2. Unit 5: The Mole 6.02 X 10 23 Mr. Blake / Mr.Gower/ Mr. Heaton

3. Review: Atomic Masses  Elements occur in nature as mixtures of isotopes.  Carbon =98.89% 12 C (6 p+, 6 n) 1.11% 13 C (6p+, 7n) <0.01% 14 C (6p+, 8n)  Carbon’s atomic mass = 12.01 amu

4. Atomic Weights are relative C: # and massH: # and massMass Ratio 1 C = 12.0 amu1 H = 1.0 amu 12 to 1 2 C = 24.0 amu 2 H = 2.0 amu 12 to 1 3 C = 36.0 amu 3 H = 3.0 amu 12 to 1 4 C = 48.0 amu 4 H = 4.0 amu 12 to 1 5 C = 60.0 amu 5 H = 5.0 amu 12 to 1 10 C = 120.1 amu10 H = 10.1 amu 12 to 1

5. What is the mole? Not this kind of mole!

6. A. What is the Mole? A counting number (like a dozen). 1 mol = 602,200,000,000,000,000,000,000 of anything. A large amount!!!!

7. Similar Words Pair: 1 pair of shoelaces = 2 shoelaces. Dozen: 1 dozen oranges = 12 oranges. Gross: 1 gross of spider rings = 144 rings. Ream: 1 ream of paper = 500 sheets of paper. Mole: 1 mole of Na atoms = 6.022 X 10 23 Na atoms.

8. The Mole So 1 mole of any element has 6.022 X 10 23 particles. This is a really big number – It’s so big because atoms are very small! Defined as the number of atoms in 12.0 grams of C-12. This is the standard! 12.0 g of C-12 has 6.022 X 10 23 atoms.

9. The Mole Amedeo ______________ (1776 – 1856)This number is named in honor of Amedeo ______________ (1776 – 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present 6.022 x 10 23 is also called _______________________. Avogadro Avogadro’s number (N A )

10. Amadeo Avogadro I didn’t discover it. Its just named after me!

11. Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

12. Everybody Knows Avogadro’s Number! But Where Did it Come From? It was NOT just picked! It was measured. One of the better methods of measuring this number was the Millikan Oil Drop Experiment. Since then we have found even better ways of measuring using x- ray technology.

13. The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.022 X 10 23 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.022 X 10 23 cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.022 X 10 23 atoms -Note that the NUMBER is always the same, but the MASS is very different! Mole is abbreviated mol

14. = 6.022 x 10 23 C atoms = 6.022 x 10 23 H 2 O molecules = 6.022 x 10 23 NaCl formula units (technically, ionics are compounds not molecules so they are called formula units) 6.022 x 10 23 Na + ions and 6.022 x 10 23 Cl – ions A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles 1 mole C 1 mole H 2 O 1 mole NaCl

15. Suppose we invented a new collection unit called a widget. One widget contains 8 objects. 1. How many paper clips in 1 widget? a) 1b) 4c) 8 2. How many oranges in 2.0 widgets? a) 4b) 8c) 16 3. How many widgets contain 40 gummy bears? a) 5b) 10c) 20 Learning Check

16. 6.022 x 10 23 particles 1 mole or 1 mole 6.022 x 10 23 particles - Note that a particle could be an atom OR a molecule! Avogadro’s Number as Conversion Factor

17. 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.022 x 10 23 Al atoms c) 3.01 x 10 23 Al atoms 2.Number of moles of S in 1.8 x 10 24 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 10 48 mole S atoms Learning Check

18. B. Molar Mass Mass of 1 mole of an element or compound. Equal to the numerical value of the average atomic mass (get from periodic table) Atomic mass tells the... – atomic mass units per atom (amu) – grams per mole (g/mol) Round to 2 decimal places (hundredths)

19. 1 mole of C atoms= 12.01 g 1 mole of Mg atoms =24.31 g 1 mole of Cu atoms =63.55 g Molar Masses of Elements (Found in Periodic Table) Carried to the hundredths place!!

20. B. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol 65.39 g/mol

21. Calculating Molecular Mass Calculate the molecular mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mol

22. 1 mole Ca = 40.08 g +2 moles Cl = 70.90 g Therefore,the mass of the entire compound is… 1 mole of CaCl 2 = 110.98 g Molar Mass of Molecules and Compounds A substance’s molecular mass (molecular weight) is the mass in grams of one mole of the compound.

