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Ch. 7 - Molar Relationships

Ch. 7 - Molar Relationships. I. The Mole (p. 170-183). VERY. A large amount!!!!. A. What is the Mole?. A counting number (like a dozen) Avogadro’s number (N A ) 1 mol = 6.02  10 23 items. A. What is the Mole?. HOW LARGE IS IT???.

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Ch. 7 - Molar Relationships

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  1. Ch. 7 - Molar Relationships I. The Mole (p. 170-183)

  2. VERY A large amount!!!! A. What is the Mole? • A counting number (like a dozen) • Avogadro’s number (NA) • 1 mol = 6.02  1023 items

  3. A. What is the Mole? HOW LARGE IS IT??? • 1 mole of pennies would cover the Earth 1/4 mile deep! • 1 mole of hockey pucks would equal the mass of the moon! • 1 mole of basketballs would fill a bag the size of the earth!

  4. B. Molar Mass • Mass of 1 mole of an element or compound. • Atomic mass tells the... • atomic mass units per atom (u) • grams per mole (g/mol) • Round to 2 decimal places

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

  6. 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

  7. B. Molar Mass Examples • sodium bicarbonate • sucrose • 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

  8. MASS IN GRAMS MOLES NUMBER OF PARTICLES C. Molar Conversions molar mass 6.02  1023 (g/mol) (particles/mol)

  9. C. Molar Conversion Examples • 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

  10. C. Molar Conversion Examples • How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules 1 mol 2.50 mol = 1.51  1024 molecules C12H22O11

  11. C. Molar Conversion Examples • 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

  12. Molar Relationships II. Molarity

  13. substance being dissolved total combined volume A. Molarity • Concentration of a solution.

  14. A. Molarity 2M HCl What does this mean?

  15. MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION B. Molarity Calculations 6.02  1023 (particles/mol) molar mass (g/mol) Molarity (mol/L)

  16. M = B. Molarity Calculations • Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g 1 mol 41.99 g = 0.238 mol NaF 0.238 mol 0.25 L = 0.95M NaF

  17. B. Molarity Calculations • How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L 0.25 mol 1 L 58.44 g 1 mol = 7.3 g NaCl

  18. Ch. 7 - Molar Relationships III. Percent Composition

  19. A. Definition • the percent by mass of each element in a compound • to distinguish between similar compounds • to determine elemental mass in a sample WHY?

  20. 55.85 g 71.85 g 16.00 g 71.85 g B. Calculation of % Comp • Find the % composition of FeO.  100 = 77.73% Fe %Fe =  100 = 22.27% O %O =

  21. 111.70 g 159.70 g 48.00 g 159.70 g B. Calculation of % Comp • Find the % composition of Fe2O3. 100= 69.944% Fe %Fe =  100 = 30.06% O %O =

  22. 28 g 36 g C. Compound Identification • A compound contains 28 g Fe and 8.0 g O. Is it FeO or Fe2O3?  100 = 78% Fe %Fe = The compound is FeO.

  23. D. Elemental Mass in a Sample • How many grams of iron are in a 38.0-gram sample of iron(III) oxide? Fe2O3 is 69.944% Fe 38.0 g Fe2O3 69.944 g Fe 100 g Fe2O3 = 26.6 g Fe - or - (38.0 g Fe2O3)(0.69944) = 26.6 g Fe

  24. CHALLENGE PROBLEM • A mining company needs to choose a mine site. Which would provide a better source of iron? Ore containing 32% FeO or ore containing 48% Fe2O3? FeO is 77.73% Fe Fe2O3 is 69.944% Fe SOLUTION

  25. 36.04 g 147.02 g E. % Water in a Hydrate • CaCl2·2H2O • calcium chloride dihydrate  100 = 24.51% H2O %H2O =

  26. Ch. 7 - Molar Relationships IV. Formula Calculations

  27. CH3 B. Empirical Formulas • Smallest whole number ratio of atoms in a compound C2H6

  28. B. Empirical Formulas 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

  29. 1.85 mol 1.85 mol B. Empirical Formulas • Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O

  30. N2O5 B. Empirical Formulas N1O2.5 Need to make the subscripts whole numbers  multiply by 2

  31. C2H6 C. Molecular Formulas • “True Formula” - the actual number of atoms in a compound CH3 empirical formula molecular formula

  32. C. Molecular Formulas 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

  33. 28.06 g/mol 14.03 g/mol C. Molecular Formulas • The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.06 g/mol? empirical mass = 14.03 g/mol = 2 (CH2)2  C2H4

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