1. Chapter 11 By: Kathryn Johnson 2008 & BuggLady 2006
The Mole
Chapter 11
By: Kathryn Johnson 2008
& BuggLady 2006

2. What is a mole?
Not that one….

3. A Mole is….. (p.310) … an SI unit of measure used for
large numbers of particles
Like the following:
1 dozen = 12
1 gross = 144
1 ream = 500
1 thousand = 1,000
1 billion = 1,000,000,000

4. A Mole (mol) = = 602,213,670,000,000,000,000,000 particles
= 6.02x1023 particles (scientific notation)
= 6.02E23 particles (calculator notation)
Particles - atoms, molecules, formula units, ions,…
This number is also called ‘Avogadro's number’

5. Why is the number so large?
Because atoms are so small!

6. Why do we use the mole?
Because it is easier to talk of 4 moles than to say we have 2,410,000,000,000,000,000,000,000 or 2.41x1024 molecules of something
We use it the same way we talk of
miles instead of feet when we travel

7. Remember… Conversion Factors (Chapter 2)
They allow us to decide whether to divide or multiply by a value
Also called:
Dimensional Analysis
Unit Cancellation
Factor Label

8. How to set one up… Set up a fraction Place the units first
the unit you have on the bottom
the other unit from the relationship on top
Scratch off (Cancel) units that are on both the top and the bottom
Repeat until you only have the units you want
Then place the numbers from the relationships into the fractions

9. Last…Do the Math Multiply all the numbers on the top together
Divide by all the numbers
on the bottom
Round off to correct
significant figures
USE YOUR
CALCULATOR!

10. Mole ↔ Molecules (p.311) Set up a conversion factor!
Moles x Molecules = Molecules
Moles
Molecules x Moles = Moles
Molecules

11. 1 mol = 6.02E23 molecules 6.02E23 molecules 1 mol multiplication
division

12. This will be easy at first and you will want to skip steps….
Be Careful!
This will be easy at first and you will want to skip steps….
But we will be setting up lots of these fractions and you have to keep them straight

13. Now, lets add another step!
Molar Mass is
the mass (g)
of one mole (mol)
of a substance

14. Molar Mass of Elements (p.313)
The mass (g) of a mole (mol) of a particular element is equal to its Atomic Mass on the Periodic Table
Use two decimal places
Carbon: 1 mol C = g C
12.01 g/mol

15. Molar Mass of Compounds (p.322)
The mass (g) of a mole (mol) of a molecule is found by adding the atomic masses of all the elements, multiply by the subscripts
Water… H2O
2(1.01) = g/mol
or mol H2O = g

16. Be Careful…. Polyatomic ions with parenthesis complicate the math!
Ammonium Phosphate… (NH4)3PO4
= 3( (1.01)) (16.00)
= 3( )
= 3(18.05) =
= g/mol

17. Another way to do it… Ammonium Phosphate… (NH4)3PO4
Molar Mass = g/mol

18. Mass ↔ Mole (p.314) Now we have a new conversion factor!
Mass x Mole = Moles
Mass
Moles x Mass = Mass
Moles

19. 1 mol = (Molar Mass) g (Molar Mass) g 1 mol multiplication 1 mol .
division

20. Mass ↔ Molecules (p.316) (the 2 step!)
Now we set up two conversion factors!
You always have to go to Moles first!
Mass x Mole x Molecules = Molecules
Mass Mole
Molecules x Mole x Mass = Mass
Molecules Mole

21. Ratio of Moles of an Element in 1 mole of a Compound (p. 320)
equal to the subscript beside the element
Water… H2O
2 mol H and mol O .
1 mol H2O mol H2O

22. It gets complicated… Watch out for those polyatomic parenthesis!
Ammonium Phosphate… (NH4)3PO4
3 mol N mol H .
1 mol (NH4)3PO mol (NH4)3PO4
and
1 mol P mol O .
1 mol (NH4)3PO mol (NH4)3PO4

23. Mass Compound ↔ Element (the 3 step!)
Now we set up many conversion factors!
Always go to moles!
Mass compound x Mole compound x Mole element… Mass compound Mole compound
… x Mass element = Mass element
Mole element

24. Yes… it keeps going! ... Don’t Skip Steps!
You can set up these conversion factors in infinite combinations to transform from one value to another
The only trick is… to go from one ‘thing’ (compound) to another ‘thing’ (element in that compound) you have to go through the mole ratio based on the subscript
... Don’t Skip Steps!

