3.1: The Mole
3.2: Determining the Formula of an Unknown
Compound
3.3: Writing and Balancing Chemical Equations
3.4: Calculating the amounts of Reactant and Product
3.5: Fundamentals of Solution Stoichiometry

MOLE
• The Mole is based upon the definition: The amount of substance that contains as many elementary particles (atoms, molecules, ions, or other?) as there are atoms in exactly 12 grams of carbon -12.
1 Mole = 6.022045 x 10 23 particles (atoms, molecules, ions, electrons, or…) = N
A particles
~100 million x 100 million x 100 million

( N
A
)
N
A
23

Fig 3.1 (P 90) Counting objects of fixed relative mass
12 red marbles @ 7g each = 84g
12 yellow marbles @4e each=48g
55.85g Fe = 6.022 x 10 23 atoms Fe
32.07g S = 6.022 x 10 23 atoms S

Mole - Mass Relationships of Elements
Element Atomic Mass Molar Mass Number of Atoms
1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms
1 atom of S =
1 atom of O = amu amu
1 mole of S =
1 mole of O = g g
= atoms
= atoms
Molecular mass:
1 molecule of O
2
1 molecule of S
8
=
1 mole of O
2
=
= g amu
= molecules amu
1 mole of S
8
= g = molecules

Mole - Mass Relationships of Elements
Element Atomic Mass Molar Mass Number of Atoms
1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms
1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 10 23 atoms
1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 10 23 atoms
Molecular mass:
1 molecule of O
2
1 molecule of S
8
= 16.00 x 2 = 32.00 amu
1 mole of O
2
= 32.00 g
= 32.07 x 8 = 256.56 amu
= 6.022 x 10 23 molecule
1 mole of S
8
= 256.56 g = 6.022 x 10 23 molecules

Molecular Mass Molar Mass ( M )
The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams , called its molar mass.
For water: H
2
O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( amu) + amu = amu
Mass of one molecules of water = amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
= 2 ( g ) + g = g g H
2
O = 6.022 x 10 23 molecules of water = 1 mole H
2
O

Molecular Mass Molar Mass ( M )
The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams , called its molar mass.
For water: H
2
O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecules of water = 18.02 amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
= 2 ( 1.008 g ) + 16.00 g = 18.02 g
18.02 g H
2
O = 6.022 x 10 23 molecules of water = 1 mole H
2
O

Fig 3.2
(P 87)
One mole of common sbustances
CaCO
3
100.09 g
Oxygen
32.00 g
Copper
63.55 g
Water
18.02 g
(This balloon volume is not really big enough. Need
~10-20 liters, depending on pressure inside.)

Fig. 3.3

Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element
3680 o C. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number!
Solution: Converting from mass of W to moles:
Moles of W =
No. of W atoms =

Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element
3680 o C. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number!
Solution: Converting from mass of W to moles:
Moles of W = 35.0 mg W x = 0.00019032 mol
183.9 g W
1.90 x 10 - 4 mol
6.022 x 10 23 atoms
NO. of W atoms = 1.90 x 10 - 4 mol W x =
1 mole of W
= 1.15 x 10 20 atoms of Tungsten

Calculating the Moles and Number of
Formula Units in a given Mass of Cpd.
Problem: Trisodium phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na
3
PO
4
. Calculating the molar mass:
M =
Converting mass to moles:
# Formula units =

Calculating the Moles and Number of
Formula Units in a given Mass of Cpd.
Problem: Trisodium Phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na
3
PO
4
. Calculating the molar mass:
M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen =
= 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol
= 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol
Converting mass to moles:
Moles Na
3
PO
4
= 38.6 g Na
3
PO
4
= 1.46 x 10 23 formula units x (1 mol Na
3
PO
4
)
163.94 g Na
3
PO
4
= 0.23545 mol Na
3
PO
4
Formula units = 0.23545 mol Na
3
PO
4 x 6.022 x 10 23 formula units
1 mol Na
3
PO
4

Flow Chart of Mass Percentage Calculation
Moles of X in one mole of Compound
Multiply by M (g / mol of X)
Mass (g) of X in one mole of compound
Divide by mass (g) of one mole of compound
Mass fraction of X
Multiply by 100 %
Mass % of X

Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C
12
H
22
O
11
) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element: mass of C per mole sucrose = mass of H / mol = mass of O / mol = total mass per mole =
Finding the mass fraction of C in Sucrose & % C :
=
To find mass % of C = mass of 1 mole sucrose

Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C
12
H
22
O
11
) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element: mass of C per mole sucrose = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H / mol = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O / mol = 11 x 16.00 g O/mol = 176.00 g O/mol total mass per mole = 342.296 g/mol
Finding the mass fraction of C in Sucrose & % C :
= 0.4210
mass of 1 mole sucrose 342.30 g Cpd/mol
To find mass % of C = 0.4210 x 100% = 42.10%

Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued mass of 1 mol sucrose
Mass % of O = x 100% = mass of 1 mol sucrose
(b) Determining the mass of carbon :
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C =

Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
= 6.479% H mass of 1 mol sucrose 342.30 g
Mass % of O = x 100% = x 100% mass of 1 mol sucrose 342.30 g
= 51.417% O
(b) Determining the mass of carbon :
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C = 24.35 g sucrose x = 10.25 g C
1 g sucrose

Calculate M and % composition of NH
4
NO
3
.
• 2 mol N x
• 4 mol H x
• 3 mol O x
Molar mass = M =
%N = x 100% = 35.00%
80.05g
%H = x 100% = 5.037%
80.05g
%O = x 100% = 59.96%
80.05g
99.997%

Calculate M and % composition of NH
4
NO
3
.
• 2 mol N x 14.01 g/mol = 28.02 g N
• 4 mol H x 1.008 g/mol = 4.032 g H
• 3 mol O x 15.999 g/mol = 48.00 g O
80.05 g/mol
%N = x 100% = 35.00%
80.05g
%H = x 100% = 5.037%
80.05g
%O = x 100% = 59.96%
80.05g
99.997%

Calculate the Percent Composition of Sulfuric Acid H
2
SO
4
Molar Mass of Sulfuric acid =
2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol
%H = x 100% =
98.09g
2
)
2.06% H
%S = x 100% = 32.69% S
98.09g
%O = x 100% = 65.25% O
98.09 g
Check = 100.00%

Empirical and Molecular Formulas
Empirical Formula The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms.
Molecular Formula The formula of the compound as it exists, it may be a multiple of the Empirical formula.

Some Examples of Compounds with the same
Elemental Ratio’s
Empirical Formula Molecular Formula
CH
2
(unsaturated Hydrocarbons) C
2
H
4
, C
3
H
6
, C
4
H
8
OH or HO H
2
O
2
S S
8
P P
4
Cl Cl
2
CH
2
O (carbohydrates) C
6
H
12
O
6

Steps to Determine Empirical Formulas
Mass (g) of Element
÷ M (g/mol )
Moles of Element
Use no. of moles as subscripts.
Preliminary Formula
Empirical Formula
Change to integer subscripts:
÷ smallest, conv. to whole #.

Determining Empirical Formulas from
Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.
Solution: Finding the moles of the elements:
Moles of Na =
Moles of Cr =
Moles of O =

Determining Empirical Formulas from
Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.
Solution: Finding the moles of the elements:
1 mol Na
Moles of Na = 5.678 g Na x = 0.2469 mol Na
22.99 g Na
Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr
52.00 g Cr
1 mol O
Moles of O = 7.902 g O x = 0.4939 mol O
16.00 g O

Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Converting to integer subscripts (dividing all by smallest subscript):
Rounding off to whole numbers:

Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Na
0.2469
Cr
0.1235
O
0.4939
Converting to integer subscripts (dividing all by smallest subscript):
Na
1.99
Cr
1.00
O
4.02
Rounding off to whole numbers:
Na
2
CrO
4
Sodium Chromate

Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
( M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements.
Solution:
Mass Carbon =
Mass Hydrogen =
Mass Oxygen =

Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
( M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements.
Solution:
Mass Carbon = 40.00% x 100g/100% = 40.00 g C
Mass Hydrogen = 6.719% x 100g/100% = 6.719g H
Mass Oxygen = 53.27% x 100g/100% = 53.27 g O
99.989 g Cpd

Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C =
Moles of H =
Moles of O =
Constructing the preliminary formula:
Converting to integer subscripts, ÷ all subscripts by the smallest:

Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C = Mass of C x = 3.3306 moles C
12.01 g C
1 mol H
Moles of H = Mass of H x = 6.6657 moles H
1.008 g H
Moles of O = Mass of O x = 3.3294 moles O
16.00 g O
Constructing the preliminary formula C
3.33
H
6.67
O
3.33
Converting to integer subscripts, ÷ all subscripts by the smallest:
C
3.33/3.33
H
6.667 / 3.33
O
3.33 / 3.33
= CH
2
O

Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula is:
Whole-number multiple = = empirical formula mass
Therefore the Molecular Formula is:

Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula is:
1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 g/mol
Whole-number multiple = = empirical formula mass
180.16
= = 6.00 = 6
30.03
Therefore the Molecular Formula is:
C
1 x 6
H
2 x 6
O
1 x 6
= C
6
H
12
O
6

•
•
•
•
• Analysis gives :
C = 56.8 %
H = 6.50 %
O = 28.4 %
N = 8.28 %
• Calculate the
Empirical Formula !

• Assume 100g!
• C =
• H =
•
O =
• N =
Divide by smallest (0.591) =>
• C =
• H =
• O =
• N =

• Assume 100g!
• C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C
• H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H
• O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O
• N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N
•
Divide by smallest (0.591) =>
• C = 8.00 mol C = 8.0 mol C or
• H = 10.9 mol H = 11.0 mol H
• O = 3.01 mol O = 3.0 mol O C
8
H
11
O
3
N
• N = 1.00 mol N = 1.0 mol N