075M solution of NaCl? How would you make 100. mL of a 0.075M solution of NaCl? moles solute = M(V) moles NaCl = (0.075mol/L) (.100L) moles NaCl = 0.0075 mol mass NaCl = 0.0075 mol (58.5 g/mol) mass NaCl = 0.439 g = 0.44 g NaCl

47. The Preparation of a Standard Aqueous Solution

48. Dilution Most solutions are purchased in highly concentrated form. They often need to be diluted before they can be used. When a solution is diluted, the moles of solute remain unchanged. Dilution involves the addition of more solvent (water), while the amount of solute stays the same.

49. Dilution Dilution problems can be easily solved using the following relationships. mols concentrated solute = mols diluted solute M conc (V conc ) = M dil (V dil )

50. Dilution How many milliliters of 6.00M H 2 SO 4 are needed to make 30.0 mL of 1.25M H 2 SO 4 ? How many milliliters of 6.00M H 2 SO 4 are needed to make 30.0 mL of 1.25M H 2 SO 4 ? M conc (V conc )= M dil (V dil ) ________

51. Dilution How many milliliters of 6.00M H 2 SO 4 are needed to make 30.0 mL of 1.25M H 2 SO 4 ? How many milliliters of 6.00M H 2 SO 4 are needed to make 30.0 mL of 1.25M H 2 SO 4 ? V conc = 1.25M(30.0 mL) (6.00M) M conc (V conc )= M dil (V dil ) ________

52. Dilution V = (0.0375 mol) V = 0.00625 L = 6.25 mL The solution requires 6.25 mL of 6.00M H 2 SO 4 diluted to a final volume of 30.0 mL. (6.00 mol/L)

53. Solution Stoichiometry How much 0.100M AgNO 3 is required to precipitate all of the chloride from 25.00 mL of 0.173M CaCl 2 ? How much 0.100M AgNO 3 is required to precipitate all of the chloride from 25.00 mL of 0.173M CaCl 2 ? 1. Write a balanced chemical equation. 2AgNO 3 CaCl 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl(s) 2AgNO 3 (aq) + CaCl 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl(s)

54. Solution Stoichiometry How much 0.100M AgNO 3 is required to precipitate all of the chloride from 25.00 mL of 0.173M CaCl 2 ? How much 0.100M AgNO 3 is required to precipitate all of the chloride from 25.00 mL of 0.173M CaCl 2 ? 2. Map out the problem. 2AgNO 3 CaCl 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl(s) 2AgNO 3 (aq) + CaCl 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl(s) V CaCl2 M CaCl2 =moles CaCl 2  moles AgNO 3  V AgNO3 coefficientsV=(mol)/(M)

55. Solution Stoichiometry 3. Solve the problem. 2AgNO 3 CaCl 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl(s) 2AgNO 3 (aq) + CaCl 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl(s) V CaCl2 M CaCl2 =moles CaCl 2  moles AgNO 3  V AgNO3 (25.00mL CaCl 2 ) (0.173 mol/L CaCl 2 ) = 4.325 mmol CaCl 2 (4.325 mmol CaCl 2 )(2 mmol AgNO 3 /1 mmol CaCl 2 ) = 8.65 mmol AgNO 3 V AgNO3 = 8.65mmol AgNO 3 = 86.5 mL AgNO 3 0.100mmol AgNO 3 /mL coefficientsV=(mol)/(M)

56. Titrations In a titration, a specific volume of a solution of unknown concentration is reacted with a solution of known concentration. The volume needed for complete reaction is determined using an indicator and a buret to measure the precise volume needed. Titrations may be used for many types of reactions, and they are often used in acid-base neutralization.

57. Acid-Base Titrations In this titration, NaOH is in the buret, and HCl and the indicator are in the flask. The NaOH solution is added until the first permanent pale pink color is formed in the flask.

58. Titration Stoichiometry 10.00 mL of a sulfuric acid solution requires 17.83 mL of a 0.100M NaOH solution for complete neutralization. Determine the concentration of the sulfuric acid. 10.00 mL of a sulfuric acid solution requires 17.83 mL of a 0.100M NaOH solution for complete neutralization. Determine the concentration of the sulfuric acid. 1. Write the balanced chemical reaction. H 2 SO 4 (aq) +2 NaOH(aq)  Na 2 SO 4 (aq) +2H 2 O(l)

59. Titration Stoichiometry 10.00 mL of a sulfuric acid solution requires 17.83 mL of a 0.100M NaOH solution for complete neutralization. Determine the concentration of the sulfuric acid. 10.00 mL of a sulfuric acid solution requires 17.83 mL of a 0.100M NaOH solution for complete neutralization. Determine the concentration of the sulfuric acid. 2. Map out the solution. H 2 SO 4 (aq) + 2 NaOH(aq)  Na 2 SO 4 (aq) + 2H 2 O(l) V NaOH M NaOH =moles NaOH  moles H 2 SO 4  M H2SO4 coefficients M=mol/liter

