1. Mr. Matthew Totaro Legacy High School Honors Chemistry
The Mole Concept
Mr. Matthew Totaro
Legacy High School
Honors Chemistry

2. Why Is Knowledge of Amounts so Important?
Everything in nature is either chemically or physically combined with other substances.
To know the amount of a material in a sample, you need to know what fraction of the sample it is.
Some Applications:
The amount of sodium in sodium chloride for a diet.
The amount of iron in iron ore for steel production.
The amount of hydrogen in water for a hydrogen fuel cell.
The amount of chlorine in freon to estimate ozone depletion.

3. Stoichiometry
Stoichiometry = comes from the Greek words ‘stoicheon’ meaning ‘element’ and ‘metron’ meaning ‘to measure’.
System used to measure quantities in chemistry
Two Branches of Stoichiometry
Composition Stoichiometry = describes the quantitative relationships among elements in compounds.
Reaction Stoichiometry = describes the quantitative relationships among substances as they participate in chemical reactions

4. Calculation of Molecular Mass

5. Formula (Molecular) Mass
The mass of an individual molecule or formula unit.
Also known as molecular mass or molecular weight.
Sum of the masses of the atoms in a single molecule or formula unit.
Whole = Sum of the parts.
Ex: Mass of 1 molecule of H2O
= 2(1.01 amu H) amu O = amu.

6. Practice—Calculate the Formula Mass of Al2(SO4)3.

7. Practice—Calculate the Formula Mass of Al2(SO4)3, Continued.

8. Counting Atoms by Moles
If we can find the mass of a particular number of atoms, we can use this information to convert the mass of an element sample to the number of atoms in the sample.
The number of atoms we use in chemistry is 6.02 x 1023 and we call this a mole.
1 mole = 6.02 x 1023 things.
Like 1 dozen = 12 things.
Avogadro’s number.

9. The mole is named in honor of him. ‘Mole’ in Latin means ‘mass’.
Amadeo Avogadro
The mole is named in honor of him. ‘Mole’ in Latin means ‘mass’.

10. The Mole - Avogadro’s Number

11. How Big is a Mole?

12. Chemical Packages—Moles
Mole = Number of things equal to the number of atoms in 12 g of C-12.
1 atom of C-12 weighs exactly 12 amu.
1 mole of C-12 weighs exactly 12 g.
In 12 g of C-12 there are 6.02 x1023 C-12 atoms.
1 mole
carbon
12.01 g

13. One Step Stoichiometry Conversions

14. Example—A Silver Ring Contains 1. 1 x 1022 Silver Atoms
Example—A Silver Ring Contains 1.1 x 1022 Silver Atoms. How Many Moles of Silver Are in the Ring?
Given:
Find:
1.1 x 1022 atoms Ag
moles Ag
Solution Map:
Relationships:
1 mol = x 1023 atoms
atoms Ag
mol Ag
Solution:
Check:
Since the number of atoms given is less than Avogadro’s number, the answer makes sense.

15. Practice—Calculate the Number of Atoms in 2.45 mol of Copper.

16. Relationship Between Moles and Mass
The mass of one mole of atoms is called the molar mass
The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu
The lighter the atom, the less a mole weighs
The lighter the atom, the more atoms there are in 1 g
1 mole
sulfur
32.06 g

17. Mole and Mass Relationships
sulfur
32.06 g
1 mole
carbon
12.01 g

18. Example—Calculate the Moles of Sulfur in 57.8 g of Sulfur.
Given:
Find:
57.8 g S
mol S
Solution Map:
Relationships:
1 mol S = g
g S
mol S
Solution:
Check:
Since the given amount is much less than 1 mol S, the number makes sense.

19. Practice—Calculate the Grams of Carbon in 0
Practice—Calculate the Grams of Carbon in mols of Pencil Lead, Continued.
Given:
Find:
g C
mol C
Solution Map:
Relationships:
1 mol C = g
mol C
g C
Solution:
Check:
Since the given amount is much less than 1 mol C, the number makes sense.

20. Molar Volume of a Gas There is so much empty space between molecules
in the gas state that
the volume of the
gas is not effected
by the size of the
molecules (under
ideal conditions).

21. Molar Volume

22. Practice—Calculate the Volume of 2.79 mols of He gas.

23. Two Step Stoichiometry Conversions

24. Example — How Many Aluminum Atoms Are in a Can Weighing 16.2 g?
Given:
Find:
16.2 g Al
atoms Al
Solution Map:
Relationships:
1 mol Al = g, 1 mol = x 1023
g Al
mol Al
atoms Al
Solution:
Check:
Since the given amount is much less than 1 mol Cu, the number makes sense.

25. Molar Mass of Compounds
The relative weights of molecules can be calculated from atomic weights.
Formula mass = 1 molecule of H2O
= 2(1.01 amu H) amu O = amu.
Since 1 mole of H2O contains 2 moles of H and 1 mole of O.
Molar mass = 1 mole H2O
= 2(1.01 g H) g O = g.

26. Practice — How Many CO2 molecules Are in a tank weighing 3.10 g?

27. Example — What Is the Mass of 4.78 x 1024 NO2 Molecules?
Given:
Find:
4.78 x 1024 NO2 molecules
g NO2
Solution Map:
Relationships:
molecules
mol NO2
g NO2
1 mol NO2 = g, 1 mol = 6.02 x 1023
Solution:
Check:
Since the given amount is more than Avogadro’s number, the mass > 46 g makes sense.

