2. The Mole and Stoichiometry Chemistry gets Real…. Tough that is

3. The “mole” A term for a certain number of something. A term for a certain number of something. Brainstorm other counting words! Brainstorm other counting words! Dozen = Dozen = Pair = Pair = Gross = Gross = A “mole” of something is 6.02 x 10 23 of something. A “mole” of something is 6.02 x 10 23 of something. 602, 000, 000, 000, 000, 000, 000, 000 602, 000, 000, 000, 000, 000, 000, 000 12 2 144

4. Molecular Weight M.W. = the weight (in grams) of a mole of substance M.W. = the weight (in grams) of a mole of substance On your periodic tables On your periodic tables Round to the nearest tenth Round to the nearest tenth Hydrogen is 1.00797  1.0 g/mol Hydrogen is 1.00797  1.0 g/mol Mole Weight is an INTENSIVE property— doesn’t depend on amount

5. Try these MW’s Ca Ca 40.1 40.1 H 2 H 2 2 (1.0) = 2.0 2 (1.0) = 2.0 BaF 2 BaF 2 [137.3 + 2(19.0)] = 175.3 [137.3 + 2(19.0)] = 175.3 MW of 2BaF 2 is still 175.3 MW of 2BaF 2 is still 175.3 1 mole of Ca weighs 40.1 grams 1 mole of H 2 weighs 2.0 grams

6. 1 penny = 2.68 1 penny = 2.68 6 pennies = g 6 pennies = g 6 pennies = 6 pennies = 2.68 2.68 X 6 = 16.08

7. Avogadro’s Number N A = 6.02 x 10 23 of anything N A = 6.02 x 10 23 of anything Avogadro (1776-1856)

8. 1 mole K = grams 1 mole K = grams MW of potassium = MW of potassium = 1 dozen K = atoms 1 dozen K = atoms 1 mole K = atoms 1 mole K = atoms 39.1 39.10 12 6.02 x 10 23

9. MW of CO 2 = MW of CO 2 = 3 moles of CO 2 = 3 moles of CO 2 = 3 moles of CO 2 = ____ g/mol 3 moles of CO 2 = ____ g/mol MW of nitrogen gas = MW of nitrogen gas = N 2 (g) = N 2 (g) = 44.0 2 ( 14.0) = 28.0 44.0

10. Which elements exist as diatomic molecules? H 2 N 2 O 2 F 2 Cl 2 Br 2 I 2 H 2 N 2 O 2 F 2 Cl 2 Br 2 I 2

11. How Big Is The Mole? One mole of marbles would cover the entire Earth to a depth of fifty miles One mole of marbles would cover the entire Earth to a depth of fifty miles One mole of hockey pucks would equal the mass of the moon. One mole of hockey pucks would equal the mass of the moon. One mole of rice grains is more than the number of grains of all crops grown since the beginning of time. One mole of rice grains is more than the number of grains of all crops grown since the beginning of time. If one mole of pennies was divided up equally between all the people on Earth, you would have enough money to spend a million dollars every hour, 24 hours a day, for your entire life. When you died, you would have spent less than half of your riches. If one mole of pennies was divided up equally between all the people on Earth, you would have enough money to spend a million dollars every hour, 24 hours a day, for your entire life. When you died, you would have spent less than half of your riches.

13. Lab 11 Formula Mass (g) MWMolesMoleculesAtoms AgNO 3 15.00169.90.088 5.3 x 10 22 2.7 x 10 23 KOH18.9856.10.34 2.0 x 10 23 6.0 x 10 23 CO47.0028.01.68 1.0 x 10 24 2.0 x 10 24 Na 2 SO 4 185.00142.11.30 7.8x 10 23 5.4 x 10 24 H2H2H2H21.002.00.50 3.0 x 10 23 6 x 10 23 Na 2 S 13.7678.10.18 1.1 x 10 23 3.3 x 10 23 NaOH214.7540.05.37 3.2 x 10 24 9.6 x 10 24 Pb6.50207.20.031 1.8 x 10 22 Pb 3 (PO 4 ) 2 18.75811.60.023 1.4 x 10 22 1.8 x 10 23 NaCl62.2558.51.06 6.6 x 10 23 1.3 x 10 24 ÷MW X N A X#atoms

14. Percent Composition A. Determined from Formulas (“Accepted Value”) Is NaCl 50.0% Na by weight? No, Na is 23.0 g/mole and Cl is 35.5 g/mole To Prove, % Na =

15. Percent Composition from Formula % oxygen in CaCO 3 ? % oxygen in CaCO 3 ? Grams of Mg in 4.00 grams of MgO? Grams of Mg in 4.00 grams of MgO? First: Calculate % Mg in MgO Second: Calculate g Mg in 4.00 g MgO

16. B. Experimental Percent Composition FROM DATA (“Experimental Value”) FROM DATA (“Experimental Value”) 4.00g of Ag 2 O is decomposed to yield 3.65g Ag. 4.00g of Ag 2 O is decomposed to yield 3.65g Ag. The Experimental % = ? The Experimental % = ? The Accepted % = ? The Accepted % = ? Your Experimental Error? Your Experimental Error?

