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All Roads Lead to Mole City

All Roads Lead to Mole City. What Have We Learned Thus Far (10.1). How to convert from molecules to moles/moles to molecules. Molecules to Mole …. If you have 7.36 x 10 22 molecules of barium sulfate, how many moles of BaSO 4 do you have?. 7.36 x 10 22 molecules BaSO 4. 1 mole BaSO 4.

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All Roads Lead to Mole City

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  1. All Roads Lead to Mole City

  2. What Have We Learned Thus Far (10.1) How to convert from molecules to moles/moles to molecules

  3. Molecules to Mole …. If you have 7.36 x 1022molecules ofbarium sulfate, how many moles of BaSO4 do you have? 7.36 x 1022 molecules BaSO4 1 mole BaSO4 = 0.122 mole BaSO4 6.02 x 1023 molecules BaSO4 Molecules / Avogadro’s # = Moles

  4. Moles to Molecules…. If you have 0.632 moles ofbarium sulfate, how many molecules of BaSO4 do you have? 0.632 moles BaSO4 6.02 x 1023 molecules BaSO4 = 3.80 x 1023 molecules BaSO4 1 mole BaSO4 Moles x Avogadro’s # = Molecules

  5. What Have We Learned Thus Far (10.1) How to calculate the molar mass of a compound

  6. For instance …. What is the molar mass of CuHAsO3? Cu = 1(63.5) = 63.5 g/mol H = 1(1.0) = 1.0 g/mol As = 1(74.9) = 74.9 g/mol O = 3(16.0) = 48.0 g/mol = 187.4 g/mol

  7. What Have We Learned Thus Far (10.2) How to convert from grams to moles/moles to grams

  8. Grams to Moles …. If you have 12.212 g ofCuHAsO3.How many moles do you have? 12.212 g CuHAsO3 1 mole CuHAsO3 = .065165421 mole CuHAsO3 187.4 g CuHAsO3 Cu = 1(63.5) = 63.5 H = 1(1.0) = 1.0 As = 1(74.9) = 74.9 O = 3(16.0) = 48.0 = .06517 mole CuHAsO3 = 6.517 x 10-2 mole CuHAsO3 = 187.4 g/mol Grams / Molar Mass = Moles

  9. Moles to Grams …. If you have 1.54 moles ofCuHAsO3.How many grams do you have? 1.54 mol CuHAsO3 187.4 g CuHAsO3 = 288.596 g CuHAsO3 1 mole CuHAsO3 Cu = 1(63.5) = 63.5 H = 1(1.0) = 1.0 As = 1(74.9) = 74.9 O = 3(16.0) = 48.0 = 289 g CuHAsO3 (only 3 sig figs) = 187.4 g/mol Moles x Molar Mass = Grams

  10. Delayed Objective (10.2) Learn how to convert between moles and volume (L, gas) We will learn this when we derive the Ideal Gas Law in Chapter 14.3, but …… just multiply or divide by 22.4 L

  11. All Roads Go Through Mole City Divide by 6.02 x 1023 Molecule Town Multiply by 6.02 x 1023 Divide by molar mass Multiply by molar mass Divide by 22.4 L Gramville Mole City Multiply by 22.4 L Gas Station

  12. Sample mixed mole problem: Molecules to Grams If you have 3.45 x 1025 molecules ofNH3, how many grams do you have? 3.45 x 1025 molecules NH3 1 mol NH3 17.0 g NH3 = 974.252 g NH3 6.02 x 1023 molecules NH3 1 mol NH3 = 974 g NH3 N = 1(14.0) = 14.0 g/mol H = 3(1.0) = 3.0 g/mol molar mass NH3 = 17.0 g/mol

  13. Sample mixed mole problem: Grams to Molecules/Atoms If you have 66.23 g ofCO2, how many oxygen atoms do you have? 6.02 x 1023 molecules CO2 66.23 g CO2 1 mol CO2 2 atoms oxygen 44.0 g CO2 1 mol CO2 1 molecule CO2 = 1.8123 x 1024oxygen atoms = 1.81 x 1024oxygen atoms C = 1(12.0) = 12.0 g/mol O = 2(16.0) = 32.0 g/mol molar mass CO2 = 44.0 g/mol

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