The Mathematics of Chemistry

1
The Mathematics of Chemistry

• Mole Concept

Kenneth E. Schnobrich
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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
Avogadros 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 Avogadros
  Number (it is unitless).

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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.

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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
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Doing the Conversion

• Always start with what you are given

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Using Table T
On Table T a formula is given for calculating the
number of moles - lets 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
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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
NaHSO4 Total Mass (1 mole) 120
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Determining Formula Mass
Element Atomic Mass atoms Total Mass
C 12 6 72
H 1 12 12
O 16 6 96

C6H12O6 Total Mass (1 mole) 180
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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
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Mixing It Up A Bit
Substance Mass (g) moles
CaCO3 300.
(NH4)2SO4 0.50
Cr(NO3)3 150.
CuSO45H2O 2.0
Sr(OH)2 50.
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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
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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
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Composition Examples
Compound C H O
C6H12O6
CH4 No Oxygen
C2H5OH
CH3COOH
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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 - BaCl22H2O
CuSO45H2O.
PROBLEM Calculate the by mass of water in one
mole of BaCl22H2O
H2O (mass of H2O in one Mol/mass of
BaCl22H2O) x 100 H2O (36 grams/243 grams) x
100 14.8 H2O
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Empirical MolecularFormulas

• 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)

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Empirical MolecularFormulas

• 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

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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

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Some Problems
Empirical Formula Molecular Mass (amu) Molecular Formula
CH2 70
CH 78
CH2 28
CH3 30
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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 -gt 4CO2 6H2O
• The coefficients represent the of moles of
  each substance.

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Moles in Equations

• Regardless of the number of moles of any reactant
  or product, the ratio must always remain -
  2746 for this reaction
• 2C2H6 7O2 -gt 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 - 17/223
• Therefore we will produce 3 moles of H2O starting
  with 1 mole of C2H6

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Other Mole Relationships

• Lets 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

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Mole Relationships(cont.)
3.5 moles
X

• 2KClO3 2KCl 3O2

2 moles
3 moles
Now setup a proportion
X 5.25 moles O2
24
Mole Relationships(cont.)
35 grams
X

• 2KClO3 2KCl 3O2

2 (122.5 grams)
3 moles
Now setup a proportion
K 39 Cl 35.5 O
16 (3) 48
122.5
X 0.43 moles O2
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Mole Relationships(cont.)
35 grams
X

• 2KClO3 2KCl 3O2

2 (122.5 grams)
3 (32 grams)
Now setup a proportion
K 39 Cl 35.5 O
16 (3) 48
122.5
X 6.86 grams O2
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Same Mole Problem Using The Factor Label Method

• 2KClO3 2KCl 3O2

35 g KClO3 mol KClO3 3 mol O2
32 g O2
1 122.5 g 2 mol KClO3
mol O2 KClO3
Answer 6.86 gram of O2