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Molar Conversions

The Mole. Molar Conversions. VERY. A large amount!!!!. What is the Mole?. A counting number (like a dozen or a pair) Avogadro’s number 6.02  10 23 1 mole = 6.02  10 23 things. (things can be atoms, molecules, formula units). A. What is the Mole?. HOW LARGE IS IT???.

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Molar Conversions

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  1. The Mole Molar Conversions

  2. VERY A large amount!!!! What is the Mole? A counting number (like a dozen or a pair) • Avogadro’s number 6.02  1023 • 1 mole = 6.02  1023 things. (things can be atoms, molecules, formula units)

  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. Example • How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules 1 mol 2.50 mol = 1.51  1024 molecules C12H22O11

  5. Mole- Volume • At the same temperature: • Equal volumes of gases contain equal numbers of moles • STP • Standard temperature and pressure • 1 mole of any gas at STP has a volume of : 22.4 L

  6. Example • How many moles of neon are in 44.8L of neon at STP? 1 mol 22.4 L 44.8 L = 2 mol

  7. Molar Mass How much do I weigh?? 

  8. Molar Mass • Mass of 1 mole (6.02 X 1023 atoms or molecules)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 when calculating

  9. Molar Mass ExamplesThis is easy! Just look at the periodic table! 30.97 g/mol 26.98 g/mol 4.00 g/mol 107.87 g/mol • Phosphorus • Aluminum • Helium • Silver

  10. Molar Mass Examples WATER: • H2O • Ca(OH)2 Ca 40.08 + O2(16) + H 2(1.01) = 74.10 g/mol x H → O → 2 1.01 g/mol = 2.02 g/mol x 1 16.0 g/mol = 16.0 g/mol 18.02 g/mol

  11. Molar Mass • sodium bicarbonate • sucrose • NaHCO3 • 84.01 g/mol • C12H22O11 • 342.34 g/mol

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

  15. 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

  16. 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 =

  17. 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

  18. Empirical/Molecular Formulas The Mole

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

  20. Empirical Formula Find mass (or %) of each element. (may be given) 2. Find moles of each element. (g to moles) 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

  21. 1.85 mol 1.85 mol 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

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

  23. Examples • An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. Determine the empirical formula. • Empirical Formula: CH • What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? • Empirical Formula: Fe2S3

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

  25. C. Molecular Formula 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.

  26. 28.1 g/mol 14.03 g/mol Molecular Formula • The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? empirical mass = 14.03 g/mol = 2.00 (CH2)2  C2H4

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