Chapter 6
Chemical
Composition
Why Is Knowledge of
Chemical Composition Important?
• Everything in nature is either chemically or
physically combined with other substances.
• To know the amount of a material in a sample,
you need to know what fraction of the sample
it is.
• Some Applications:
 The amount of sodium in sodium chloride for diet.
 The amount of sugar in a can of pop.
 The amount of hydrogen in water for hydrogen
fuel.
 The amount of chlorine in freon to estimate ozone
depletion.
2
Counting Nails by the Pound
• People usually buy thousands
of nails for construction
• Hardware stores sell nails in
the pound.
• How do I know how many
nails I am buying when I buy a
pound of nails?
3
Counting Nails by the Pound,
Continued
A hardware store customer buys 2.60 pounds of
nails. A dozen nails has a mass of 0.150 pounds.
How many nails did the customer buy?
1 dozen nails = 0.150 lbs.
12 nails = 1 dozen nails
Solution map:
4
Counting Nails by the Pound,
Continued
1 doz. nails 12 nails
2.60 lbs. 

 208 nails
0.150 lbs.
1 doz.
• The customer bought 2.60 lbs of nails and
received 208 nails. He counted the nails by
weighing them!
5
Counting Nails by the Pound,
Continued
• What if he bought a different size nail?
Would the mass of a dozen be 0.150 lbs?
Would there still be 12 nails in a dozen?
Would there be 208 nails in 2.60 lbs?
How would this effect the conversion factors?
6
Counting Atoms by Moles
• Just as we can use the dozen to count nails, we
can use the mole to count atoms in a sample.
• 1 mole = 6.022 x 1023 things.
1 mole of atoms = 6.022 x 1023 atoms
1 mole of apple = 6.022 x 1023 apples
Note that just like the “dozen” the mole is a constant
number and this number is called the Avogadro’s
number.
7
Using Moles as Conversion Factors
• Since 1 mole atoms = 6.022 x1023 atoms
6.022  10 atoms
1 mole
23
OR
1 mole
6.022  10 23 atoms
8
Example 6.1—A silver ring contains 1.1 x 1022 silver atoms.
How many moles of silver are in the ring?
1.1 x 1022 atoms Ag
moles Ag
Given:
Find:
Solution Map:
atoms Ag
mol Ag
1 mol
6.022  1023 atoms
1 mol = 6.022 x 1023 atoms
Relationships:
Solution:
1.1 10
22
1 mol
atoms Ag 
23
6.022  10 atoms
 1.8266  10- 2 mol Ag  1.8  10- 2 mol Ag
Check:
Since the number of atoms given is less than
Avogadro’s number, the answer makes sense.
Tro's "Introductory Chemistry",
Chapter 6
9
Example 6.1a—Calculate the number of
atoms in 2.45 mol of copper.
Given:
Find:
Solution Map:
2.45 mol Cu
atoms Cu
mol Cu
atoms Cu
6.022 10 23 atoms
1 mol
Relationships:
Solution:
1 mol = 6.022 x 1023 atoms
6.022  10 23 atoms
2.45 mol Cu 
1 mol
 1.48  10 24 atoms Cu
Tro's "Introductory Chemistry",
Chapter 6
10
Practice—Calculate the number of atoms in
1.45 mol of Sodium.
Answer: 8.73 x 1023 Na atoms
11
Relationship Between
Moles and Mass
• The mass of one mole of a substance is called the
molar mass.
• The molar mass of an element, in grams, is
numerically equal to the element’s atomic mass, in
amu
 E.g., 1 Carbon atom weighs 12.00 amu, therefore the
molar mass of Carbon is 12.00 g
• The lighter the atom, the less a mole weighs.
 Carbon (C) is lighter than Sodium (Na); 1 mol C
weighs 12.00 g and 1 mol Na weighs 22.99 g.
12
Mole and Mass Relationships
Substance Pieces in 1 mole
Weight of 1 mole
Hydrogen
6.022 x 1023 atoms
1.008 g
Carbon
6.022 x 1023 atoms
12.01 g
Oxygen
6.022 x 1023 atoms
16.00 g
Sulfur
6.022 x 1023 atoms
32.06 g
Calcium
6.022 x 1023 atoms
40.08 g
Chlorine
6.022 x 1023 atoms
35.45 g
Copper
6.022 x 1023 atoms
63.55 g
1 mole
sulfur
32.06 g
1 mole
carbon
12.01 g
13
Example 6.2a—Calculate the moles of sulfur
in 57.8 g of sulfur.
