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Chapter 3 Stoichiometry

Chapter 3 Stoichiometry. 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical Equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products

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Chapter 3 Stoichiometry

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  1. Chapter 3Stoichiometry 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical Equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products 3.9 Calculations Involving a Limiting Reactant

  2. Atomic Mass • We measure mass, volume, pressure, etc. • We want number of atoms (or number of moles) • grams x (grams/mole)-1 = moles

  3. Atoms • Avogadro’s Number is the number of 12C atoms in exactly 12 grams of carbon N0 = 6.02 X 1023 • The mass, in grams, of Avogadro's number of atoms of an element is numerically equal to the relative atomic mass of that element (relative to carbon) • The mass of 6.02 X 1023 atoms of 12C = 12.00 g (defn) • The mass of 1 atom of 12C = 12.00 amu (12.00 Daltons) • The mass of 6.02 X 1023 atoms of C = 12.01 g • The mass of 6.02 X 1023 atoms of 35Cl = 35.0 g • The mass of 6.02 X 1023 atoms of Cl = 35.4 g

  4. Molar Mass • Molecular Mass (aka the molecular weight) of a molecule equals the sum of the atomic masses of all of the atoms making up the molecule. • Examples Fructose (C6H12O6) Benzene (C6H6) Water (H20) Sodium Chloride (NaCl) Hemoglobin (best to look that one up)

  5. Moles • The # of moles of chemical is its amount. • One mole of a substance equals the amount that contains Avogadro's number of atoms or molecules. • One mole of an element or molecule has a Molecular Weight (MWt) of that element or molecule, expressed in grams • For example, the Molecular Weight of Glucose (Glucose C6H12O6) is MWt = 6(12.0 g/mol) + 12(1.00 g/mol) + 6(16.0 g/mol) = 180 g/mol Zumdahl Chapter 3

  6. Isoamyl acetate have the formula C7H14O2. Calculate (a) how many moles and (b) how many molecules are contained in 0.250 grams of isoamyl acetate.Strategy: Use the units. You are given grams. You need moles. The molecular weight is given in g / mol). Then you need molecules. Avo’s # is molecules / mole. • Calculate MWt of C7H14O2 • Calculate the number of moles in 0.250 grams • Using Avogadro’s number to calculate the number of molecules in the number of moles of C7H14O2

  7. Percentage Composition (g/g) from Empirical or Molecular Formula Tetrodotoxin, a potent poison found in the ovaries and liver of the globefish, has the empirical formulaC11H17N3O8. Calculate the mass percentages of the four element in this compound. Strategy: • Calculate molar mass of C11H17N3O8, by finding the mass contributed by each element. • Assume you have one mole of the compound. • Divide the mass of each element by the total mass of the compound. Zumdahl Chapter 3

  8. Tetrodotoxin has the empirical formula C11H17N3O8. Calculate the mass percentages (g/g) of the four element in this compound. Solution: • Calculate molar mass of C11H17N3O8, by finding the mass contributed by each elementC : 11x12 = 132 g / molH : 1x17 = 17N : 3x14 = 42O : 8x16 = 126tot = 317 g / mol • Calculate the mass of one mole of tetrodotoxin317 g / mol x 1 mol = 317 g • Divide the mass of each element by the mass of the compound.Example C: 132 g / 317 g = 0.42 g/g (42 %) Zumdahl Chapter 3

  9. Milli, Micro, etc 1 mmol = 1 millimole =1x10-3 mol 1mg = 1 milligram =1x10-3 g 1msec = 1 millisec =1x10-3 sec 1 u mol = 1 micromole =1x10-6 mol 1u g = 1 microgram =1x10-6 g Also nano (10-9) pico (10-12) femto (10-15) An Angstrom (Å)= 10-15 meters = 0.1 nm

  10. The Law of Conservation of Mass The mass/matter of a closed system is constant. Mass can be rearranged but not created or /destroyed. In a chemical reaction the mass of the reactants equals the mass of the products. Mass is conserved during a change of state (solid to liquid to gas)(This law holds in this class, but maybe not in your physics class)

  11. Using Conservation of Mass to determine an Empirical Formula Moderate heating of 97.4 mg of a compound containing nickel, carbon and oxygen and no other elements drives off all of the carbon and oxygen in the form of carbon monoxide (CO) and leaves 33.5 mg of metallic nickel (Ni) behind. Determine the empirical formula of the compound.

