1. The Mole
Molar Conversions

2. A. What is the Mole? A counting unit
Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000
6.02 X 1023 (in scientific notation)
This number is named in honor of
Amedeo Avogadro (1776 – 1856),

3. A. What is the Mole? HOW LARGE IS IT???
1 mole of hockey pucks would equal the mass of the moon!
1 mole of basketballs would fill a bag the size of the earth!
1 mole of pennies would cover the Earth 1/4 mile deep!

4. How does a Mole relate? 6.02 x 1023 cookies 6.02 x 1023 cars
1 Mol = 6.02 x 1023 particles
If I had 1 mol of cookies, how many cookies would I have?
6.02 x 1023 cookies
If I had 1 mol of cars, how many cars would I have?
6.02 x 1023 cars
If I had 1 mol of Aluminum, how many atoms I have?
6.02 x 1023 atoms of Al
Note that the NUMBER is always the same, but the MASS is very different!

5. A Mole of Anything Contains 6.02 x 1023 particles
= x 1023 C atoms
= x 1023 H2O molecules
= x 1023 NaCl
“formula units”
6.02 x 1023 Na+ ions and
6.02 x 1023 Cl– ions
1 mole C
1 mole H2O
1 mole NaCl

6. Learning Check 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms
b) x 1023 Al atoms
c) x 1023 Al atoms
2.Number of moles of S in 1.8 x 1024 S atoms
a) mole S atoms
b) mole S atoms
c) x 1048 mole S atoms

7. B. Molar Mass
Describes the mass (how heavy) 1 mole of an element or compound is.
Atomic mass tells the...
molar mass (g/mol)
the mass of one mole of an element or atom
formula mass (g/mol)
the mass of one mole of an ionic compound
molecular mass (g/mol)
the mass of one mole of a covalent compound

8. B. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol

9. Learning Check!
Find the molar mass – Round to the hundredths position. It is important for significant digits!
= g/mol
1 mole of Br atoms
1 mole of Sn atoms
= g/mol

10. B. Molar Mass Examples water H2O 2(1.01) + 16.00 = 18.02 g/mol
sodium chloride
H2O
2(1.01) = g/mol
NaCl
= g/mol

11. B. Molar Mass Examples sodium bicarbonate NaHCO3
sucrose
NaHCO3
(16.00) = g/mol
C12H22O11
12(12.01) + 22(1.01) + 11(16.00) = g/mol

12. B. Molar Mass of antacid Al(OH)3 = ?
Learning Check!
Molar Mass of K2O = ? g/mol
B. Molar Mass of antacid Al(OH)3 = ?

13. C. Molar Conversions molar mass 6.02  1023 MOLES (g/mol)IN
GRAMS
MOLES
NUMBER OF
PARTICLES
(g/mol)
(particles/mol)

14. C. Molar Conversion Examples
How many moles of carbon are in 26 g of carbon?
26 g C
1 mol C
12.01 g C
= 2.2 mol C

15. C. Molar Conversion Examples
How many moles of oxygen are in 85 g of Oxygen gas?
85 g O2
1 mol O2
32.00 g O2
= 2.6 mol O2

16. C. Reverse!
How many grams of nitrogen are in 0.25 moles of nitrogen gas?
0.25 mol N2
28.02 g N2
1 mol N2
= 7.0 grams N2

17. C. Reverse! How many grams of water are in 3.45 moles of water?
3.45 mol H2O
18.02 g H2O 1 mol H2O
= 62.2 g H2O

18. C. Moles to Molecules Ex
How many molecules are in 2.50 moles of C12H22O11?
6.02  1023
molecules
1 mol
2.50 mol
= 1.50  1024
molecules
C12H22O11

19. C. Molecules to Moles Ex How many moles are in
1.505 x 1023 molecules of HCl?
1.505 1023
molecules
1 mol
= .25 Moles HCl
6.02  1023
molecules

20. Learning Check!
The artificial sweetener aspartame (Nutra-Sweet) formula C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

21. Percent Composition

22. When you are asked to determine the percentage composition of a compound, it should be understood that this refers to the percentage by mass. 
In other words, the percentage composition of water shows what % (by mass) is Hydrogen, and what % is Oxygen.
Percentage composition by mass does not tell us the ratio by which the atoms combine. (Does not tell us the subscripts)

23. Example 1
Homer has a dozen assorted doughnuts.  3 of these doughnuts are chocolate. 
What percentage of Homer's doughnuts are chocolate? (note, we are talking about total number, not mass, in this case.)

24. Example 2
Jim goes to a school that has 258 total students.  If 164 of the students are boys, what percentage of the student body is made up of girls?

