1. THE Mole
Chapter 10

2. 10.1 MeasURING mATTER
Chemists use the mole to count atoms, molecules, and ions.
Atoms – individual atoms
Molecules – two or more atoms covalently bonded
ions – positively or negatively charged particles
Formula Units – Ionic compounds

3. A mole is used to compare large amounts of atoms, particles, or # of molecules
1 mole = 6.02 x atoms/particles/molecules

4. Avogadro Avagadro’s Number = 6.02 x 1023 Named in Honor of 
In 1811 he determined the volume
of 1 mol of a gas

5. How was Avogodro’s number first calculated?
The first person to have calculated the number of molecules in any mass of substance seems to have been Josef Loschmidt, ( ), an Austrian high school teacher, who in 1865, using the new Kinetic Molecular Theory (KMT) calculated the number of molecules in one cubic centimeter of gaseous substance under oridnary conditions of temperature of pressure, to be somewhere around 2.6 x 1020 molecules. This has always been known as the"Loschmidt Number."

6. How was Avogodro’s number first calculated?
Use the following conversions to calculate Avagodro’s number
100 ml /1litre
l/1mole

7. How was Avogodro’s number first calculated?
(2.68 x molecules) x (1000 ml) x ( Liters ) =
1cm3 1 litre 1 mole
(2.68 x 10 19molecules) x (1000 ml) x ( Liters ) = molecules
1cm3 1 litre mole 1 mole

8. The mole is independent of mass and size
The mole is independent of mass and size. It only measures number of particles

9. A mole comparison

10. How many Roses?

11. Converting between real world units
To convert use a conversion factor (aka a ratio in math) to convert or switch from one unit to another unit.
Ex. How many roses are in 3 dozen?
The conversion factor is 1 dozen / 12 roses or 12 roses / 1 dozen
The set up
What you are given is written first.
Then use the correct conversion factor
3 dozen x 12 roses = 36 roses
1 dozen
Cross out = the units that cancel = one on top and one on bottom

12. Converting between moles and particles
Ex. How many molecules are in 3.50 moles of sucrose?
3.50 mol sucrose x 6.02 x 1023 molecules sucrose = mol sucrose
=2.11 x 1024 molecules sucrose
Do questions 1& 2 pg 307, 4&5 pg 309

13. Converting between moles and particles
3.50 mol sucrose x 6.02 x 1023 molecules sucrose = mol sucrose
=2.11 x 1024 molecules sucrose
Do questions 1& 2 pg 307, 4&5 pg 309

14. Converting between moles and particles
Practice 1. How many particles are in 2.5 moles of Zn?
2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn
1 mole Zn

15. Converting between moles and particles
2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn
1 mole Zn

16. Converting between moles and particles
2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn
1 mole Zn

17. More Practice
Silver nitrate (AgNO3) is used to make several different silver halides used in photographic films. How many formula units of AgNO3 are there in 3.25 moles of AgNO3.

18. Converting Particles to Moles
Start with the number of Representative Particles (aka atoms, molecules, ion, or formula units)
Use the correct conversion factor to cancel out unwanted units.
How many moles in 2.11 x 1024 molecules of sucrose?
Ex x 1024 molecules of sucrose x 1 mole sucrose =
6.02x1023 mol sucrose
= 3.50 mols sucrose

19. Practice particles to moles
How many moles are in 4.50 x 1024 atoms Zn?
How many moles in 2.50 x 1020 atoms Fe?

20. 10.2 Mass and the mole
A mole always contains the same number of particles; however, moles of different elements/compounds have different masses
Molar mass: the mass in grams of one mole of any pure substance: units=g/mol
Atomic Mass = g/mol

21. Real World Example Was is the mass of 5 dozen jelly beans
1 dozen jelly beans = 35 g jelly beans
5 dozen jelly beans x 35 g jelly beans = 175 g jelly beans
1 dozen jelly beans

