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Chapters 8-10

Chapters 8-10. The mole, equations and stoichiometry. The Structure of Atoms. Atomic Mass Unit 1 amu = 1/12 of the mass of one atom of Carbon-12 1 amu = 1.6605 x 10 -24 g. Atomic and Molecular Mass. Mass: proton = 1.00728 amu neutron = 1.0086 amu electron = 0.0005486 amu

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Chapters 8-10

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  1. Chapters 8-10 The mole, equations and stoichiometry

  2. The Structure of Atoms Atomic Mass Unit 1 amu = 1/12 of the mass of one atom of Carbon-12 1 amu = 1.6605 x 10-24 g

  3. Atomic and Molecular Mass Mass: proton = 1.00728 amu neutron = 1.0086 amu electron = 0.0005486 amu 12C atom = 12.00000 amu 13C atom = 13.00335 amu

  4. Atomic and Molecular Mass • The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes. • Mass of C = average of 12C and 13C • = 0.9889 x 12 amu + 0.0111 x 13.0034 amu • = 12.011 amu

  5. Atomic and Molecular Mass The mass of a molecule is just the sum of the masses of the atoms making up the molecule. m(C2H4O2) = 2·mC + 4·mH + 2·mO • = 2·(12.01) + 4·(1.01) + 2·(16.00) • = 60.06 amu

  6. Avogadro and the Mole • One mole of a substance is the gram mass value equal to the amu mass of the substance. • One mole of any substance contains 6.02 x 1023 units of that substance. • Avogadro’s Number (NA,6.022 x 1023) is the numerical value assigned to the unit, 1 mole.

  7. Avogadro and the Mole

  8. Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H) Avogadro and the Mole

  9. The Mole: Allows us to make comparisons between substances that have different masses. Avogadro and the Mole

  10. Counting atoms Amount in moles can be converted to number of atoms: the conversion is similar to changing 20 dozen eggs into the equivalent number of individual eggs:

  11. Convert 2.66 mol of an element to its equivalent number of atoms:

  12. How many moles in 2.54 x 1024 Fe atoms? 2.54 x 1024 atoms x

  13. # particles The mole

  14. # Particles Avogadro’s number X / The mole

  15. Practice problems p 278 • How many atoms are present in 3.7 mol of sodium? • How many atoms are present in 155 mol of arsenic? • How many moles of xenon is 5.66 x 1026 atoms? • How many moles of silver is 2.888 x 1015 atoms? • Recall that 1 mol = 6.02 x 1023 atoms

  16. Balancing Chemical Equations • A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved. reactants  products limestone  quicklime + gas Calcium carbonate  calcium oxide + carbon dioxide CaCO3(s)  CaO(s) + CO2(g)

  17. Molar mass relates moles to grams Molar mass- is numerically equal to the element’s atomic mass and has the unit grams per mole (g/mol) - molar masses are used as conversion factors in the relationship between moles and the mass in grams of a substance.

  18. # of Particles Avogadro’s number X / (x 1mol/6.02EE23) 6.02EE23 x mass x 1 mol 1 mol mass The Mole Mass (g)

  19. Balancing Chemical Equations CaCO3(s)  CaO(s) + CO2(g) The letters in parentheses following each substance are called State Symbols (g) → gas (l) → liquid (s) → solid (aq) → aqueous

  20. Balancing Chemical Equations A balanced equation MUST have the same number of atoms of each element on both sides of the equation. H2 + O2→ H2O Not Balanced H2 + ½O2→ H2O Balanced 2H2 + O2→ 2H2O Balanced

  21. Balancing Chemical Equations The numbers multiplying chemical formulas in a chemical equation are called: Stoichiometric Coefficients (S.C.) 2H2 + O2→ 2H2O Balanced Here 2, 1, and 2 are stoichiometric coefficients.

  22. Balancing Chemical Equations Hints for Balancing Chemical Equations: • Save single element molecules for last. • Try not to change the S.C. of a molecule containing an element that is already balanced. • If possible, begin with the most complex molecule that has no elements balanced.

  23. Balancing Chemical Equations Hints for Balancing Chemical Equations: 4) Otherwise, trial and error!!

  24. Balancing Chemical Equations Example 1: CH4 + O2→ CO2 + H2O Balance O2 last C is already balanced Start by changing S.C. of H2O to balance H CH4 + O2→ CO2 + 2H2O

  25. Balancing Chemical Equations Example 1: CH4 + O2→ CO2 + 2H2O Now C and H are balanced Balance O by changing the S.C. of O2 CH4 + 2O2→ CO2 + 2H2O BALANCED!

  26. Balancing Chemical Equations Example 2: B2H6 + O2→ B2O3 + H2O Balance O last B is already balanced Start by changing S.C. of H2O: B2H6 + O2→ B2O3 + 3H2O

  27. Balancing Chemical Equations Example 2: B2H6 + O2→ B2O3 + 3H2O B and H are balanced Balance O by changing S.C. of O2 B2H6 + 3O2→ B2O3 + 3H2O BALANCED!

