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Chemistry Chapter 7

Chemistry Chapter 7. Part 2 Molar mass, conversions, %composition, empirical formula. Molar Mass. H 2 O. = 2.0 g. H. 2. x 1.0 g. = 16.0 g. 1. O. x 16.0 g. 18.0 g. Molar Mass. C 6 H 12 O 6. C. 6. x 12.0 g. = 72.0 g. = 12.0 g. H. 12. x 1.0 g. = 96.0 g. 6. O.

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Chemistry Chapter 7

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  1. Chemistry Chapter 7 Part 2 Molar mass, conversions, %composition, empirical formula

  2. Molar Mass H2O = 2.0 g H 2 x 1.0 g = 16.0 g 1 O x 16.0 g 18.0 g

  3. Molar Mass C6H12O6 C 6 x 12.0 g = 72.0 g = 12.0 g H 12 x 1.0 g = 96.0 g 6 O x 16.0 g 180.0 g

  4. Percent Compositon part X 100 = % whole H 2 x 1.0 g = 2.0 g H2O O 1 x 16.0 g = 16.0 g 18.0 g

  5. H2O part % = X 100 whole 2.0 g H % = 11.1 H % = x 100 18.0 g 16.0 g O % = 88.9 O % = x 100 18.0 g

  6. part % = X 100 C6H12O6 whole 72.0 g C % = 40.0 C % = X 100 180.0 g 12.0 g H % = X 100 H % = 6.7 180.0 g 96.0 g O % = X 100 O % = 53.3 180.0 g

  7. Conversions Convert 10.0 g C6H12O6 to moles. 1 mol C6H12O6 x = 180.0 g C6H12O6 10.0 g C6H12O6 x = 0.0555556 mol C6H12O6 x = 0.0556 mol C6H12O6

  8. Conversions Convert 10.0 g NaOH to formula units. 40.0 g NaOH 10.0 g NaOH = 6.02 x 1023 fu NaOH x x = 1.505 x 1023 fu NaOH x = 1.51 x 1023 fu NaOH

  9. Conversions Convert 1.50 mole NaCl to formula units. 1 mol NaCl 1.50 mol NaCl = 6.02 x 1023 fu NaCl x x = 9.03 x 1023 fu NaCl

  10. Conversions Convert 90.0 g C6H12O6 to moles. 1 mol C6H12O6 x = 180.0 g C6H12O6 90.0 g C6H12O6 x = 0.5 mol C6H12O6 x = 0.500 mol C6H12O6

  11. Conversions Convert 15.0 g CaSO4 to moles. 1 mol CaSO4 x = 136.2 g CaSO4 15.0 g CaSO4 x = 0.0073421 mol CaSO4 x = 0.000734 mol CaSO4

  12. Conversions Convert 2.50 mol Ca(OH)2 to grams. 1 mol Ca(OH)2 2.50 mole Ca(OH)2 = 74.1 g Ca(OH)2 x x = 185.25 g Ca(OH)2 x = 185 g Ca(OH)2

  13. Empirical Formula Ag % = 63.5 N % = 8.2 O % = 28.3 Assume 100g, convert to moles. x = 0.588507 mol Ag x 1 mol Ag 1 = x = 0.589 mol Ag 63.5 g Ag 107.9 g Ag x 1 mol N x = 0.587142 mol N 1 = 14.0 g N 8.2 g N x = 0.59 mol N x 1 mol O x = 1.76875 mol O 3 = 28.3 g 0 16.0 g O x = 1.77 mol O AgNO3

  14. Empirical Formula C % = 40.0 H % = 6.7 O % = 53.3 Assume 100g, convert to moles. x = 3.3333 mol C x 1 mol C 1 = x = 3.33 mol C 40.0 g C 12.0 g C 2 x 1 mol H = x = 6.7 mol H 1.0 g H 6.7 g H x 1 mol O x = 3.33125 mol O 1 = 53.3 g 0 16.0 g O x = 3.33 mol O CH2O

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