The Mole I
Mathematical Relationships with
Chemical Formulas
mole (mol) - the amount of a substance that
contains the same number of entities as
there are atoms in exactly 12 g of carbon12.
This amount is 6.022x1023. The number is
called Avogadro’s number and is
abbreviated as N.
One mole (1 mol) contains 6.022x1023
entities (to four significant figures)
One mole
of common
substances.
CaCO3
100.1 g
Oxygen
32.0 g
Water
18.0 g
Copper
63.5 g
molar mass (MM) = mass of one
mole of a substance
• 1 mol Ca = 40.1 g = 6.02 x 1023 Ca atoms
• 1 mol Cu = 63.5 g = 6.02 x 1023 Cu atoms
• 1 mol Hg = 200.6 g = 6.02 x 1023 Hg atoms
Molar Mass
• 1 mol Cl2 = 71.0 g
= 6.02 x 1023 Cl2 molecules
= 2(6.02 x 1023) Cl atoms
• 1 mol (NH4)2SO4
= 6.02 x 1023 units of (NH4)2SO4
= 15(6.02 x 1023) atoms
= [2(14.0g) + 8(1.0g) + 1(32.1g) +
4(16.0g)] = 132.1 g
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Carbon (C)
Hydrogen (H)
Oxygen (O)
6 atoms
12 atoms
6 atoms
12 moles
of atoms
Atoms/mole of
compound
6 moles
of
atoms
6(6.022 x
1023) atoms
12(6.022 x
1023) atoms
6 moles
of
atoms
6(6.022 x 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
12.10 g
96.00 g
Atoms/molecule
of compound
Moles of atoms/
mole of compound
Mass/mole of
compound
72.06 g
Calculating molesA from gramsA
• How many moles of Ca are in 22 g of Ca?
40.1 g
1 mol Ca
or
1 mol Ca = 40.1 g so
40.1 g
1 mol Ca
Calculating gramsA from molesA
• How many g of (NH4)2SO4 are in 0.0335
mol of (NH4)2SO4 ?
1 mol (NH4)2SO4 = 132.1 g so
132.1 g
1 mol (NH4 )2 SO4
or
132.1 g
1 mol (NH4 )2 SO4
132.1 g
0.0335 mol (NH4 )2 SO4 x
= 4.43 g (NH4 )2 SO4
1 mol (NH4 )2 SO4
Mass percent from the chemical formula
Mass % of element X =
moles of X in formula x molar mass of X
molecular (or formula) mass of compound
x 100
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
Glucose (C6H12O6) is the most important
nutrient in the living cell for generating
chemical potential energy. What is the mass
percent of each element in glucose?
Per mole glucose there are:
6 moles of C
12 moles H
6 moles O
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
6 mol C x
6 mol O x
12.01 g C
1 mol C
16.00 g O
1 mol O
= 72.06 g C
= 96.00 g O
mass percent of C =
12 mol H x
1.008 g H
= 12.096 g H
1 mol H
M = 180.16 g/mol
72.06 g C
180.16 g glucose
x 100 = 39.99 mass %C
12.096 g H
mass percent of H =
x 100 = 6.714 mass %H
180.16 g glucose
96.00 g
mass percent of O =
x 100 = 53.29 mass %O
180.16 g glucose
Empirical and Molecular Formulas
Empirical Formula The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest
set of whole numbers of atoms.
C1H2O1
Molecular Formula The formula of the compound as it exists, it may be
a multiple of the empirical formula.
C6H12O6
Determining the Empirical Formula from Masses of Elements
Elemental analysis of a sample of an ionic compound
showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O.
What are the empirical formula and name of the
compound?
2.82 g Na
4.35 g Cl
7.83 g O
1 mol Na
22.99 g Na
1 mol Cl
35.45 g Cl
1 mol O
16.00 g O
= 0.123 mol Na
Divide each mole value by the
smallest value to get the whole
number subscripts.
= 0.123 mol Cl
= 0.489 mol O
Na1 Cl1 O3.98
NaClO4
NaClO4 is sodium perchlorate.
Determining a Molecular Formula from Elemental Analysis and Molar
Mass
During physical activity, lactic acid (MM=90.08 g/mol) forms in muscle tissue
and is responsible for muscle soreness. Elemental analysis shows that this
compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
Determining a Molecular Formula from Elemental Analysis and Molar
Mass
Assuming there are 100. g of lactic acid, the constituents are:
40.0 g C 1 mol C
6.71 g H 1 mol H
12.01g C
53.3 g O 1 mol O
1.008 g H
16.00 g O
3.33 mol C
6.66 mol H
Divide each mole amount by the smallest value…
C 3.33 H6.66
3.33
3.33
O3.33
3.33
CH2O
3.33 mol O
empirical
formula
Divide the molar mass by the mass of the empirical formula
molar mass of lactic acid
mass of CH2O
90.08 g
30.03 g
3
C3H6O3 is the
molecular formula
Calculating moleB <-> moleA
• How many moles of hydrogen are contained
in 2.8 mol of (NH4)2SO4?
