The Mole and Stoichiometry

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

2. The “mole” • A term for a certain number of something. • Brainstorm other counting words! • Dozen = • Pair = • Gross = • A “mole” of something is 6.02 x 1023 of something. • 602, 000, 000, 000, 000, 000, 000, 000 12 2 144

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

4. Try these MW’s • Ca • 40.1 • H2 • 2 (1.0) = 2.0 • BaF2 • [137.3 + 2(19.0)] = 175.3 • MW of 2BaF2 is still 175.3 1 mole of Ca weighs 40.1 grams 1 mole of H2 weighs 2.0 grams

5. 1 penny = 2.68 • 6 pennies = g • 6 pennies = 2.68 X 6 = 16.08 2.68

6. Avogadro’s Number Avogadro (1776-1856) • NA = 6.02 x 1023 of anything

7. 1 mole K = grams • MW of potassium = • 1 dozen K = atoms • 1 mole K = atoms 39.10 39.1 12 6.02 x 1023

8. MW of CO2 = • 3 moles of CO2 = • 3 moles of CO2 = ____ g/mol • MW of nitrogen gas = • N2 (g) = 44.0 44.0 2 ( 14.0) = 28.0

9. Which elements exist as diatomic molecules? • H2 N2 O2 F2 Cl2 Br2 I2

10. 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 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. How Big Is The Mole?

12. ÷MW X NA X#atoms Lab 11

13. 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 =

14. Percent Compositionfrom Formula • % oxygen in CaCO3? • Grams of Mg in 4.00 grams of MgO? First: Calculate % Mg in MgO Second: Calculate g Mg in 4.00 g MgO

15. B. Experimental Percent Composition • FROM DATA (“Experimental Value”) • 4.00g of Ag2O is decomposed to yield 3.65g Ag. • The Experimental % = ? • The Accepted % = ? • Your Experimental Error?

16. Experimental % Ag Equation: Ag2O  Experimental % Ag: 2 4 Ag + O2 4.00 g 3.65 g

17. True (Accepted) %Ag Percent Error?

18. Empirical Formula • Definition:The simplest formula indicating the mole ratio of elements in a compound • Examples: • H2O2 HO • C6H6CH • N2O4? • CO2 ? NO2 CO2

19. Empirical Formula STEPS • Change grams to moles • Divide by the least # moles for a RATIO • Apply ratio to the formula

20. Solving Empirical Formula—determined from gram composition • A compound contains 0.90 g Ca and 1.60 g Cl CaCl2

21. CH2 Try This… • 0.556 g Carbon and 0.0933 g Hydrogen

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

23. What formula would this ratio give? • K = 0.26 moles • N = 0.25 moles • O = 0.78 moles • Yields an Empirical Formula of… • KNO3

24. X 2 Try This: 70.5 % Fe and 29.5 % O • ___ moles Fe, ___ moles O • Assume 100g of substance: 1.26 1.84 1.26 moles 1.26 moles FeO1.5 Fe2O3

25. Try This: 40.0% C 6.7% H 53.3% O • ___ moles C, ___ moles H, ___ moles O • Assume 100g of substance: 3.3 6.7 3.3 After dividing by the least moles yields: CH2O

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

27. Molecular Formulas • Definiton: Formula of an actual compound as it exists in molecules. • Benzene exists as C6H6 not CH. • Hydrogen Peroxide exists as H2O2 not HO.

28. X 2 X 2 17 g/mol H2O2 100 g/mol C2H2O2Cl4 75.5 g/mol C3Cl3N6

29. 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.

30. 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

31. C. Coefficients yield the mole ratio!!! 2 H2 + O2 2 H2O 2 : 1 : 2

32. D. Example 4 Fe + 3 O2 2 Fe2O3 4 : 3 : 2 If 4 moles of iron rust, moles of Fe2O3 will form If 8 moles of iron rust, moles of Fe2O3 will form 2 4

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

34. 28.00 g of iron yields ? g of rust? 2:1 0.50 moles 0.25 moles 4 Fe + 3 O2 2 Fe2O3 28.00 g Fe 40.00 g Fe2O3 ? g Fe2O3

36. 36.00 g of water resulted from ? g of methane? 1:2 1.00 moles 2.00 moles CH4 + O2 CO2 + H2O Balance 2 2 ? g CH4 16.00g of methane 36.00 g H2O

37. Variation:12 moles of oxygen combusting will yield how many grams of CO2? 2:1 12 moles 6 moles CH4 + O2 CO2 + H2O 2 2 ? g CO2 264.00 g CO2