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DO NOW: (Refresh your memory)

DO NOW: (Refresh your memory). Classify the following as an atom, molecule, ion, or formula unit: 1.Fe _________ 2.F 2 _________ 3.H 2 O _________ 4.Na _________ 5.NaCl _________ 6.Na +1 _________. DO NOW: (Refresh your memory).

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DO NOW: (Refresh your memory)

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  1. DO NOW: (Refresh your memory) Classify the following as an atom, molecule, ion, or formula unit: 1.Fe _________ 2.F2 _________ 3.H2O _________ 4.Na _________ 5.NaCl _________ 6.Na+1 _________

  2. DO NOW: (Refresh your memory) Classify the following as an atom, molecule, ion, or formula unit: 1.Fe ATOM 2.F2 MOLECULE 3.H2O MOLECULE 4.Na ATOM 5.NaCl FORMULA UNIT 6.Na+1 ION

  3. Chapter 10 THE MOLE!!!

  4. What is a mole?

  5. What is a mole? • It is a unit of measure. • 6.02 x 1023 • If we write this out (standard notation), we get: • 602,000,000,000,000,000,000,000

  6. A mole is equal to: • 6.02 x 1023 Representative Particles • Atoms, molecules, formula units, and ions Fe F2 NaCl Na+1 • Now we have a conversion unit to use: • 1mole 6.02 x 1023 r.p. (any of the four choices above)

  7. Lets do some problems • 2.1 moles of Fe = how many r.p. of Fe? • First what r.p. is Fe? (atom) 2.1 moles Fe 6.02 x 1023 atoms Fe 1mole Fe • 2.1moles Fe = 1.3x 1024 atoms Fe

  8. 140 moles of MgCl2 = how many r.p. of Cl-? • First what r.p. is MgCl2 and then Cl-? (formula unit, ion) 140 moles MgCl2 6.02 x 1023 f.u. MgCl2 2 ions Cl- 1mole MgCl2 1 f.u. MgCl2 140 moles MgCl2 = 1.7 x 1026 ions Cl-

  9. Gram Formula Mass & Gram Molar Mass (molar mass) • Use individual atomic masses to determine overall mass • In 1 mole of NaCl, there are 58 g of NaCl 23 + 35 = 58 • In 1 mole of H2O, there are 18 g of H20 2(1) + 16 = 18

  10. A mole is equal to: • The molar mass of a substance • Now we have a conversion unit to use: • 1mole molar mass of a substance EX: what is the gfm of C6H12O6 180g/mol So, in 1 mole of C6H12O6, there are 180 g 6(12) + 12(1) + 6(16) = 180

  11. Lets do some problems • 11.3 moles of C6H12O6 = how many grams? 11.3 moles C6H12O6 180 g C6H12O6 1mole C6H12O6 11.3 moles C6H12O6 = 2030 g C6H12O6 = 2.03 x 103 g C6H12O6

  12. A mole is equal to: • The 22.4L of gas at STP • STP= Standard Temperature and Pressure • = 0o C or 273 K and • (101.3 kPa or 1 atm or 760 mmHg or 760 torr) • Now we have a conversion unit to use: • 1mole 22.4L of a gas @ STP

  13. Molar Massgfm or gmm of a substance 6.02x1023 r.p. (Avogadro’s #) 1 mole of a substance 22.4 Lof gas per mole @ STP THE MOLE ISLAND!!!!

  14. Lets do some problems • 5.3x1024 molecules of CH4 = how many grams CH4? !!!!! You must go from the r.p. island then to the mole island, and then to the molar mass island!!!!! 5.3x1024 molecules of CH4 1 mole CH4 16g CH4 6.02x1023r.p. CH4 1 mole CH4 5.3x1024 molecules of CH4= 140 g CH4 = 1.4 x 102 g CH4

  15. Chapter 10 Percent Composition

  16. Calculating % Composition Chemists use this calculation when new compounds are created in the lab and they would have to determine the formula of the cmpd.

  17. Formula % mass of element = grams of element x 100 grams of cmpd

  18. Lets do some problems • An 8.20g piece of magnesium combines completely with 5.40g of oxygen to form a compound. What is the % composition of Magnesium and Oxygen the cmpd?

  19. Step 1 : Add masses to get total 8.20g + 5.40g = 13.60g • Step 2 : Find the % of each element. Mg: (8.20g/13.60g)*100 = 60.3% O: (5.40g/13.60g)*100 = 39.7% • Step 3 : Make sure your %s add up to 100.

  20. Chapter 10 Empirical Formulas

  21. Calculating Empirical Formulas Gives the lowest whole number ratio of the atoms of the elements in a cmpd.

  22. Remember the Poem % to mass Mass to mole Divide by the smaller # And multiply until whole

  23. Lets do some problems • What is the empirical formula of the cmpd that is 25.9% N and 74.1% O? 1. 25.9% N = 25.9g N 74.1% O = 74.1g O 2. 25.9g N 1mole N =1.85 mole N 14 g N 74.1 g O 1 mole O =4.63 mole O 16 g O

  24. 1.85 = 1.00 mole for N 4.63 = 2.50 mole for O 1.85 1.85 Make into whole #s by multiplying by 2 1 mole N x 2 = 2 moles N 2.5 mole O x 2 = 5 moles O So, the empirical formula is N2O5

  25. Empirical vs. Molecular Formulas (lowest whole # ratio) vs (multiple of empirical)

  26. Calculating Molecular Formulas 1. Go through all of the steps of an empirical formula problem. (poem) Ex: CH3 2. Add up the mass of the empirical formula. Ex: 15g/mol (efm)

  27. 3. Divide the mass of the molecular formula (gfm), which will be given in the problem (30g/mol), by the mass of the empirical formula (efm). (gfm/efm) Ex: 30g/mol = 2 15g/mol

  28. 4. Multiply all of the subscripts in the empirical formula by the 2. This will be the new molecular formula. CH3 (empirical) x 2=C2H6 (molecular formula) • * Try #38 and #39 p. 312.

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