17. Experimental % Ag Equation: Ag 2 O  Ag 2 O  Experimental % Ag: 4.00 g3.65 g Ag + O 2 2 4

18. True (Accepted) %Ag Percent Error?

19. Empirical Formula Definition: The simplest formula indicating the Definition: The simplest formula indicating the mole ratio of elements in a compound Examples: Examples: H 2 O 2  HO H 2 O 2  HO C 6 H 6  CH C 6 H 6  CH N 2 O 4  ? N 2 O 4  ? CO 2  ? CO 2  ? NO 2 CO 2

20. Empirical Formula STEPS 1.Change grams to moles 2.Divide by the least # moles for a RATIO 3.Apply ratio to the formula

21. Solving Empirical Formula— determined from gram composition A compound contains 0.90 g Ca and 1.60 g Cl A compound contains 0.90 g Ca and 1.60 g Cl CaCl 2

22. Try This… 0.556 g Carbon and 0.0933 g Hydrogen 0.556 g Carbon and 0.0933 g Hydrogen CH 2

23. Why is the Empirical Formula a ratio of small WHOLE numbers? Can’t have half of an atom Can’t have half of an atom Atoms combine as whole units Atoms combine as whole units Shows the simplest way that atoms can pair Shows the simplest way that atoms can pair

24. What formula would this ratio give? K = 0.26 moles K = 0.26 moles N = 0.25 moles N = 0.25 moles O = 0.78 moles O = 0.78 moles Yields an Empirical Formula of… Yields an Empirical Formula of… KNO 3 KNO 3

25. Try This: 70.5 % Fe and 29.5 % O ___ moles Fe, ___ moles O Assume 100g of substance: 1.261.84 Fe 2 O 3 1.26 moles FeO 1.5 X 2

26. Try This: 40.0% C6.7% H 53.3% O ___ moles C, ___ moles H, ___ moles O Assume 100g of substance: 3.36.73.3 After dividing by the least moles yields: CH 2 O

27. Why doesn’t the ratio of the % give the empirical formula? Why doesn’t the ratio of the % give the empirical formula? Must account for differing masses of elements. Must account for differing masses of elements. Why does the ratio of the moles give the empirical formula? Why does the ratio of the moles give the empirical formula? The ratio of the # of atoms normalizes for mass differences. The ratio of the # of atoms normalizes for mass differences.

28. Molecular Formulas Definiton: Formula of an actual compound Definiton: Formula of an actual compound as it exists in molecules. as it exists in molecules. Benzene exists as C 6 H 6 not CH. Benzene exists as C 6 H 6 not CH. Hydrogen Peroxide exists as H 2 O 2 not HO. Hydrogen Peroxide exists as H 2 O 2 not HO.

29. Empirical Formula MW Empirical MW Molecular Molecular Formula HO 34 g/mol CHOCl 2 200 g/mol CClN 2 226.5 g/mol H2O2H2O2 17 g/mol X 2 100 g/mol 75.5 g/mol C 2 H 2 O 2 Cl 4 C 3 Cl 3 N 6 X 2

30. Why is the M.W. needed to determine the molecular formula? Why is the M.W. needed to determine the molecular formula? Need M.W. of actual compound to find how many each type of atom is in a molecule. Need M.W. of actual compound to find how many each type of atom is in a molecule.

31. Stoichiometry – The Big Leagues A. Define: Problem Solving involving mass-mass relationships in chemical changes Ex. How many grams of rust are formed when 12.00 g of Fe reacts with oxygen. B. Must use balanced equations for the correct mole ratios

32. C. Coefficients yield the mole ratio!!! 2 H 2 + O 2  2 H 2 O 2 : 1 : 2 2 : 1 : 2

33. D. Example 4 Fe + 3 O 2  2 Fe 2 O 3 4 : 3 : 2 4 : 3 : 2 If 4 moles of iron rust, moles of Fe 2 O 3 will form If 8 moles of iron rust, moles of Fe 2 O 3 will form moles of Fe 2 O 3 will form 4 2

34. Solving Mass-Mass Problems 1) Determine the mole ratio (from the balanced equation) 2) Convert grams to moles 3) Apply the mole ratio 4) Convert moles to grams

35. 28.00 g of iron yields ? g of rust? 4 Fe + 3 O 2  2 Fe 2 O 3 28.00 g Fe 0.50 moles 2:1 0.25 moles 40.00 g Fe 2 O 3 ? g Fe 2 O 3

37. 36.00 g of water resulted from ? g of methane? CH 4 + O 2  CO 2 + H 2 O 36.00 g H 2 O 1.00 moles 1:2 2.00 moles 16.00g of methane 2 2 Balance ? g CH 4

38. Variation: 12 moles of oxygen combusting will yield how many grams of CO 2 ? 12 moles 2:1 6 moles 264.00 g CO 2 CH 4 + O 2  CO 2 + H 2 O 2 2 ? g CO 2