22. Example What is the mass of 3.00 moles of Cu?
Identify the Molar mass of Cu from the periodic table(round to 4 sig figs)
Cu = g/mol
Calculation
3.00 moles Cu x 63.55g Cu = 191 g Cu
1 mol Cu

23. Practice Pg. 328 # 15b Determine the mass in grams of 42.6 mol Si # 16
2.45 x 10-2 mol Zn (answer in scientific notation)

24. Mass to Mole Conversion
Ex. Convert 525 g of Ca to mol
525 g Ca x 1 mol Ca = mol Ca
40.08 g Ca
Do questions 7 & 8 pg ,12-15

25. Practice Pg. 329 # 18b. How many moles in 300.0 g of S
#19b. How many moles in 1.00 Kg Fe

26. 10.3 Moles of Compounds Identify the molar mass of a compound.
Add up the molar mass of the atoms.
Ex. CCl4
C = g/mol
Cl = g/mol x 4 = g/mol
141.8 g/mol g/mol = g/mol = g/mol CCl4
Ex. How many moles are in g of CCl4
500.0 g CCl4 x 1 mol CCl = mol CCl4
153.8 g CCl4

27. Practice What is the Molar Mass of H2SO4?
How many grams are in 3.25 moles of H2SO4?

28. Mass to Mole Ex. How many moles are in 35.0g of HCl?
35.0 g HCl x 1 mol HCl = mol HCl
36.46 g HCl

29. Practice
How many moles are in 145 g of AgNO3?

30. Mass to particles Mass in g  moles  particles Two conversions
Ex. How many particles are in 35.0 g of AlCl3
1st Identify the molar mass of AlCl3.
Al = g/mol
Cl = g/mol x 3 = g/mol
AlCl3 = g/mol g/mol = g/mol

31. To convert from Mass to Particles
The given value goes in the upper left spot
35.6 g of AlCl3 x 1 mol AlCl3 = mol AlCl3
133.3 g AlCl3
0.267 mol AlCl3 x 6.02 x 1023 formula units = 1.61 x 1023 formula units AlCl3
1 mol AlCl3
OR 35.6 g of AlCl3 x 1 mol AlCl x 6.02 x 1023 formula units = 1.61 x g AlCl mol AlCl Formula Units
AlCl3

32. How many ions? 1.61 x 1023 formula units AlCl3
Identify the number of ions that make up the compound and multiply by that number.
1.61 x 1023 formula units AlCl3 x 3 Cl- Ions = 4.83 x1023 Cl- ions
1 AlCl3 formula unit
1.61 x 1023 formula units AlCl3 x 1 Al3+ Ions = 1.61 x1023 Al3+ ions

33. Mass of 1 Formula Unit/molecule
g/mol AlCl3 (molar mass) divide by Avogadro’s number to get the mass of 1 Formula unit
g/1 mol AlCl3 x 1 mol/6.02 x 1023 formula units
= 2.21 x g AlCl3/formula unit

34. 10.4 Empirical and Molecular Formulas
Percent composition
Synthetic chemists makes the new compounds
Analytical chemists identifies what its composition is
Percent composition from experimental data
Identify the percent by mass of a compound
A 100 g sample of XY contains 55 g of X and 45 g of Y.

35. Calculating percent by mass
Percent by mass (element) = (mass of element/mass of compound) x 100
Ex (55g X/ 100g XY) x 100 = 55% X
Ex (45g Y/ 100g XY) x 100 = 45% Y
Percent by mass from the chemical formula
Percent by mass = mass of element in 1 mol of compound x 100
molar mass of compound

36. Example Compound NaHCO3 Na = 22.99 g/mol H = 1.008 g/mol
C = g/mol
O = g/mol x 3 = g/mol
NaHCO3 = g/mol
Find the percent of Na in NaHCO3.
Percent Na = (22.99 g/mol /84.01 g/mol) x 100 = 27.37%

