Welcome to the MOLE
What is a mole?
This is not such a bad
mole, but not what we
need to discuss. . .
This little stinker is just plain mean
and ugly. . .
and the Mole People are in the dark and clueless. . . . .
We are discussing the Mole used in
Chemistry:
Avogadro’s Number (NA)
or 6.02214179 x 10 23
shortened to
6.022 x 1023
This amount = 1 mole
The mole is a way to describe the number of
something without writing a huge number.
• It is similar to common terms like dozen,
gross or even π in geometry
• 1 mole of anything – atoms, molecules,
cockroaches or even galaxies will number
6.022 x 1023
• We need a number like this since atoms
and molecules are extremely small and so
many take up such a small space
Key Equations – KNOW THESE!
Divide by Molar Mass
Grams
(mass) of
Substance
x by Molar Mass
x by NA
Number
of
Particles
Moles of
Substance
Divide by NA
Molar Mass = Atomic Weight in Grams per Mole
(g/mol)
• Molar Mass is the mass (g) of 1 mole of
an element
– For example – Na is 22.989 amu which is
22.989 g, and 1 mole of Na = 22.989 g
– CO2 is 1 Carbon at 12.011 g and 2 Oxygens
at 15.999 each; therefore the molar mass
of CO2 = Σ of 1 C + 2 O ≈ 44 g
• Mass (m) = # mols x # g
• # mols = m / g
• # Mols = m / Molar Mass
• % Composition by Mass = m element x 100
m compound
• # Mols = Concentration (Molar) x Volume (L)
• Volume = # mols / Concentration
Here’s a trick – to find the needed equation, just cover up
the wanted result and what is left is the equation!
Mass
Atomic X Mols
Mass
Mols
Concentration X Volume
The Mole Concept
• Atomic Mass (or atomic weight) is the
mass of the element in amu (μ)
– It is the number under the elemental symbol
– Simply make this number into grams (g)
– This represents the mass (m) of 1 mole of that
element; and/or the m of 6.022 x 1023 atoms
of that substance!
– 1 mole of any gas = 22.4 L
• Mole (mol) – is the # of atoms, ions,
molecules that is equal to NA
• Molar Mass (M) – this is the m in g of 1
mol of a substance (g/mol)
– Example – Manganese = 54.94 μ, thus its
M = 54.94 g/mol;
and 54.94 g of Mn will contain 6.022 x 1023 atoms
and this is equal to 1 mol of Mn
Mass to Mole Calculations
• Remember – each
element has a different
amu and thus, 1 mol of
each will differ in mass
The Mass of a Mole
–
–
–
–
Uses the C 12 isotope as its standard
H = μ of 1; or 1/12 of 1 atom of C 12
He has μ of 4 or 4/12 (1/3) of a C 12 atom
Remember – atomic masses use isotopes and their %
abundance in nature to calculate – and the closer to a whole
number – the fewer the isotopes
I. Molar Mass of Substance = Grams Substance
1 mol Substance
Therefore:
1. Mols of A = grams of A given x 1 mol A
gram A
2. Mass of B = Mols of B given x gram B
1 mol B
Examples:
• 3 mol Mn = ? Grams
3 mol Mn X 54.9 g Mn
1 mol Mn
• 25 g Au = ? Mols
25 g Au x 1 mol Au =
196.97 g Au
=
165 g Mn
25
196.97
=
0.127 mol Au
• 0.127 mols Au = ? Atoms
0.127 mol Au x 6.022 x 1023 = 7.65 x 1022 atoms Au
1 mol Au
II. Moles to Mass
• Mols (given) x # grams = mass
1 mol
Example:
– 2.5 mol of (C3H5)2S has what mass?
M = 1 mol S = 32.07 g
6 mol C = 6 x 12.01 = 72 g
10 mol H = 10 x 1 = 10 g
Σ 114.07 g/mol
2.5 mol x 114.07 g = ≈ 286 g
1 mol
We just used the bottom left of the diagram!
Divide by Molar Mass
Grams
(mass) of
Substance
x by Molar Mass
x by NA
Number
of
Particles
Moles of
Substance
Divide by NA
Molar Mass = Atomic Weight in Grams per Mole
(g/mol)
III. Mass to Moles with Compounds
• Example:
m of Ca(OH)2 = 325 g (rounded off)
M = ?
# mols = ?
M = 1 mol Ca = 40.08 g
2 mol O = 2 x 16 = 32 g
2 mol H = 2 x 1 = 2 g
Σ 74.096 g/mol
Given m of 325 g Ca(OH)2 x 1 mol Ca(OH)2
M of 74.096 g
=
4.3 mol
IV. Mass (g) to Particles
• Mols x NA = # Particles
• Example:
– m = 35.6 g of AlCl3
What is the number of Al+3 and Cl- ions?
M = Al 26.981 g/mol
Cl 35.452 g/mol x 3 = 106.356
Σ 133.337
Mols Al =
m given
= _____ mols x NA = _____ Al ions
26.981 g/mol
Mols Cl =
m given
= _____ mols x NA = _____ Cl ions
106.356 g/mol
Continued
• And:
35.6 g AlCl3
=
133.337 g/mol AlCl3
0.267 mol
• 0.267 mol AlCl3 x (6.022 x 1023) =
1.6 x 10 23 molecules
V. Percent Composition of Compounds
• Mass Element (m) x
Mass Cmpd (M)
100 = % by mass
Example:
H2O; what percent is H and what percent is O?