Given:
Find:
Solution Map:
57.8 g S
mol S
gS
1 mol S
32.07 g
mol S
Relationships: 1 mol S = 32.07 g
Solution:
1 mol
57.8 g S 
32.07 g
 1.80 mol S
14
Example 6.2b—Calculate the moles of carbon
in 0.0265 g of pencil lead.
Given:
Find:
Solution Map:
0.0265 g C
mol C
gC
1 mol
12.01 g
mol C
Relationships: 1 mol C = 12.01 g
Solution:
1 mol
0.0265 g C 
12.01 g
 2.21 10-3 mol C
15
Example 6.3—How Many Aluminum Atoms
Are in a Can Weighing 16.2 g?
Given:
Find:
Solution Map:
16.2 g Al
atoms Al
g Al
mol Al
1 mol
26.98 g
Relationships:
Solution:
atoms Al
6.022 10 23 atoms
1 mol
1 mol Al = 26.98 g, 1 mol = 6.022 x 1023
1 mol Al 6.022  10 23 atoms
16.2 g Al 

26.98 g Al
1 mol
 3.62  10 23 atoms Al
16
Practice
1) Calculate the grams of carbon in
2.21 x 10-3 mol of lead pencil.
2) How many copper atoms are in a
penny weighing 3.10 g?
Tro's "Introductory Chemistry",
Chapter 6
17
Practice
1) Calculate the grams of carbon in 2.21 x 10-3
mol of lead pencil.
Answer: 0.0265 g Carbon
18
Practice
2) How many copper atoms are in a penny
weighing 3.10 g?
Answer: 2.94 x 1022 Copper atoms
19
Molar Mass of Compounds
• The molar mass of a compound can be calculated
by adding up the molar mass of all the atoms that
the compound is compose of.
Example: Calculate the molar mass of water (H2O)
• Since 1 mole of H2O contains 2 moles of H and 1
mole of O.
Molar mass = 2(1.01 g H) + 16.00 g O = 18.02 g.
20
Example 6.4—Calculate the mass of
1.75 mol of H2O.
Given:
Find:
Solution Map:
1.75 mol H2O
g H2O
mol H2O
18.02 g
1 mol H 2 O
g H2O
Relationships: 1 mol H2O = 18.02 g
Solution:
18.02 g
1.75 mol H 2 O 
1 mol
 31.535 g  31.5 g H 2 O
Check: Since the given amount is more than 1 mol, the mass
being > 18 g makes sense.
21
Example 6.4a—How many moles are in 50.0 g of
PbO2? (Pb = 207.2, O = 16.00)
Given:
Find:
Solution Map:
50.0 g mol PbO2
moles PbO2
g PbO2
1 mol PbO 2
239.2 g
mol PbO2
Relationships: 1 mol PbO2 = 239.2 g
Solution:
Check:
1 mol
50.0 g PbO 2 
239.2 g
 0.20903 mol  0.209 mol PbO 2
Since the given amount is less than 239.2 g, the
moles being < 1 makes sense.
22
Practice—What is the mass of 4.78 x 1024 NO2
molecules?
Answer: 365 g
23
Example 6.5—How many formula units are in
50.0 g of PbO2? (PbO2 = 239.2)
Given:
Find:
Solution Map:
50.0 g PbO2
formula units PbO2
g PbO2
units PbO2
6.022 1023 molecules
239.2 g
Relationships:
Solution:
239.2 g = 6.022 x 1023
6.022 1023 units PbO 2
50.0 g PbO 2 
239.2 g PbO 2
 1.26 1023 units PbO 2
24
Mole Relationships in
Chemical Formulas
• Examples
Moles of compound
1 mol NaCl
Moles of constituents
1 mol Na, 1 mol Cl
1 mol H2O
2 mol H, 1 mol O
1 mol CaCO3
1 mol C6H12O6
1 mol Ca, 1 mol C, 3 mol O
6 mol C, 12 mol H, 6 mol O
Fact check: How many moles of H atoms are in
1 mole of CH3OCH2CH3
Tro's "Introductory Chemistry",
Chapter 6
25
Example 6.6—Calculate the moles of oxygen
in 1.7 moles of CaCO3.
