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Molar Conversions

Topic 9 The Mole. Molar Conversions. A. What is the Mole?. A counting number (like a dozen) Avogadro’s number (N A ) 1 mol = 6.02  10 23 representative particles (atoms, ions, formula units, or molecules.

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Molar Conversions

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  1. Topic 9 The Mole Molar Conversions

  2. A. What is the Mole? • A counting number (like a dozen) • Avogadro’s number (NA) • 1 mol = 6.02  1023 representative particles (atoms, ions, formula units, or molecules If you could spend 1 million dollars per second, how many years will it take to spend a mole of dollars, given there are 31536000 seconds per year. 1.91 x 1010 years or 19.1 billion years

  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. A. Moles • Used as a Conversion Factor: • 1 mole = 6.02 x 1023 so… • Moles converted to Particles = # mol 6.02  1023 Particles 1 mol

  5. A. Moles • Rep. Particles converted to moles = # particles 1 mol 6.02  1023 Particles

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

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

  8. 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

  9. B. Molar Mass Examples • sodium hydrogen carbonate • 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

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

  11. 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

  12. 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

  13. 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

  14. Topic 9 – The Mole II. Formula Calculations

  15. A. Percentage Composition • the percentage by mass of each element in a compound

  16. 127.10 g Cu 159.17 g Cu2S 32.07 g S 159.17 g Cu2S A. Percentage Composition • Find the % composition of Cu2S.  100 = %Cu = 79.852% Cu %S =  100 = 20.15% S

  17. 28 g 36 g 8.0 g 36 g A. Percentage Composition • Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.  100 = 78% Fe %Fe =  100 = 22% O %O =

  18. A. Percentage Composition • How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.852% Cu (38.0 g Cu2S)(0.79852) = 30.3 g Cu

  19. 36.04 g 147.02 g A. Percentage Composition • Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 24.51% H2O %H2O =  100 =

  20. CH3 B. Empirical Formula • Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts

  21. B. Empirical Formula Calculation 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.

  22. 1.85 mol 1.85 mol B. Empirical Formula • 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

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

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

  25. C. Molecular Formula Calculation 1. Find the empirical formula. 2. Calculate the empirical formula molar mass. 3. Divide the molecular molar mass by the empirical molar mass. 4. Multiply each subscript by the answer from step 3.

  26. 28.06 g/mol 14.03 g/mol C. Molecular Formula • 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 = 2.00 (CH2)2  C2H4

  27. Topic 9 The Mole III. Molarity

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

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

  30. 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) Aqueous solution

  31. 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

  32. 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

  33. StandardTemperature&Pressure 0°C and 1 atm C. Molar Volume at STP 1 mol of any gas=22.4 L at STP

  34. C. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

  35. C. Molar Conversion Examples • How many Liters are in 2.50 moles of of Helium gas at STP? 22.4 L 1 mol 2.50 mol = 56.0 L He

  36. C. Molar Conversion Examples • Find the mass of 137 L of SO2 gas at STP. 137 L SO2 1 mol 22.4 L SO2 64.07 g 1 mol = 392 g SO2

  37. Find the number of Liters of CO2 at STP if it has a mass of 132 grams.

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