3. Mass Spectrometer Mass Spectrometers are used to determine the mass of an atom. Click for an explanation of Mass SpectrometerMass Spectrometer

4. Atomic Mass l Scientists use Atomic Mass Units (amu) where the mass of 12 C is defined as 12 amu. l The standard was established in 1961 l Value of any elements mass is relative to 12 C l The decimal numbers on Periodic table are average atomic masses.

5. Example l There are two isotopes of carbon which make up the largest percentage of carbon: 1. 12 C with a mass of 12.00000 amu (98.89%) 2. 13 C with a mass of 13.0034 amu (1.11%) To calculate the average atomic mass (Isotope Mass)(abundance) + (Isotope Mass)(abundance) Yields Average Atomic Mass for Element Must use decimal equivalent of abundance value

6. They are not whole numbers l Based on averages of known isotopes and their relative abundance. Lets try one p.117 # 28 Ans. 47.88 amu Ti Answer is behind orange

7. The Mole l The mole is a counting number. l 6.022 x 10 23 of anything is a mole USE 6.022 l A large dozen. l The number of atoms in exactly 12 grams of carbon-12.

8. Molar mass l Mass of 1 mole of a substance. l Often called molecular weight. l To determine the molar mass of an element, look on the periodic table. l The molar mass of an element is equal to its atomic mass units in grams.

9. Lets Try Some #33 p. 118 l Your Turn –#38 –HFC-134a = 102.04 g/mol –HCFC-124 = 136.48 g/mol

10. Percent Composition l Percent of each element in compound l Find the mass of each element, divide by the total mass, multiply by a 100. l Calc. Percent Composition of CO 2 l Carbon 12.01 g + Oxygen 16.00 (2) = 44.01g

11. Lets Try Some – p.119 –#59 – the first element You finish it! The next one is all yours –#63=1360 g/mol

12. Empirical Formulas l From percent composition, you can determine the empirical formula. l Empirical Formula - lowest whole number ratio of atoms in a molecule. l Based on mole ratios l #69 p. 119 KEY: Start by assuming 100.00g sample.

13. Out of 100.00 g of adrenaline, there are Divide each mol value by the smallest number Empirical Formula is C 8 H 11 O 3 N

14. How do we get to Carnegie Hall? l Try these on p. 119-120 –#70C 6 H 11 NO –#72N 2 H 4 CO

15. Empirical To Molecular Formulas l Empirical is lowest ratio. l Molecular is actual molecule. l Need Molar mass. l Ratio of empirical to molar mass will tell you the molecular formula. l Must be a whole number

16. #74 p. 120 Assuming 100.00 g of compound Divide by lowest mol value gives 1 mol P, 1 mol N, & 2 mol Cl for an Empirical formula PNCl 2 Empirical formula mass = 80.43 Because 580/80.43 ≈ 7 the molecular formula is P 7 N 7 Cl 14

17. Answer: Practice Here we come Carnegie Hall l Determine the empirical and molecular formula for a compound. l Try these on p. 120 –#75Emp. C 3 H 5 O 2 Molec. C 6 H 10 O 4 –#76Emp. CHOMolecular C 4 H 4 O 4

18. Chemical Equations l Are sentences. l Describe what happens in a chemical reaction. Reactants  Products l Equations must be balanced.

19. Balancing equations CH 4 + O 2  CO 2 + H 2 O ReactantsProducts C11 O23 H42

20. Balancing equations CH 4 + O 2  CO 2 + 2 H 2 O ReactantsProducts C11 O23 H424

21. Balancing equations CH 4 + O 2  CO 2 + 2 H 2 O ReactantsProducts C11 O23 H424 4

22. Balancing equations CH 4 + 2O 2  CO 2 + 2 H 2 O ReactantsProducts C11 O23 H424 4 4

23. Abbreviations ( s )  Solid ( g )  Gas l ( aq ) Aqueous l heat  heat l catalyst

24. Meaning l A balanced equation can be used to describe a reaction in moles, molecules and atoms. l Not grams.

25. Lets Just Do One Problem l #84 on p. 120 2KO 2(s) + 2H 2 O (l) → 2KOH (aq) + O 2(g) + H 2 O 2(aq) l Fe 2 O 3(s) + 6HNO 3(aq)  2Fe(NO 3 ) 3(aq) + 3H 2 O (l)

26. Stoichiometry l Given an amount of either starting material or product, determine the other quantities. l Use conversion factors from –molar mass (g - mole) –balanced equation (mole – mole) –Mole – volume –Mole – number of particles (atoms, molecules, formula units)

27. Remember Flux Capacitor on Steroids

28. I’ll get you started l #89 p. 121

29. Your Turn l #92 Ba(OH) 2. 8H 2 O (s) + 2NH 4 SCN (s)  Ba(SCN) 2(s) + 10H 2 O (l) + 2NH 3(g) 3.2 g NH 4 (SCN) l #95 See instructor for correct answer.

30. Limiting Reagent l Reactant that determines the amount of product formed. l The one you run out of first. l Makes the least product. l Book shows you a ratio method p.106. l It works. l So does mine

31. Limiting reagent l To determine the limiting reagent requires that you do multiple stoichiometry problems. l Figure out how much product each reactant makes. l The one that makes the least is the limiting reagent. l p. 122 #101

32. The reactant that produces the smallest amount will run out 1 st and is limiting reagent = O 2

33. Check Out p.122 –#102 See instructor for correct answer.

34. Excess Reagent l The reactant you don’t run out of. l The amount of stuff you make is the yield. l The theoretical yield is the amount you would make if everything went perfect in a lab. Stoichiometrically calculated. l The actual yield is what you make in the lab.

35. Percent Yield l % yield = Actual x 100% Theoretical l % yield = what you got in lab x 100% calculated yield l Problem #103 p.122

36. Find limiting reactant C 7 H 6 O 3 is limiting reagent Theoretical yield is

37. Practice, Practice, Practice l #104 l #106 See instructor for correct answer.