4.2 Mole Quantities
Thursday, September 22, 2011
5:34 PM
If we look at a simple compound like water we can determine that the mass
percentages are as follows:
Oxygen = 88%
Hydrogen = 12%
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Total = 100%
While, this tells us the complete composition of the compound, it does not tell
us the formula of the compound, why?
We are not accounting for differing relative atomic masses of the
elements!
Consider the following
Element Percentage Rel. AM
H
12
1.008
O
88
16.00
Do we have scales that measure in 'amu?' NO
We need to express the mass of a collection of atoms, therefore we need a
standard number of atoms to count. Atoms are so small that this number will
have to be very large to equal to a significant mass. The solution is a unit called
a mole.
Mole = a collection of atoms, molecules, ions, or formula units equal to 6.022×
1023 units.
Why 6.022×1023!?
This number, called Avagadro's Number, is the number of atoms of 12C
needed to mass at 12 g (remember how we standardized relative atomic mass
to this isotope).
You use this to calculate the number of atoms in a sample.
EXAMPLE
In a mole of Water, H2O, how many atoms of hydrogen are present? Of
oxygen?
1 mole water = 1 mole of Oxygen and 2 moles of Hydrogen
Chapter 4 Page 1
1 mole water = 1 mole of Oxygen and 2 moles of Hydrogen
Therefore
1 mole oxygen = 6.022×1023 atoms of oxygen
2 moles hydrogen = 2 X 6.022×1023 = 12.04×1024 atoms of hydrogen
Molar Mass = Since the definition of a mole is based off the basis of the
relative atomic mass, a mole can also be extended to be used to define mass in
grams. A mole of a substance has a mass in grams equal to its relative atomic
mass in atomic mass units.
EXAMPLE
1 mole of 12C = 12 g (exact)
1 mole of hydrogen = 1.008 g
We can use this to determine ratios in formula units or chemical formula.
EXAMPLE
Take our example of Water.
Say we have a 1 g of water… this means we have 0.12 g Hydrogen and 0.88 g
Oxygen.
Let us calculate the moles of each.
Moles of H = 0.12g H X (1 mole H/1.008g H) = 0.11 moles H
Moles of O = 0.88g O X (1 mole O/16.00 g O) = 0.055 moles O
Moles H
0.11 moles H
----------- = ----------------------- = 2:1 hydrogen:oxygen = H2O
Moles O
0.055 moles O
PROBLEM
Given what we determined about the mass percentages of Emeralds last
section, can you calculate the formula ratios?
Beryllium = 5.04%
Chapter 4 Page 2
section, can you calculate the formula ratios?
Beryllium = 5.04%
Aluminum = 10.04%
Silicon = 31.35%
Oxygen = 53.56%
Chapter 4 Page 3