1. The Mole Chapter 11 Chemistry RiverDell High School Ms. C. Militano
What is a mole in chemistry?
What conversion factors are associated with the mole?
Types of conversions involving mole equalities

2. I. What is a Mole? SI base unit that measures amount of a substance
B. 1 mol = Avogadro Number of particles (particles can be atoms, molecules or ions) x 1023 is Avogadro's Number
Molar Mass – mass of one mole of atoms
of an element
1. ex. C = 12.0amu N = 14.0 amu
D. Mole Equalities
- 1 mole = molar mass
- 1 mole = x 1023 particles

3. II. Mole Conversions [mass-mole-atoms] A. Type of Problem Equality
1. MOLES  MASS
2. MASS  MOLES mole= molar mass (g)
3. MOLES  ATOMS
4. ATOMS  MOLES 1 mole = 6.02 X 1023 atoms
5. MASS  ATOMS mole = 6.02 x 1023 atoms
6. ATOMS MASS mole = molar mass (g)

4. Making Conversions – use quantities in the Atom and Mass box as conversion factors
6.02x molar mass
ATOMS MOLE MASS(g)
1 mole mole

5. B. Using Factor Label to Make Mole Conversions
1 mole  molar mass
6.02 x mole
PARTICLES <----> MOLES <----> MASS
6.02 x  mole
1 mole molar mass

6. C. Solving Mole Problems
EXAMPLES
1.00 mole of He = 4.00 g.
2.00 mole of He = _____g
2.00 mol He X 4.00g He =
mole He
8.00 g He

7. 2.00 mole He = ________atoms He
EXAMPLES
1.00 mole He = X 1023 atoms
2.00 mole He = ________atoms He
2.00 mole He x x 1023 atoms He = x 1023
mole He
1.20 x 1024 atoms He
16.00g He = _____ moles He
16.0 g He x 1 mole He =
g He g He

8. 3.01 X 1023 atoms He = _____ moles 8.00g He =______atoms He EXAMPLES
3.01 x 1023 atoms He x mole He =
x 1023 atoms He
.500 mol He
8.00g He =______atoms He
8.00 g He x 1 mole He x x 1023 atoms He =
g He mole He
12.04 x 1023 atoms He = 1.20 x 1024 atoms He

9. Sample Problems – More Practice
Moles to mass.
Find the mass of 3.50 moles of carbon.
Mass to Moles
How many moles of carbon are contained in
60.0 g of carbon?
Moles to Atoms
How many atoms of carbon are found in 4.00 moles of carbon?

10. Sample Problems More Practice
Atoms to Moles
How many moles of carbon are represented by
1.806 x 1024 atoms of carbon?
Mass to Atoms
How many carbon atoms are found in 36.0g of
carbon?
Atoms to Mass
What is the mass of x 1024 atoms of carbon?

11. Answers to Sample Problems
g C
mol C
x 1023 = 2.41 x 1024 atoms C
x 10-1 = mol C
x = x 1024 atoms C
g C

12. More Sample Problems 2.00 moles of Cu = atoms of Cu
60.0 grams of C = moles of C
3.00 x 1023 atoms He = moles of He
2.50 moles Al = grams of Al
28.0 grams N = atoms of N
1.80 x 1023 atoms Mg = grams of Mg

13. Answers to More Sample Problems
12.04 x 1023 = 1.20 x 1024 atoms Cu
5.00 mol C
.498 mol He or x 10-1 mol He
67.5 g Al
12.04 x = x 1024 atoms N
7.27 g Mg

14. Compounds and Diatomic Molecules Convert 4.00 mol NaOH to grams.
a. Na = O = H = 1.00
formula mass = 40.0
b mol NaOH x 40.0g NaOH = 160.g NaOH
mol NaOH

15. 2. How many molecules are found in 2.00 mol of H2SO4? (sulfuric acid)
2.00 mol H2SO4 x x 1023 molecules H2SO4
mol H2SO4
12.04 x 1023 =
1.20 x molecules H2SO2

16. How many moles are found in 64.0 g of oxygen gas (O2)?
64.0g O2 x 1mol O2 = mol O2
g O2
4. How many formula units are found in 117.0g of sodium chloride (NaCl)?
117.0g NaCl x 1 mol NaCl x 6.02x1023 units NaCl
g NaCl mol NaCl
12.04 x 1023 = x 1024 formula units NaCl

17. III. Percent Composition
Procedure
1. Determine total mass for each element
2. Determine the molar mass (formula
mass)
3. Divide mass of each element by the
molar mass (formula mass)
4. Multiply by 100%

18. B. Problem Solving
Determine the percent composition of each element in carbon dioxide (CO2)
C = O = 2 x 16.0 = 32.0
Sum = (formula mass)
Carbon 12.0/ = .273 = %
Oxygen 32.0/44.0 = %

19. B. Problem Solving
Determine the percent composition of each element in calcium hydroxide Ca(OH)2
Ca = 40.1
O = 16.0 x 2 = 32.0
H = x 2 = sum is 74.1(molar mass)
% Ca = 40.1/74.1 = 54.1%
% O = 32.0/74.1 = 43.2%
% H = /74.1 = %

20. B. Problem Solving
Determine the precent composition of each element in TNT (trinotrotoluene) C7H5(NO2)3
7C = 7(12.0) = 84.0
5H = 5(1.00) =
3N = 3(14.0) = 42.0
6O = 6(16.0) = sum = 227 (molar mass)
% C = /227 = %
% H = /227 = %
% N = /227 = %
% O = /227 = %

21. IV. Hydrates
Definitions
1. hydrate – compound that has a specific number of water molecules in its crystal (solid state)
2. water of hydration – water molecules that are part of the crystal (solid state)
3. anhydride – compound without water of
hydration