Given:
Find:
Solution Map:
1.7 mol CaCO3
mol O
mol CaCO3
3 mol O
1 mol CaCO 3
mol O
Relationships: 1 mol CaCO3 = 3 mol O
Solution:
3 mol O
1.7 mol CaCO3 
1 mol CaCO3
 5.1 mol O
Tro's "Introductory Chemistry",
Chapter 6
26
Example 6.7—Find the mass of carbon in
55.4 g C10H14O.
Given:
Find:
Solution Map:
55.4 g C10H14O
gC
g C10H14O
mol C10H14O
1 mol
150.2 g
Relationships:
mol C
10 mol C
1 mol C10 H14O
12.01 g
1 mol
gC
1 mol C10H14O = 150.2 g, 1 mol C = 12.01 g,
10 mol C : 1 mol C10H14O
Solution:
1 mol C10 H14O
10 mol C
12.01 g C
55.4 g C10 H14O 


 44.3 g C
150.2 g
1 mol C10 H14O 1 mol C
27
Practice—Find the Mass of Sodium in 6.2 g of NaCl
(MM: Na = 22.99, Cl = 35.45)
Answer: 2.4 g
28
Percent Composition of Compounds
•
Percentage of each element in a compound by
mass can be determined from:
 The formula of the compound.
 The experimental mass analysis of the compound.
mass of element X in 1 mol
Percentage 
 100%
mass of 1 mol of the compound
Tro's "Introductory Chemistry",
Chapter 6
29
Example 6.9—Find the mass percent of
Cl in C2Cl4F2.
Given:
Find:
Solution Map:
Relationships:
C2Cl4F2
% Cl by mass
4  molar mass Cl
Mass % Cl 
100%
molar mass C 2Cl4 F2
mass element X in 1 mol
Mass % element X 
100%
mass 1 mol of compound
Solution: 4  molar mass Cl  4(35.45 g/mol)  141.8 g/mol
molar mass C 2Cl4 F2  2(12.01)  4(35.45)  2(19.00)  203.8 g/mol
Mass % Cl 
141.8 g/mol
100%  69.58%
203.8 g/mol
Practice—Determine the mass percent composition
of Ca and Cl in CaCl2:
(Ca = 40.08, Cl = 35.45)
Answer: 36.11 % Ca and 63.86 % Cl
31
Mass Percent as a
Conversion Factor
• The mass percent tells you the mass of a
constituent element in 100 g of the
compound.
The fact that NaCl is 39% Na by mass means that
100 g of NaCl contains 39 g Na.
• This can be used as a conversion factor.
100 g NaCl = 39 g Na
g NaCl 
39 g Na
 g Na
100 g NaCl
g Na 
Tro's "Introductory Chemistry",
Chapter 6
100 g NaCl
 g NaCl
39 g Na
32
**Example**: The FDA recommends that you consume
2.4 g of sodium a day. How many grams of NaCl must
you consume a day in order to keep up with your sodium
intake. NaCl is 39 % sodium by mass
Answer: 6.2 g NaCl
33
Empirical Formulas
• The simplest, whole-number ratio of atoms in a
molecule is called the empirical formula.
Can be determined from percent composition or
combining masses.
• The molecular formula is a multiple of the
empirical formula.
34
Empirical Formulas, Continued
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HO
Benzene
Molecular formula = C6H6
Empirical formula = CH
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O
35
Example—Determine the Empirical Formula of
Benzopyrene, C20H12,
•
Find the greatest common factor (GCF) of
the subscripts.
20 and 12
GCF = 4
•
Divide each subscript by the GCF to get
the empirical formula.
C20/4H12/4 = C5H3
Empirical formula = C5H3
36
Finding an Empirical Formula
From % Mass
1. Convert the percentages to grams.
a.
Skip if already grams.
2. Convert grams to moles.
a.
Use molar mass of each element.
3. Write a pseudoformula using moles as subscripts.
4. Divide all by smallest number of moles.
5. Multiply all mole ratios by number to make all
whole numbers, if necessary.
a.
b.
-If ratio is 0.5, multiply all by 2
-if ratio is 0.33 or 0.67, multiply all by 3
-if ratio is 0.25 multiply by 4
Skip if already whole numbers after Step 4.
