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Ch. 3 Stoichiometry

Its Back!!. Ch. 3 Stoichiometry. Atomic Mass. How heavy is an atom of oxygen? There are different kinds of oxygen atoms. More concerned with average atomic mass. Based on abundance of each element in nature. Don’t use grams because the numbers would be too small. Measuring Atomic Mass.

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Ch. 3 Stoichiometry

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  1. Its Back!! Ch. 3 Stoichiometry

  2. Atomic Mass • How heavy is an atom of oxygen? • There are different kinds of oxygen atoms. • More concerned with average atomic mass. • Based on abundance of each element in nature. • Don’t use grams because the numbers would be too small

  3. Measuring Atomic Mass • Unit is the Atomic Mass Unit (amu) • One twelfth the mass of a carbon-12 atom. • = 1.660540 * 10 -27 • Amu is the number on the periodic table • Refers to the mass of one atom of an element • Each isotope has its own atomic mass we need the average from percent abundance.

  4. Calculating averages • You have five rocks, four with a mass of 50 g, and one with a mass of 60 g. What is the average mass of the rocks? • Total mass = 4 x 50 + 1 x 60 = 260 g • Average mass = 4 x 50 + 1 x 60 = 260 g 5 5 • Average mass = 4 x 50 + 1 x 60 = 260 g 5 5 5

  5. Calculating averages • Average mass = (4/5) x 50 + (1/5) x 60 = 260 g • Average mass = .8 x 50 + .2 x 60 • 80% of the rocks were 50 grams • 20% of the rocks were 60 grams • Average = % as decimal x mass + % as decimal x mass + % as decimal x mass

  6. Atomic Mass • Is not a whole number because it is an average. • These are responsible for the decimal numbers on the periodic table.

  7. The Mole 6.02 X 1023 http://www.teachertube.com/viewVideo.php?title=Happy_Mole_Day_to_You_Chemistry_Song&video_id=41557

  8. Molar Mass • Molecular Mass/Molecular Weight: • The molecular mass/weight is the amount of mass in 1 mole of the molecule.

  9. Molar Mass • Molar mass of atoms: • 1 mole of Br atom = 79.9 g / mole. • 1 mole of Sn atom = 118.7 g / mole. • Molar masses for compounds or Molecules are the sum of molar masses of each atom. • 1 Mole of CaCl2 = 111.1 g / mole {40.1 + 2(35.5)}

  10. Percent composition • Percent of each element in a compound. TO Calculate: • Find the mass of each element • Divide that by the molar mass • Multiply by a 100.

  11. Percent Composition • Find the percent composition of CH4 • C= 1x 12g= 12g • H= 4 x 1g = 4g 16g is the molar mass of CH4 • % of C = 12/16 = 75% • % of H = 4/16 = 25%

  12. Empirical Formula

  13. Empirical Formulas • Empirical Formula = the lowest ratio of atoms in a molecule. • Ex. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula?

  14. Formula from percentage • Assume 100 g of sample • Divide by the molar mass for each element. • Find the smallest approximate whole number ratio of smallest number to each of the other element • Determine formula of the compound.

  15. Example 1. Determine the empirical formula of a compound that contains 36.5% sodium, 25.4% sulfur, and 38.1% oxygen.

  16. Assume 100 g of sample • Solution:  By assuming that we can study a 100g sample of the compound, we can change % to grams. so: • So the problem now reads: Determine the empirical formula of a compound that contains 36.5g sodium, 25.4g sulfur, and 38.1g oxygen.

  17. Divide by the molar mass for each element • Now we can solve them by finding the molar ratio by which the elements combine. • mass of that element in the sample moles of an element = --------------------- Molar mass of the element

  18. Divide by the molar mass for each element •     36.5 g# of moles of sodium = ------------- = 1.59 moles                                     23.0g/mole •                                      25.4g# of moles of sulfur = --------------- = 0.791 moles                                    32.1 g/mole •                                        38.1 g# of moles of oxygen = --------------- = 2.38 moles                                       16.0 g/mole         

  19. Find the smallest approximate whole number ratio of smallest number to each of the other element • # of moles of Na  = 1.59 moles                              -------------- = 2.01                              0.791 moles         • # of moles of S   = 0.791 moles                              -------------- = 1                              0.791 moles • # of moles of O   = 2.38  moles                              -------------- = 3.01                              0.791 moles

  20. Determine formula of the compound • The ratio shows that 2 atoms of sodium combine with 1 atom of sulfur and 3 atoms of oxygen, so our answer is Na2SO3. • Answer: Na2SO3

  21. Mass to moles(mass is already given) • Determine mass of each of the elements. • Divide by the molar mass for each element • Find the smallest approximate whole number ratio of smallest number to each of the other element • Determine formula of the compound.

