We must think of the mole in three ways:

1. As the molecular mass
in grams, e.g. for O_{2}, 1 mole = 2x16 =32g.,

for CO_{2}, 1 mole = 12+(2x16) = 44g.

and for H_{2}SO_{4}, 1 mole = (2x1) +32 + (4x16) = 98g.

2. As the amount that contains
6x10^{23} particles:

i.e. 1 mole, or 32g. of oxygen contains 6x10^{23} molecules.

3. If the substance is
a gas, then 1 mole of it has a volume of 22.4 dm^{3} (22,400 cm^{3})
at S.T.P.

so 32g. of oxygen has a volume of 22.4 dm^{3} at S.T.P.

**Worked Examples**

(i)How many molecules are there in 64g. of oxygen?

32g.
(of oxygen) = 1 mole

So 64g. = 2 moles

1 mole contains 6 x 10^{23}
molecules

So 2 moles contain 12 x
10^{23} molecules

i.e.
1.2 x 10^{24} molecules. = Ans.

(ii) What is the volume
in dm^{3} of 8.5g. of ammonia gas (NH_{3}) at S.T.P?

17g. (of NH_{3})
= 1 mole

so 8.5g. = 0.5 mole

so 0.5 mole has a volume
of 22.4 / 2 = 11.2 dm^{3}. = Ans.

(iii) How many molecules
are there in 4.48 dm^{3} of sulphur dioxide?

(The volume is measured at S.T.P.)

22.4 dm^{3} = 1 mole

so 4.48 dm^{3} = 4.48 / 22.4 mole

= 0.2 mole

1 mole
contains 6x10^{23} molecules

so 0.2
mole contains 0.2 x 6 x 10^{23} molecules

= 1.2 x 10^{23} molecules = Ans.

(iv) What is the mass in
grams at S.T.P. of 44,800 cm^{3} of CO_{2}?

22400
cm^{3} of CO_{2} at S.T.P. = 1 mole

so 44800 cm^{3}
of CO_{2} at S.T.P. = 2 moles

1 mole
of CO_{2} = 44g.

so 2 moles of CO_{2}
= 88g. = Ans.

(v) What is the relative
molecular mass of a gas if 0.36g. of the gas has a volume of 500 cm^{3}
at S.T.P?

Suggest what the gas might
be.

At S.T.P. 500 cm^{3}
of the gas weigh 0.36g.

so 1 cm^{3} weighs 0.36 / 500

so 22400 cm^{3} weigh 22400 x 0.36 / 500 = 16.128 g.

The gas is methane, CH_{4},whose
M_{r} = 16.

Remember the three ways
of looking at the mole:

1. 6 X 10^{23}
Particles
2. Volume = 22.4 dm^{3} (if a gas!) |

(Avogadro Number) |

3. 1 mole contains the Atomic / Molecular Mass in grams |