The Mathematics of
Chemistry
Mole Concept
Kenneth E. Schnobrich
Counting Things

It is relatively easy to count the number of things
in:
A pair of soxs
 A dozen oranges
 A gross of paper clips
Atoms, molecules, and ions are extremely small units of
matter and visually it would be impossible to count them
in any sample we are given.



Experimentally, we can determine the number of
atoms in a given sample indirectly
Avogadro’s Number


It has been determined that the number of
atoms contained in a mass equivalent to
the atomic mass of an element, expressed
in grams = 6.02 x 1023 atoms of that
element.
The term used for 6.02 x 1023 “things” is a
MOLE (or MOL). It is also referred to as
Avogadro’s Number (it is unitless).
An Example



There are 1 x 1028 atoms in the average
sized person (an Octillion)
If the atom were the size of a pea, an
Octillion peas would cover the complete
surface of 250,000 planets, each the size
of Earth, to a depth of 4 feet
The atom is an incredibly small unit of
matter.
Some Mole Relationships
AMOUNT
# OF MOLES
7 grams of Li
1 mole of Li
28 grams of Fe
0.5 moles of Fe
8 grams of Mg
0.25 moles of Mg
1.204 x 1024 Na atoms
2 moles of Na
3.01 x 1023 S atoms
0.5 moles of S
Doing the Conversion

Always start with what you are given:
28 g Fe 1 mole Fe
= 0.5 mole Fe
56 g Fe
1.204 x 1024 atoms Na
1 mole Na
= 2 mole Na
6.02 x 1023 atoms Na
Using Table T*
On Table T a formula is given for calculating the number of
moles - let’s use it for a calculation:
# of moles (n) = given mass/gram formula mass*
# of moles (n) = given mass/gram atomic mass**
*Gram formula mass - mass of the compound in grams
**Gram atomic mass - mass of the element in grams
Given 8.0 grams of Mg how many moles do you have?
#moles(n) = 8.0 grams/24.0 grams
#moles(n) = 0.25
*NYS Reference Tables for Chemistry
Determining Formula Mass
Element
Atomic
Mass
# atoms
Total Mass
Na
23
1
23
H
1
1
1
S
32
1
32
O4
16
4
64
Total Mass
(1 mole)
120
NaHSO4
Determining Formula Mass
Element
Atomic
Mass
# atoms
Total Mass
C
12
6
72
H
1
12
12
O
16
6
96
Total Mass
(1 mole)
180
C6H12O6
Using Formula Mass
Given 90. grams of C6H12O6 how many moles in the sample?
#moles (n) = given mass/ gram formula mass
#moles (n) = 90. grams/ 180 grams/mole = 0.50 moles
Mixing It Up A Bit
Substance
Mass (g)
CaCO3
300.
(NH4)2SO4
Cr(NO3)3
0.50
150.
CuSO4•5H2O
Sr(OH)2
#moles
2.0
50.
Mols and Gases
At STP one mol of any gas occupies a volume of 22.4 L
So, if we had 2 mols of H2 gas at STP it would occupy 44.8 Liters
PROBLEM: If a sample of N2 gas at STP occupies a volume of 67.2
Liters, how many moles of the gas do you have in the sample?
# mols = volume given/volume of one mol at STP
# mols = 67.2 Liters/22.4 Liters/mol = 3 mols
PROBLEM:
How many grams of N2 gas are there in the sample?
#g = #mols x #g/one mol
# g = 3 mol (28 g/mol) = 84 grams
Percent Composition




These problems are worked like any
percentage - if you miss 25 out of 50
questions you have a 50%.
% = (mass of part/mass of whole) x 100
This equation can be found on Table T*
If I wanted to know the percent by mass of
hydrogen in water

