1. Stoichiometry! The heart of chemistry

2. The Mole The mole is the SI unit chemists use to represent an amount of substance. 1 mole of any substance = 6.02 x 10 23 representative particles. This is known as Avogadro’s number.

3. Types of Representative Particles 1.The representative particle of most elements is the atom. Ex. 1 mol Ca = 6.02 x 10 23 atoms Ca 2.The representative particle of covalent compounds (including diatomic elements) is the molecule. Ex. 1 mol SO 3 = 6.02 x 10 23 molecules SO 3. 3.The representative particle of ionic compound is the formula unit. Ex. 1 mol NaCl = 6.02 x 10 23 formula units NaCl.

4. Remember A mole represents number of particles, not mass. Each element has a unique atomic mass. Substances with the same number of moles can have different masses.

5. Molar Mass Molar mass represents how much mass there is in one mole of a substance. Molar mass is expressed in units of g/mol. atomic mass Ex) Molar mass of Sr 87.62 g/mol

6. How do I calculate molar mass? 1.Molar mass of an element = atomic mass of an element (periodic table) Ex. 1 mol He = 4.00 g He. The molar mass of He is 4.00 g/mol. 1 mol S = 32.07 g S. The molar mass of S is 32.07 g/mol.

7. Calculating molar mass cont. 2.Molar mass of a compound = sum of the masses of the elements in the compound a)What is the molar mass of water? H 2 O H: 2 x 1.01 = 2.02 O: 1 x 16.00 = 16.00 = 18.02 g/mol b)What is the molar mass of oxygen gas? O 2 O: 2 x 16.00 = 32.00 g/mol

8. Target Check What is the molar mass of lithium? 6.94 g/mol What is the molar mass of carbon? 12.01 g/mol What is the molar mass of calcium nitrate? Ca(NO 3 ) 2 Ca: 1 x 40.08 = 40.08 N: 2 x 14.01 = 28.02 O: 6 x 16.00 = 96.00 = 164.10 g/mol

9. What if I have more/less than one mole of a substance?! 1.Calculate the number of moles of NaCl in 175.5 g of the salt. 2.How many atoms are in 2.5 moles of sodium metal? 2.A sample of nitrogen gas contains 1.20 x 10 25 molecules. Determine the number of grams of nitrogen in the sample.

10. Target Check 1.Calculate the mass of 0.800 moles of H 2 SO 4. 2.How many atoms are in a 10.0 g sample of calcium metal?

11. Molar Volume The volume of a gas is usually measured at a standard temperature and pressure (STP). ▫Standard temperature is 0°C. ▫Standard pressure is 1 atm. A mole of any gas at STP occupies a volume of 22.4 L. This is the molar volume of a gas.

12. Molar Volume Examples Determine the volume, in liters, of 0.600 moles of sulfur dioxide gas at STP. Determine the number of molecules in 33.6 L of hydrogen gas at STP.

13. Target Check Determine the number of moles of 58.6 L of CO 2 gas at STP. Determine the number of liters of 76.0 g of oxygen gas at STP.

14. Percent Composition Percentage composition is the percent by mass of each element in a compound. % of element in compound =

15. Percentage Composition Example Find the percentage composition of aluminum sulfate. Al 2 (SO 4 ) 3 molar mass = 342.17 g/mol Al: 2 x 26.98 g = 53.96 g Al % Al = (53.96 g/342.17 g) x 100 = 15.8% S: 3 x 32.07 g = 96.21 g S % S = (96.21 g/342.17 g) x 100 = 28.1% O: 12 x 16.00 g = 192.00 g O % O = (192.00 g/342.17 g) x 100 = 56.1%

16. Target Check 1.Find the percent of carbon in carbon dioxide. CO 2 molar mass = 44.01 g/mol C: 1 x 12.01 g = 12.01 g C % C = (12.01 g/44.01 g) x 100 = 27.3%

17. Empirical Formulas An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. Examples: CH 4, NH 3, CO, LiClO 3 Non-examples: N 2 H 4 or C 2 H 6

18. Empirical Formulas Example Problems What is the empirical formula of a compound that is 79.8% carbon and 20.2% hydrogen? 79.8 g C x (1 mol C/12.01 g C) = 6.64 mol C 6.64 mol C/6.64 = 1 20.2 g H x (1 mol H/1.01 g H) = 20.0 mol H 20.0 mol H/6.64 = 3 The empirical formula is CH 3.

19. Empirical Formulas Example Problems What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? 25.9 g N x (1 mol N/14.01 g N) = 1.85 mol N 1.85 mol N/1.85 = 1 x 2 = 2 74.1 g O x (1 mol O/16.00 g O) = 4.63 mol O 4.63 mol O/1.85 = 2.5 x 2 = 5 The empirical formula is N 2 O 5.

