1. CHAPTER 10
THE MOLE

2. DIMENSIONAL ANALYSIS Also known as factor label method
Problem solving method that focuses on the UNITS that are used to describe matter
Unit labels are treated as FACTORS that can be divided out
Uses conversion factors
Example: 12 inches = 1 foot

3. Problem:
5 ft - _____________ inches
5 ft
12 in
1 ft
5 X 12 1
60 in

4. 10 mm = 1 cm 100 cm = 1 m 14 m = ___________ mm 14 m 100 cm 10 mm
14,000 mm
1 m
1 cm
14 x 100 x 10 1

5. 1000 L= 1 kL 1000 mL = 1 L 4 kL = ____________ mL 4kL 1000 L 1000 mL

6. 4 moles bananas = ___________ bananas
6.02 x bananas = 1 mole
4 moles bananas = ___________ bananas
6.02 x 1023 bananas
4 mole bananas
1 mole banana
4 x (6.02 x 1023) 1
2 x 1024 bananas

7. AVOGADRO’S CONSTANT The number of particles in a mole
Equal to 6.02 x 1023
Mole = a unit for the chemical quantity of a substance
A term that represents a certain # of particles
Need a large number because atoms and molecules are so small
1 mole = 6.02 x 1023 of anything (particles, atoms, molecules etc.)

8. MASS AND THE MOLE Atomic mass = mass of one atom of a substance
Measure in atomic mass unit (amu)
1 hydrogen atom = amu

9. MOLAR MASS Mass in grams of a mole of any pure substance
Use the atomic mass (off the periodic table); change unit to grams/mole
1 atom hydrogen = amu
1 mole hydrogen = grams

10. M0LAR MASS OF COMPOUNDS AND MOLECULES
Sum of the masses of all of the atoms of a substance
Example: H2O
H 2 x 1.0 =
O 1 x 16.0 = 16.0
18.0 g/mole
masses

11. KMnO4
K 1 x = 39.1
Mn 1 x 54.9 =
O 4 x 16.0 =
158.0 g/mole
DON’T FORGET SIG DIGS!!

12. Molar Mass of Hydrates (added H2O)
CoCl2·6H2O
Co x = 58.9
Cl x = 71.0
H2O x 18.0 = 108.0
237.9 g/mole

13. PERCENT COMPOSITION
A way to determine the formula of newly invented compounds (ones that you don’t know the oxidation numbers of)
%composition is a calculation of the % by mass of each element in a compound (part from whole)
Use the formula of the compound and the molar mass of each part

14. K2CrO4
K 2 x 39.1 = 78.2
Cr 1 x 52.0 = 52.0
O 4 x 16.0 = 64.0
molar mass = g /mole
% K = 78.2 / x 100 = 40.3 %
%Cr = 52.0 / x 100 = 26.8%
%O = 64.0 / x 100= %

16. CONVERSION OF THE MOLE Remember: Know: Grams moles particles,etc
1 mole = 6.02 x 1023 particles, atoms etc
Molar mass = _________ g/mole
( g= 1 mole)
Know:
Grams moles particles,etc
Molar mass
g/mole
6.02 x 1023
Particles etc./mole
To go from: grams to moles use molar mass
To go from: moles to molecules use Avogadro’s #
To go from: grams to molecules use both

17. GRAMS TO MOLES Calculate the moles of 1.5 g of sodium
Use molar mass sodium = 23.0 g/mole
1.5 g Na
1 mole Na
.065 moles Na
23.0 g Na

18. MOLES TO GRAMS Calculate the mass in grams of 50.0 moles of KCl
Use molar mass KCl  74.6 g/mole
(K= 39.1; Cl = 35.5)
50.0 mole KCl
74.6 g KCl
3730 g KCl
1 mole KCl

20. MOLES TO PARTICLES(ATOMS, MOLECULES, FORMULA UNITS)
How many formula units are in 4.9 moles of MgSO4?
Use: x 1023 formula units/mole
4.9 moles MgSO4
1 mole MgSO4
6.02 x 1023 formula units
2.9 x 1024 formula units of
MgSO4

21. PARTICLES TO MOLES How many moles are in 2.3 x 1025 molecules of CO2?
Use: x 1023 molecules/mole
2.3 x 1025 molecules CO2
1 mole CO2
6.02 x 1023 molecules CO2
= 38 moles CO2

23. GRAMS TO PARTICLES How many atoms are in 14.0 g of barium?
Use: molar mass Ba = g/mole
Use: x 1023 atoms/mole
14.0 g Ba
1 mole Ba
137.3 g Ba
6.02 x 1023 atoms Ba
= 6.14 x 1022 atoms Ba

24. PARTICLES TO GRAMS How many grams are in 4.5 x 1024 molecules of H2O?
Use: x 1023 molecules/mole
Use: molar mass H2O  g/mole
4.5 x 1024 molc. H2O
6.02 x 1023 molecules
H2O
1 mole H2O
18.0 g H2O
= or g H2O

