1. Quick Review: The Mole A very large counting number = 6.022 x 10 23 (6.022 followed by 23 zeros) Also known as Avogadro’s number The number shown under an element on the periodic table is the mass of 1 mole of that element (from the weighted average of naturally occurring isotopes). 1 mole of Li has a mass of 6.941 grams. 6.022 x 10 23 atoms of Li has a total mass of 6.941 grams.

2. A mole of Cl has a mass of 35.45 grams. The molar mass of chlorine is 35.45 g/mole What is the mass of 6.022 x 10 23 atoms of Al?

3. From the periodic table we can see that the molar mass of Al is 26.98 g/mol

4. Use conversion factors to convert 10.0 g Al to the correct number of moles Grams to moles Given End Unit Conversion factor

5. 10.0 g Al x 1 mol Al = 0.371 mol Al 26.98 g Al GivenEnd Unit Periodic Table

6. How many grams of Al are there in 10.0 mol of Al? Invert the conversion factor (Mol to Gram) 10.0 mol Al x 26.98g Al= 269.8 g Al mol Al How many atoms of Al are in the sample? (Mol to atoms) 10 mol Al x 6.022 x 10 23 atoms Al mol Al = 6.022 x 10 24 atoms of Al.

7. Review: Molar Mass of compounds (Molecular Weight) Compounds are made up of two or more elements The molar mass of the compound is the sum of the molar masses of each element in the compound. Molar mass of O 2 (the element, oxygen) is 2 x molar mass of the O atom. 2 x 16.00 grams/mol = 32.00 grams/mol The molar mass for the element O 2 is 32.00 g/mol 6.022 x 10 23 molecules of O 2 weigh 32.00 g

8. In the molecule CO 2, there is one atom of carbon. The subscript on the symbol for oxygen tells us that there are 2 atoms of oxygen in the molecule. There is one mole of carbon in CO 2 There are two moles of oxygen in CO 2 The molar mass of CO 2 is 1 x 12.01 g C / mol CO 2 2 x 16.00 g O / mol CO 2 44.01 g / mol CO 2

9. What is the molar mass of SO 2 ? a.how many moles of S? b.how many moles of O?

10. 1 mol S x 32.06 g/mol S = 32.06 g/mol 2 mol O x 16.00 g/mol of O = 32.00 g/mol 64.06 g/mol 1 mol SO 2 = 64.06 g This is an equivalence statement 1 mol SO 2 =164.06 g SO 2 = 1 64.06 g SO 2 1 mol SO 2 This is the same type of equivalence statement as 2.54 cm = 1 in and can be used as conversion factors in calculations.

11. Calculate the number of moles of CO 2 in 2.25 x 10 2 g of the gas. (Grams to moles)

12. Calculate the molar mass of CO 2. 1 x 12.01 g/mol C 2 x 16.00 g/mol O 44.01 g/mol CO 2 1 mol CO 2 = 44.01 g Second use conversion factors to convert grams to moles. 2.25 x 10 2 g CO 2 x 1 mol CO 2 = 5.11 mol CO 2 44.01 g CO 2

13. Mass percent composition of compounds (molecules or ionic compounds) Expressed by identifying the elements present and giving the mass percent of each. mass fraction, = mass of element in 1 mol compound given element mass 1 mol of compound Mass % element A = molar mass of element A x 100 molar mass of compound

14. Mass percent of elements in ethanol, C 2 H 5 OH The molar mass of ethanol is 46.08 g/mol In each mole of ethanol there are 2 moles of carbon The mass percent of carbon in ethanol is 2 x 12.01 = 24.02 g C in each mole of ethanol 24.02 g C x 100 = 52.13% 46.08 g Ethanol 52.13% of the mass of ethanol is from the mass of carbon in the molecule.

