1. 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 _________

2. DO NOW: (Refresh your memory) Classify the following as an atom, molecule, ion, or formula unit: 1.FeATOM 2.F 2 MOLECULE 3.H 2 OMOLECULE 4.NaATOM 5.NaClFORMULA UNIT 6.Na +1 ION

3. Chapter 10 THE MOLE!!!

4. What is a mole?

6. It is a unit of measure. 6.02 x 10 23 If we write this out (standard notation), we get: 602,000,000,000,000,000,000,000

7. A mole is equal to: 6.02 x 10 23 Representative Particles Atoms, molecules, formula units, and ions Fe F 2 NaCl Na +1 Now we have a conversion unit to use: 1mole 6.02 x 10 23 r.p. (any of the four choices above)

8. 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 Fe6.02 x 10 23 atoms Fe 1mole Fe 2.1moles Fe = 1.3x 10 24 atoms Fe

9. 140 moles of MgCl 2 = how many r.p. of Cl - ? First what r.p. is MgCl 2 and then Cl - ? (formula unit, ion) 140 moles MgCl 2 6.02 x 10 23 f.u. MgCl 2 2 ions Cl - 1mole MgCl 2 1 f.u. MgCl 2 140 moles MgCl 2 = 1.7 x 10 26 ions Cl -

10. 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 H 2 O, there are 18 g of H 2 0 2(1) + 16 = 18

11. 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 C 6 H 12 O 6 180g/mol So, in 1 mole of C 6 H 12 O 6, there are 180 g 6(12) + 12(1) + 6(16) = 180

12. Lets do some problems 11.3 moles of C 6 H 12 O 6 = how many grams? 11.3 moles C 6 H 12 O 6 180 g C 6 H 12 O 6 1mole C 6 H 12 O 6 11.3 moles C 6 H 12 O 6 = 2030 g C 6 H 12 O 6 = 2.03 x 10 3 g C 6 H 12 O 6

13. A mole is equal to: The 22.4L of gas at STP STP= Standard Temperature and Pressure = 0 o 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

14. THE MOLE ISLAND!!!! 1 mole of a substance Molar Mass gfm or gmm of a substance 6.02x10 23 r.p. (Avogadro’s #) 22.4 Lof gas per mole @ STP

15. Lets do some problems 5.3x10 24 molecules of CH 4 = how many grams CH 4 ? !!!!! You must go from the r.p. island then to the mole island, and then to the molar mass island!!!!! 5.3x10 24 molecules of CH 4 1 mole CH 4 16g CH 4 6.02x10 23 r.p. CH 4 1 mole CH 4 5.3x10 24 molecules of CH 4 = 140 g CH 4 = 1.4 x 10 2 g CH 4

16. Chapter 10 Percent Composition

17. 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.

18. Formula % mass of element = grams of elementx 100 grams of cmpd

19. 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?

20. 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.

21. Chapter 10 Empirical Formulas

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

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

24. 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

25. 1.85 = 1.00 mole for N 4.63 = 2.50 mole for O 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 N 2 O 5

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

27. Calculating Molecular Formulas 1. Go through all of the steps of an empirical formula problem. (poem)Ex: CH 3 2. Add up the mass of the empirical formula. Ex: 15g/mol (efm)

28. 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

29. 4. Multiply all of the subscripts in the empirical formula by the 2. This will be the new molecular formula. CH 3 (empirical) x 2=C 2 H 6 (molecular formula) * Try #38 and #39 p. 312.