% H = 2 x 1 (the molar mass of H) = 2 x 100 = 11.2%
(the molar mass of H2O)
18
Thus, all compounds equal 100%, so 100 – 11.2 = 88.8 % for O
• What is the % of C and O in CO2?
gC
x 100 = 12.01 C
total g CO2
32 g O
x
44.01 g CO2
x 100 = 27.29%
44.01 g CO2
100
= 72.71 %
Example:
• H3PO4 (aq) (Phosphoric Acid)
%H
=
%P
=
%O
=
3 g H
M = 98 g
31 g P
98 g
64 g O
98 g
x 100
=
3%
x 100
=
32%
x 100
=
65%
m Compound = H (3 x 1) + P (1 x 31) + O (4 x 16) = 98 g
VI. Mole Ratios
• Given Vitamin C (ascorbic acid) with the
following percentages, determine formula:
40.92 %
4.58 %
54.5 %
Set up with unknown moles (n):
nC = 40.92 g C / 12.01 g C
nH = 4.58 g H / 1.00 g H
nO = 54.5 g O / 16 g O
C
H
O
=
=
=
This is the Mole Ratio
3.4 mol C
4.5 mol H
3.4 mol O
• Set Mole Ratio values as subscripts
• Divide each by the lowest value:
C 3.4 / 3.4
H 4.5 / 3.4
O 3.4 / 3.4
=
C 1
H 1.33
O 1
• The 1.33 on H needs to be Δ’d into integer
– Do this by multiplying until closest to a whole
number
• 1.33 x 2 = 2.66
• 1.33 x 3 = 3.99 which can be rounded off to 4
– So the “magic” number is 3 – must multiply all
subscripts by 3
• Result is C1 x 3H1.33 x 3O1 x 3 or C3H4O3!
What is the molecular formula that has 92.2% C and
7.8%H. The molar mass is 52.1.
• First – assume a 100 g sample of the substance
• The element’s percentages are assumed to be masses
(g)
• Determine the moles of elements in compound:
92.2 g C x 1 mol C
12.01 g C
7.8 g H x 1 mol H
1.01 g H
=
7.68 mol C
=
7.72 mol H
• Divide all mols by the lowest value
7.68 C
7.68
=
1 mol C
7.72 H
7.68
=
Continued
1.01 mol H
• The Empirical Formula is C1H1 (this is the basic form)
• To get the Molecular Formula:
– Molar mass of the Empirical Formula is 12.01 +
1.01 = 13.02 g/mol
– Molar mass of unknown is 52.1 g/mol
• So:
Molar Mass Compound
=
Molar Mass Emp. Formula
52.1 g/mol = 4
13.02 g/mol
Whole # to multiply
subscripts to get
molecular formula
thus, (CH) x 4 = C4H4
Once again, a molecular formula calculation:
Given 38.7% C, 9.7% H, and 51.6% O with a
molecular formula mass of 62.0 g. What is the
true molecular formula?
First - find the empirical formula (this is CH3O)
Find the formula mass:
C 1 x 12 = 12
H 3 x 1.01 = 3.03
O 1 x 16 = 16.0
Σ 31.0
• Divide the molecular mass by the
empirical formula mass:
62 g (given) / 31 g (mass of emp. form.) = 2
• Multiply each subscript by (n) or 2 in
this example. . .
Thus, the molecular formula:
(CH3O)(n)  (CH3O)(2)  C2H6O2
VII. Hydrates
• Hydrates are
substances that include
H2O in their
formulas, but are
not wet!
– Hydration – adding H2O
– Dehydration – removing it
– Anhydrous – no H2O present
• Methane Hydrate is found on the ocean’s
floor
– The methane will burn – but the water in it
keeps the skin from burning!
• The methane molecule (CH4) is in a cage
of water molecules
• There is 1 mole CH4 per 5.75 mols H2O
• It is found at depths of 300 meters or more
• There is an estimated 1,300 trillion cubic
feet of methane hydrate in the oceans
• However – the problem is that methane is
one of the major greenhouse gases which
contributes to global warming – so more
study on retrieval and use is needed
Example:
• Barium Chloride Hydrate
– Mass = 5 g. How many H2O per molecule?
– BaCl2•____H2O
• The sample is heated and the result is 4.26 g
anhydrous BaCl2
– The difference between the 5 g hydrate and the
4.26 anhydrate is .74 g H2O
So. . . . . . . . .
A common hydrate is. . .
MgSo4•7H2O
Magnesium Sulfate
Heptahydrate
4.26 g BaCl2
=
0.0205 mols BaCl2
=
0.041 mol H2
M = 208.23 g/mol
0.74 g H2O
18.02 g/mol
# H2O = x = mols H2O = 0.041 = 2
mols cmpd
0.0205
Thus: BaCl2•2H2O or barium chloride dihydrate
Summary