2. 8/1/20151 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 3.9 Calculations Involving a Limiting Reactant

3. Atomic Mass We measure mass, volume, pressure, etc. We want number of atoms (or number of moles) grams x (grams/mole) -1 = moles 8/1/20152

4. 3 Atoms Avogadro’s Number is the number of 12 C atoms in exactly 12 grams of carbon N 0 = 6.02 X 10 23 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 10 23 atoms of 12 C = 12.00 g (def n ) The mass of 1 atom of 12 C = 12.00 amu (12.00 Daltons) The mass of 6.02 X 10 23 atoms of C = 12.01 g The mass of 6.02 X 10 23 atoms of 35 Cl = 35.0 g The mass of 6.02 X 10 23 atoms of Cl = 35.4 g

5. 8/1/20154 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 (C 6 H 12 O 6 ) Benzene (C 6 H 6 ) Water (H 2 0) Sodium Chloride (NaCl) Hemoglobin (best to look that one up)

6. 8/1/2015Zumdahl Chapter 35 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 C 6 H 12 O 6 ) is MWt = 6(12.0 g/mol) + 12(1.00 g/mol) + 6(16.0 g/mol) = 180 g/mol

7. 8/1/20156 Isoamyl acetate have the formula C 7 H 14 O 2. 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. 1.Calculate MWt of C 7 H 14 O 2 2.Calculate the number of moles in 0.250 grams 3.Using Avogadro’s number to calculate the number of molecules in the number of moles of C 7 H 14 O 2

8. 8/1/2015Zumdahl Chapter 37 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 formula C 11 H 17 N 3 O 8. Calculate the mass percentages of the four element in this compound. Strategy: 1.Calculate molar mass of C 11 H 17 N 3 O 8, by finding the mass contributed by each element. 2.Assume you have one mole of the compound. 3.Divide the mass of each element by the total mass of the compound.

9. 8/1/2015Zumdahl Chapter 38 Tetrodotoxin has the empirical formula C 11 H 17 N 3 O 8. Calculate the mass percentages (g/g) of the four element in this compound. Solution: 1.Calculate molar mass of C 11 H 17 N 3 O 8, by finding the mass contributed by each element C : 11x12 = 132 g / mol H : 1x17 = 17 N : 3x14 = 42 O : 8x16 = 126 tot = 317 g / mol 2.Calculate the mass of one mole of tetrodotoxin 317 g / mol x 1 mol = 317 g 3.Divide the mass of each element by the mass of the compound. Example C: 132 g / 317 g = 0.42 g/g (42 %)

10. 8/1/20159 1 mmol = 1 millimole =1x10 -3 mol 1mg = 1 milligram =1x10 -3 g 1msec = 1 millisec =1x10 -3 sec 1  mol = 1 micromole =1x10 -6 mol 1  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 Milli, Micro, etc

11. 8/1/201510 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)

12. 8/1/201511 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.

13. 8/1/201512 Using Conservation of Mass to determine an Empirical Formula 1.Write the reaction.Ni x C y O y -> X Ni + Y CO 2.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) 3.Find the number of moles of CO and of Ni. CO : 63.9 mg / (12.0+16.0 g/mol) = 2.28 mmol Ni : 33.5 mg / 58.7 g / mol) = 0.57 mmol 4.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 5.Y/X = 4: answer is NiC 4 O 4

14. 8/1/2015Zumdahl Chapter 313 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

15. 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! H 2 + O 2  H 2 O (not balanced) 2H 2 + O 2  2H 2 O (balanced) Chemical Equations

16. 4 Al(s) + 3 O 2 (g) → 2 Al 2 O 3 (s) This equation means 4 Al atoms + 3 O 2 molecules ---produces---> 2 molecules of Al 2 O 3 or 4 moles of Al + 3 moles of O 2 ---produces---> 2 moles of Al 2 O 3 Br 2 (l) + ___ Al (s) → __Al 2 Br 6 (s)

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

18. 8/1/201517 Step 1: Set the stoichimetric coefficient of the most complicated molecule (with the largest number of different elements ) to 1. Step 4: Eliminate fractional coefficients. To Balance an Equation Step 2: Balance as many atoms as possible in the second most complicated molecule. Ignore atoms that show up elsewhere in homonuclear species like O 2 and H 2. Step 3: Balance the atoms in the in the homonuclear species (O 2 and H 2 ). Step 5: Count the each atom type on each side of the equation.

