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Unit 08 – Moles and Stoichiometry

Unit 08 – Moles and Stoichiometry. Molar Conversions. A. What is the Mole?. VERY. A large amount!!!!. A counting number (like a dozen) Avogadro’s number ( 6.02  10 23 particles ) (SI unit) 1 mol = molar mass. A. What is the Mole?. HOW LARGE IS IT???.

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Unit 08 – Moles and Stoichiometry

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  1. Unit 08 – Moles and Stoichiometry • Molar Conversions

  2. A. What is the Mole? VERY A large amount!!!! • A counting number (like a dozen) • Avogadro’s number (6.02  1023 particles) (SI unit) • 1 mol = molar mass

  3. A. What is the Mole? HOW LARGE IS IT??? • 1 mole of pennies would cover the Earth 1/4 mile deep! • 1 mole of hockey pucks would equal the mass of the moon! • 1 mole of basketballs would fill a bag the size of the earth!

  4. B. Molar Mass • Molar Mass- the mass of a mole of any element or compound (in grams) • Round to 2 decimal places • Also called: • Gram Formula mass – sum of the atomic masses of all the atoms in a formula of a compound • Formula weight

  5. B. Molar Mass Examples • water • sodium chloride • H2O • (1.01g x 2) + 16.00g = 18.02 g • NaCl • 22.99g + 35.45g = 58.44 g

  6. B. Molar Mass Examples • sodium hydrogen carbonate • sucrose • NaHCO3 • 22.99g + 1.01g + 12.01g + (16.00g x 3) = 84.01 g • C12H22O11 • (12.01g x12) + (1.01g x 22) + (16.00g x11)= 342.34 g

  7. C. Number of Particles in a Mole 1 mole = 6.02 × 10 23representative particles (also called Avogadro’s Number) • Atom- rep. particle for most elements • Ions – if atom is charged • Molecule- rep. particle for covalent compounds and diatomic molecules “BrINCl HOF” • Formula unit- rep. particle for ionic compounds What is a representative particle? How the substance normally exists:

  8. D. Volume of a Mole of Gas • The Volume of a gas varies with a change in temperature or pressure. • Measured at standard temperature and pressure (STP) • 0°C at 1 atmosphere (atm) • 1 mole of any gas occupies a volume of 22.4L

  9. The MoleRoad Map Molecule Atoms (ions) Formula unit

  10. E. Molar Conversion Examples • How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

  11. E. Molar Conversion Examples • How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules 1 mol 2.50 mol = 1.51  1024 molecules C12H22O11

  12. E. Molar Conversion Examples • Find the number of molecules of 12.00 L of O2 gas at STP. 6.02 x 1023 molecules 1 mol 1 mol 22.4 L 12.00 L = 3.225 x 1023 molecules

  13. II. Stoichiometric Calculations

  14. A. Proportional Relationships Ratio of eggs to cookies • I have 5 eggs. How many cookies can I make? 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

  15. A. Proportional Relationships • Stoichiometry • mass relationships between substances in a chemical reaction • based on the mole ratio • Mole Ratio • indicated by coefficients in a balanced equation 2 Mg + O2 2 MgO

  16. B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Convert known to moles (IF NECESSARY) Line up conversion factors. 4. Use Mole ratio – from equation 5. Convert moles to unknown unit (IF NECESSARY) 6. Calculate and write units.

  17. C. Stoichiometry Problems • Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. N2 + 3H2→ 2NH3

  18. III. Stoichiometry in the Real World

  19. A. Limiting Reactants • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly • Limiting Reactant • bread • Excess Reactants • peanut butter and jelly

  20. A. Limiting Reactants • Limiting Reactant • used up in a reaction • determines the amount of product • Excess Reactant • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  21. A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: • limiting reactant • amount of product

  22. A. Limiting Reactants Using the following equation identify the limiting reagent. • How many moles of ammonia (NH3) can be produced from the reaction of 28.2 L of nitrogen and 25.3 L of hydrogen? N2 + 3H2 2NH3 28.2 L ? mol 25.3 L

  23. A. Limiting Reactants N2 + 3H2 2NH3 28.2 L ? mol 25.3 L 28.2 L N2 1 mol N2 22.4 L N2 2 mol NH3 1 mol N2 = 2.5 mol NH3

  24. A. Limiting Reactants N2 + 3H2 2NH3 28.2 L ? mol 25.3 L 25.3 L H2 1 mol H2 22.4 L H2 2 mol NH3 3 mol H2 = 0.753 mol NH3

  25. A. Limiting Reactants N2: 2.5 mol NH3 H2: 0.753 mol NH3 Limiting reactant: H2 Excess reactant: N2 Product Formed: 0.753 mol NH3

  26. Limiting Reactants Mg + 2HCl → MgCl2 + H2 How many grams of magnesium chloride are produced from the reaction of 2.08 mol of Mg and 2.08 mol of HCl?

  27. B. Percent Yield measured in lab calculated on paper

  28. B. Percent Yield • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  29. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

  30. B. Percent Yield 46.3 g 49.4 g K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.4 g KCl 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield =

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