• 1 mol (NH4)2SO4 = 2 mol N, 8 mol H, 1 mol S,
4 mol O
8 mol H
2.8 mol (NH4 )2 SO4 x
= 22 mol H
1 mol (NH4 )2 SO4
Calculating particles from moles
• Particles can be molecules, formula units
(ionic compounds), atoms or ions.
• 1 mol particles = 6.02 x 1023 units
Ne atom
NH3 molecule
NaCl formula unit
O2 molecule
Calculating particles <-> moles
• How many atoms of oxygen are in 0.580
mol of iron (II) nitrate?
6 mol O
6.02 x 1023 O atoms
0.580 mol Fe(NO3 )2 x
x
= 2.09 x 1024 O atoms
1 mol Fe(NO3 )2
1 mol O
• How many moles of sulfur trioxide can
be made from 4.0 x 1023 oxygen atoms?
4.0 x 1023 O atoms x
1 mol SO3
1 mol O
X
= 0.22 mol SO3
23
6.02 x 10 O atoms
3 mol O
Calculating particles <-> g
• How many atoms of potassium are in 23.4 g of
potassium carbonate?
1 mol K2CO3
2 mol K
6.02 x 1023 atoms K
23.4 g K2CO3 x
x
x
= 2.04 x 1023 atoms K
138.2 g K2CO3
1 mol K2CO3
1 mol K
• How many g of sulfur hexachloride can be
made from 3.33 x 1024 chlorine atoms?
3.33 x 1024 Cl atoms x
245.1 g SCl6
1 mol SCl6
1 mol Cl atoms
x
x
= 226 g SCl6
23
6.02 x 10 Cl atoms
6 mol Cl
1 mol SCl6
GramA <-> GramB with Formulas
• How many g of oxygen are contained in
458 g of C12H22O11?
458 g C12H22O11 x
16.0 g O
1 mol C12H22O11
11 mol O
x
x
= 236 g O
342 g C12H22O11
1 mol C12H22O11
1 mol O
Summary of the mass-mole-number relationships
in a chemical formula.
MASS(g)
of substance A
MM (g/mol)
of compound
A
molar ratio from
AMOUNT(mol)
of substance A
formula subscripts
Avogadro’s number
(particles/mol)
MOLECULES or ATOMS
(or formula units)
of substance A
Modified from Silberberg, Principles of Chemistry
MASS(g)
of substance B
MM (g/mol)
of compound
B
AMOUNT(mol)
of substance B
Avogadro’s number
(particles/mol)
MOLECULES or ATOMS
(or formula units)
of substance B
Questions to Ask When Solving a
Problem
• Are grams mentioned?
If yes you will need to use a molar mass
conversion.
• Are you comparing two different substances?
If yes you will need a conversion using the
subscripts in a formula.
• Are particles (atoms or molecules) involved?
If yes you will need a conversion with Avogadro’s
number.
You may need any number of these conversions
depending on the problem.
How many atoms of oxygen are required
to form 6.8 grams of aluminum nitrate?
•
•
•
•
Grams? Yes, need molar mass
Two substances? Yes, need subscripts
Particles? Yes, need Avogadro’s #
So at least 3 conversions are needed…
1 mol Al(NO3 )3
9 mol O
6.02 x 1023 O atoms
6.8 g Al(NO3 )3 x
x
x
= 1.7 x 1023 O atoms
213.0 g Al(NO3 )3 1 mol Al(NO3 )3
1 mol O
How many moles of nitrogen are contained
in 45.0 grams of aluminum nitrate?
•
•
•
•
Grams? Yes, need molar mass
Two substances? Yes, need subscripts
Particles? No
So, at least 2 conversions are needed…
1 mol Al(NO3 )3
3 mol N
45.0 g Al(NO3 )3 x
x
= 0.634 mol N
213.0 g Al(NO3 )3 1 mol Al(NO3 )3
Molarity
• Because many reactions happen between
substances in solution it is helpful to
have a concentration based on moles.
moles of solute (dissolved substance)
Molarity (M) =
1 liter of solution
http://fuelcell.com
Calculating the Molarity of a Solution
Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity
of an aqueous solution that contains 0.715 mol of glycine in 495 mL?
Molarity is the number of moles of solute per liter of solution.
0.715 mol glycine 1000mL
495 mL soln
1L
= 1.44 M glycine
Summary of
mass-mole-number-volume
relationships in solution.
MASS (g)
of compound
in solution
MM (g/mol)
AMOUNT
(mol)
of compound
in solution
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound
in solution
M (mol/L)
VOLUME (L)
of solution