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Exploring The Mole Concept: Avogadro's Number, Atomic Masses, and Molar Conversions

Understand the concept of the mole, Avogadro's number, and molar masses with examples and calculations. Learn about atomic masses, elements, and converting moles to grams. Discover the importance of Avogadro's number in chemistry.

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Exploring The Mole Concept: Avogadro's Number, Atomic Masses, and Molar Conversions

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  1. Unit 6: The Mole Mr. Blake 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 = 602,000,000,000,000,000,000,000 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 = 6.02 X 1023 Na atoms.

  7. The Mole • So 1 mole of any element has6.02 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 • 6.02 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 = 6.02 X 1023 cookies • 1 dozen cars = 12 cars • 1 mole of cars = 6.02 X 1023 cars • 1 dozen Al atoms = 12 Al atoms • 1 mole of Al atoms = 6.02 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 = 6.02 x 1023 C atoms =6.02 x 1023 H2O molecules = 6.02 x 1023 NaCl “molecules” (technically, ionics are compounds not molecules so they are called formula units) 6.02 x 1023 Na+ ions and 6.02 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 6.02 x 1023 particles 1 mole or 1 mole 6.02 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) 6.02 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 x 40.08 g/mol= 40.08 g/mol +2 moles Cl x 35.45 g/mol=70.90 g/mol Therefore,the mass of the entire compound is… 1 mole of CaCl2 = 110.98 g/mol

  22. B. Molar Mass Examples • water • 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 • C12H22O11 • 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol • sodium bicarbonate • sucrose

  24. molar mass 6.02  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 Converting moles to grams Ex. #1 How many grams of lithium are in 3.50 moles of lithium? = g Li 24.3

  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 =81.0 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? 6.02  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 molecules of NaHCO3. 2.1  1024 molecules 1 mol 6.02  1023 molecules 84.01 g 1 mol = 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 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu =3.4 X 1023 atoms Cu

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

  41. ( ) ( ) 49.32g C = 4.107 molC ( ) 12.01 gC ( ) ( ) 6.851 gH 1 molH = 6.78 molH ( ) 1.01 gH ( ) ( ) 43.84 g O 1molO = 2.74 molO ( ) 16.00 gO Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1 mol C

  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 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. Hydrated salt 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. _ ____________ 2.60 g H2O Step 1: Mass of water? 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. IQ #2 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.

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