60. Titration Stoichiometry 3. Solve the problem. H 2 SO 4 (aq) + 2 NaOH(aq)  Na 2 SO 4 (aq) + 2H 2 O(l) 10.00 mL 17.83mL; 0.100M V NaOH M NaOH =moles NaOH  moles H 2 SO 4  M H2SO4 (17.83 mL)(0.100 mol NaOH/L) = 1.783 mmol NaOH (1.783 mmol NaOH) (1mmol H 2 SO 4 ) = 0.892 mmol (2 mmol NaOH) H 2 SO 4 (2 mmol NaOH) H 2 SO 4 coefficientsM=mol/liter

61. Titration Stoichiometry 3. Solve the problem. H 2 SO 4 (aq) + 2 NaOH(aq)  Na 2 SO 4 (aq) + 2H 2 O(l) 10.00 mL; 0.892mmol 1.78 mmol V NaOH M NaOH =moles NaOH  moles H 2 SO 4  M H2SO4 (0.892 mmol H 2 SO 4 )/10.00 mL = 0.0892 mol/L = 0.0892 M coefficientsM=mol/liter

62. Chemical Composition Chemical composition can be expressed in several ways, including percentages by mass, or chemical formulas. For example, water contains 11.2% hydrogen and 88.8% oxygen by mass. This information must be consistent with the chemical formula for water, H 2 O. For example, water contains 11.2% hydrogen and 88.8% oxygen by mass. This information must be consistent with the chemical formula for water, H 2 O.

63. Chemical Composition For example, water contains 11.2% hydrogen and 88.8% oxygen by mass. This information must be consistent with the chemical formula for water, H 2 O. For example, water contains 11.2% hydrogen and 88.8% oxygen by mass. This information must be consistent with the chemical formula for water, H 2 O. 2 H atoms = 2(1.008 amu) = 2.016 amu 1 O atom =1( 16.00 amu) =16.00 amu molecular mass of water = 18.02 amu % H = (2.016/18.02) x 100% = 11.19%H % O = (16.00/18.02) x 100% = 88.79%O

64. Chemical Composition The early scientists analyzed new chemical compounds to determine their composition and chemical formulas. Modern analytical laboratories still provide this service.

65. Chemical Composition Usually, the compound is combusted in the presence of oxygen. Any carbon in the compound is collected as carbon dioxide (CO 2 ), and any hydrogen is collected as water (H 2 O).

66. Chemical Composition Similar techniques exist to analyze for other elements. The formula obtained for the compound is the simplest whole number ratio of the elements in the compound, or the empirical formula. It may differ from the actual formula. For example, hydrogen peroxide is H 2 O 2, but chemical analysis will provide an empirical formula of HO.

67. Determining Empirical Formulas If given % composition: 1. Assume a quantity of 100 grams of the compound. 2. Determine the number of moles of each element in the compound by dividing the grams of each element by the appropriate atomic mass. 3. To simplify the formula into small whole numbers, divide the moles of each element by the smallest number of moles.

68. Determining Empirical Formulas 4. If necessary, multiply each number of moles by a factor that produces whole number subscripts. 5. If you know the approximate molar mass of the compound, determine the molecular formula.

69. % Composition Problem An oxide of titanium contains 59.9% titanium. Determine the empirical formula of the compound. An oxide of titanium contains 59.9% titanium. Determine the empirical formula of the compound.

70. Determining Empirical Formulas If given combustion data: The ultimate goal is to get the simplest whole number ratio of the elements in the compound. Usually the compound contains carbon, hydrogen and perhaps oxygen or nitrogen. 1. Use the information about CO 2 to determine the moles and mass of carbon in the compound.

71. Determining Empirical Formulas If given combustion data: 2. Use the information about H 2 O to determine the moles and mass of hydrogen in the compound. 3. The mass and moles of oxygen (or a third element) can be obtained by difference. 4. Once moles of each element is obtained, find the relative number of moles and empirical formula (as with % composition).

72. Formulas from Combustion Data This method assumes the compound contains only C and H.

73. Empirical Formula using Combustion Data- Problem A compound, which contains C, H and O, is analyzed by combustion. If 10.68 mg of the compound produces 16.01 mg of carbon dioxide and 4.37 mg of water, determine the empirical formula of the compound. A compound, which contains C, H and O, is analyzed by combustion. If 10.68 mg of the compound produces 16.01 mg of carbon dioxide and 4.37 mg of water, determine the empirical formula of the compound. If the compound has a molar mass of 176.1 g/mol, determine the molecular formula of the compound. If the compound has a molar mass of 176.1 g/mol, determine the molecular formula of the compound.