1. The Mole
Chapter 8.2

2. What is a mole? It is "that equal number" of atoms arbitrarily chosen.
the number of atoms in the atomic weight in g of any element.
the number of atoms in 16 g of oxygen, in 4 grams of He, in 32 g of sulfur, etc.

3. the number of molecules in the molecular weight, in grams, of any compound.
the number of molecules in 18 g of water, in 40 g of lithium carbide.
Much later, a mole, that equal number, was found to be x 1023.

4. What Can Be Expressed Using Moles?
A mole can be used to represent 6.02 x 1023:
-atoms
-molecules
-ions (But not Grams)
-compounds
-particles
-items (EX: students, homework assignments)

5. Test your understanding with the following questions:
1. How many bananas are in a mole of bananas?
2. How many kisses is a half mole of kisses? x ÷ 2 =
6.02 x 1023
3.01 x 1023

6. Calculating Molar Mass
Molar mass is the mass of 1 mole of a pure substance.
Molar mass may also be referred to as:
-gram formula mass (ionic compounds)
-formula mass (ionic compounds)
-molecular mass (non metals)

7. Calculating Molar Mass
The mass of one mole (6.02 x 1023) of atoms is based on the mass of one mole of Carbon-12, which is exactly 12.0 grams.
The molar masses of elements can then be obtained from their respective atomic masses (rounded to 2 places after the decimal).
Use the unit g/mol to denote molar mass.

8. Calculating Molar Mass
EX: Find the molar masses for the following:
Silver
Argon
Potassium
Oxygen
Hydrogen

9. Calculating Molar Mass
The molar mass of a molecule or formula can be determined by adding the molar masses of each of the elements present.
EX: H2O
Elements Moles of each Atomic mass
H x = 2.02
O x =16.00
Molar mass of H20  g/mol

10. Calculate molar mass of the following:
Al2S3
Al (2 x )
S (3 x )
g/mol
Add hydrate example: CuCl2 · 2H2O

11. Mole Conversions
The key to converting between any units is to label numeric values with appropriate units and work with these until the only one that will not cancel is the one you want.

12. Converting MolesMolecules
Use Avogadro’s constant (6.02 x 1023) when converting between moles and molecules.
EX: How many atoms are in 3.4 moles of iron atoms?
6.02 x 1023 small unit = mole
molec, atoms, ions, particles,
formula units, ionic compounds

13. Calculation 1. Start with known value divided by 1.
2. Then line up units so that only the desired units are not cancelled.
3. Divide the products of the numerators by the products of the denominators.
3.4 moles x 1023 atoms = 2.1 x 1024
1 1 mole atoms

14. Practice 2.54 x 1021 molecules of CO2 is equal to ___________ moles.
0.010 moles of HCl is equal to ____________ molecules of HCl.
2 moles of M & Ms is equal to __________ pieces.
4.22 x 10-3 mol
6.0 x 1021 molecules
1 x 1024 pieces

15. Converting Moles  Mass
Molar mass is the conversion bridge needed to convert between mass and mole of any substance.
EX: g/mol = g = 1 mole
1 mole g

16. Mole Conversions What is the mass in grams of 2.5 mol of O2?
2.5 mol (32.00 g) =
( 1 mol )
80. g of O2
*Divide notepacket area into 2 sections for 2nd problem.
Determine the number of moles in 5.00 g of H2O.
5.00 g ( 1 mol ) =
(18.02 g)
0.277 mol of H2O

17. Convert Molec  Mass
1 mole of any substance is equivalent to 1 mole of another substance. 1 mole = 6.02 x 1023 particles.
Molar masses of substances are not equivalent. EX: NaCl = 58.44g/mol ; NaOH = g/mol
There is no direct conversion between molecules and grams.
So… Molec MolesMass

18. Mole Conversions
Determine the number of molecules there are in a 5.45 g sample of CaCl2.
5.45 g CaCl2 x ( 1 mol ) ( 6.02 x 1023 molecules)
( g ) ( 1 mol )
2.96 x 1022 molecules of CaCl2

19. Percent Composition
Percent composition of a compound is a statement of the relative mass each element contributes to the mass of the compound as a whole.
Chemists often compare the percent compositions of unknown compounds to those of known compounds to identify the unknown.

20. Calculating % Composition
Find the % composition of the elements in NaCl.
Step 1: calculate the molar mass of the substance.
Step 2: divide the molar mass of each of the
elements by the molar mass of the substance.
Step 3: multiply by 100 to get into percent.

21. Molar mass of NaCl is
58.44 g/mol
Na = g/mol x =
58.44g/mol
Cl = g/mol x =
39.34%
60.66%

22. Na = %
Cl = % %
Each formula unit of table salt is about 40 % sodium cation and 60 % chlorine anion by mass.
Refer to page 206 in text for additional info.

23. Empirical Formulas
Remember that the elements in a compound combine in whole number ratios such as 1:1, 1:2, 2:3, and so forth.
If elements combine in these whole number ratios, we can predict the same applies for moles of each atom.
We can use this principle to find the empirical (lowest whole number) formula for compounds based on the relative masses of each of the elements in the compound.

24. Calculating Empirical Formulas
What is the empirical formula for a compound if a 2.50 gram sample contains g of calcium and 1.60 g of chlorine?
Step 1: convert grams of each of the elements into moles.
Step 2: obtain the simplest ratio by dividing the moles by the smallest number of moles.
Step 3: round or multiply to express the ratio using whole numbers.

25. 0.900 g Ca 1 mol
g =
mol
1.60 g Cl mol
g =
mol
The smallest mole is so we divide each by this number.

26. Ca = moles
moles = 1
Cl = moles
moles =
(round to 2)

27. Based on our calculations, there is 1 mole of calcium for every 2 moles of chlorine.
The formula is then:
CaCl2

28. Practice Problem
What is the empirical formula of a compound that is 66.0% Ca and 34.0% P?
*When given only the percentages of each element, assume that there is 100 g of the compound.

29. Sometimes dividing by the smallest number of moles will not lead to a whole number.
EX:
66.0 g Ca 1 mole
g = mol
34.0 g P 1 mole
g = mol

30. Ca = mole
1.10 mole =
P = mole
1.10 mole = 1
1.5 can not be rounded to the next highest number.
We must multiply each by 2 in order to turn 1.5 into a whole number.

31. Ca = x 2 = 3
P = x 2 = 2
The empirical formula for this compound is:
Ca3P2

32. Hints for rounding or multiplying until a whole number is achieved:
If <0.2, or >.8 then round to whole number.
If 1.5, then multiply each substance in the problem by 2 to get 3. 
If 1.25, then multiply each substance in the problem by 4 to get 5.
If 1.33, then multiply by 3 to get 4.

33. Empirical Formula Poetry
% to Mass
Mass to Mole
Divide by Small
Multiply til Whole

34. Molecular Formulas Molecular formulas are true formulas for compounds.
Steps: 1. Determine empirical formula.
2. Determine molar mass of empirical formula.
3. Divide molecular mass that was given by the empirical molecular mass.
4. Distribute whole number as subscript to empirical formula.

35. molecular mass = n
empirical formula mass
xnyn
Find the molecular formula for CH2 if the molecular mass is g/mol.
42.09 g/mol = 3 = C3H6
14.03 g/mol

36. Practice:
1. Vitamin C (ascorbic acid) contains % C, 4.58 % H, and % O, by mass. The experimentally determined molecular mass is 176 g/mol. What is the empirical and molecular formula for ascorbic acid?
2. NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.  Calculate the empirical formula of NutraSweet and find the molecular formula.  (The molar mass of NutraSweet is g/mol)