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Formulas, Equations, and Moles

Formulas, Equations, and Moles. 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

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Formulas, Equations, and Moles

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  1. Formulas, Equations, and Moles

  2. 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)

  3. 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

  4. 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

  5. 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.

  6. 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.

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

  8. 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

  9. 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!

  10. 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

  11. 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!

  12. 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

  13. 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

  14. 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

  15. 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!

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

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

  18. 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

  19. 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

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

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

  22. 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

  23. 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

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

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

  26. 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

  27. 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

  28. 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.

  29. Avogadro and the Mole

  30. 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

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

  32. Avogadro and the Mole

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

  34. 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?

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

  36. 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

  37. 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

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

  39. Stoichiometry

  40. 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?

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

  42. Stoichiometry 4.50 g Salicylic acid (C7H6O3) = ? moles MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00 = 138.12 g/mole

  43. Stoichiometry Since all compounds have the same S.C., there must be 0.0326 moles of all 4 of them involved in the reaction. g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin = .0326 x [9x12.01 + 8x1.008 + 4x16.00] =.0326 mole x 180.15 g/mole 5.87 g Aspirin

  44. Yields of Chemical Reactions:If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate apercentage yield. Stoichiometry

  45. Stoichiometry • Dichloromethane (CH2Cl2) is prepared by reaction of methane (CH4) with chlorine (Cl2) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%?

  46. Stoichiometry CH4 + Cl2→ CH2Cl2 + HCl Balance CH4 + 2Cl2→ CH2Cl2 + 2HCl 1.85 kg CH4 = ? moles CH4

  47. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl 1.85 kg CH4 = ? moles CH4 MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole

  48. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl 115 moles CH4 in theory we should produce: 115 moles of CH2Cl2 and 230 moles of HCl And use up 230 moles of Cl2

  49. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl 115 moles of CH2Cl2 = ? g MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93 115 moles x (84.03 g/mole) = 9770 g

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