  22. Solution:  Find the mass of the empirical formula (CH2): • C = 12.0 x 1 atom = 12.0 gH = 1.01 x 2 atoms = 2.02 g                               ----------14.0 g

  23. next, divide that number into the molecular mass: • 56.0 g-------  = 414.0 g • Now use that number, 4, as a multiplier for the subscripts in the empirical formula: • CH2 x 4 = C4H8 • Answer = C4H8

  24. Another Way to Determine Molecular Formulas Caffeine a stimulant found in coffee, contains, 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol determine molecular formula • Assume 100 total grams the element for each 100 grams of compound • Multiply by molar mass of the compound. • Divide by molar mass of the element.

  25. 49.48 g C x 194.2g x 1 mol C = 8.01 mol C 100g caffeine 1 mol 12.01g C • 5.15 g H x 194.2g x 1 mol H = 9.92 mol H 100g caffeine 1 mol 1.01g H • 28.87g N x 194.2g x 1 mol N = 4.002 mol N 100g caffeine 1 mol 14.01g N • 16.49 g O x 194.2g x 1 mol O = 2.001 mol O 100g caffeine 1 mol 16.0g O Molecular Formula C8H10N4O2

  26. Law of Conservation of Mass and Balancing Chemical Equations • Matter is neither created nor destroyed during a chemical reaction. Therefore, all the atoms that were present at the start of the reaction must be present at the end of the reaction.

  27. Steps to Balance Chemical Equations • Write the formula equation with the correct symbols and formulas. Na + Cl2 NaCl • Count the number of atoms of each element on each side of the arrow. • Balance atoms by using coefficients. 2Na + Cl2 2NaCl • Check your work by counting atoms of each element.

  28. Mole/Mole Problems • A balanced chemical equation • Determine the mole ratio • Set-Up problem….start with the given

  29. Li + O2 Li2O • Balance equation: 4 Li + O2 2 Li2O • Mole ratio of lithium to lithium oxide: 4 mol lithium to 2 mol lithium oxide • Set-Up problem, start with given.

  30. 2 mol Li x 2 mol Li2O = 4 mol Li Answer: 1 mol Li2O

  31. Mass/Mole and Mole/Mass Problems • Must have 3 pieces of information: • Balanced chemical equation • Mole ratio • Molar mass

  32. Problem 1 • What mass in grams of magnesium oxide is produced from 2.00 mol of magnesium? • Mg + O2 MgO • 2 Mg + O2 2 MgO • Molar Mass of MgO • 40.31 g/mol 2.00 mol Mg x 2 mol MgO x 40.31g MgO = 2 mol Mg mol MgO

  33. Mass/Mass Problems • Balanced equation • Mole ratio • Molar Masses

  34. Problem 1 • Tin (II) fluoride, SnF2 is used in toothpaste. How many grams of tin (II) fluoride are produced from the reaction of 30.0 g of HF with Sn? (Hydrogen is also a product.) • Sn + HF  SnF2 + H2 • Sn + 2 HF  SnF2 + H2

  35. Sn + 2 HF  SnF2 + H2 • How many grams of tin (II) fluoride are produced from the reaction of 30.0 g of HF with Sn? • Mole ratio? • 2 mole HF : 1 mole SnF2 • Molar masses of HF and SnF2? • HF: 20.01 g/mol • SnF2: 156.71 g/mol

  36. How many grams of SnF2 are produced from the reaction of 30.0 g of HF with Sn? 30.0gHF x mol HF x 1 mol SnF2 x 156.71gSnF2 20.01gHF 2 mol HF mol SnF2 Answer: 117 g SnF2

  37. Limiting Reactant Problems • Balanced chemical equation. • Determine the number of moles of each reactant. • Divide each number of moles of each reactant by the coefficient in the balanced equation. The smallest number is the limiting reactant. • Proceed with solving problem with original numbers of limiting reactant.

  38. Limiting Reactant Problem If 20.5 g of chlorine is reacted with 20.5 g of sodium, which is the limiting reactant? How much salt is formed? • Cl2 + Na  NaCl • Cl2 + 2 Na  2 NaCl

  39. Cl2 + 2 Na  2 NaCl • If 20.5 g of chlorine is reacted with 20.5 g of sodium, which is the limiting reactant? 20.5 g Cl2 x 1 mol Cl2 = 0.289 mol 70.9 g 20.5 g Na x 1 mol Na = 0.892 mol 22.99 g Cl2 : 0.289 ÷ 1 = 0.289 Na : 0.892 ÷ 2 = 0.445

  40. How much salt is formed? • Limiting Reactant? • Chlorine 0.289 mol Cl2 x 2 mol NaCl x 58.44 g NaCl 1 mol Cl2 1 mol NaCl 33.8 g NaCl

  41. Percent Yield • Theoretical Yield: the maximum amount of product that can be produced from a given amount of reactant. • Actual Yield: the measured amount of product obtained from a reaction.

  42. Percent Yield = Actual yield x 100 Theoretical yield

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