1. Formulas and the Mole

2. Mass of one atom of an element In a.m.u. Remember amu?
Atomic Mass
Mass of one atom of an element
In a.m.u.
Remember amu?
Mass of atom relative to C-12 (the standard)
1 amu = 1/12 mass of C-12 atom

3. Atomic Mass ionic covalent
The sum of the atomic masses of all of the elements present in the formula
For compounds (_____) it is called the FORMULA MASS
For molecules (_______) it is called the MOLECULAR MASS
ionic
covalent

4. Determining Formula Mass
Try Na2CO3
2 Na + 1 C + 3 O
2 (22.99) + 1 (12.01) + 3 (16.00) =
Formula unit
amu
Go to the 100’s place!

5. Practice
What is the molecular mass of oxygen (__)?
What is the molecular mass of CH3OH?
What is the formula mass of Ca(OH)2? O2
2 x = amu
1 C + 4 H + 1 O
1 (12.01) + 4 (1.01) + 1 (16.00) =
32.05 amu
PRACTICE
1 Ca + 2O + 2H
1 (40.08) + 2(16.00) + 2(1.01) =
74.10 amu

6. Gram-Atomic Mass (GAM)
Atomic mass of an element, expressed in grams
The GAM is 1 mole of atoms of the element
Remember the mole?
1 mole contains ____________ atoms
1 mole = GAM (from the Periodic Table)
1 mole = 6.02 x 1023 atoms
6.02 x 1023

7. Let’s try it
1 mole of Oxygen = ____ grams (from Periodic Table) = ________ atoms
1 mole of S = _____ grams (from Periodic Table) = ________ atoms
6.02 x 1023 atoms = __ mole of F = _____ grams of F
grams Cl = __ mole of Cl = ________ atoms of Cl
16. 00
6.02 x 1023
32.06
6.02 x 1023 1
19. 00 1
6.02 x 1023

8. Gram-Formula Mass (GFM)
Formula mass of a compound is expressed in grams
The GFM is the mass of 1 mole of formula units or molecules
GFM Na2CO3: g = 1 mole Na2CO3 formula units (amu)
GFM CH3OH: g = 1 mole CH3OH molecules

9. Molar Mass
The molar mass is the same as the gram-atomic mass (GAM) or gram-formula mass (GFM)
For example….
Number of items in a mole of a substance is always the same, the mass of that mole varies

10. Hydrogen Peroxide (H2O2) Molecule 34.02 amu 34.02 g
Substance
Type of Substance
Atomic or Formula Mass
Molar Mass
(GAM or GFM)
Calcium (Ca)
Atomic element
40.08 amu
40.08 g
Hydrogen (H2)
Diatomic element
2.02 amu
2.02 g
Hydrogen Peroxide (H2O2)
Molecule
34.02 amu
34.02 g
Calcium Chloride (CaCl2)
Ionic compound
amu g
PRACTICE

11. g The Mole Map # particles g  mole particles (atoms or molecules)
(atoms & molecules)
6.02 x 1023 atoms
1 mol
GAM
MOLE

12. Mole Conversions 4.0 moles F x 19.00 g F 76 = ___ g F 1 mole F
1. Find the mass of 4.0 moles of fluorine atoms.
4.0 moles F x
19.00 g F 76
= ___ g F
1 mole F

13. Mole Conversions 0.30 moles N2 x 28.02 g N2 8.4 = ___ g N2 1 mole N2
2. Calculate the mass of 0.30 moles of nitrogen (N2).
0.30 moles N2 x
28.02 g N2
8.4
= ___ g N2
1 mole N2
1 mol = 2(14.01) = g

14. Mole Conversions 35.0 g CaCO3 x 1 mol CaCO3 .350 = ___ mol CaCO3
3. Find the total number of moles represented by 35.0 g CaCO3 .
35.0 g CaCO3 x
1 mol CaCO3
.350
= ___ mol CaCO3
g CaCO3
1 Ca + 1 C + 3 O
40.08 g g + 3 (16.00 g) = g
1 mol

15. + Mole Conversions ( ) 1 mol 85.o g x .348 = _____ mol BaCl2*2H2O
How many moles are present in 85.o g of
BaCl2 * 2 H2O ? +
( )
1 mol
BaCl2 * 2H2O
85.o g x
BaCl2*2H2O
.348
= _____ mol BaCl2*2H2O g
BaCl2 * 2H2O
1 Ba + 2 Cl + 4 H + 2 O
(35.45) + 4(1.01) + 2(16.00) = g
1 mol

16. Mole Conversions 1 mol H2SO4 32 g H2SO4 x .3262642741 = ____________
5. Given 32 g of H2SO4, how many molecules are present in the sample?
1 mol H2SO4
32 g H2SO4 x
= ____________
mol H2SO4
98.08 g H2SO4
2 H+ 1 S + 4 O
2(1.01) (16.00) = g
1 mol
HOW MANY MOLECULES? ……

17. Question 5 Continues 2.0 x 1023 1 mole = 6.02 x 1023 molecules
5.How many molecules are present in .33 mol of H2SO4?
6.02 x 1023
molecules H2SO4
2.0 x 1023
.33 mol x
H2SO4
= ____________
molecules H2SO4
1 mol H2SO4
1 mole = 6.02 x 1023 molecules

