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Chapter 1

Chapter 1. Be familiar with the metric system and its prefixes (conversion). Apply the rules for significant figures (and do not forget to use them after chapter 1). Chapter 1. Do not resist using conversion factors. They will be used over and over again throughout the course.

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Chapter 1

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  1. Chapter 1 • Be familiar with the metric system and its prefixes (conversion). • Apply the rules for significant figures (and do not forget to use them after chapter 1).

  2. Chapter 1 • Do not resist using conversion factors. They will be used over and over again throughout the course. • Note: solubility calculations involve conversion factors (relates grams of solute to grams of solvent).

  3. Chapter 1 • Element vs. compound vs. mixture (heterogeneous and homogeneous). • Separation techniques. • Distinguish between intensive and extensive properties. • Distinguish between chemical and physical properties.

  4. Chapter 1 • Example: 1. How much water is required to dissolve 80g of lead nitrate at 100oC? • 57g H2O (show all work)

  5. Chapter 1 • Example: 2. Cool to 10oC. How much lead nitrate remains in solution? How much crystallizes? 28g lead nitrate, 52g crystallizes.

  6. Chapter 2 • Conceptually this chapter is easy to assimilate… BUT…there is a lot of memorizing. • It is mostly CHEM T material.

  7. Chapter 2 • You must be able to: • Learn the molecular formulas of the elements (fig 2.8), charges of the transition metal cations (fig. 2.10), names and formulas of polyatomic ions (table 2.2), and the nomenclature system for oxoacids.

  8. Chapter 2 • Naming compounds requires that you recognize ionic versus covalent compounds.

  9. Chapter 3 • Students have little trouble calculating atomic masses from isotopic data. The reverse process is tricky due to the algebra and significant figures.

  10. Chapter 3 • Note that mole-gram conversions will be required in almost all future chapters, usually as the first step in solving a complex problem.

  11. Chapter 3 • Formulas: subscripts not only give atom ratios, but also mole ratios (important for obtaining simplest formula from mass percents).

  12. Chapter 3 • Students normally do well calculating formulas from mass percents, but struggle when do it from analytical data.

  13. Chapter 3 • A chemical equation describes what happens in the laboratory. Physical states should be included when possible (not needed on the AP exam equation writing section, however).

  14. Chapter 3 • USE CONVERSION FACTORS (not ratios-proportions). • There are many ways to do limiting reagent and theoretical yield problems. The approach in 3.4 is recommended.

  15. Chapter 3 • Meaning of atomic masses - gives relative masses of atoms. Based on C-12 scale; most common isotope of carbon is assigned a mass of exactly 12 amu.

  16. Chapter 3 • Ca = 40.08 amu • Ni = 58.69 amu • A nickel atom is 58.69/40.08 = 1.464 times as heavy as a calcium atom.

  17. Chapter 3 • Atomic masses from isotopic composition: • A.M. = (mass of isotope1)(%/100) + (mass of isotope2)(%/100)+….

  18. Chapter 3 • Ne-20: 20.00u(90.92%), Ne-21: 21.00u(0.26%), Ne-22: 22.00u(8.82%) • What is the average atomic mass of neon? • (20.18 amu)

  19. Chapter 3 • Masses of atoms:Since the atomic masses of H, Cl and Ni are 1.008amu, 35.45amu, and 58.69amu, it follows that 1.008g H, 35.45g Cl, and 58.69g Ni all contain the same number of atoms, NA(avagadro’s number).

  20. Chapter 3 • Find the mass of a hydrgen atom in grams. 1.674 x 10-24 g. • Find the number of atoms in one gram of nickel. 1.026 x 1022 atoms.

  21. Chapter 3 • Formula Mass = sum of atomic masses in a formula • FM of H2O = 18.02 amu • If the formula represents a molecule, formula mass = molecular mass

  22. Chapter 3 • The MOLE • Meaning: 1 mol = 6.022 x 1023 items • 1 mol H = 6.022 x 1023 H atoms; mass = 1.008g • 1 mol Cl = 6.022 x 1023 Cl atoms; mass = 35.45g

  23. Chapter 3 • Find the molar mass of C6H12O6. 180.18 amu, 180.18 g/mol

  24. The Mole • Calculate the mass in grams of 13.2 mol of calcium chloride. • 1.47 x 103 g

  25. The Mole • Calculate the number of moles of molecules in 16.4 g of C6H12O6. Calculate the moles of H atoms. • 0.0910 mol of molecules, 1.09 mols of H atoms

  26. The Mole • Calculate the percent composition of each element in potassium chromate. • %K = 40.27%, %Cr = 26.78%, %O = 32.96%

  27. The Mole • Simplest Formula from Analytical Data: A sample of acetic acid (C, H, O atoms) weighing 1.000g burns to give 1.466g CO2 and 0.6001g H2O. Find its empirical formula. • CH2O

  28. The Mole • Simplest Formula from Analytical Data: Find the mass of C in the sample (from CO2), then the mass of H (from H2O) and finally the mass of O from the difference. • CH2O

  29. The Mole • Molecular Formula from Simplest Formula : If the molar mass of acetic acid is 60.0g/mol, and its empirical formula is CH2O, then what is the molecular formula? • C2H4O2

  30. Chemical Equations • Balancing: • Conservation of mass - must have same types and number of atoms on both sides of the equation. You may adjust coefficients but you can’t change the chemical formulas.

  31. Chemical Equations • Balancing: • Guess, check your guess, keep guessing. • Ex. Combustion of propane. Combustion of butane. Ferrous nitrate with potassium iodide.

  32. Chemical Equations • Stoichiometry (stoiching) : How many moles of carbon dioxide are produced when 1.65 mol of propane burns? • 4.95 mol CO2

  33. Chemical Equations • Stoichiometry (stoiching) : What is the mass of O2 required to react with 12.0g of propane? • 43.6g O2

  34. Chemical Equations • Product yield: • Limiting reactants, theoretical yield: ordinarily reactants are not present in the exact ratio required for the reaction. One is used up completely, while the other is in excess.

  35. Chemical Equations • The consumed reactant (limiting reagent) limits the amount of product and gives the theoretical yield of the reaction.

  36. Chemical Equations • 1. Calculate the yield expected if the first reagent is limiting. • 2. Repeat for the other reagents. • 3. The theoretical yield is the smaller of these yields and identifies the limiting reagent.

  37. Chemical Equations • 1.00g of silver is reacted with 1.00g of iodine to synthesize silver iodide. What is the limiting reagent and theoretical yield of this reaction? • I2 is limiting, theoretical yield = 1.85g AgI

  38. Chemical Equations • Theoretical vs. actual yield. • In the previous problem, only 1.50g is collected. What is the percent yield? • % yield = actual yield/theoretical yield x 100% = 81.1%

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