New Measurement Unit: The Mole
3 mathematical definitions:
1) 1 mol= atomic mass of atom
= molecular mass of a mlc
= formula mass of an ionic
compound
MOLAR MASS
These values come from the periodic table.
The atomic mass #’s are in grams/1 mol
Calculating Formula Molecular mass
Need periodic table (keep the decimal
numbers)
Count the number of atoms in a substance
and multiply it by it’s mass number then
add.
Ex 1) K (atom) 39.098 g = atomic mass
1 mol
Ex 2) P (atom) 30.974 g = atomic mass
1 mol
Ex 3) Potassium Chloride - KCl (formula unit)
1 K x 39.098
1 Cl x 35.453
= 74.551 g/mol
74.551
Ex 4) Magnesium Chloride - MgCl2
1 Mg x 24.305 = 24.305
2 Cl x 35.453 = 70.906
95.211 g/mol
Ex 5) Ammonium Carbonate - (NH4)2CO3
2 N = 28.02
8 H = 8.08
1 C = 12.01
3O = 48.00
= 96.11 g/mol
2) 1 mol= 6.02 x 1023 particles (avagadro’s
number)
where a “particle” can be:
a) molecule
b) atom
c) ion
d) formula unit of ionic cmpd
3) 1 mol= 22.4 L for gas molecules
Volume
of gas at
STP
formula mass,
atomic mass, or
molecular mass
Mass
22.4 dm3 / mole at STP
(gases only)
Moles
6.02 x 1023 particles/
mole (Avogadro
Number)
number of
particles
(atoms, ions,
molecules,
formula units)
One-Step Conversion Problems
Use road maps to convert between
one unit of chemical quantity to
another.
Dimensional Analysis – like metrics
Use the three definitions of a mole; an
extension of the definitions in
numerical form.
One-Step Conversion Problems
These are your conversions factors!
1) 1 mole
or
x grams
x grams
1 mole
2) 1 mol
or
6.02 x 1023 particles
3) 1 mol
22.4 L
or
6.02 x 1023 particles
1 mole
22.4 L
1 mol
GASES ONLY!
One- Step Conversions
Ex 1) Convert 4.3 grams of NaCl to moles.
Mass  mol
Given:
Unk:
4.3 g NaCl x 1 mol NaCl = 7.4 x 10-2 mol NaCl
58.45 g NaCl
Ex 2) Convert 0.00563 mol NH3 to grams.
Mol -> mass
G:
U:
0.00563 mol NH3 x 17 g NH3 =9.57 x 10 –2 g NH3
1 mol NH3
More One-Steppers
Ex)
Given: 0.91 Mol NaClO3
Unk: ? Formula Units NaClO3
Ex)
Given 22.92 x 1018 mlcs Cl2
Unk: ? Mol Cl2
More One-Steppers
Ex)
Given: 460. Mol Cl2 gas
Unk: ? L Cl2
Ex)
Given 84.56 L Cl2
Unk: ? Mol Cl2
TWO –STEP CONVERSIONS
Ex 1) Given: 8.631 x 1021 atoms Na
Unkn: ? g Na
TWO –STEP CONVERSIONS
Ex 2) Given: 1.5 x 1022 f.u. MgCl2
Unkn: ? g MgCl2
TWO –STEP CONVERSIONS
Ex 3) Given: 2.63 L O2
Unkn: ? mg O2
% Composition
Definition: the % (in mass) of each element in a
compound
% mass element = g element x 100%
g compound
Ex An 8.40 g sample of flourine completely combines with a
4.90 g sample of sodium. Calculate the % composition of the
compound that forms.
mass element 1 + mass element 2 = total;
8.40 g F2 + 4.90 g Na = 13.30 g NaF
%F: 8.40 g x 100 = 63.2% F
13.30 g
%Na: 4.90g x 100 = 36.8% Na
13.30 g
% mass CuSO4
1 Cu x 63.546 g = 63.546 g
1 S x 32.066 g = 32.066 g
4 O x 15.999 g = 63.996 g x
159.608 g
% Cu= 63.546 x 100 = 39.8%
159.608
% S= 32.066 x 100 = 20.1%
159.608
%O= 63.996 x 100 = 40.1%
159.608
HYDRATES
Compounds with water molecules chemically
attached to ionic crystalline structure
formula – CuSO4 5H2O
name – copper (II) sulfate pentahydrate
CaSO4 4H2O BaCl2 9H2O –
H2O mlcs can be removed by heating the compound.
This is not evaporation. Therefore, it is NOT a
physical change. (Decomposition reaction)
HYDRATES
Compounds with water molecules chemically
attached to ionic crystalline structure
formula – CuSO4 5H2O
name – copper (II) sulfate pentahydrate
CaSO4 4H2O BaCl2 9H2O –
H2O mlcs can be removed by heating the compound.
This is not evaporation. Therefore, it is NOT a
physical change. (Decomposition reaction)
Mole-to-Mole Relationships
Decomposition of water:
2H20 2H2 + O2
2 mol of H20 yields 2 mol H2 and 1 mol O2
Ex) You have 4 mol of water. If you decompose 4 mol
of water, how many mols of products do you get?
2[2H2O 2H2 + O2]
4H2O 4H2 + 2O2
4 mol of H2O yields 4 mol of H2 plus 2 mol of O2