23. –H2O–H2O –2(1.01) + 16.00 = 18.02 g/mol –NaCl –22.99 + 35.45 = 58.44 g/mol B. Molar Mass Examples 1) water 2) sodium chloride

24. –NaHCO 3 –22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol –C 6 H 12 O 6 –6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol B. Molar Mass Examples 3) sodium bicarbonate 4) glucose

25. C. Molar Conversions molar mass (g/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES 6.022  10 23 (particles/mol)

26. molar mass Grams Moles Calculations with Molar Mass

27. Converting moles to grams Ex. #1 How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 1 mol Li 6.94 g Li = g Li24.3

28. Ex. #2 Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? Goal: 3.00 moles Al ? g Al Converting Moles and Grams

29. 1. Molar mass of Al 1 mole Al = 26.98 g Al 2. Conversion factors for Al 26.98 g Al or 1 mol Al 1 mol Al 26.98 g Al 3. Setup 3.00 moles Al x 26.98 g Al 1 mole Al Answer = 80.9 g Al

30. Molar Conversion Examples 3. How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

31. 4) How many molecules are in 2.50 moles of C 12 H 22 O 11 ? 2.50 mol 6.022  10 23 molecules 1 mol = 1.51  10 24 molecules C 12 H 22 O 11

32. molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

33. Ex. #1 Find the mass of 2.1  10 24 formula units of NaHCO 3. 2.1  10 24 f.u.s 1 mol 6.022  10 23 f.u.s = 290 g NaHCO 3 84.01 g 1 mol NaHCO 3.

34. Ex. #2 How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.022 X 10 23 atoms Cu 63.55 g Cu 1 mol Cu = 3.35 X 10 23 atoms Cu

35. V. Percent Composition: The percent by ______ of each _________ in a compound. masselement What is the % composition of Ba(CN) 2 ? 1(137.33 g Ba) + 2(12.01 g C) + 2(14.01 g N) = 189.37 g Ba(CN) 2 72.52 % Ba 12.68 % C 14.80 % N  100 x g Ba137.33  Ba% gBa(CN)189.37 2

36. What is the % composition of Iron III sulfate, _________? Fe 2 (SO 4 ) 3 2(55.85 g Fe) + 3(32.07 g S) + 12 (16.00 g O) = 399.91 g Fe 2 (SO 4 ) 3 24.06 % S 48.01 % O 27.93 % Fe  100 x g Fe2(55.85)  Fe% 399.91 g Fe 2 (SO 4 ) 3

37. Percent Composition (Hydrates): 1. Ba(CN) 2 ∙ 2 H 2 O is what % water by mass? 1(Ba) + 2(C) + 2(N) + 2(H 2 O) = 225.41 g Ba(CN) 2 2 H 2 O 15.99 % H 2 O

38. Formulas  molecular formula = (empirical formula)n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

39. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

40. Formulas (continued) Formulas for molecular compounds might be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

41. Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers. 100% = 100 g

42. Empirical Formula Determination Adipic acid contains 49.32% C, 6.85% H, & 43.84% O by mass. What is the empirical formula of adipic acid? 49.32g C 12.01 g C 4.107 mol C = 6.85 g H 1.01 g H 6.78 mol H = 1 mol C x x 1 mol H 43.84 g O 16.00 g O 2.74 mol O = x 1 mol O

43. Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen: 4.107 1.50 2.74 molC molO = 6.78 2.47 2.74 molH molO = 2.74 1.00 2.74 molO =

44. Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

45. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the empirical mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

46. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the mass given by the emipirical formula. 146 2 73 

47. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4 146 2 73 =

48. Hydrates A. An _______ compound that has _______ trapped in the crystal lattice structure. 1. ________________: A compound with water (i.e. BaI 2 2 H 2 O). 2. __________________: A compound without water (i.e. BaI 2 ). B. Determination of a hydrate: 1. Determine the mass of ________. 2. Determine the mass of the _________________. 3. Find the __________ of water and anhydrous salt. 4. Find the mol ratio. X =* Whole number 5. Write the formula. H2OH2O Hydrated salt Anhydrous salt moles ionic H2OH2O anhydrous salt salt anhydrous mol OH 2

49. C. Example: 1.A barium iodide hydrate is heated in a crucible. Determine the formula of the hydrate given the following data: Before heating: 10.407 g BaI 2 X H 2 O (Hydrated Salt) After heating: 9.520 g BaI 2 (Anhydrous Salt) Step 1: Mass of water? ____________ _ 0.887 g H 2 O Step 2: Mass of Anhydrous salt? Step 3: Calculate the mol of the water & the anhydrous salt. Step 4: Calculate the number of waters. Step 5: 9.520 g BaI 2 = 0.0492 mol H 2 O 1 mol H 2 O 18.02 g H 2 O 0.887 g H 2 O 1 mol BaI 2 391.13 g BaI 2 9.520 g BaI 2 = 0.0243 mol BaI 2 = 2 BaI 2 2 H 2 O salt anhydrous mol OH 2 = x 2 2 BaI mol 0.0243 OH mol 0.0492 =

50. 2. Data: Before heating: 5.20 g ZnSO 3 X H 2 O (Hydrated salt) After heating:2.60 g ZnSO 3 (Anhydrous salt) ____________ _ Step 1: Mass of water?2.60 g H 2 O Step 2: Mass of Anhydrous salt? Step 3: Calculate the mol of the water & the anhydrous salt. Step 4: Calculate the number of waters. Step 5: 2.60 g ZnSO 3 = 0.144 mol H 2 O 1 mol H 2 O = 8 ZnSO 3 8 H 2 O 18.02 g H2O 2.60 g H 2 O = 0.0179 mol ZnSO 3 1 mol ZnSO 3 145.43 g ZnSO 3 2.60 g ZnSO 3