18. Mole Conversions = _________molecules of Cl2 8.49 x 1022 1 mol Cl2
g  mole  atoms/molecules
Mole Conversions
6. A sample contains 10.0 g of chlorine (Cl2) molecules. How many molecules are present? X =
1 mol Cl2
10.0 g x
Cl2
6.02 x 1023 molec Cl2
70.90 g Cl2
1 mol Cl2
_________molecules of Cl2
8.49 x 1022
1 mole = g
1 mole = 6.02 x 1023 molecules

19. Mole Conversions = 3.7 x 1023 _________molecules of N2O4 57 g x
g  mole  atoms/molecules
Mole Conversions
7. A 57 g sample of dinitrogen tetroxide (N2O4) contains how many molecules? X =
6.02 x 1023 molec N2O4
57 g x
N2O4
1 mol N2O4
92.02 g N2O4
1 mol N2O4
3.7 x 1023
_________molecules of N2O4
1 mole = g
1 mole = 6.02 x 1023 molecules

20. Mole Conversions = 1.8 x 1024 _________atoms of O 48 g O x 1 mol O
g  mole  atoms/molecules
8. How many atoms does a sample of oxygen contain if its mass is 48 g? X =
48 g O x
1 mol O
6.02 x 1023 atoms O
16.00 g O
1 mol O
1.8 x 1024
_________atoms of O
1 mole = g
1 mole = 6.02 x 1023 molecules

21. Mole Conversions = 160 ____ g of SO3 1 mol SO3 1 mol SO3 1 S + 3 O
g  mole  atoms/molecules
Mole Conversions
9. There are 12.0 x 1023 molecules of sulfur trioxide (SO3) present in a sample. What is the mass of the sample? X =
1 mol SO3
80.07 g SO3
12.0 x 1023 molec x
SO3
6.02 x 1023 molec
SO3
1 mol SO3
160
____ g of SO3
1 S + 3 O
(16.00) = g
1 mole = g

22. Mole Conversions = 9.0 ____g of H2O 1 mol H2O 1 mol H2O 2 H + 1 O
g  mole  atoms/molecules
Mole Conversions
10.If a sample contains 3.0 x 1023 molecules of water, what is the mass? X =
1 mol H2O
18.02 g H2O
3.0 x 1023 molec x
H2O
6.02 x 1023 molec
H2O
1 mol H2O
PRACTICE!
____g of H2O
9.0
2 H + 1 O
2 (1.01) = g
1 mole = g

23. Percentage Composition
If 11 of 15 students in a class are wearing jeans, what percentage of the students are wearing jeans?
part
whole
X 100 = % or
part
whole = %
100 11
Percentage composition of a compound tells you the percent of the mass made up by each element in the compound
X 100
= 73 % 15

24. Practice
Example 1: What is the percent of iron in iron ore, Fe3O4?
First find the GFM (gram formula mass).
3 (55.85) + 4 (16.00) =
g of Fe3O4
3 (Fe) (O) g
% Fe:
X 100
= % g

25. multiply the parenthesis!
Practice
Example 2: What is the percentage composition of
iron(III)sulfate, Fe2(SO4)3?
2 (55.85) +
3(32.06) +
12 (16.00)
= g of Fe2(SO4)3
*Don’t forget to
multiply the parenthesis!
If they don’t tell you which part to find the percent composition, DO THEM ALL! g g
% Fe:
% O:
96.18 g
X 100
X 100
% S:
X 100 g g g
= %
= %
= 24.05%

26. multiply the parenthesis!
Practice
Example 3: What is the percentage of water in hydrated sodium carbonate; Na2CO3 * 10 H2O ? +
( )
2(22.99) +
1(12.01) +
3(16.00) +
20(1.01)+
10(16.00) Na C O H O
*Don’t forget to
multiply the parenthesis!
Total Na2CO H2O = g
Part that is H2O = g g
% H2O:
X 100
= % g

27. H2O 2 : 1 ratio! 1 Determining Formulas
Gives atom ratio AND mole ratio
2 Hydrogen atoms : 1 Oxygen atom
2 * 6.02 x 1023 : 1 * 6.02 x 1023 atoms
2 moles of H atoms : 1 mole O atoms
1 mole H2O1
2 : 1 ratio!

28. Practice Consider percentage is out of a 100 g.
A compound has the following percentage composition: Na = 21.60%; Cl = 33.33%; O = 45.07%. Determine the compound’s simplest (empirical) formula.
Consider percentage is out of a 100 g.
Trying to find the mole ratio…..

