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Feb. 4. Announcements

Feb. 4. Announcements. 1. EXAM 1 is on Feb. 12, Tuesday. Help Session on Friday and Monday. 2. HW2 is due on Feb. 7; HW3 is due on Feb.12. 3. Guidelines and Samples for EXAM 1 are posted on Aris.

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Feb. 4. Announcements

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  1. Feb. 4. Announcements 1. EXAM 1 is on Feb. 12, Tuesday. Help Session on Friday and Monday. 2. HW2 is due on Feb. 7; HW3 is due on Feb.12. 3. Guidelines and Samples for EXAM 1 are posted on Aris. 4. Please come to class with ppt slides downloaded/printed. They are on Aris by 9pm on the day before class.

  2. Weight and molar percentage of elements in compounds A dry-cleaning solvent is suspected to by a cancer-causing agent. Elemental analysis of this compound produces 1.23 g of C, 0.069 g of H and 1.20 g of Cl. Determine the molecular formula for this solvent if it has a molar mass of 146.99 g/mol. The same empirical formula corresponds to many different molecular formulas : CH2Cl, C2H4Cl4 etc. Also, the same molecular formula may correspond to several different structural formulas (isomers): Cl-CH2-CH2-Cl or H3C-CHCl2

  3. m 2 m 2 CnHm + (n+ ) O2 = n CO(g) + H2O(g) Figure 3.4 Combustion train for the determination of the chemical composition of organic compounds.

  4. PROBLEM: Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion =85.35g mass of CO2 absorber before combustion =83.85g mass of H2O absorber after combustion =37.96g mass of H2O absorber before combustion =37.55g What is the molecular formula of vitamin C? Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis PLAN: difference (after-before) = mass of oxidized element find the mass of each element in its combustion product preliminary formula empirical formula molecular formula find the mols

  5. CO2 85.35 g-83.85 g = 1.50 g H2O 37.96 g-37.55 g = 0.41 g 2.016 g H2O 12.01 g CO2 44.01 g CO2 18.02 g H2O 12.01 g C 1.008 g H 16.00 g O 0.409 g C 0.046 g H 0.545 g O 176.12 g/mol 88.06 g Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis continued SOLUTION: There are 12.01 g C per mol CO2 1.50 g CO2 = 0.409 g C There are 2.016 g H per mol H2O. 0.41 g H2O = 0.046 g H O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 = 0.0341 mol C = 0.0456 mol H = 0.0341 mol O C1H1.3O1 C3H4O3 = 2.000 C6H8O6

  6. Table 3.3 Some Compounds with Empirical Formula CH2O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) Molecular Formula M (g/mol) Whole-Number Multiple Name Use or Function 30.03 1 CH2O formaldehyde disinfectant; biological preservative 60.05 2 C2H4O2 acetic acid acetate polymers; vinegar(5% soln) 3 90.09 C3H6O3 lactic acid sour milk; forms in exercising muscle 4 120.10 C4H8O4 erythrose part of sugar metabolism 5 150.13 C5H10O5 ribose component of nucleic acids and B2 6 180.16 C6H12O6 glucose major energy source of the cell CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6

  7. C4H10 C2H6O Table 3.4 Two Pairs of Constitutional Isomers Property Butane 2-Methylpropane Ethanol Dimethyl Ether M(g/mol) 58.12 58.12 46.07 46.07 Boiling Point -0.50C -11.060C 78.50C -250C Density at 200C 0.579 g/mL (gas) 0.549 g/mL (gas) 0.789 g/mL (liquid) 0.00195 g/mL (gas) Structural formulas Space-filling models

  8. Stoichiometry Calculations from balanced equations

  9. Figure 3.6 The formation of HF gas on the macroscopic and molecular levels.

  10. Chemical Equations • reaction of H2 and F2 • 2.016 g of H2 + 38.00 g F2→ 40.02 g HF • mass conserved • no information about actual particles • convert to moles: • 1 mole H2 + 1 mole F2→ 2 moles HF • reflects number of reacting particles

  11. Writing a balanced equation • atoms cannot be created or destroyed • atoms which appear in the reactants must be present in the products • number must be the same • usually specify the states of matter if known