  12. Using Conservation of Mass to determine an Empirical Formula • Write the reaction. NixCyOy -> X Ni + Y CO • Use conservation of mass to find the mass of CO.97.4 mg (mass tot) – 33.5 mg (mass Ni) = 63.9 g (mass CO) • Find the number of moles of CO and of Ni.CO : 63.9 mg / (12.0+16.0 g/mol) = 2.28 mmolNi : 33.5 mg / 58.7 g / mol) = 0.57 mmol • Find the ratios of the moles by dividing each by the smallest one, i.e., normalize to the smallest.2.28 mmol CO /.57 mmol Ni = 4 • Y/X = 4: answer is NiC4O4

  13. Chemical Equations • Chemical Reactions tell us three things • What atoms or molecules react together to form what products. • How much reactant how much product. • The state of each species aA (l) + bB (s)  cC (s) + dD (g) Reactants Products Zumdahl Chapter 3

  14. Chemical Equations The Law of Conservation of Mass says that a chemical equation must have the same number of atoms of a given kind on each side (a chemical reaction cannot create or destroy carbon, or oxygen, or hydrogen, or etc.) Chemical equations must be balanced! H2 + O2 H2O (not balanced) 2H2 + O2 2H2O (balanced)

  15. Chemical Equations 4 Al(s) + 3 O2(g) → 2 Al2O3(s) This equation means 4 Al atoms + 3 O2 molecules ---produces---> 2 molecules of Al2O3 or 4 moles of Al + 3 moles of O2 ---produces---> 2 moles of Al2O3 Br2 (l) + ___Al (s)→ __Al2Br6 (s)

  16. Balancing Chemical Equations • The same atoms are present in a reaction at the beginning and at the end. • KClO3 (s)  KCl (s) + O2 (g) no • KClO3 (s)  KCl (s) + 3/2O2 (g) better • 2KClO3 (s)  2KCl (s) + 3O2 (g) best

  17. To Balance an Equation Step 1: Set the stoichimetric coefficient of the most complicated molecule (with the largest number of different elements ) to 1. Step 2: Balance as many atoms as possible in the second most complicated molecule. Ignore atoms that show up elsewhere in homonuclear species like O2 and H2. Step 3: Balance the atoms in the in the homonuclear species (O2 and H2). Step 4: Eliminate fractional coefficients. Step 5: Count the each atom type on each side of the equation.

  18. Balancing Equations C3H8(g) + O2(g) CO2(g) + H2O(g) Combustion of Propane

  19. Balancing Equations Combustion of Propane • 1C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

  20. Writing Balanced Chemical Equations PbO2 + Pb + H2SO4→ PbSO4 + H2O  PbO2 + Pb + 2 H2SO4→ 2 PbSO4 + 2 H2O 2 Pb Check for balance 2 Pb 10 O 10 O 4 H 4 H Balanced Problem: Suppose we have 1.45 grams of Pb in the presence of excess lead oxide and sulfuric acid. How many grams of Lead Sulfate are produced? Zumdahl Chapter 3

  21. Limiting Reactants 3 Br2 (l) + 2 Al (s)→ 1 Al2Br6 (s) excess limiting Given the amounts below

  22. Limiting Reagent • Balance the reaction. • Convert reactant masses to moles [g(g/mol)-1=mol] • Normalize the moles of each reactant by its stoichiometric coefficient. • Find the smallest normalized number of moles • Use the ratio of stoichiometric coefficients to find the moles of product. • Convert to mass (if necessary).

  23. Example Calculation Involving a Limiting Reactant Suppose that 1.00 g of sodium and 1.00 g of chlorine react to form sodium chloride (NaCl). Which of these is limiting, and what is the mass of product. 2 Na + Cl2→ 2 NaCl nNa = 1.00 g x (1 mol Na / 23.0 g Na) = 0.0435 mol Na nCl2 = 1.00 g × (1 mol / 70.9 g Cl2) = 0.0141 mol Cl2 → nNa/2 = 0.022 (normalized) nCl2/1 = 0.0141 (normalized) Cl2 is the limiting reagent 0.0141 moles of Cl2 gives 0.0282 moles of NaCl. Use the molecular weight of NaCl to find the mass of NaCl produced. →

  24. What mass (in grams) of xenon tetrafluoride would be required to react completely with 1.000 g of water? XeF4 + 2 H2O → Xe + 4 HF + O2 =

  25. At one point in the purification of silicon, gaseous SiHCl3 reacts with gaseous H2 to give gaseous HCl and solid Si. (a) Determine the chemical amount (in moles) of H2 required to react with 160.4 mol of SiHCl3. (b) Determine the chemical amount of HCl that is produced. (c) Determine the mass (in grams) of Si that is produced. SiHCl3 (g) + H2 (g) → 3 HCl (g) + Si (s) (a) (b) Zumdahl Chapter 3

  26. At one point in the purification of silicon, gaseous SiHCl3 reacts with gaseous H2 to give gaseous HCl and solid Si. (a) Determine the chemical amount (in moles) of H2 required to react with 160.4 mol of SiHCl3. (b) Determine the chemical amount of HCl that is produced. (c) Determine the mass (in grams) of Si that is produced. 1 mol SiHCl3 1 mol H2 3 mol HCl 1 mol Si SiHCl3 (g) + H2 (g) → 3 HCl (g) + Si (s) (c) Zumdahl Chapter 3

  27. Isotopes of Cl: 35Cl contains 17 protons and 18 neutrons 37Cl contains 17 protons and 20 neutrons Zumdahl Chapter 3

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