29. Must be very close to whole number!
Na = 21.60%; Cl = 33.33%; O = 45.07%.
Element g
mole
Mole Ratio Na Cl O
.9395 1
21.60 g
1 mole
.9395 mol X
22.99 g
.9395
.9402
33.33 g
.9402 mol 1
35.45 g
.9395
45.07 g
2.82 3
2.82 mol
16.00 g
.9395
Must be very close to whole number!
(within 0.05)
Na1Cl1O3 or NaClO3

30. Fe2O3 Fe 1 O 1.49 Practice 1.49 x 2 = 3. That works! 70.00 g 55.85 g
A certain compound was analyzed and found to have the following percentage composition: Fe = 70% and O = 30%. Find the simplest formula of the compound.
Element g
mole
Mole Ratio Fe
70.00 g
1 mole
1.2533 1
mol X
55.85 g
1.2533
1.875 O
30.00 g
1.875 mol
1.49
16.00 g
Because we can’t have a decimal, we multiply by the smallest
integer to get whole numbers. Let’s try 2….
1.49 x 2 = 3. That works!
Fe2O3
We have to now multiply everything by 2 so 1 x 2 = 2.
Therefore, our formula becomes:

31. ZnS Zn 1 1 S Practice 21.66 g 65.39 g 32.32 -21.66 10.66 g 1 mole
21.66 grams of zinc combine with sulfur to form g of a compound. What is its simplest formula?
Element g
mole
Mole Ratio Zn
1 mole
.3313
21.66 g
.3313 mol 1
65.39 g
.3313 X
32.32
-21.66
10.66 g
.3325 1 S
.3325mol
32.06 g
ZnS

32. Find the molecular formula
C1H2O1
C6H12O6 X6
1(12.01)
+ 2 (1.01)
1(16.00)
30.03 g
6(12.01)
+ 12(1.01)
6(16.00) g
180.18
30.03
= 6
Molecular formula is always whole # multiple of empirical formula.
What is the empirical formula?

33. Find the molecular formula
If C3H7 is the empirical formula for a compound whose molecular mass is 86 g, find the molecular formula of the compound.
Step 1: 3 C + 7 H
3 (12.01) + 7(1.01) = total g
Whole Number
Step 2: 86
43.10
= 1.99
= 2
C6H14
Step 3: 2 x C3H7 =

34. C2H5…..not done 1 2.5 2.5 x 2 = 5 That works! C H 82.9 g 12.01 g
Calculate the molecular formula for the compound containing 82.9% carbon and 17.1% hydrogen. It’s molecular mass is 58 g.
Element g
mole
Mole Ratio 1
1 mole
6.902 mol
6.902 mol C H
82.9 g
12.01 g
6.902
2.5
mol
mol
17.1 g
1.01 g
Because we can’t have a decimal, we multiply by the smallest
integer to get whole numbers. Let’s try 2….
2.5 x 2 = 5 That works!
C2H5…..not done
We have to now multiply everything by 2 so 1 x 2 = 2.
Therefore, our formula becomes:

35. C2H5 is the empirical formula. We want the molecular formula.
Remember that the molecular mass is 58 g.
Step 1: 2 C + 5 H
2 (12.01) + 5(1.01) = total g
Whole Number
Step 2: 58
29.07
= 1.99
= 2
Step 3: 2 x C2H5 =
C4H10

36. More Percentage Problems
part
whole
X 100 = % or = %
100

37. More practice 8.20 + 5.40 = 13.60 g total 8.20 Mg: X 100 = 60.3% 13.60
part
whole
X 100 = %
part
whole = %
100
An 8.20 g piece of magnesium combines completely with 5.40 grams of oxygen to form a compound. What is the percent composition of this compound?
= g total
8.20
Mg:
X 100
= 60.3%
13.60
5.40 O:
X 100
= 39.7%
13.60

38. = 39.7% = 60.3% X X More practice 8.20 + 5.40 = 13.60 g total Mg: =
part
whole
X 100 = %
part
whole = %
100
An 8.20 g piece of magnesium combines completely with 5.40 grams of oxygen to form a compound. What is the percent composition of this compound?
= g total
Mg: =
100 X
8.20 =
100 X
5.40 O:
13.60
13.60
(X) (13.60) = (8.20) (100)
(X) (13.60) = (5.40) (100)
13.60
13.6
13.60
13.60
= 39.7%
= 60.3%

39. Let’s try
part
whole
X 100 = %
part
whole = %
100
A strip of pure copper, mass of 6.36 grams, is heated with oxygen until it is completely converted to a compound of copper and oxygen with a mass of 7.95 grams. What is the percentage of oxygen in the compound?
1.59 % X
100 =
g = 1.59 g O
7.95
1.59
X 100 =
20.0%
(100)(1.59)
= (7.95)X
7.95
7.95
7.95
X= 20.0%

40. 30.0% O X 30.0 = 425 100 X = 128 g O More Practice
part
whole
X 100 = %
part
whole = %
100
70.0% of an iron oxide compound is iron. How many grams of oxygen could you extract from a 425 gram sample of the iron oxide?
30.0% O X
30.0
(X) (100) = (30.0) (425) =
425
100
100
100
X = 128 g O

41. More practice
part
whole = %
100
part
whole
X 100 = %
A copper sulfide compound contains 56.9% copper. What mass of copper sulfide contains 40.0 g of copper?
40.0
56.9 =
100
4000 = X X
56.9
(40.0) (100) = (x) (56.9)
X = 70.3 g
56.9
56.9