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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.

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Presentation on theme: "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."— Presentation transcript:

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. 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 divide by 0.0130 C4H5N2OC4H5N2O

6 Molecular Formula The simplest formula is C 4 H 5 N 2 O 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 C 4 H 5 N 2 O 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 C 4 H 5 N 2 O units and molecular formula is C 8 H 10 N 4 O 2

8 Reaction Yield Ethanol, C 2 H 5 OH, is produced industrially by the following reaction: C 2 H 4 + H 2 O -----> C 2 H 5 OH If the reaction proceeds with 100% yield, how much ethanol can be produced from 500 tonne of ethene ( C 2 H 4 )?

9 Reaction Yield Ethanol, C 2 H 5 OH, is produced industrially by the following reaction: C 2 H 4 + H 2 O -----> C 2 H 5 OH If the reaction proceeds with 100% yield, how much ethanol can be produced from 500 tonne of ethene ( C 2 H 4 )? 500 tonne = 500 tonne x 1000 kg tonne -1 = 5.00 x 10 5 kg / 0.0280 kg mol -1 = 1.79 x 10 7 mol ethene

10 Reaction Yield Ethanol, C 2 H 5 OH, is produced industrially by the following reaction: C 2 H 4 + H 2 O -----> C 2 H 5 OH If the reaction proceeds with 100% yield, how much ethanol can be produced from 500 tonne of ethene ( C 2 H 4 )? 500 tonne = 500 tonne x 1000 kg tonne -1 = 5.00 x 10 5 kg / 0.0280 kg mol -1 = 1.79 x 10 7 mol ethene should get 1.79 x 10 7 mol of ethanol 1.79 x 10 7 mol x 0.0420 kg mol -1 = 8.75 x 10 5 kg = 875 t

11 Ethene or Ethylene Ethanol or Ethyl Alcohol CC H H H H CC HH HH HO H

12 Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO 3 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 KClO 3.

13 Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO 3 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 KClO 3. KClO 3 ----> KCl + O 2

14 Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO 3 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 KClO 3. KClO 3 ----> KCl + O 2 2KClO 3 ----> 2KCl + 3O 2

15 Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO 3 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 KClO 3. 2KClO 3 ----> 2KCl + 3O 2 This means 3 mol O 2 / 2 mol KClO 3 0.0532 kg KClO 3 / 0.122 kg mol -1 = 0.436 mol KClO 3 0.436 mol KClO 3 x 3 mol O 2 / 2 mol KClO 3 = 0.654 mol O 2

16 Gas Yield Oxygen gas can be produced on a small scale from the thermal decomposition of potassium chlorate, KClO 3 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 KClO 3. n = 0.654 mol O 2 and V = nRT / P V = 0.654 mol x 8.31 Nm mol -1 K -1 x 298 K 1.00x10 5 Nm 2 = 0.0155 m 3

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 H 2 ?

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 H 2 ? n He = 1.00 g / 4.00 g mol -1 = 0.250 mol n H 2 = 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 P total = 1.86 x 10 6 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 P total = 0.332 x 1.86 MPa = 0.617 MPa MPa = megapascal = 10 6 Pa


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