1. The Mathematics of Chemistry
Mole Concept
Kenneth E. Schnobrich

2. 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

3. 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).

4. 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.

5. 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

6. Doing the Conversion Always start with what you are given: 28 g Fe
1 mole Fe
56 g Fe
0.5 mole Fe =
1.204 x 1024 atoms Na
6.02 x 1023 atoms Na
1 mole Na
= 2 mole Na

7. 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

8. Determining Formula Mass
Element
Atomic Mass
# atoms
Total Mass Na 23 1 H S 32 O4 16 4 64
NaHSO4
Total Mass (1 mole)
120

9. Determining Formula Mass
Element
Atomic Mass
# atoms
Total Mass C 12 6 72 H 1 O 16 96
C6H12O6
Total Mass (1 mole)
180

10. 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

11. Mixing It Up A Bit Substance Mass (g) #moles CaCO3 300. (NH4)2SO4 0.50
Cr(NO3)3
150.
CuSO4•5H2O
2.0
Sr(OH)2
50.

12. At STP one mol of any gas occupies a volume of 22.4 L
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

13. *NYS Reference Tables for Chemistry
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

14. % Composition Examples
Compound %C %H %O
C6H12O6
CH4
No Oxygen
C2H5OH
CH3COOH

15. % 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

16. 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)

17. 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

18. 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

19. Some Problems Empirical Formula Molecular Mass (amu) Molecular Formula
CH2 70 CH 78 28
CH3 30

20. 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.

21. Moles in Equations
Regardless of the number of moles of any reactant or product, the ratio must always remain :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

22. 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

23. Mole Relationships(cont.)
3.5 moles X
2KClO3 = 2KCl + 3O2
2 moles
3 moles
Now setup a proportion
3.5 moles
2 moles
3 moles X =
X = 5.25 moles O2

24. Mole Relationships(cont.)
35 grams X
2KClO3 = 2KCl + 3O2
2 (122.5 grams)
3 moles
K =
Cl =
O = 16 (3) = 48
Now setup a proportion
35 grams
245 grams
3 moles X =
122.5
X = 0.43 moles O2

25. Mole Relationships(cont.)
35 grams X
2KClO3 = 2KCl + 3O2
2 (122.5 grams)
3 (32 grams)
K =
Cl =
O = 16 (3) = 48
Now setup a proportion
35 grams
245 grams
48 grams X =
122.5
X = 6.86 grams O2

26. Same Mole Problem Using The Factor Label Method
2KClO3 = 2KCl + 3O2
35 g KClO mol KClO mol O g O2
g mol KClO mol O2
KClO3
Answer = 6.86 gram of O2