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Chapter 3

Chapter 3. Mass Relations: Stoichiometry. Atomic number. # of p + in nucleus. Mass number. # of p + and n 0 in nucleus. Mass # v. Atomic mass. Avg. Atomic Mass (atomic weight). Weighted avg. of atomic masses of naturally occurring isotopes of an element. Isotopes.

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Chapter 3

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  1. Chapter 3 Mass Relations: Stoichiometry

  2. Atomic number • # of p+ in nucleus

  3. Mass number • # of p+and n0 in nucleus

  4. Mass # v. Atomic mass

  5. Avg. Atomic Mass (atomic weight) • Weighted avg. of atomic masses of naturally occurring isotopes of an element

  6. Isotopes • Different forms of the same element with different mass • Same # p+ • Diff. # n0

  7. e.g. : naturally occurring Cu • 69.17% Cu-63 (atomic mass 62.939) • 30.83% Cu-65 (atomic mass 64.927) • (0.6917)(62.939)+(0.3083)(64.927) = 63.546 • Found on periodic table

  8. Masses of individual atoms

  9. The Mole (mol)

  10. Mole (mol) • Amt. of a substance that contains same # of particles as # of atoms in exactly 12 g of carbon-12

  11. Avogadro’s number • Number of particles in exactly one mole of a pure substance (6.02 x 1023) • Named for Amedeo Avogadro [Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e di Cerreto (1776 - 1856) ]

  12. Molar mass • Mass in grams of one mole of a pure substance • Used as a conversion factor, number is taken from the periodic table

  13. What is the mass (in g) of 2.5 mol of cobalt Molar mass Co = 58.933g 2.5 mol Co x _________ 2.5 mol Co x 58.933 g Co / 1 mol Co = 147.3325 g Co = 150 g Co

  14. How many atoms are in 0.000820 g of platinum? 0.000820 g Pt x ________ 0.000820 g Pt x I mol Pt/ 195.08 g Pt = 0.000004203 mol Pt 0.000004203 mol Pt x 6.02 x 1023 atoms Pt / 1 mol Pt = 2.530449047 x 1018 atoms 2.53 x 1018 atoms

  15. Percent composition (Mass of element / molar mass cmpd. X 100%) • e.g. find % composition of Cu2S • 2 mol Cu and 1 mol S • 2 mol Cu x 63.546g Cu / 1 mol Cu = 127.09 g Cu • 1 mol S x 32.06g S / 1 mol S = 32.06 g S molar mass Cu2S = 159.15 g • 127.09g Cu/ 159.15 g Cu2S x 100% = 79.855% Cu • 32.06g S/ 159.15g Cu2S x 100% = 20.14% S

  16. Determining simplest formula (Formula showing smallest whole number ratio of atoms) • e.g. find the simplest formula for a cmpd. containing 26.56% K, 35.41% Cr, and 38.03% O • If 100g of cmpd., then: K = 26.56 g, Cr = 35.41 g, O = 38.03 g

  17. (cont.) • 26.56g K x 1 mol K / 39.098g K = 0.6793 mol K • 35.41g Cr x 1 mol Cr / 51.996g Cr = 0.6810mol Cr • 38.03g O x 1 mol O / 15.999g O = 2.377 mol O divide by smallest number  0.6793 mol K / 0.6793 = 1.000 mol K 0.6810 mol Cr / 0.6793 = 1.003 mol Cr 2.377 mol O / 0.6793 = 3.499 mol O • 1.000 :1.003 : 3.499  2 : 2 : 7 K2Cr2O7

  18. Chemical Equations and Chemical Reactions

  19. Chemical equation • Represents (w/ symbols and formulas) the reactions and products in a chemical reaction • The same # of atoms of each element must appear on each side of the equation • Use coefficients to balance equation

  20. Word equations e.g. methane + oxygen  carbon dioxide + water

  21. Formula equations • CH4(g) + O2(g)  CO2(g) + H2O(g) (unbalanced) • CH4(g) + O2(g)  CO2(g) + 2H2O(g) • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

  22. Rules • H2, N2, O2, F2, Cl2, Br2, I2 (ClIF H BrON)

  23. Writing equations • e.g., Write a formula equation for the reaction between hydrogen gas and fluorine gas to produce hydrogen fluoride gas • H2(g) + F2(g)  HF(g) • H2(g) + F2(g)  2HF(g)

  24. Balancing equations • Balance: Al + Fe2O3 Al2O3 + Fe • 2Al + Fe2O3  Al2O3 + Fe • 2Al + Fe2O3  Al2O3 + 2Fe

  25. Balance: • NH3(g) + O2(g)  N2(g) + H2O(g) • 2NH3(g) + O2(g)  N2(g) + H2O(g) • 2NH3(g) + O2(g)  N2(g) + 3H2O(g) • 2NH3(g) + 2O2(g)  N2(g) + 4H2O(g)  wrong • 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)