28. Practice — How Many Formula Units Are in 50. 0 g of PbO2. (PbO2 = 239

29. Percent Composition

30. Percent Composition Percentage of each element in a compound. By mass.
Can be determined from:
The formula of the compound.
The experimental mass analysis of the compound.
The percentages may not always total to 100% due to rounding.

31. Example—Find the Percent Composition of C2Cl4F2.
Given:
Find:
C2Cl4F2
% Cl by mass
Solution Map:
Relationships:
Solution:
Check:
Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense.

32. Empirical Formula

33. Empirical Formulas
The simplest, whole-number ratio of atoms in a molecule is called the empirical formula.
Can be determined from percent composition or combining masses.
The molecular formula is a multiple of the empirical formula.
% A mass A (g) moles A
100g MMA
% B mass B (g) moles B
100g MMB
moles A
moles B

34. Empirical Formulas, Continued
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HO
Benzene
Molecular formula = C6H6
Empirical formula = CH
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O

35. Practice—Determine the Empirical Formula of Benzopyrene, C20H12.

36. Practice—Determine the Empirical Formula of Benzopyrene, C20H12, Continued
Find the greatest common factor (GCF) of the subscripts.
20 factors = (10 x 2), (5 x 4)
12 factors = (6 x 2), (4 x 3)
GCF = 4
Divide each subscript by the GCF to get the empirical formula.
C20H12 = (C5H3)4
Empirical formula = C5H3

37. Finding an Empirical Formula from Experimental Data

38. Example:
A laboratory analysis of aspirin determined the following mass percent composition.
C = 60.00%
H = 4.48%
O = 35.53%
Find the empirical formula.

39. Example: Find the empirical formula of aspirin with the given mass percent composition.
Write down the given quantity and its units.
Given: C = 60.00%
H = 4.48%
O = 35.53%
Therefore, in 100 g of aspirin there are g C, g H, and g O.

40. Find: empirical formula, CxHyOz
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Write down the quantity to find and/or its units.
Find: empirical formula, CxHyOz

41. 1 mole C = 12.01 g C 1 mole H = 1.01 g H 1 mole O = 16.00 g O
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Collect needed conversion factors:
1 mole C = g C
1 mole H = 1.01 g H
1 mole O = g O

42. Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g; 1 mol H = 1.01 g; 1 mol O = g
Write a solution map:
g C
mol C
whole
number
ratio
mole
ratio
empirical
formula
pseudo-
formula
g H
mol H
g O
mol O

43. Apply the solution map:
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Calculate the moles of each element.

44. Apply the solution map:
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: mol C, 4.44 mol H,
2.221 mol O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Write a pseudoformula.
C4.996H4.44O2.221

45. Apply the solution map:
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: C4.996H4.44O2.221
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Find the mole ratio by dividing by the smallest number of moles.

46. Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: C2.25H2O1
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Multiply subscripts by factor to give whole number.
{ } x 4
C9H8O4

47. Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00).

48. All These Molecules Have the Same Empirical Formula
All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?
Name
Molecular
Formula
Empirical
Glyceraldehyde
C3H6O3
CH2O
Erythrose
C4H8O4
Arabinose
C5H10O5
Glucose
C6H12O6

49. All These Molecules Have the Same Empirical Formula
All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?, Continued
Name
Molecular
Formula
Empirical
Molar
Mass, g
Glyceraldehyde
C3H6O3
CH2O 90
Erythrose
C4H8O4
120
Arabinose
C5H10O5
150
Glucose
C6H12O6
180

50. Molecular Formula

51. Molecular Formulas
The molecular formula is a multiple of the empirical formula.
To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound.
Molar massreal formula
Molar massempirical formula
= Factor used to multiply subscripts

52. Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8.
Determine the empirical formula.
May need to calculate it as previous.
C5H8
Determine the molar mass of the empirical formula.
5 C = 60.05, 8 H = 8.064
C5H8 = g/mol

53. Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.
Divide the given molar mass of the compound by the molar mass of the empirical formula.
Round to the nearest whole number.

54. Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.
Multiply the empirical formula by the factor above to give the molecular formula.
(C5H8)3 = C15H24

55. Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01)

56. Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01), Continued
C5 = 5(12.01 g) = g
H3 = 3(1.01 g) = g
C5H3 = g
Molecular formula = {C5H3} x 4 = C20H12

57. Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N. (C=12.01, H=1.01, N=14.01)

58. Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued
Given: 74.0% C, 8.7% H, {100 – ( )} = 17.3% N 
in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N.
Find: CxHyNz
Conversion Factors:
1 mol C = g; 1 mol H = 1.01 g; 1 mol N = g
Solution Map:
g C
mol C
whole
number
ratio
mole
ratio
pseudo-
formula
empirical
formula
g H
mol H
g N
mol N

59. Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued.
Apply solution map:
C5 = 5(12.01 g) = g
N1 = 1(14.01 g) = g
H7 = 7(1.01 g) = g
C5H7N = g
C6.16H8.6N1.23
{C5H7N} x 2 = C10H14N2