25. How many grams are in 3.29 x 1024 molecules of chlorine gas?
Use: x 1023 molecules/mole
Use: molar mass of chlorine 71.0 g/mole
3.29 x 1024 molecule Cl2
1 mole Cl2
6.02 x 1023 moleculeCl2
71.0 g Cl2
= 388 g Cl2

26. EMPIRICAL FORMULAS
Definition: the simplest whole # ratio for a formula
You can calculate this if you know the % composition of each element of a newly made compound

27. STEPS FOR WRITING EMPIRICAL FORMULAS
1. Given the grams or % composition (cross off the % and change unit to grams)
2. Convert to moles for each element (use molar mass)
3.Find ratio of moles compared to other elements (divide by smallest # of moles)
4.From ratios, write the empirical formula

28. Example:
What is the empirical formula for a compound that contains .900 g of Ca and 1.60 g of Cl?
1. Determine moles from molar mass for each element
.900g Ca mole Ca
40.1 g Ca = mole Ca
1.60 g Cl mole Cl
35.5 g Cl = mole Cl

29. 3. Write empirical formula using ratios as subscripts CaCl2
2. Find simplest ratio (take smallest # of moles found in step 1 and divide each by that)
Ca = =
Cl = =
3. Write empirical formula using ratios as subscripts
CaCl2

30. FRACTIONS: if the ratios divide out to be fractions, do the following:
1.5:1 ratio difference is .5  multiply both by 2
If the difference is .33 or .67 multiply both by 3
If the difference is .25 or .75 multiply both by 4

31. What is the empirical formula of a compound that is 66. 0 % Ca and 34
What is the empirical formula of a compound that is 66.0 % Ca and 34.0 % P?
40.1 g Ca
66.0g Ca
1 mole Ca
1.65 mole
34.0 g P
31.0 g P
1 mole P
1.10 mole

32. Ca /1.10 = 1.5
P / = 1
Since there is a fraction, multiply both by 2
Ca 1.5 x 2 = 3
P 1 x 2 = 2
The formula is then Ca3P2 a 3:2 ratio

33. MOLECULAR FORMULA Shows the actual # of atoms of each element
Example: Empirical = HO
Molecular = H2O2
To solve, you need one more piece of information  the molar mass of the molecular formula
Use that to find the whole # multiplier

34. The empirical formula of a compound is found to be CH2O
The empirical formula of a compound is found to be CH2O. The molar mass of the molecular formula is g/mole. What is the molecular formula of the compound.

35. Example:
Calculate the empirical formula and molecular formula for a compound that is 56.4% P and 43.7% O. The molar mass of the molecular formula is 220 g/mole.
56.4 g P
1 mole
31.0 g P
= moles
43.7 g O
1 mole
16.0 g 0
= moles

36. Since 1.82 is the smallest amount of moles, we divide both by that
P 1.82 / = 1
O / = 1.5
Since we have a fraction, multiply both by 2
P 1 x 2 = 2
O 1.5 x 2 = 3
The empirical formula is then P2O3

37. To find the molecular formula:
Get molar mass of Empirical:
P2O = 110g/mole
Divide that into molar mass of molecular which is given in the problem
220/110 = 2 which equal the whole # multiplier
Empirical = P2O3
Molecular = P4O6

38. HYDRATES Crystals that contain water
Forms when crystals form in water solution (water molecules stick to the crystal in a specific ratio)
Ex: CuSO4 · 5H2O
CuSO4
H2O
Copper (II) sulfate
pentahydrate

39. Hydrate problems
Write the formula for the hydrate with 89.2% BaBr2 and 10.8% H2O.
89.2 g BaBr2
1 mole BaBr2
297.0 g BaBr2
.300 1
.300
10.8 g H2O
1 mole H2O
18.0 g H2O
.600
.300 2
.600
BaBr2 · 2 H20
Barium bromide dihydrate

40. ANHYDROUS
Without water

41. Determine the hydrate formula for:
.391 g of Li2SiF6 and g H20

42. Determine the hydrate formula for:
76.9% CaSO3 and 23.1% H20

43. Ch. 10 Test
15 multiple choice: definitions, conversions
9 calculations
Molar mass
Percent composition
Grams<-> moles, moles<-> particles, grams<-> particles
Empirical and molecular formulas
Hydrate formulas
Periodic table and conversion “cheat” formula will be provided

44. I am a Mole
- Videos, more practice and tutorials on webpage – Chapter Practice section

45. Popcorn lab (video slow mo) (how its made)
Calculations (show work-both brands)
Mass of unpopped kernels= Data #2 - Data #1
Mass of popped kernels = Data #4 - Data #1
Difference in volume = Data #5 - Data#3
Mass of water in popcorn= Calc. #1 – Calc #2
% composition of water= Calc #4 / Calc#1 x 100
Lab prep
Data (2 brands- get another groups info)
Calculations (show work for both groups)
Questions 1-6 (use articles
Conclusion (full- what learned, error, application)