15. What is the mass percent of hydrogen in ethanol, C 2 H 5 OH? Total molar mass = 46.08 g. (1 mol ethanol = 46.08 g)

16. 6 moles of hydrogen in each mole of ethanol x 1.01 g/mol = 6.06 g/mol 6.06 g/mol hydrogen /46.08 g/mol ethanol x 100 = 13.2% Do the two add up to 100%??? Where is the difference? 16.00 g/mol O / 46.08 g/mol ethanol x 100 = 34.7 %

17. Percent composition makes it possible to determine a compound’s EMPIRICAL formula, the smallest whole number ratio of one element to others in the compound that agrees with the percent composition. CH 2 O is the empirical formula for formic acid, acetic acid (C 2 H 4 O 2 ) and glucose (C 6 H 12 O 6 ). What is the molar mass of each compound?

18. C 6 H 12 O 6 192.3 g / mol CH 2 O 32.05 g / mol C 2 H 4 O 2 64.06 g / mol

19. Steps - empirical forumula Assume 100 g of compound Divide the mass percent of the element by the molar mass of the element to get the number of moles of each element. Divide the # moles of each element by the smallest # moles to get the ratio of the elements, one to another. If any result is far from a whole number multiply through by a common factor that converts each number of moles to integers (or close) If each is close to a whole # round off each to nearest integer

20. An unknown clear solution was analyzed and found to contain 84.1% carbon and 15.9% hydrogen by mass. If we assume grams of the substance, the compound would contain 84.1 g carbon atoms and 15.9 grams of hydrogen atoms. 100

21. Taking 84.1 g carbon and using the atomic mass and dimensional analysis, we can determine the number of moles of the two elements in the substance. 84.1 g C x 1 mol C = 7.00 mol C 12.01 g C 15.9 g H x 1 mol H = 15.8 mol H 1.01 g H divide moles of each component by smallest number of moles 15.8 mol H = 2.26 mol H 7.00 mol C 1 mol C

22. The C-H mole ratio is 1:2.26 We can write C 1 H 2.26 as a temporary formula. This is obviously not a true chemical formula (cannot have 2.26 atoms – violates Dalton’s law). We must multiply the formula by a number which converts the fractional number of atoms of hydrogen into a whole number. In the above instance, if we multiply by four, we get C 4 H 9.

23. Work Time Finish warm-up sheet 11.4 notes Section assessment questions page 337 (58- 61)

24. C 4 H 9 This is an empirical formula. The true chemical formula could be a whole number multiple of the empirical formula. The chemical formula of the substance analyzed could be C 8 H 18 or any other whole number multiple. Molar mass of empirical formula 4 moles C x 12.01 g/mol C = 48.04 g C 9 moles of H x 1.01 g/mol H = 9.09 g H 57.13 g total

25. Vitamin C (ascorbic acid) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid? Assume 100 g. 40.92 g C x 1 mol C= 3.41 mol C 12.01 g 4.58 g H x 1 mol H= 4.53 mol H 1.01 g 54.50 g O x 1 mol O= 3.41 mol O 16.00 g

26. 40.92 g C x 1 mol C= 3.41 mol C 12.01 g 4.58 g H x 1 mol H= 4.53 mol H 1.01 g 54.50 g O x 1 mol O= 3.41 mol O 16.00 g To get empirical formula divide the all by the smallest # moles 3.41 mol C / 3.41 = 1 4.53 mol H / 3.41 = 1.33 3.41 mol O / 3.41 = 1 C 1 H 1.33 O 1 ???

27. Must multiply subscripts of the derived formula, C 1 H 1.33 O 1, by some factor to convert the fractional atom into a whole number. This is done by trial and error, but the multiples are usually small, whole numbers. x 2 = C 2 H 2.66 O 2 x 3 = C 3 H 4 O 3 molar mass = 88.07 g / mol This is the empirical formula, which may or may not be the molecular formula. It tells us only the relative numbers of atoms. What is the molecular formula for ascorbic acid whose molar mass is 176.13

28. C 3 H 4 O 3 molar mass = 88.07 g / mol molar mass compound = 176.14 g / mol 176.14 / 88.07 = 2 molecular formula ascorbic acid = C 6 H 8 O 6

29. The composition of ibuprofen is 75.7% C, 8.8% H and 15.5% O, by mass. Determine the mass percentage of each element and from that data determine the empirical formula. Start with 100 grams

