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Unit 5: The Mole

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

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Unit 5: The Mole

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  1. Unit 5: The Mole Mr. Blake / Mr.Gower/ Mr. Heaton 6.02 X 1023

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

  3. Atomic Weights are relative

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

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

  6. 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 1023 Na atoms.

  7. The Mole • So 1 mole of any element has6.022 X 1023 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 1023 atoms.

  8. The Mole • 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 1023 is also called _______________________. Avogadro Avogadro’s number (NA)

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

  10. 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.

  11. 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.

  12. The Mole • 1 dozen cookies = 12 cookies • 1 mole of cookies = 6.022 X 1023 cookies • 1 dozen cars = 12 cars • 1 mole of cars = 6.022 X 1023 cars • 1 dozen Al atoms = 12 Al atoms • 1 mole of Al atoms = 6.022 X 1023 atoms - Note that the NUMBER is always the same, but the MASS is very different! • Mole is abbreviated mol

  13. A Mole of Particles Contains 6.02 x 1023 particles = 6.022 x 1023 C atoms =6.022 x 1023 H2O molecules = 6.022 x 1023 NaCl formula units (technically, ionics are compounds not molecules so they are called formula units) 6.022 x 1023 Na+ ions and 6.022 x 1023 Cl– ions 1 mole C 1 mole H2O 1 mole NaCl

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

  15. Avogadro’s Number as Conversion Factor 6.022 x 1023 particles 1 mole or 1 mole 6.022 x 1023 particles - Note that a particle could be an atom OR a molecule!

  16. Learning Check 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.022 x 1023 Al atoms c) 3.01 x 1023 Alatoms 2.Number of moles of S in 1.8 x 1024 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 1048 mole S atoms

  17. 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)

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

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

  20. Calculating Molecular Mass Calculate the molecular mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mol

  21. Molar Mass of Molecules and Compounds A substance’s molecular mass(molecular weight) is the mass in grams of one mole of the compound. 1 mole Ca = 40.08 g + 2 moles Cl =70.90 g Therefore,the mass of the entire compound is… 1 mole of CaCl2 = 110.98 g

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

  23. B. Molar Mass Examples • NaHCO3 • 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol • C6H12O6 • 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol 3) sodium bicarbonate 4) glucose

  24. molar mass 6.022  1023 MASS IN GRAMS MOLES NUMBER OF PARTICLES (g/mol) (particles/mol) C. Molar Conversions

  25. Calculations with Molar Mass molar mass Grams Moles

  26. 3.50 mol Li 6.94 g Li 1 mol Li = g Li 24.3 Converting moles to grams Ex. #1 How many grams of lithium are in 3.50 moles of lithium?

  27. Converting Moles and Grams Ex. #2Aluminum 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

  28. 1. Molar mass of Al1 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

  29. 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

  30. 4) How many molecules are in 2.50 moles of C12H22O11? 6.022  1023 molecules 1 mol 2.50 mol = 1.51  1024 molecules C12H22O11

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

  32. Ex. #1 Find the mass of 2.1  1024 formula units of NaHCO3. 2.1  1024 f.u.s 1 mol 6.022  1023 f.u.s 84.01 g 1 mol NaHCO3. = 290 g NaHCO3

  33. 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 1023 atoms Cu 63.55 g Cu 1 mol Cu =3.35 X 1023 atoms Cu

  34. V. Percent Composition: The percent by ______ of each _________ in a compound. mass element 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)2 137.33 g Ba 72.52 % Ba = = % Ba x 100 189.37 g Ba(CN) 2 12.68 % C 14.80 % N

  35. Fe2(SO4)3 2(55.85 g Fe) + 3(32.07 g S) + 12 (16.00 g O) = 399.91 g Fe2(SO4)3 What is the % composition of Iron III sulfate, _________? 27.93 % Fe 399.91 g Fe2(SO4)3 2(55.85) g Fe = = % Fe x 100 24.06 % S 48.01 % O

  36. Percent Composition (Hydrates): 1. Ba(CN)2 ∙ 2 H2O is what % water by mass? 1(Ba) + 2(C) + 2(N) + 2(H2O) = 225.41 g Ba(CN)2• 2 H2O 15.99 % H2O

  37. Formulas • Empirical formula: the lowest whole number • ratio of atoms in a compound. • molecular formula = (empirical formula)n [n = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH • Molecular formula: the true number of atoms • of each element in the formula of a compound.

  38. Formulas(continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: K2CO3 NaCl MgCl2 Al2(SO4)3

  39. Formulas(continued) Formulas for molecular compounds might be empirical (lowest whole number ratio). Molecular: C6H12O6 H2O C12H22O11 Empirical: H2O CH2O C12H22O11

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

  41. x x x = 4.107 mol C = 6.78 mol H = 2.74 mol O 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? 1 mol C 49.32g C 12.01 g C 1 mol H 6.85 g H 1.01 g H 1 mol O 43.84 g O 16.00 g O

  42. 4.107 molC = 1.50 2.74 molO 6.78 molH = 2.47 2.74 molO 2.74 molO = 1.00 2.74 molO Empirical Formula Determination(part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

  43. 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 x 2 x 2 3 5 2 Empirical formula: C3H5O2

  44. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the empirical mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  45. 146 = 2 73 Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  46. 146 = 2 73 Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

  47. Hydrates ionic H2O Hydrated salt A. An _______ compound that has _______ trapped in the crystal lattice structure. 1. ________________: A compound with water (i.e. BaI2 •2 H2O). 2. __________________: A compound without water (i.e. BaI2). 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. Anhydrous salt H2O anhydrous salt moles mol H O 2 mol anhydrous salt

  48. _ ____________ 0.0492 mol H O mol H O 2 2 = = x mol anhydrous salt 0.0243 mol BaI 2 C. Example: • A barium iodide hydrate is heated in a crucible. Determine the formula of the hydrate given the following data: Before heating: 10.407 g BaI2 • X H2O (Hydrated Salt) After heating: 9.520 g BaI2 (Anhydrous Salt) Step 1: Mass of water? 0.887 g H2O Step 2: Mass of Anhydrous salt? 9.520 g BaI2 Step 3: Calculate the mol of the water & the anhydrous salt. 0.887 g H2O 1 mol H2O = 0.0492 mol H2O 18.02 g H2O 9.520 g BaI2 1 mol BaI2 = 0.0243 mol BaI2 391.13 g BaI2 Step 4: Calculate the number of waters. = 2 Step 5: BaI2• 2 H2O

  49. _ ____________ Step 1: Mass of water? 2.60 g H2O 2. Data: Before heating: 5.20 g ZnSO3 • X H2O (Hydrated salt) After heating: 2.60 g ZnSO3 (Anhydrous salt) Step 2: Mass of Anhydrous salt? 2.60 g ZnSO3 Step 3: Calculate the mol of the water & the anhydrous salt. 2.60 g H2O 1 mol H2O = 0.144 mol H2O 18.02 g H2O 2.60 g ZnSO3 1 mol ZnSO3 = 0.0179 mol ZnSO3 145.43 g ZnSO3 Step 4: Calculate the number of waters. = 8 Step 5: ZnSO3• 8 H2O

  50. x x x x x = 123 mol Hg 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 C6H14 and that the density of gasoline is approximately 0.85 grams/mL. 0.85 g C6H14 1 mol Hg 1 L 4 qt 1 mL 1 gal 1 gal 1.057 qt 10-3 L mL 200.59 g

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