20. Target Check A compound is found to contain 46.0 g sodium, 52.0 g chromium, and 64.0 g oxygen. What is the empirical formula of the compound? 46.0 g Na x (1 mol Na/22.99 g Na) = 2.00 mol Na 2.00 mol Na/1.00 = 2 52.0 g Cr x (1 mol Cr/52.00 g Cr) = 1.00 mol Cr 1.00 mol Cr/1.00 = 1 64.0 g O x (1 mol O/16.00 g O) = 4.00 mol O 4.00 mol O/1.00 = 4 The empirical formula is Na 2 CrO 4.

21. Molecular Formulas The molecular formula of a compound is the actual formula of molecular compound. It is either the same as the experimentally determined empirical formula, or it is some whole-number multiple of it. In order to find the molecular formula of a compound, you must know the empirical formula first.

22. Molecular Formulas Example Problem The molar mass of a compound is 92 g/mol. Analysis of a sample of the compound diciates that it contains 0.606 g N and 1.390 g O. Find its molecular formula. 0.606 g N x (1 mol N/14.01 g N) = 0.0433 mol N 0.0433 mol N/0.0433 = 1 1.390 g O x (1 mol O/16.00 g O) = 0.0869 mol O 0.0869 mol O/0.0433 = 2 The empirical formula is NO 2 (46.01 g/mol). 92 g/mol/46.01 g/mol = 2 The molecular formula is N 2 O 4.

23. Target Check 1. The percent composition of a compound is 40.0% C, 6.7% H, and 53.3 % O. The molar mass of the compound is 90.0 g/mol. What is the molecular formula of the compound? 40.0 g C x (1 mol C/12.01 g C) = 3.33 mol C 3.33 mol C/3.33 = 1 6.7 g H x (1 mol H/1.01 g H) = 6.63 mol H 6.63 mol H/3.33 = 2 53.3 g O x (1 mol O/16.00 g O) = 3.33 mol O 3.33 mol O/3.33 = 1 The empirical formula is CH 2 O (30.0 g/mol). 90.0 g/mol / 30.0 g/mol = 3 The molecular formula is C 3 H 6 O 2.

24. Target Check 2. Calculate the molecular formula of CH 4 N. The molar mass is 120.0 g/mol. CH 4 N = 30.0 g/mol 120 g/mol / 30.0 g/mol = 4 The molecular formula is C 4 H 16 N 4.

25. Stoichiometry Stoichiometry is the study of the amount of substances produced and consumed in chemical reactions. Calculations are carried out step-by-step using dimensional analysis.

26. Molecular Relationships in Balanced Equations A balanced equation shows the smallest whole number of moles of each component involved in the reaction. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) 1 molecule2 molecules1 molecules 2 molecules 1 mole2 moles1 mole2 moles

27. Molecular Relationships Example Problems C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) 1.For everyone one mole of C 3 H 8 that completely reacts, how many moles of CO 2 are produced? 3 moles CO 2 2.For every 3 moles of CO 2 produced by the reaction, how many moles of H 2 O will be produced? 4 moles H 2 O 3.What is the ratio of moles of O 2 to moles of CO 2 ? 5:3

28. Mole A  Mole B Calculations In a mole-mole problem, the number of moles of one substance in a given reaction is given and you are asked to calculate the number of moles of another substance. Steps in solving mole-mole calculations. ▫Write a balanced chemical equation. ▫Use the mole ratio to convert moles of given to moles of unknown.

29. Mole A  Mole B Calculations Example How many moles of oxygen gas would result from the decomposition of 13.4 moles of water? 2H 2 O  2H 2 + O 2

30. Target Check How many moles of nitrogen gas are required to completely react with 4.5 moles of hydrogen gas to produce ammonia? N 2 + 3H 2  2NH 3

31. Mole A  Mass B Calculations In a mole-mass problem, the number of moles of one substance in a given reaction is given and you are asked to calculate the number of grams of another substance, or vice versa. Information needed to solve a mole-mass calculation: ▫A balanced chemical equation. ▫Mole ratio. ▫Molar mass of substance being converted to or from grams.

32. Mole A  Mass B Calculations Example Problem How many grams of zinc chloride would be produced in the reaction of 4.5 moles of hydrochloric acid with excess zinc? Zn + 2HCl  ZnCl 2 + H 2

33. Mass A  Mass B Calculations In a mass-mass problem, the mass of one substance in a reaction is given and you are asked to calculate the mass of one or more of the other substances involved in the reaction. Steps in solving mass-mass calculations: ▫Write a balanced equation. ▫Convert the mass of the given substance to moles. ▫Use the mole ratio to calculate the moles of the second substance needed to react with the given substance. ▫Convert the moles of the second substance to grams.