25. Mass of Element Atoms of Element Moles of Element Moles of Compound
(atomic mass) g = 1 mol
1 mol = (6.02E23) atoms
Moles of Element
(subscript) mol of Element = 1 mol of Compound
Moles of Compound
(molar mass) g = 1 mol
1 mol = (6.02E23) molecules
Mass of Compound
Molecules of Compound

26. And we still have it easy…
wait until we
start dealing with
Chemical Reactions!
(Chapter 12)

27. Percent Composition (p.328)
percent (%) by mass of each element in a compound
Review: Name → Formula (Chapter 8&9)
Watch out for those polyatomic parenthesis!
% Element = (subscript)x(Atomic Mass)x100
Molar Mass Compound
All the % should add up to 100!

28. Example: Water…. H2O % H = 2(1.01) g H x 100 = 11.2% 18.02 g H2O
% O = g O x 100 = 88.8%
Check your work:
11.2% % = 100%

29. Another way to do it… Ammonium Phosphate… (NH4)3PO4
Molar Mass = g/mol

30. Practice
Determine the percent by mass of each element in calcium chloride.
Calculate the percent composition of sodium sulfate.
Which has the larger percent by mass of sulfur: H2SO4 or H2S2O8?

31. Empirical Formula (p.331)
The smallest whole-number mole ratio of the elements in a compound
Derived from %comp of the elements
Turn Percent Composition (%) into Mass (g)
Convert Mass→Mole with the Molar Mass
Divide by the smallest value to get the whole number ratios
Like a %comp in reverse

32. Example: 59.95% O & 40.05% S 0: 59.95 g x 1 mol = 3.747 mol = 3
16.00 g mol
S: g x 1 mol = mol = 1
32.07 g mol
Therefore: SO3… Sulfur trioxide!

33. If the ratios are still decimals, multiply them by a small factor to make them whole
(EX: 0.5 = ½, multiply by 2)
(EX: 0.33 = 1/3, multiply by 3)
(EX: 0.25 =1/4, multiply by 4)

34. Example: % C, 8.16% H, & 43.20% O
0: g x 1 mol = mol = 1 x 2 = 2
16.00 g mol
C: g x 1 mol = mol = 1.5 x 2 = 3
12.01 g mol
H: 8.16 g x 1 mol = mol = 3 x 2 = 6
1.01 g mol
Therefore: C3H6O2

35. Two or more substances with different properties can have the same % composition and empirical formula!
The simplest ratio in the empirical formula does not indicate the actual number of elements in the compound

36. Molecular Formula (p.333) The actual numbers of atoms in a compound
Identifies the compound
A multiple of the empirical formula
Molecular Formula = (Empirical Formula) n
n = Measured molar mass of compound_ _.
Calculated molar mass of empirical formula

37. Example: 46.68% N & 53.32% O Measured Molar Mass = 60.01 g/mol
0: g x 1 mol = mol = 1
16.00 g mol
N: g x 1 mol = mol = 1
14.01 g mol
Empirical Formula: NO … nitrogen monoxide
Molar Mass = = g/mol
But, g/mol = 2 therefore…
30.01 g/mol
Molecular Formula: N2O2 … dinitrogen dioxide!

38. Review Mol ↔ Molecules/Atoms (Avogadro's #) Mol ↔ Mass (Molar Mass)
Mass ↔ Molecules/Atoms (The 2 step)
Element ↔ Compound (Subscript, 3 step)
% Composition
Empirical/Molecular Formula

39. Practice…

40. You have to be fast enough to finish the test!
Work Smart, Not Hard!
You have to be
fast enough
to finish the test!

41. Mole Ratios Coefficients indicate the number of moles
Mole Ratios – the ratio between the numbers of moles of any 2 substances in a balanced equation
Ex: 2 Al Br2  2 AlBr3
All possible mole ratios:
2 mole Al : 3 mole Br2 2 mole Al : 2 mole AlBr3
3 mole Br2 : 2 mole Al 3 mole Br2 : 2 mole AlBr3
2 mole AlBr3 : 2 mole Al 2 mole AlBr3 : 3 mole Br2

42. Interpreting Mole Conversions
4 Fe O2  2 Fe2O3
You can interpret any balanced equation in terms of moles, mass, and molecules
Molecules: 4 atoms of iron react with 6 atoms of oxygen to produce 2 molecules of iron (III) oxide
Moles: 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron (III) oxide
Mass: convert to grams first