37. Practice
Find the % composition for H, C, and O in NaHCO3.

38. Empirical formula The opposite of % composition.
When you know the % you can determine the empirical formula
Empirical Formula: the formula with the smallest whole- number mole ratio of the elements
May or may not be the same as the molecular formula (aka the actual chemical formula)

39. Solving for Empirical Formula
How to solve for Empirical formula
1. Turn % into mass (g) by assuming you have 100% and therefore g
2. Convert all masses in g to mole using the molar mass of the compound.
3. Divide each mole value by the smallest mole value (this gives the molar ratio)
4. If decimals are present multiply the all mole values by the smallest number to equal all whole numbers
5 Write the empirical formula with symbols and numbers of atoms present

40. Example Pg. 345 Example Problem 10.11
Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: % C, 8.16% H, and 43.20% Oxygen

41. Practice empirical formula
# 58 pg. 346
The circle graph at the right gives the percent composition for a blue solid. What is the empirical formula for this solid?
63.16% O
36.84% N

42. Warm Up
Identify the % composition of H and O is H2O.

43. Molecular FOrmula
Empirical Formula = smallest whole number ratio. (many compounds have the same empirical formula
Molecular formula = actual formula
1. Begin with the steps to identify the empirical formula!
2. Identify the molar mass of the empirical formula by adding up all the molar masses
3. Compare that to the molar mass of the compound given. Divide the given molar mass/empirical formula molar mass
4. Distribute the value through your empirical formula to create the molecular formula

44. Example of Molecular formulaPg
Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% C, 5.08% H, and % O and has a molar mass of g/mol. Determine the empirical then the molecular formula.

45. Practice Molecular Formula
Pg. 350 # 62
A compound was found to contain g of C and g of H. The molar mass is g/mol. What is the empirical formula and the molecular formula?

46. Warm Up
Pg. 361 #164
Vitamin D3: Your body’s ability to absorb calcium is aided by vitamin D3. Chemical analysis yields the data that D3 is made of 84.31% C, % H, and 4.16% O. Find the Empirical and Molecular Formulas
Molar mass of D3 = 384 g/mol
Pg. 361 # 186
A g sample of a hydrate of magnesium iodide is heated until its mass is reduced to g and all water has been removed. What is the formula of the hydrate?

47. Chapter 10.5 Hydrates
What is a Hydrate? a compound that has a specific number of water molecules bound to its atoms

48. Naming Hydrates
The number of water molecules is written following a dot for example
Na2CO3 * 10H2O = sodium carbonate decahydrate
Na = Sodium
CO3 = Carbonate
10 = Deca
H2O = Hydrate
The first compound is named based on Ionic or covalent naming.

49. Analyzing a Hydrate
When a hydrate is heated it loses the water molecules and becomes the anhydrous form.
How do you determine the formula of a hydrate? (how many waters it has?)
Find the number of moles of water associated with 1 mole of the hydrate.
Dehydrate your sample
Calculate the water lost and convert to the value to moles
Calculate the mass of the anhydrate and convert to moles
Divide the moles of water by the moles of anhydrate

50. Example
Pg. 352
How many water molecules are in the hydrate BaCl2 * xH2O?
A 5.00g sample of BaCl2 * xH2O is dehydrated. The resulting mass is 4.26 g of anhydrous BaCl2.
Subtract: 5.00g – 4.26g = 0.74 g H2O lost.
Convert to moles:
4.26g x 1mol BaCl2/208.2g BaCl2 = mol BaCl2
0.740g x 1 mol H2O/ g H2O =
mol H2O
Divide the water by barium chloride
.0411mol H2O/ mol BaCl2 =2
So 2 moles of water to every 1 mole barium chloride = BaCl2 * 2H20

51. Practice
A mass of 2.50 g of blue, hydrated copper (II) sulfate is placed in a crucible and heated. After heating, 1.59g of white anhydrous copper sulfate remains. What is the formula for the hydrate? (How many waters?) Name the hydrate.