19. Balancing Equations C 3 H 8 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) C 3 H 8 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) Combustion of Propane

20. Balancing Equations Combustion of Propane 1C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) 1C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g)

21. 8/1/2015Zumdahl Chapter 320 Writing Balanced Chemical Equations PbO 2 + Pb + H 2 SO 4 → PbSO 4 + H 2 O  PbO 2 + Pb + 2 H 2 SO 4 → 2 PbSO 4 + 2 H 2 O 2 Pb 10 O 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? Check for balance

22. 8/1/201521 3 Br 2 (l) + 2 Al (s) → 1 Al 2 Br 6 (s) Limiting Reactants excesslimitingGiven the amounts below

23. Limiting Reagent 1.Balance the reaction. 2.Convert reactant masses to moles [g(g/mol) - 1 =mol] 3.Normalize the moles of each reactant by its stoichiometric coefficient. 4.Find the smallest normalized number of moles 5.Use the ratio of stoichiometric coefficients to find the moles of product. 6.Convert to mass (if necessary). 8/1/201522

24. 8/1/201523 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 + Cl 2 → 2 NaCl n Na = 1.00 g x (1 mol Na / 23.0 g Na) = 0.0435 mol Na n Cl 2 = 1.00 g × (1 mol / 70.9 g Cl 2 ) = 0.0141 mol Cl 2 → → n Na /2 = 0.022 (normalized) n Cl 2 /1 = 0.0141 (normalized) Cl 2 is the limiting reagent 0.0141 moles of C l2 gives 0.0282 moles of NaCl. Use the molecular weight of NaCl to find the mass of NaCl produced.

25. 8/1/201524 What mass (in grams) of xenon tetrafluoride would be required to react completely with 1.000 g of water? XeF 4 + 2 H 2 O → Xe + 4 HF + O 2 =

26. 8/1/2015Zumdahl Chapter 325 At one point in the purification of silicon, gaseous SiHCl 3 reacts with gaseous H 2 to give gaseous HCl and solid Si. (a) Determine the chemical amount (in moles) of H 2 required to react with 160.4 mol of SiHCl 3. (b) Determine the chemical amount of HCl that is produced. (c) Determine the mass (in grams) of Si that is produced. SiHCl 3 (g) + H 2 (g) → 3 HCl (g) + Si (s) (a) (b)

27. 8/1/2015Zumdahl Chapter 326 At one point in the purification of silicon, gaseous SiHCl 3 reacts with gaseous H 2 to give gaseous HCl and solid Si. (a) Determine the chemical amount (in moles) of H 2 required to react with 160.4 mol of SiHCl 3. (b) Determine the chemical amount of HCl that is produced. (c) Determine the mass (in grams) of Si that is produced. SiHCl 3 (g) + H 2 (g) → 3 HCl (g) + Si (s) 1 mol SiHCl 3 1 mol H 2 3 mol HCl 1 mol Si (c)

28. 8/1/2015Zumdahl Chapter 327 Isotopes of Cl: 031 150 msec EC/ECp,11.98 0 S-31/P-30 30.9924 S-31P-30 032 298 msec EC/ECa/ECp,12.6 85 S-32/Si-28/P-31 31.985688 S-32Si-28P-31 033 2.511 sec EC,5.583 MeV S-33 32.97745 S-33 034 1.5264 sec EC,5.492 MeV S-34 33.97376 S-34 035 75.77% Stable 34.9688 036 3.01E+5 yr B- /EC,10.413 Ar-36/S-36 35.9683 Ar-36S-36 037 24.23% Stable 36.9659 038 37.24 min B-,4.917 MeV Ar-38 37.968010 Ar-38 039 55.6 min B-,3.442 MeV Ar-39 38.968008 Ar-39 040 1.35 min B-,7.480 MeV Ar-40 39.970413 Ar-40 041 38.4 sec B-,5.730 MeV Ar-41 40.970649 Ar-41 042 6.8 sec B-,9.430 MeV Ar-42 41.973172 Ar-42 043 3.3 sec B-,7.950 MeV Ar-43 42.974202 Ar-43 044 0.43 sec B-/B- n,3.920 Ar-44/Ar-43 43.978539 Ar-44Ar-43 045 400 msec B-/B-n,10.800 Ar-45/Ar-44 44.979710 Ar-45Ar-44 046 0.22 sec B-/B-n,6.900 Ar-46/Ar-45 45.984111 Ar-46Ar-45 047 200 nsec B-/B- n,14.700 Ar-47/Ar-46 46.98797647 Ar-47Ar-46 MASS abund. Halflife Particle, Energy Decay Product(s) Isotopic Mass 35 Cl contains 17 protons and 18 neutrons 37 Cl contains 17 protons and 20 neutrons