22. B. Naming Hydrates 1. CaCl2  2H2O calcium chloride dihydrate
2. NaC2H3O2  3H2O sodium acetate trihydrate
3. CuSO4  5H2O copper sulfate pentahydrate
4. MgSO4  7H2O magnesium sulfate heptahydrate
5. Na2CO3  10H2O sodium carbonate decahydrate

23. C. Heating a Hydrate ∆
Hydrate  anhydride + water
Na2CO3  10H2O  Na2CO H2O
MgSO4  7H2O  MgSO H2O
Na2CO3  10H2O  Na2CO H2O

24. D. Problem Solving - % Water in Hydrates
Procedure
a. determine formula mass of the compound
b. divide mass of only water molecules
by the formula mass of the compound
c. multiply answer by 100%

25. Examples – Find % water in the hydrate 2a. CaCl2  2 H2O Ca = 40.1
Total mass is 147.1
Mass of water 2(18.0) = 36.0
% H2O mass water = = =
mass of compound %

26. 2b. Find the % water in the hydrate
MgSO4 7H2O
Mg = 24.3
S =
O(4) = 64.0
H2O(7) = 18(7) = sum = 246.4
% water = 126/246.4 = 51.1%

27. V. Empirical Formulas A. lowest whole number ratio of subscripts
B. How to Determine the Empirical Formula
1. if given % composition write % as a
number of grams without the % sign
2. divide # grams by molar mass to get the
number of moles
3. Divide # moles for each element by smallest
4. Round to nearest whole number when possible
5. Multiply by 2,3,or 4 to get whole numbers-
if necessary
6. Use resulting numbers as subscripts in the
formula

28. C. Sample Problems Formula is K4O
1. Find the empirical formula of a compound
containing g of potassium (39.08) and
2.00g of oxygen (16.00).
Determine # moles of each element
# moles Oxygen – / = .125
# moles Potassium / = .500
Divide # moles of each by the smallest
.125/.125 = /.125 = 4.00
Formula is K4O

29. Find the empirical formula of a compound
containing 5.41g Fe (55.85), 4.64g Ti (47.88), and
4.65g O (16.00).
# moles Iron(Fe) /55.85 =
# moles Titanium(Ti) / = .0969
# moles Oxygen(O) /16.00 =
Fe = /.0968 = Ti = .0969/.0968 = 1.00
O = / = 3.01
Formula is FeTiO3

30. 3. Find the empirical formula of methyl acetate which contains 48
3. Find the empirical formula of methyl acetate which contains 48.64% C, 8.16% H, 43.20% O.
# moles C = /12.00 = 4.053
# moles H = / = 8.16
# moles O = /16.00 = 2.700
Oxygen – /2.700 = x 2 = 2.000
Hydrogen – 8.16/ = x 2 = 6.04
Carbon / = x 2 = 3.000
Formula is C3H6O2

31. VI. Determining Molecular Formulas
Procedure
1. Determine empirical formula
2. Calculate empirical formula mass
3. Divide actual formula mass to get “X”
empirical formula mass
4. Multiply subscripts in the empirical
formula by “X”

32. B. Determine Molecular Formula
1.Determine the molecular formula of succinic acid.
(Molar mass is 118.1g) C = 40.68% H = 5.05%, O =54.24%
Carbon = /12.0 =
Hydrogen = 5.08/1.00 = # of moles
Oxygen = /16.0 = 3.39
C = 3.39/3.39 = 1.0
O = 3.39/3.39 = mole ratio
H = 5.08 / 3.39 = 1.49
Multiply by 2 to get the smallest whole number ratio

33. Empirical formula = C2H3O2 Empirical formula mass =
Calculate empirical formula mass
Empirical formula mass =
2(12.00) + 3(1.00) + 2(16.00) = g
Divide formula mass by empirical formula mass
118.1/59.0 = = “X”
Multiply each subscript in the empirical formula by X
The molecular formula is C4H6O4

34. 2. Determine the molecular formula for styrene.
C = 92.25%, H = 7.75%. (Molar mass – g)
92.25/ 12.0 = 7.69 mol C /1.00 = 7.75 mol H
7.69/7.69 = /7.69 = 1.01
Empirical formula is CH
Empirical formula mass is = 13.0
104.00/13.0 = 8 ( “X”)
Formula is C8H8

35. 3. Determine the molecular formula for ibuprofen.
C=75.7%, H=8.80%, O=15.5% Molar Mass g
Determine the # of moles
75.7/12.0 = 6.31 mol C / 1.00 = 8.80 mol H
15.5/16.0 = .969 mol O
Divide by the smallest number of moles
.969/.969 = /.969 = /.969 = 9.08
Multiply by 2 to get whole number ratio
Empirical Formula is C13H18O
Empirical Formula Mass 13(12.0)+18(1.00)+2(16.0) = 206.0
Divide formula mass by empirical formula mass to get X
206/206 = (“X”)
Molecular Formula is C13H18O2

36. Given the molecular formula, determine the empirical formula for the following.
a. C6H6 (benzene)
b. C2H6 (ethane)
c. C10H8 (naphthalene)
d. C8H10N4O2 (caffeine)
e. C14H18N2O5 (aspartame)
Answers a) CH b) CH3 c) C5H4
d) C4H5N2O e) C14H18N2O5

37. 5. Determine the molecular formula if the empirical
formula is CH and the molar mass is 78.00g.
Empirical formula mass is = 13.O
78.00/13.0 = (“X”)
The molecular formula is C6H6
Determine the molecular formula for butane if
the empirical formula is C2H5 and the molar mass
is 58.00g.
2(12.0) + 5(1.00) = 29.0 (empirical formula mass)
58.0/29.0 = (“X”)
The molecular formula is C4H10