  12. Balance each of the following equations: • KClO3→ KCl + O2 • Cl2 + HBr → HCl + Br2 • Fe2S + O2 → Fe2O3 + SO2

  13. Write balanced equations to represent the following: • In a flash-bulb, magnesium wire and oxygen gas react to produce powdery magnesium oxide. • Within the cylinders of a car engine, octane (C8H18) mixes with the oxygen in the air and burns to form carbon dioxide and water vapor. • Sodium reacts violently with water to form hydrogen gas and sodium hydroxide solution [ LAB DEMO]

  14. Aqueous nitric acid reacts with calcium carbonate to form calcium nitrate, carbon dioxide and water. • Phosphorus trfluoride is prepared by the reaction of phosphorus trichloride and hydrogen fluoride; hydrogen chloride is the other product. • Liquid nitroglycerine (C3H5N3O9) explodes to produce a mixture of carbon dioxide, water vapor, nitrogen and oxygen

  15. Balancing Chemical Equations translate the statement balance the atoms adjust the coefficients check the atom balance specify states of matter

  16. PLAN: SOLUTION: 2C8H18 + 25O216CO2 + 18H2O C8H18 + O2 CO2 + H2O C8H18 + O2 CO2 + H2O 2C8H18 + 25O216CO2 + 18H2O balance the atoms adjust the coefficients check the atom balance 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) specify states of matter Sample Problem 3.7 Balancing Chemical Equations PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. translate the statement 25/2 8 9

  17. Summary of the mass-mole-number relationships in a chemical reaction. Figure 3.8 MASS(g) of compound A MASS(g) of compound B M (g/mol) of compound A M (g/mol) of compound B molar ratio from balanced equation AMOUNT(mol) of compound A AMOUNT(mol) of compound B Avogadro’s number (molecules/mol) Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound B

  18. In the combustion of propane, how many moles of O2 are consumed when 10.0 mol of H2O are produced? • C3H8 (g) + 5O2 (g) → 3 CO2 (g) + 4 H2O (g) • 5 mol O2 produce 4 mol H2O • this gives a conversion factor: • 12.5 mol O2 consumed

  19. General Steps • write a balanced equation for the reaction • convert mass (or #) to moles • use molar ratio as conversion factor • convert moles back to mass (or #)

  20. Copper is obtained from sulfide ores, such as chalcocite, or copper (I) sulfide, by a multistage process. After an initial grinding step, the first stage is to “roast” the ore (heat it strongly in oxygen) to form powdered copper (I) oxide and gaseous sulfur dioxide.How many moles of oxygen are required to roast 10.0 mol of copper (I) sulfide? • 2Cu2S + 3O2→ 2Cu2O + 2SO2 • n of O2 = 15.0 mol

  21. What mass (in grams) of sulfur dioxide are formed when 10.0 mol of copper (I) sulfide is roasted? • n SO2 = 10 mol; M = 64.07; m = 641 g What mass (in kg) of oxygen is required to form 2.86 kg of copper (I) oxide? M of Cu2O = 143.10 g; moles = 20.0 mass of O2 = 0.960 kg

  22. Thermite is a mixture of iron (III) oxide and aluminum powders that was once used to weld railroad tracks. It undergoes a spectacular reaction to yield solid aluminum oxide and molten iron.What mass (in grams) of iron form when 135 g of aluminum reacts? • Fe2O3 + 2Al → 2Fe + Al2O3 • n of Al = 5.004; mass of Fe = 279 g How many atoms of Al are required to produce each 1.00 g of aluminum oxide? M of Al2O3 = 101.96; n = 0.00981 atoms Al = 1.18 x 1027 atoms

  23. Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water.How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide? • 2 LiOH (s) + CO2 (g) → Li2CO3 (s) + H2O (l) • M; LiOH = 23.95; CO2 = 44.01 • 0.919 g CO2

  24. The decomposition of KClO3 is commonly used to prepare small amounts of oxygen in the laboratory, according to the reaction 2KClO3→ 2KCl + 3O2Calculate the mass of O2 which can be prepared from a 4.50 g sample of KClO3 • M of KClO3 = 122.55 • m = 1.77 g

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