1 / 64

CHAPTER 3

CHAPTER 3. STOICHIOMETRY. Determination of quantities of materials consumed and produced in a chemical reaction. CHEMICAL REACTION. A + B Product Reactants . Periodic Table. Atomic Mass number below the element

sara-valenzuela
Download Presentation

CHAPTER 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 3

  2. STOICHIOMETRY • Determination of quantities of materials consumed and produced in a chemical reaction.

  3. CHEMICAL REACTION • A + B Product • Reactants

  4. Periodic Table • Atomic Mass • number below the element • not whole numbers because the masses are averages of the masses of the different isotopes of the elements

  5. STOICHIOMETRY • For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.

  6. Determination of Aver. Mass • Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

  7. Take Note: • If there are more than 2 isotopes, then formula has to be re-adjusted

  8. Sample Problem 1 • Assume that element Uus is synthesized and that it has the following stable isotopes: • 284Uus (283.4 a.m.u.) 34.6 % • 285Uus (284.7 a.m.u.) 21.2 % • 288Uus (287.8 a.m.u.) 44.20 %

  9. Solution • Ave. Mass of Uus = • [284Uus] (283.4 a.m.u.)(0.346) • [285Uus] +(284.7 a.m.u.)(0.212) • [288Uus] +(287.8 a.m.u.)(0.4420) • = 97.92 + 60.36 + 127.21 • = 285.49 a.m.u (FINAL ANS.)

  10. For Your Benefit • Do Problem 24 on page 123 of Zumdahl text.

  11. The MOLE • Amount of substance that contains as many entities as there are in exactly 12 grams of carbon-12.

  12. The MOLE • The mass of 1 mole = the atomic mass of the element in grams

  13. Formula for Mole Mole = mass of element atomic mass of element

  14. Sample Mole Calculations 1 mole of C = 12.011 grams • 12.011 gm/mol • 0.5 mole of C = 6.055 grams • 12.011 gm/mol

  15. Avogadro’s Number • Way of counting atoms • Avogadro’s number = 6.02 x 1023

  16. Point to Remember One mole of anything is 6.02 x 1023 units of that substance.

  17. Avogadro’s Number and the Mole • Ifone mole of anything is 6.02 x 1023 units of that substance, then: • 1 mole of oranges = 6.02 x 1023 oranges

  18. And…….. • 1 mole of C has the same number of atoms as one mole of any element

  19. Also….. • 1 mole of sand = 6.02 x 1023 particles

  20. An Even Better Analogy….. • 1 dozen = 12 entities • a dozen apples has the same number of entities as a dozen oranges

  21. Summary • Avogadro’s Number gives the number of particles or atoms in a given number of moles • 1 mole of anything = 6.02 x 10 23 atoms or particles

  22. Sample Problem 2 • Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.

  23. Solution • PART I: • Formula for Mole: • Mole = mass of element atomic mass of element

  24. Solution (cont.) • Part II: To determine # of atoms • # atoms = moles x Avogadro’s number

  25. Problem # 2 • A diamond contains 5.0 x 1021 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?

  26. Molar Mass • Often referred to as molecular mass • Definition: • mass in grams of 1 mole of the compound

  27. Example Problem • Determine the Molar Mass of C6H12O6

  28. Solution • Mass of 6 mole C = 6 x 12.01 = 72.06 g • Mass of 12 mole H = 12 x 1.008 = 12.096 g • Mass of 6 mole O = 6 x 16 = 96 g • Mass of 1 mole C6H12O6 = 180.156 g

  29. Problem # 2 • A diamond contains 5.0 x 1021 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?

  30. Problem #3 • What is the molar mass of (NH4)3(PO4)?

  31. % Mass Determination Step I. Total % Masses of atoms = 100 %

  32. % Mass Determination Step II. If formula is given, break the compound down and get total atomic masses of each element.

  33. % Mass Determination Step III. Divide total atomic masses of each element by total molar mass to determine element contribution

  34. % Mass Determination Step IV. Multiply by 100 to get percent

  35. Sample Problem • Find the % Mass of: - FeO (% Fe = ? and % O = ?) • Fe2O3 (% Fe = ? and % O = ?)

  36. Composition of Compounds How many grams of silicon are there in 217.00 grams of SiO2? [Hint: Determine % composition first.]

  37. Empirical Formula • Only gives the types of elements in the compound and the simplest ratio of the elements in the formula

  38. Empirical Formula • Does not tell exactly how many of the elements are in the compound

  39. Molecular Formula • Gives the exact number of elements in the compound as it exists. • Gives you the exact elemental composition of the compound • Formula of the compound as it would actually exist.

  40. EF vs. MF Sucrose or table sugar: Molecular Formula = C6H12O6 Empirical Formula = CH2O

  41. Empirical Formula • EF Determination when % Masses are given

  42. Steps in Determining EF • Step 1. Sum up all given percentages. • If sum of percentages = 100 % or very close to it, proceed to Step 2. • If sum is < 100 %, the missing percentage is often due to oxygen or the missing element present in the elemental analysis.

  43. Step 2. Convert Mass % to grams. • Step 3. Convert all grams to moles using the equation: mole = gram of element atomic mass of element

  44. Step 4. Divide all calculated moles by the smallest calculated mole to get a simplest ratio of 1.

  45. Step 5. If the ratios are whole numbers, you now have the Empirical Formula. The ratios are the subscripts of the elements in the empirical formula. • If the ratios are not whole numbers, follow the rule of rounding.

  46. Rule of Rounding Molar Ratios • Mole ratios can only be rounded to the nearest whole number if they are < 0.2 away from the nearest whole number. For ex: 1.95 = 2; 3.18 = 3 and 4. 13 = 4. • If the mole ratio is > 0.2 away from the nearest whole number, multiply the mole ratio by a certain integer to get it close to the nearest whole number. For ex: 3.5 x “2” = 7; 6.33 x “3” = 18.99 = 19; 4.25 x “4” = 11.

  47. Please Remember • If you have to multiply a mole ratio by an integer to get close to a whole number, you MUST multiply all the other mole ratios by the same integer. • “In short, what you do to one mole ratio, you also do to the rest.” • The ratios give you the subscripts in the EF.

  48. Steps To Determine the Molecular Formula • Step 1. Now that you have the empirical formula, get the ratio of the “given” molar mass to the empirical formula mass. Ratio = Given Molar Mass Empirical Formula Mass * Round ratio to the nearest whole number. • Please note that the Empirical formula Mass is the sum of the atomic masses of all the elements in the Empirical Formula.

  49. Step 2. Once the ratio has been determined, multiply all the subscripts in the empirical formula by the ratio. This gives you the Molecular Formula.

  50. Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas D = heat (aq) = aqueous solution

More Related