  26. Mole relationships • H2(g) + Cl2(g)  2HCl(g) • 1 molecule of hydrogen reacts with 1 molecule of chlorine to yield 2 molecules of hydrogen chloride or • 1 mol H2 reacts with 1 mol Cl2 to yield 2 mol HCl or

  27. (cont.) • 2g H2 (1 x molar mass) reacts with 71g Cl2 (1 x molar mass) to yield 73g HCl (2 x molar mass) • I mol H2 : 1 mol Cl2 : 2 mol HCl • 2g H2 : 71g Cl2 : 73g HCl

  28. Types of chemical reactions • Synthesis, 2Mg(s) + O2(g)  2MgO(s) • Decomposition, 2H2O(l)  2H2(g) + O2(g) • Single replacement, Mg(s) + 2HCl(aq)  H2(g) + MgCl2 • Double replacement, Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2 KNO3(aq)

  29. (cont.) 5. Combustion, C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

  30. Specific examples • Decomposition of metal hydroxides, Ca(OH)2(s)  CaO(s) + H2O(g) • Decomposition of metal chlorates 2KClO3(s)  2 KCl(s) + 3O2(g) • Replacement of hydrogen in water by a metal, 2K(s) + 2H2O(l)  2 KOH(aq) + H2(g)

  31. Stoichiometry

  32. Molar mass • Mass in grams of one mole of a pure substance • Used as a conversion factor, number is taken from the periodic table

  33. Stoichiometry is the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions

  34. Stoichiometric air-fuel ratios of common fuels Fuel By weight Percent fuel Gasoline 14.7 : 1 6.8% Natural Gas 17.2 : 1 5.8% Ethanol 9 : 1 11.1% Diesel 14.6 : 1 6.8%

  35. Stoichiometry(mass relationships) • Start with a balanced equation •  mole ratio • 2Al2O3  4Al + 3O2 • 2 mol Al2O3 : 4 mol Al : 3 mol O2 • e.g. 2 mol Al2O3/ 4 mol Al

  36. ? Mol Al produced f/ 15.0 mol Al2O3 15.0 mol Al2O3 x 4 mol Al / 2 mol Al2O3 = 30.0 mol Al

  37. 4 types of stoichiometry problems • Mole – mole • Mole – mass • Mass – mole • Mass - mass

  38. mole - mole • CO2 + 2LiOH  Li2CO3 + H2O • How many of LiOH are required to react with 30 mol of CO2

  39. (cont.) • 30 mol CO2 x 2 mol LiOH / 1 mol CO2 = 60 mol LiOH

  40. Mole - mass 6 CO2 + 6H2O  C6H12O6 + 6O2

  41. Given 3.00 mol of water and an excess of carbon dioxide how many grams of glucose will be produced? • 3.00 mol H2O x 1 mol C6H12O6/ 6 mol H2O x 180 g C6H12O6/ 1 mol C6H12O6 = 90.0 g C6H12O6

  42. Mass-mole • C + SO2 CS2 + CO • If 8.00 g of SO2 reacts with an excess of carbon how many moles of CS2 are formed? • 5C + 2SO2 CS2 + 4CO • 8.00 g SO2 x 1 mol SO2/ 64.1 g SO2 x 1 mol CS2/ 2 mol SO2 = 0.0624 mol CS2

  43. Mass -mass • Sn(s) + 2HF(g)  SnF2(s) + H2(g) • How many g of SnF2 is produced from the reaction of 30.00g of HF with an excess of Sn?

  44. 30.00g HF x 1 mol HF/ 20.01g HF x 1 mol SnF2/ 2 mol HF x 156.7g SnF2/ 1 mol SnF2 = 117.5 g SnF2

  45. Limiting Reactant

  46. Limiting reactant • Reactant that limits the amt. of the other reactants that can combine, and the amt. of product formed

  47. Limiting reactant • Silicon dioxide (quartz) reacts with hydrogen fluoride according to the following reaction: SiO2 (s) + 4HF(g)  SiF4(g) + 2H2O(l)

  48. (cont.) • If 2.0 mol of HF is combined with 4.5 mol of SiO2, which is the limiting reactant? 2.0 mol HF x 1 mol SiO2/ 4 mol HF = 0.50 mol SiO2 • Therefore 2.0 mol HF requires 0.50 mol of SiO2 to completely react, 4.5 mol SiO2 is more than 0.50 mol, therefore HF is the limiting reactant

  49. Percent yield • Percent yield = actual yield/ theoretical yieldx 100% • C6H6 + Cl2 C6H5Cl + HCl • If 36.8 g of C6H6 reacts with an excess of Cl2 the actual yield is 38.8 g of C6H5Cl. What is the percent yield?

  50. (cont.) • 36.8 g C6H6 x 1 mol C6H6/ 78.1 g C6H6 x 1 mol C6H5Cl/ 1 mol C6H6 x 113 g C6H5Cl/ 1 mol C6H5Cl = 53.2 g C6H5Cl (theoretical yield)

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