25. In both cases, you found the percent composition by dividing one part by the total and multiplying by 100.
               part
% composition = x 100        whole    

26. Finding the percentage composition by mass works the same way
Finding the percentage composition by mass works the same way.  You divide the mass that one element contributes to the compound by the mass of the entire compound.                                                                                                    mass from element
% Composition = x 100                          total mass of the compound

27. Example 3
What percentage of the mass of carbon dioxide (CO2) is made up by the carbon?
Mass of Carbon = x 1 = 12.01
Mass of Oxygen = x 2 = 32.00

28. Continued part % comp. of Carbon = --------------- x 100 whole 12.01
44.01
Answer = 27.29%

29. Example 4 What is the percentage composition of glucose (C6H12O6) ?
Start by determining the molar mass!
C = x 6 = 72.06
H = x 12 = 12.12
O = x 6 = 96.00
180.18

30. Remember!!! part % comp. of Element = --------------- x 100 whole
% for Carbon = x 100
                        
40.00% C

31. 12.12 % for Hydrogen = ------------- x 100
% for Oxygen = x 100
                         % O

32. Check your Answer!
One way to check your answer is to make sure that all of the percentages add up to approximately 100%.  (i.e. 40.0% + 6.7% % = 100%)  
Your total may be off by a few tenths of a percent, due to rounding.

33. Empirical and Molecular Formulas

34. Empirical Formula What are we talking about???
Empirical Formula represents the smallest ratio of atoms in a formula.
In other words it represents the simplest chemical formula for a particular compound.

35. Steps to Solve!
You need to follow these basic steps to determine your answer.
1) Make sure you are starting in grams
2) Convert all grams to moles (Molar Mass)
3) Divide each by the smallest number of moles.
4) Create whole number ratio of subscripts
5) Multiply if necessary

36. Example 1
In an unknown molecule, there are 4.15 grams of Carbon and 1.38 grams of Hydrogen. Determine the empirical formula for the substance!
Step 1) Make sure you are starting in grams
Step 2) Convert each mass in grams to moles.

38. Now What?
Step 3) Divide each by the smallest number of moles. Set up the Ratio of elements

39. Solution!
The 1 and the 4 represent the whole number ratio of the elements in the chemical formula.
The 1 represents that there is only one Carbon atom
The 4 represents that there are four Hydrogen atoms
The Solution: The empirical formula is CH4

40. Phenol, a general disinfectant, is 76. 57 % Carbon, 6
Phenol, a general disinfectant, is % Carbon, 6.43 % Hydrogen, and % Oxygen. Determine it’s empirical formula.
Step 1) Make sure you are starting in grams…….
How can we go from percent to grams?
Easy! Assume we have a 100 gram sample so there would be how much of each element?

41. Step 2) Convert each mass in grams to moles.

43. Try this!
Element “A” is 78.1% abundant and has a molar mass of g/mol. Element “B” is 21.9% abundant and has a molar mass of 1.01 g/mol. Determine the empirical formula.
78.1g A X 1mol/10.81g = 7.22 mol B
21.9g B X 1mol/1.01g = mol H
7.22 mol ÷ 7.22mol =1
21.7 mol ÷ 7.22mol = 3.01
Ratio of 1:3
AB3

44. Molecular Formula
The molecular formula is related, but different to the empirical formula.
Remember that Empirical Formula represents the lowest ratio of the atoms in a compound.
The Molecular Formula is the actual, or true ratio of the elements in the compound.

45. Needs and Steps!
1. In order to solve you need the molar mass of the molecular formula for a compound in g/mol.
2. Determine the empirical formula.
3. Calculate the molar mass of the empirical formula
4. Divide the molar mass of the molecular formula by the molar mass of the empirical formula. Apply ratio to all subscripts.

46. Example
The empirical formula for hydroquinone, a chemical used in photography, is C3H3O. The molecular weight of the compound Is 110 g/mol. Determine the molecular formula.
Step 1) Determine the molecular weight of the compound: Given at 110 g/mol.

47. Step 3) Determine the molar mass of the empirical formula.
Step 2) Determine the empirical formula: Given C3H3O.
Step 3) Determine the molar mass of the empirical formula.

48. Think…I Know it’s hard…
Step 4) Think about the molar mass of the molecular formula compared to the molar mass of the empirical formula.
Mol. Formula Emp. Formula
110 g/mol g/mol
Divide and get the ratio
110 g/mol / g/mol = 2

49. So….
That 2 represents what you will multiply all the subscripts by to determine the correct molecular formula.
C3H3O x 2 = C6H6O2
Molecular Formula

50. Try this
The empirical formula for a compound is C3H7. If the molecular weight is 86 g/mol, then what is the molecular formula?
1. The molecular weight is given:
86 g/mol
2. The empirical formula is given:
- C3H7

51. 3. Calculate the molar mass for the empirical formula.

52. Compare the two…. The Molecular weight was 86 g/mol
The Empirical weight was 43.1 g/mol
Divide Molecular by Empirical to determine the ratio!
86 g/mol / 43.1 g/mol = 2

53. So 2 is the ratio, multiply all the subscripts by 2
C3H7 x 2 = C6H14 is the Mol. Formula

54. Practice!
A) Empirical Formula is S
Molecular weight is 256 grams per mol
B) Empirical Formula is NO2
Molecular Weight is 46 g/mol