  28. Balancing Chemical Equations Example 3: MnO2 + KOH + O2→ K2MnO4 + H2O Balance O last Mn is already balanced Change S.C. of KOH to balance K MnO2 + 2KOH + O2→ K2MnO4 + H2O

  29. Balancing Chemical Equations Example 3: MnO2 + 2KOH + O2→ K2MnO4 + H2O Mn, K, and H are balanced (H was balanced by chance) Balance O MnO2 + 2KOH + ½O2→ K2MnO4 + H2O or 2MnO2 + 4KOH + O2→ 2K2MnO4 + 2H2O

  30. Balancing Chemical Equations Example 4: NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4 Hard one (no single element molecules) S is balanced Start with NaNO2 to balance Na 2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4

  31. Balancing Chemical Equations Example 4: 2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4 S, Na, and N are balanced Cannot balance H without changing S.C. for H2SO4! Boo! Option 1: trial and error Option 2: Go on to next problem!

  32. Balance the following equations: C6H12O6→ C2H6O + CO2 Fe + O2→ Fe2O3 NH3 + Cl2→ N2H4 + NH4Cl KClO3 + C12H22O11→ KCl + CO2 + H2O Balancing Chemical Equations

  33. Balance the following equations: C6H12O6→ 2C2H6O + 2CO2 Fe + O2→ Fe2O3 NH3 + Cl2→ N2H4 + NH4Cl KClO3 + C12H22O11→ KCl + CO2 + H2O Balancing Chemical Equations

  34. Balance the following equations: C6H12O6→ 2C2H6O + 2CO2 4Fe + 3O2→ 2Fe2O3 (balance O first) NH3 + Cl2→ N2H4 + NH4Cl KClO3 + C12H22O11→ KCl + CO2 + H2O Balancing Chemical Equations

  35. Balance the following equations: C6H12O6→ 2C2H6O + 2CO2 4Fe + 3O2→ 2Fe2O3 (balance O first) NH3 + Cl2→ N2H4 + NH4Cl N:H is 1:3 on left, must get 1:3 on right! Balancing Chemical Equations

  36. NH3 + Cl2→ N2H4 + NH4Cl N:H is 1:3 on left, must get 1:3 on right! 4NH3 + Cl2→ N2H4 + 2NH4Cl Balancing Chemical Equations

  37. Balance the following equations: C6H12O6→ 2C2H6O + 2CO2 4Fe + 3O2→ 2Fe2O3 4NH3 + Cl2→ N2H4 + 2NH4Cl KClO3 + C12H22O11→ KCl + CO2 + H2O (tough!) Balancing Chemical Equations

  38. Balance the following equations: KClO3 + C12H22O11→ KCl + CO2 + H2O balance C KClO3 + C12H22O11→ KCl + 12CO2 + H2O balance H KClO3 + C12H22O11→ KCl + 12CO2 + 11H2O balance O 8KClO3 + C12H22O11→ KCl + 12CO2 + 11H2O Balancing Chemical Equations

  39. Balance the following equations: 8KClO3 + C12H22O11→ KCl + 12CO2 + 11H2O balance K (and hope Cl is balanced) 8KClO3 + C12H22O11→ 8KCl + 12CO2 + 11H2O Balanced! Balancing Chemical Equations

  40. Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below: Balancing Chemical Equations

  41. Avogadro and the Mole

  42. Stoichiometry • Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

  43. Stoichiometry Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl2: • 2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l) How many grams of NaOH are needed to react with 25.0 g of Cl2?

  44. Stoichiometry 2 NaOH + Cl2→ NaOCl + NaCl + H2O 25.0 g Cl2 reacts with ? g NaOH

  45. Calculate the molar mass of the following: Fe2O3 (Rust) C6H8O7 (Citric acid) C16H18N2O4 (Penicillin G) Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe2O3. Fe2O3(s) + CO(g) Fe(s) + CO2(g) Avogadro and the Mole

  46. Fe2O3 + CO → Fe + CO2 Balance (not a simple one) Save Fe for last C is balanced, but can’t balance O In the products the ratio C:O is 1:2 and can’t change Make the ratio C:O in reactants 1:2 Fe2O3 + 3CO → 2Fe + 3CO2 Avogadro and the Mole

  47. Avogadro and the Mole Fe2O3 + 3CO → 2Fe + 3CO2

  48. Stoichiometry

  49. Stoichiometry • Aspirin is prepared by reaction of salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) to form aspirin (C9H8O4) and acetic acid (CH3CO2H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by-product?

  50. Stoichiometry Salicylic acid + Acetic anhydride → Aspirin + acetic acid C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 Balanced! Equal # moles for all

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