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Sample Calculations

Sample Calculations. Simplest Formula. A 1.261 g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains 0.624 g C, 0.065 g H, 0.364 g N, and 0.208 g O. Calculate the simplest formula of caffeine. Simplest Formula.

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Sample Calculations

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  1. Sample Calculations

  2. Simplest Formula A 1.261 g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains 0.624 g C, 0.065 g H, 0.364 g N, and 0.208 g O. Calculate the simplest formula of caffeine.

  3. Simplest Formula A 1.261 g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains 0.624 g C, 0.065 g H, 0.364 g N, and 0.208 g O. Calculate the simplest formula of caffeine. The concept of a formula implies that we can relate the numbers of atoms of each kind to each other and we must relate these masses to numbers of moles.

  4. Simplest Formula A 1.261 g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains 0.624 g C, 0.065 g H, 0.364 g N, and 0.208 g O. Calculate the simplest formula of caffeine. 0.624 / 12.0 = 0.0520 mol C 0.065 / 1.01 = 0.0650 mol H 0.364 / 14.0 = 0.0260 mol N 0.208 / 16.0 = 0.0130 mol O

  5. Simplest Formula A 1.261 g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains 0.624 g C, 0.065 g H, 0.364 g N, and 0.208 g O. Calculate the simplest formula of caffeine. divide by 0.0130 0.624 / 12.0 = 0.0520 mol C 4 0.065 / 1.01 = 0.0650 mol H 5 0.364 / 14.0 = 0.0260 mol N 2 0.208 / 16.0 = 0.0130 mol O 1 C4H5N2O

  6. Molecular Formula The simplest formula is C4H5N2O but the molar mass is found to be 0.194 kg mol-1. What is the molecular formula?

  7. Molecular Formula The simplest formula is C4H5N2O but the molar mass is found to be 0.194 kg mol-1. What is the molecular formula? The molar mass of the simplest unit is: 4 x 0.012 + 5 x 0.001 + 2 x 0.014 + 0.016 = 0.097 kg mol-1 Molecule must contain two C4H5N2O units and molecular formula is C8H10N4O2

  8. Reaction Yield Ethanol, C2H5OH, is produced industrially by the following reaction: C2H4 + H2O -----> C2H5OH If the reaction proceeds with 100% yield, how much ethanol can be produced from 500 tonne of ethene ( C2H4 )?

  9. Reaction Yield Ethanol, C2H5OH, is produced industrially by the following reaction: C2H4 + H2O -----> C2H5OH If the reaction proceeds with 100% yield, how much ethanol can be produced from 500 tonne of ethene ( C2H4 )? 500 tonne = 500 tonne x 1000 kg tonne -1 = 5.00 x 105 kg / 0.0280 kg mol -1 = 1.79 x 107mol ethene

  10. Reaction Yield Ethanol, C2H5OH, is produced industrially by the following reaction: C2H4 + H2O -----> C2H5OH If the reaction proceeds with 100% yield, how much ethanol can be produced from 500 tonne of ethene ( C2H4 )? 500 tonne = 500 tonne x 1000 kg tonne -1 = 5.00 x 105 kg / 0.0280 kg mol -1 = 1.79 x 107mol ethene should get 1.79 x 107 mol of ethanol 1.79 x 107 mol x 0.0420 kg mol -1 = 8.75 x 105 kg = 875 t

  11. Ethene or Ethylene H H C C H H Ethanol or Ethyl Alcohol H H H C C O H H H

  12. Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO3 to give oxygen and potassium chloride, KCl. Calculate the volume of oxygen at 298 K and 100 kPa that can be produced from 0.0532 kg of KClO3 .

  13. Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO3 to give oxygen and potassium chloride, KCl. Calculate the volume of oxygen at 298 K and 100 kPa that can be produced from 0.0532 kg of KClO3 . KClO3 ----> KCl + O2

  14. Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO3 to give oxygen and potassium chloride, KCl. Calculate the volume of oxygen at 298 K and 100 kPa that can be produced from 0.0532 kg of KClO3 . KClO3 ----> KCl + O2 2KClO3 ----> 2KCl + 3O2

  15. Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO3 to give oxygen and potassium chloride, KCl. Calculate the volume of oxygen at 298 K and 100 kPa that can be produced from 0.0532 kg of KClO3 . 2KClO3 ----> 2KCl + 3O2 This means 3 mol O2 / 2 mol KClO3 0.0532 kg KClO3 / 0.122 kg mol -1 = 0.436 mol KClO3 0.436 mol KClO3 x 3 mol O2 / 2 mol KClO3 = 0.654 mol O2

  16. Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO3 to give oxygen and potassium chloride, KCl. Calculate the volume of oxygen at 298 K and 100 kPa that can be produced from 0.0532 kg of KClO3 . n = 0.654 mol O2 and V = nRT / P V = 0.654 mol x 8.31 Nm mol-1 K-1 x 298 K 1.00x105 Nm2 = 0.0155 m3

  17. Gas Mixtures What is the pressure at 300K in a 1.00L flask that contains 1.00 g of He and 1.00 g of H2?

  18. Gas Mixtures What is the pressure at 300K in a 1.00L flask that contains 1.00 g of He and 1.00 g of H2? nHe = 1.00 g / 4.00 g mol -1 = 0.250 mol nH2 = 1.00 / 2.02 g mol -1 = 0.495 mol n total = 0.745 mol P = nRT / V = 0.745 x 8.31 x 300 / 1.00 x 10 -3 Ptotal = 1.86 x 106 Pa

  19. Mole fraction Partial Pressure We can express the composition in mole fraction: He = n He / n total = 0.250mol / 0.745 mol = 0.332 NO UNITS!

  20. Mole fraction Partial Pressure The partial pressure is just : P He = He x Ptotal = 0.332 x 1.86 MPa = 0.617 MPa MPa = megapascal = 106 Pa

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