1. Sample Calculations

2. 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.

3. 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.

4. 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

5. 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

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

7. 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

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 / 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 / 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

11. 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

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

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 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

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 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

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 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

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

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