51. Using your knowledge of mole calculations and unit conversions, determine how many atoms there are in 1 gallon of gasoline. Assume that the molecular formula for gasoline is C 6 H 14 and that the density of gasoline is approximately 0.85 grams/mL. x 1 gal 4 qt 1 gal x 1.057 qt 1 L x 10 -3 L 1 mL x mL 0.85 g C 6 H 14 x 200.59 g 1 mol Hg = 123 mol Hg

52. Using a conversion factor of 3785 mL per gallon, we can determine that the mass of gasoline in one gallon is 3785 mL x 0.8500 g/mL = 3217 grams. Because the molar mass of C 6 H 14 is 86 g/mole, there are 3217 / 86 moles of gasoline molecules, or 37.4 moles of molecules present. Multiplying 37.4 x 20 (the number of atoms per mole of gasoline), there are 748 moles of atoms. Finally, multiplying 748 moles of atoms by 6.02 x 10 23 atoms/mole, we can find that there are 4.50 x 10 26 atoms present in the sample.

53. Answers for Station Review Station 1 1.Aluminum Oxide2. Iodine3. Iron (II) oxide 4.Nitrogen trioxide5. Manganese (III) oxide 6. Calcium phosphide Station 2 1. 197.84 g/mol2. 164.10 g/mol 3.18.02 g/mol 4. 78.05 g/mol Station 3 591 g Br 2

54. Station 4 0.966 mol KBr Station 5 All of the same Station 6 3.04 x10 24 atoms Station 7 8.26 g Na 2 SO 4 Station 8 Na: 46.91%C: 24.51 %N: 28.59%

55. I. Determination of Empirical Formula: 1. Laboratory analysis shows that a compound is 5.80 % H and 94.2 % S by weight (mass). What is its E.F.? Step 1:Calculate the moles of each element present. (Assume 100 % = 100 g) H2SH2S 2 atoms H : 1 atom S 2 mol H : 1 mol S 5.80 g Hx 1.01 g 1 mol H = 5.74 mol H 94.2 g Sx 32.07 g 1 mol S = 2.94 mol S

56. Step 2:Determine the ratio between moles (smallest whole numbers) by dividing by the smallest value in step 1. H2SH2S 2 Name? Hydrosulfuric acid +/- 0.05 of the whole number

57. 2.Remember that these values are measurements and need not be exact. They should be close. What would be the formulas for elements “X” and “Y”. (Step 2 is done) X 1.00 Y 1.03 X 1.5 Y 1.0 X 1.00 Y 1.33 X 1.00 Y 1.25 3.(a) Laboratory analysis shows that a compound has the following composition: Zinc 52.0 %,Carbon 9.60 %, & Oxygen 38.4 %. Determine its empirical formula. 2 X3Y2X3Y2 XY 3 X3Y4X3Y4 4 X4Y5X4Y5 52.0 g Znx 65.39 g 1 mol Zn = 0.80 mol Zn 9.60 g Cx 12.01 g 1 mol C = 0.80 mol C 38.4 g Ox 16.00 g 1 mol O = 2.4 mol O 0.80 = 1 = 3 ZnCO 3 Name? Zinc carbonate +/- 0.05 of the whole number

58. (b) What is the empirical formula of a compound that is 52.9 % Al and 47.1 % oxygen? J. Determination of the molecular formula: 1. You must be given the _________ weight. 2. What is the relationship between the empirical formula weight and the molecular formula weight? 3. Molecular formula = Empirical formula x _________ 52.9 g Alx 26.98 g 1 mol Al = 1.96 mol Al 47.1 g Ox 16.00 g 1 mol O = 2.94 mol O 1.96 = 1 = 1.5 Al 2 O 3 Name? Aluminum oxide x 2 molecular Molecular formula is a multiple of empirical formula! multiple given calculate

59. 4.Analysis shows that a compound is 30.4 % Nitrogen and 69.6 % Oxygen. If it has a molecular weight of 92 g, what is its Molecular formula? Step 1: Calculate the Empirical formula. Step 2:Calculate the Empirical formula weight. 30.4 g Nx 14.01 g 1 mol N = 2.17 mol N 69.6 g Ox 16.00 g 1 mol O = 4.35 mol O 2.17 = 1 = 2 NO 2 Name? Nitrogen dioxide 1 (14.01) + 2 (16.00) = 46.01 g

60. Step 3: Divide the Empirical formula weight into molecular weight (must be given). Step 4: Determine the Molecular formula. = 2 2(NO 2 )= N 2 O 4 Name? Dinitrogen tetroxide Note: Follow normal rounding rules to round to a whole number.