34. Mass A  Mass B Calculations Example Problem If 30.0 g of lithium react with an excess of water, how many grams of lithium hydroxide will be produced?

35. Target Check 1.Sodium chloride is prepared by the reaction of sodium metal with chlorine gas. How many moles of sodium are required to produce 40.0 g of sodium chloride? 2Na + Cl 2  2NaCl

36. Target Check 2. How many grams of aluminum will be produced by the decomposition of 25.0 g of aluminum oxide? 2Al 2 O 3  2Al + 3O 2

37. Mass A  Volume B Calculations In mass-volume problems, either the given or the unknown quantity is the volume of a gas at STP.

38. Mass A  Volume B Calculations Example Problem Oxygen is prepared in the lab by the decomposition of potassium chlorate. 2KClO 3  2KCl + 3O 2 a. Calculate the volume of oxygen in liters, measured at STP that would be obtained from 183.9 g of potassium chlorate.

39. Mass A  Volume B Calculations Example Problem Oxygen is prepared in the lab by the decomposition of potassium chlorate. 2KClO 3  2KCl + 3O 2 b. Calculate the grams of potassium chlorate required to produce 2.80 L of oxygen at STP.

40. Volume A  Volume B Calculations In volume-volume problems, both the substance sought and the substance given are gases. Under the same conditions of temperature and pressure, the volume of reacting gases are proportional to the number of moles of the gases in the balanced equation.

41. Volume A  Volume B Calculations Sample Problem 2C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O(g) Calculate the volume of CO 2 results from the combustion of 6.0 L of C 2 H 6. Shortcut:

42. Target Check 1.Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. If 150L of hydrogen gas is collected at STP, how many grams of aluminum were used? 2Al + 6HCl  2AlCl 3 + 3H 2

43. Target Check 2. Calculate the volume of water vapor resulting from the combustion of sufficient C 2 H 6 to require 2.0 L of O 2. 2C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O(g)

44. Limiting Reagents The limiting reagent in a chemical reaction limits the amounts of the other reactants that can combine—and the amount of product that can form—in a chemical reaction. In other words, it is the reactant that runs out first. The reaction will stop once the limiting reagent is used up.

45. Excess Reagent The excess reagent in a chemical reaction is the substance that is not used up completely in a chemical reaction. In other words, it is the reactant that you have leftovers of.

46. Limiting Reagent Sample Problem Nitrogen gas and hydrogen gas react to form ammonia. What will occur will 6.70 moles of N 2 reacts with 3.20 moles of H 2 ?

47. N 2 + 3H 2  2NH 3 a.What is the limiting reagent? Step 1: Convert both of the reactants to the products.

48. Step 2: The reactant that produced the least amount of product ran out first, and thus, is the limiting reagent. The hydrogen gas is the limiting reactant. b. What is the maximum number of moles of NH 3 that can be produced? Based on the answer to part a, only 2.13 mol of NH 3 can be produced.

49. Target Check A student reacts 80.0 g of Copper metal with 25.0 g of Sulfur to form copper (I) sulfide. 2Cu + S  Cu 2 S a.What is the limiting reagent? Cu is the limiting reagent!

50. Target Check cont. b. What is the maximum number of grams of copper(I) sulfide that could be produced? Based on the answer from part a, only 100. g of Cu 2 S can be produced.

51. Percent Yield When a chemical reaction is used to calculate the amount of product that will form during a reaction, then a value for the theoretical yield is obtained. This is the maximum amount of product that could be formed. The amount of product that forms when the reaction is carried out in the laboratory is called the actual yield. It is often less than the theoretical yield.

52. Percent Yield The percent yield is the ratio of the actual yield to the theoretical yield. % yield =

53. Percent Yield There are many factors that cause percent yield to be less than 100%. Reactions do not always go to completion. Impure reactants and competing side reactions may cause other products to be formed. Some of the products may be lost during purification.

54. Percent Yield Example Problem A student decomposes 24.8 g of calcium carbonate in the lab. CaCO 3  CaO + CO 2 a.Calculate the theoretical yield of calcium oxide.

55. Percent Yield Example Problem b. What is the percent yield if the student only produced 13.1 g of calcium oxide in the laboratory?

56. Target Check When 50 g of silicon dioxide is heated with an excess of carbon, 32.2 g of silicon carbide is produced. SiO 2 (s) + 3C(s)  SiC(s) + 2CO(g) a.First, calculate the theoretical yield.

57. Target Check b. Second, calculate the percent yield. % yield =