The Mole and Chemical
Composition
Chapter 7
• Mole (mol) = amount of a substance that
contains as many representative particles as
there are atoms in exactly 12 g of carbon
• Avogadro’s Number (NA) = 6.02 x 1023 =
actual # of atoms in 12 g of carbon
experimentally
2
Molar Mass of an Element
• Molar Mass (M) = mass in g of 1 mole of a
substance
• Has units of g/mol
• Molar mass in g of C = atomic mass in amu
of C
• If you know the atomic mass of an element,
you also know the molar mass
3
Relationships between Mass
units
M = molar mass in g/mol
NA = Avogadro’s number
4
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
0.551 g K
1 mol K
x
39.10 g K
6.022 x 1023 atoms K
x
1 mol K
=
8.49 x 1021 atoms K
5
3.2
Molecular mass = sum of the atomic masses (in amu) in
the molecule
H2O: 2*(atomic mass of H) + atomic mass of O
2(1.008 amu) + (16.00 amu) = 18.02 amu
Molar mass (g) is numerically equal to the molecular
mass (amu)
Molecular mass of H2O = 18.02 amu
Molar mass of H2O = 18.02 g/mol (contains 6.022 x
1023 molecules)
6
Mole-mass, Mole-volume
relationships
1 mole
• Moles = mass (g) x
mass (g)
• At STP (standard temperature & pressure),
Volume of gas =
22.4 L
moles of gas x
1 mole
7
Percent Composition of
Compounds
• Tells what % of total mass is contributed by each
element
% composition of an element
n  molar mass of element

100%
molar mass of compound
n = # of moles of element in 1 mole of compound
8
Chemical Formulas
• Chemical formulas tell the # and type of atoms present in a
substance
• Empirical Formula tells the simplest whole #
ratio of the atoms present in a compound
– Not necessarily the actual # of atoms
9
Determining Empirical
Formulas
•
•
•
•
1. Convert % composition to grams
2. Convert g to mol using molar mass
3. Divide by smallest # of moles
4. Insert whole number subscripts
10
Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition: K 24.75%, Mn
34.77%, O 40.51%.
nK = 24.75 g K x
nMn = 34.77 g Mn x
nO = 40.51 g O x
1 mol K
39.10 g K
= 0.6330 mol K
1 mol Mn
= 0.6329 mol Mn
54.94 g Mn
1 mol O
16.00 g O
= 2.532 mol O
11
Percent Composition and Empirical Formulas
nK = 0.6330, nMn = 0.6329, nO = 2.532
K:
Mn :
O:
0.6330 ~
~ 1.0
0.6329
0.6329
= 1.0
0.6329
2.532 ~
~ 4.0
0.6329
KMnO4
12
Molecular Formulas
• Empirical formula is calculated from % composition
because subscripts are always reduced to smallest
whole #
• Molecular formula shows the exact # of atoms of each
element in the smallest unit of a substance; the true
formula
– Subscript indicates # of atoms of an element
present
13
Determination of Molecular
Formulas
• To determine, must know approximate
molar mass of a compound in addition to
the empirical formula
• Molar mass of molecular formula is an
integral multiple of molar mass of empirical
formula
14
A sample of a compound contains 1.52 g N
and 3.47 g O. The molar mass of this
compound is between 90 g and 95 g.
Determine the molecular formula and the
accurate molar mass of this compound.
Empirical = NO2 , Molecular = N2O4
15