%H = (2.00 g H/18.0 g H2O) x 100 = 11.2%
*NYS Reference Tables for Chemistry
% Composition Examples
Compound
%C
%H
%O
C6H12O6
CH4
C2H5OH
CH3COOH
No Oxygen
% H2O in a Hydrate
Certain substances have water molecules associated with their structure. Many salts
fall into this category and we refer to them as hydrated salts. Different salts have
different numbers of of water molecules - BaCl2•2H2O; CuSO4•5H2O.
PROBLEM: Calculate the % by mass of water in one mole of BaCl2•2H2O
%H2O = (mass of H2O in one Mol/mass of BaCl2•2H2O) x 100
%H2O = (36 grams/243 grams) x 100 = 14.8% H2O
Empirical & Molecular
Formulas

Empirical formulas represent the smallest
whole number ratio of atoms in a
compound (cannot be reduced any further)

Molecular (or True) formula represents the
actual whole number ratio of atoms in a
compound (it will be a small whole number
multiple of the Empirical formula)
Empirical & Molecular
Formulas

CH2O would be the empirical formula of a
typical monosaccharide

C6H12O6 would represent the molecular or
true formula of the monosaccharide

6 x (CH2O) - each of the subscripts would be
multiplied by 6
Simple EF/MF Problems

Vitamin C has an empirical formula of
C3H4O3 and a molecular mass of 176 amu.
What is the molecular formula?
First determine the mass of the empirical
formula: (3 x 12) + (4 x 1) + (3 x 16) = 88
 Now divide the molecular mass by the
empirical mass: 176/88 = 2
 Now multiply each of the subscripts in the
empirical formula by 2: C6H8O6

Some Problems
Empirical
Formula
Molecular Mass
(amu)
CH2
70
CH
78
CH2
28
CH3
30
Molecular
Formula
Moles in Equations




In chemical reactions it is important that
equations be balanced and the mole ratio
taken into account.
For this unit we will provide the balanced
equation
2C2H6 + 7O2 -> 4CO2 + 6H2O
The coefficients represent the # of moles
of each substance.
Moles in Equations



Regardless of the number of moles of any
reactant or product, the ratio must always
remain 2:7:4:6 for this reaction
2C2H6 + 7O2 -> 4CO2 + 6H2O
Problem: Starting with 1 mole of C2H6 how
many moles of H2O will be produced?
Divide the coefficients by 2
 The new ratio is - 1:7/2:2:3
 Therefore we will produce 3 moles of H2O
starting with 1 mole of C2H6

Other Mole Relationships
Let’s use another reaction and take a slightly different
approach:
2KClO3 = 2KCl + 3O2
Problem: 3.5 moles of KClO3 will produce how many
moles of O2?
Place the given # of moles over the substance
Place an X over the substance in question
Below the substance given indicate the # of moles
from the balanced equation
Below the substance you are looking for place the
number of moles from the balanced equation
Mole Relationships(cont.)
3.5 moles
X
2KClO3 = 2KCl + 3O2
2 moles
3 moles
Now setup a proportion
3.5 moles
2 moles
=
X
3 moles
X = 5.25 moles O2
Mole Relationships(cont.)
35 grams
X
2KClO3 = 2KCl + 3O2
2 (122.5 grams)
K=
39
Cl =
35.5
O = 16 (3) = 48
122.5
3 moles
Now setup a proportion
35 grams
245 grams
=
X
3 moles
X = 0.43 moles O2
Mole Relationships(cont.)
35 grams
X
2KClO3 = 2KCl + 3O2
2 (122.5 grams)
K=
39
Cl =
35.5
O = 16 (3) = 48
122.5
3 (32 grams)
Now setup a proportion
35 grams
245 grams
=
X
48 grams
X = 6.86 grams O2
Same Mole Problem Using
The Factor Label Method
2KClO3 = 2KCl + 3O2
35 g KClO3
mol KClO3
1
122.5 g
KClO3
3 mol O2
32 g O2
2 mol KClO3
mol O2
Answer = 6.86 gram of O2