# Sample Calculations.

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Sample Calculations

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

Simplest Formula A g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains g C, g H, g N, and 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.

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

Simplest Formula A g sample of caffeine ( reputed to be a stimulant for the central nervous system commonly used by students to avoid sleep ) contains g C, g H, g N, and g O. Calculate the simplest formula of caffeine. divide by 0.624 / = mol C 0.065 / = mol H 0.364 / = mol N 0.208 / = mol O C4H5N2O

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

Molecular Formula The simplest formula is C4H5N2O but the
molar mass is found to be kg mol-1. What is the molecular formula? The molar mass of the simplest unit is: 4 x x x = kg mol-1 Molecule must contain two C4H5N2O units and molecular formula is C8H10N4O2

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 )?

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 / kg mol -1 = 1.79 x 107mol ethene

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 / kg mol -1 = 1.79 x 107mol ethene should get 1.79 x 107 mol of ethanol 1.79 x 107 mol x kg mol -1 = x 105 kg = 875 t

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

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 kg of KClO3 .

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 kg of KClO3 . KClO > KCl O2

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 kg of KClO3 . KClO > KCl O2 2KClO > 2KCl O2

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 kg of KClO3 . 2KClO > 2KCl O2 This means 3 mol O2 / 2 mol KClO3 kg KClO3 / kg mol -1 = mol KClO3 0.436 mol KClO3 x 3 mol O2 / 2 mol KClO3 = mol O2

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 kg of KClO3 . n = mol O2 and V = nRT / P V = mol x 8.31 Nm mol-1 K-1 x 298 K 1.00x105 Nm2 = m3

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?

Gas Mixtures Ptotal = 1.86 x 106 Pa
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 = mol nH2 = / g mol -1 = mol n total = mol P = nRT / V = x 8.31 x 300 / 1.00 x 10 -3 Ptotal = 1.86 x 106 Pa

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

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