43. Practice Mole Ratios For the following equations:
Balance the equations
Interpret each in terms of moles, molecules, and mass
Determine all possible mole ratios
1. CaC2 + N2  CaNCN + C
2. NH3 + CuO  Cu + H2O + N2
3. MnO2 + HCl  MnCl2 + Cl2 + H2O
4. P4O C  P CO
5. ZrI4  Zr + I2
6. PbS + O2  PbO + SO2

44. Mole to Mole Conversions
Ex: Determine the number of moles of H2 produced when moles of K is used.
Steps for solving:
Step 1. Write the balanced equation:
2 K H2O  2 KOH + H2
Step 2. Identify the known (given) substances and the unknown substances:
Known: moles K
Unknown: moles of H2
Step 3. Determine the mole ratio of unknown substance to known substance:
1 mole H2 (use coefficients from balanced equation)
2 mole K
Step 4. Convert using the known substance and the mole ratio:
0.040 moles K x 1 mole H2 = mole H2 produced

45. Practice
Handout: Mole to Mole Stoichiometry Problems

46. Mole to Mass Conversions
Ex: Determine the mass of NaCl produced when 1.25 moles of Cl2 reacts with sodium
Steps:
1. Write the balanced equation:
2 Na + Cl2  2 NaCl
2. Identify the known (given) substances and the unknown substances:
Known: moles Cl2
Unknown: mass of NaCl
3. Determine the mole ratio of unknown substance to known substance:
2 moles NaCl (use coefficients from balanced equation)
1 mole Cl2
4. Convert using the known substance and the mole ratio:
1.25 moles Cl2 x 2 moles NaCl = 2.5 moles NaCl
5. Convert moles of unknown to grams of unknown: (use molar mass)
2.5 mole NaCl x g = 146 grams NaCl
1 mole

47. Example: Mole to Mass Reactants: 4 mole Fe x 55.85 g Fe = 223.4 g Fe
3 mole O2 x g O2 = g O2
1 mole O2
Product:
2 mole Fe2O3 x g Fe2O3 = gFe2O3
1 mole Fe2O3

48. Example, cont.
** Law of Conservation of Mass – the sum of the reactants should ALWAYS equal the sum of the product – check your work!!!
223.4 grams of Fe react with 96.0 grams of O2 to produce grams of Iron (III) oxide

49. Mass to Mass Conversions
Ex: Determine the mass of H2O produced from the decomposition of 25.0 grams of NH4NO3
Steps:
1. Write the balanced equation:
NH4NO3  N2O + 2H2O
2. Identify the known (given) substances and the unknown substances:
Known: grams of NH4NO3
Unknown: mass of H2O
3. Convert grams of known to moles of known:
25.0 g NH4NO3 x 1 mole NH4NO3 = mole NH4NO g
4. Determine the mole ratio of unknown substance to known substance:
2 moles H2O (use coefficients from balanced equation)
1 mole NH4NO3
5. Convert using the known substance and the mole ratio:
0.312 mole NH4NO3 x 2 moles H2O = mole H2O
6. Convert moles of unknown to grams of unknown: (use molar mass)
0.624 mole H2O x g = grams H2O
1 mole

50. Hydrates (p.338)
Compound that has a specific number of water molecules (H2O) bound to its atoms
Solids in which water molecules are trapped
Like a sponge, different amounts of water can be trapped

51. Formulas and Names
1 formula unit of compound with a dot and then the attached water molecules
Ex: Na2CO3∙10H2O
Name the molecule
Use the prefixes in front of the word ‘hydrate’
Ex: sodium carbonate decahydrate 1
Mono- 2
Di- 3
Tri- 4
Tetra- 5
Penta- 6
Hexa- 7
Hepta- 8
Octa- 9
Nona- 10
Deca-

52. Anhydrous “without water” Water is removed from the hydrate by heating
Desiccants are used to absorb excess moisture in packages
Calcium chloride
Calcium sulfate

53. How much water attached?
How many moles of water attached to one mole of the molecule (BaCl2∙xH2O)
compare the mass before and after heating
Mass hydrate – Mass anhydrous = Mass water
Convert the Mass water and Mass anhydrous to moles using the molar mass of each
Calculate the ratio x = mol H2O / mol BaCl2