30. 75.7 g from C, 8.8 g from hydrogen, 15.5 g from O 75.7 g C x 1 mol C = 6.30 mol C 12.01 g C 8.8 g H x 1 mol H = 8.70 mol H 1.01 g H 15.5 g O x 1 mol O = 0.969 mol 16.00 g O

31. DIVIDE EACH NUMBER OF MOLES BY THE SMALLEST TO GIVE THE FOLLOWING. 6.30 mol C / 0.969 mol O = 6.50 mol C / mol O 8.7 mol H / 0.969 mol O = 9.0 mol H / mol O 0.969 mol O / 0.969 mol O = 1 mol O / mol O Empirical formula = C 6.50 H 9 O

32. Elemental Analysis Combustion A compound of unknown composition is decomposed by heat. The elements are carefully trapped and the number of moles of each are analyzed.

33. A sample of a compound composed of carbon oxygen and hydrogen are combusted in a stream of O 2 to produce CO 2 and H 2 O. The H 2 O and CO 2 are trapped and the masses of each measured.

34. The sample has a mass of 0.255g. When the reaction is complete, 0.561 g of CO 2 and 0.306g of H 2 O are produced. What is the empirical formula of the compound? 1.Determine the mass of C in the sample. For each mole of CO 2, there is one mole of C. Convert moles of C to grams of C. 0.561 g CO 2 x (1 mol CO 2 / 44.01 g CO 2 ) x (1 mol C / 1 mol CO 2 ) (12.01 g C/ mol C) = 0.153 g C

35. There are 2 moles of hydrogen per mole of H 2 O. 0.306 g H 2 O x (1 mol H 2 O / 18.0 g H 2 O) x (2 mole H / mole H 2 O) x (1.01 g H / mol H) = 0.0343 g H Mass O = mass sample - mass H - mass C Mass sample = 0.255 g Mass O = 0.255 - 0.153 - 0.0343 = 0.068 g O

36. To get empirical formula, convert g back to moles 0.153 g C x ( 1 mol C / 12.01 g C ) = 0.0128 mol C 0.0343 g H x (1 mol H / 1.01 g H) = 0.0340 mol H 0.068 g O x (1 mol O / 16.0 g O) = 0.0043 mol O Divide each by 0.0043 to get ratio of each element to O C 0.0128 mol C / 0.0043 mol O = 2.98 ~ 3 There are 3 moles of carbon for each mole of oxygen H 0.0340 mol H / 0.0043 mol O = 7.91 ~ 8 There are 8 moles of hydrogen per mole of oxygen Empirical Formula C 3 H 8 O

37. Chemical Equations Short hand describing a chemical change using symbols and formulas to represent the elements and compounds involved in the change. Carbon reacts with molecular oxygen to yield carbon dioxide C + O 2 CO 2 The equation tells us that 1 mole of C reacts with 1 mole of O 2 to yield 1 mole of CO 2

38. Stoichiometric Calculations: Stoicheion (element) ; metron (measurement) Relating mass or moles of reactants in a chemical reaction to mass or moles of products. EQUATION MUST BE BALANCED Reaction of C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) If we have 15.0 g of C 2 H 4 how many moles of HCl are needed to carry out the reaction to completion?

39. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) The balanced equation tells us it takes 1 mole of HCl for each mole of C 2 H 4 reacted. We must begin by converting the number of grams of C 2 H 4 which has a molar mass of 28.08 g/mol, to moles 15.0 g C 2 H 4 x 1 mol C 2 H 4 = 0.534 mol C 2 H 4 28.08 g C 2 H 4

40. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) The coefficients in the balanced equation tells us that one mole of HCl is needed for each mole of C 2 H 4 (ethylene). It therefore takes 0.534 mol of HCl to consume all of the ethylene. What mass of HCl is needed to carry the reaction through to completion?

41. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) To determine the mass of HCl needed, use molar mass. Begin with what is asked. 0.534 mol HCl x 36.5 g HCl = 19.6 g HCl mol HCl

42. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) How many moles of product are made when 15.0 g of C 2 H 4 is reacted with an excess of HCl? Pay close attention to what is being asked. Note from balanced eqn that there is one mole of C 2 H 4 is needed to make one molecule of C 2 H 5 Cl. Note also that the question only refers to the reactant C 2 H 4 and product, not HCl.

43. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) To determine moles from mass use dimensional analysis. Molar mass of C 2 H 4 (ethylene) = 28.08 g/mol The problem states that you have 15.0 g C 2 H 4 15.0 g C 2 H 4 x 1mole C 2 H 4 = 0.534 mol C 2 H 4 28.06 g C 2 H 4 Since the stoichiometry of the balanced equation indicates that 1 mol of C 2 H 5 Cl is produced from each mole of C 2 H 4, the ratio is 1:1.

44. C 2 H 4 (g) + HCl(g) C 2 H 5 Cl(g) 0.534 mol 0.534 mol (molar mass C 2 H 5 Cl = 64.51 g/mol) Mass of product: 0.534 mol product x 64.51 g/mol = 34.448 g = 34.5 g in correct # sig fig (we started with 15.0 g reactant. Answer must be in 3 sig.fig).

45. Calculations of product masses when the stoichiometric coefficients are not the same. N 2 (g) + H 2 (g) NH 3 (g) Note from the balanced equation that two moles of product are yielded when 1 mole of N 2 reacts with 3 mole H 2. To determine the mass of the product, we follow the same steps, but now we must consider another factor, the mole ratio. 32

46. N 2 (g) + 3H 2 (g) 2NH 3 (g) If there is enough N 2 in the mixture and we have 4.8 mol H 2, how many moles of NH 3 are produced? The relationship is now 3:2 and we must take that into consideration. Again, begin with what you are given. You have 4.8 mol H 2.

47. N 2 (g) + 3H 2 (g)2NH 3 (g) 4.8 mol H 2 x 2 mol NH 3 = 3.2 mol NH 3 3 mol H 2 Mass of NH 3 is determined by calculating mass from moles using molar mass as before.

48. N 2 + 3H 2 2NH 3 1. Assuming there is enough hydrogen for the reaction to go to completion how many grams of nitrogen are needed to yield 0.430 moles of NH 3 ? 2. For the reverse reaction, how many grams of nitrogen are produced by the decomposition of 3.24 grams of gaseous NH 3 ?

49. Limiting Reactants Suppose you do not have enough of one reactant to carry the reaction to completion? To determine the moles or masses of products, we need to know which reactant will run out first and which is in excess. We cannot use coefficients or masses to make the determination. As usual, everything is measured in moles.

50. Limiting Reactants 1. Write the balanced equation. 2. Convert masses to moles. 3. The reactant producing the lowest moles of product will be consumed (limiting reactant).

51. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) If 0.30 mol Zn is added to 0.52 mol HCl, how many moles of H 2 will be produced?

52. Zn(s) + 2HCl(aq)ZnCl 2 (aq) + H 2 (g) 0.30 mol Zn x 1 mol H 2 = 0.30 mol H 2 1 mol Zn 0.52 mol HCl x 1 mol H 2 = 0.26 mol H 2 2 mol HCl The mole ratio conversions come directly from the stoichiometric coefficients in the balanced equation. 0.26 mol is the maximum amount of H 2 produced when 0.30 mol of Zn and 0.52 mol of HCl are mixed.

53. Grams of A Moles of A Moles of B Grams of B Use molar mass as a conversion factor Use coefficients from balanced equation as a conversion factor Use molar mass as a conversion factor

54. Percent Yield In most reactions not all reactants go forward to form products. Some are lost to byproducts and some do not react. The actual yield is usually less than the anticipated, theoretical yield. Percent yield = actual yield x 100 theoretical yield

55. Zn(s) + 2 HCl(aq)ZnCl 2 (aq) + H 2 (g) Theoretical yield is 0.26 mol H 2. If 0.19 moles are produced, what is the percent yield?

56. (0.19/0.26) x 100 = 73%

57. Balanced eqn Masses to moles Mol ratios from coefficients Compare mol ratios, which is limiting? Mole ratio of limiting reactant To products Moles product Mass product

58. You must practice these types of conversions to proceed any further in the class. We will be using these skills through the balance of the semester. Go through sample exercises while they are still fresh in your mind.