37
Example 6.11—Finding an empirical
formula from experimental data
A laboratory analysis of aspirin determined the
following mass percent composition. Find the
empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Tro's "Introductory Chemistry",
Chapter 6
38
Example:
Find the empirical formula
of aspirin with the given
mass percent composition.
Step 1: Convert percentages to grams
Given:
C = 60.00%
H = 4.48%
O = 35.53%
Therefore, in 100 g of aspirin there are 60.00 g C,
4.48 g H, and 35.53 g O.
39
Example:
Find the empirical
formula of aspirin with
the given mass percent
composition.
Information:
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g
1 mol H = 1.01 g
1 mol O = 16.00 g
• Step 2: Convert masses to moles:
1 mol C
 4.996 mol C
12.01 g C
1 mol H
4.48 g H 
 4.44 mol H
1.01 g H
1 mol O
35.53 g O 
 2.221 mol O
16.00 g O
60.00 g C 
40
Example:
Find the empirical
formula of aspirin with
the given mass percent
composition.
Information:
Given: 4.996 mol C, 4.44 mol H,
2.221 mol O
Write the empirical formula,
CxHyOz
• Step 3: Write a pseudoformula in the form CxHyOz
.
C4.996H4.44O2.221
41
Information:
Given: C4.996H4.44O2.221
Find: empirical formula, CxHyOz
Solution Map:
g C,H,O  mol C,H,O 
mol ratio  empirical
formula
Example:
Find the empirical
formula of aspirin with
the given mass percent
composition.
• Step 4: Find the mole ratio by dividing by the smallest number of
moles in C4.996H4.44O2.221
C 4.996 H 4.44 O 2.221
2.221
2.221
2.221
C 2.25 H 2O1
42
Example:
Find the empirical
formula of aspirin with
the given mass percent
composition.
Information:
Given:
C2.25H2O1
Find: empirical formula, CxHyOz
Solution Map:
g C,H,O  mol C,H,O 
mol ratio  empirical formula
• Step 5: Multiply subscripts by factor to give whole
number.
{ C2.25H 2O1 } x 4
C9H8O4
43
Practice 1—Determine the Empirical Formula
of Tin Fluoride, which Contains 75.7% Sn
and the Rest Fluorine.
Practice 2—Determine the Empirical Formula
of a compound which Contains 69.94% Fe
and the Rest Oxygen.
47
Practice 1—Determine the Empirical Formula
of Stannous Fluoride, which Contains 75.7%
Sn (118.70) and the Rest Fluorine (19.00).
Answer: SnF2
48
Answer: SnF2
Tro's "Introductory Chemistry",
Chapter 6
49
Practice 2—Determine the Empirical Formula
of a compound which Contains 69.94 % Fe
(55.85) and the Rest Oxygen (16.00).
Answer: Fe2O3
50
Answer: Fe2O3
51
Empirical Formula and Molecular
Formula.
Name
Empirical
Formula
Molecular
Formula
Molar
Mass, g
Glyceraldehyde
CH2O
C3H6O3
90
Erythrose
CH2O
C4H8O4
120
Arabinose
CH2O
C5H10O5
150
Glucose
CH2O
C6H12O6
180
52
Molecular Formulas
• The molecular formula is a multiple of the
empirical formula.
• To determine the molecular formula, you
need to know the empirical formula and the
molar mass of the compound.
Molar massreal formula = Factor used to multiply subscripts
Molar massempirical formula
53
Example 6.13—Determine the Molecular
Formula of Cadinene if it has a Molar Mass of
204 g and an Empirical Formula of C5H8.
1. Determine the empirical formula.
 May need to calculate it as previous.
C5H8
2. Determine the molar mass of the empirical
formula.
5 C = 60.05, 8 H = 8.064
C5H8 = 68.11 g/mol
54
Example 6.13—Determine the Molecular
Formula of Cadinene if it has a Molar Mass of
204 g and an Empirical Formula of C5H8,
Continued.
3. Divide the given molar mass of the
compound by the molar mass of the
empirical formula.
 Round to the nearest whole number.
204 g/mol
3
68.11 g/mol
Tro's "Introductory Chemistry",
Chapter 6
55
Example 6.13—Determine the Molecular
Formula of Cadinene if it has a Molar Mass of
204 g and an Empirical Formula of C5H8,
Continued.
4. Multiply the empirical formula by the
factor above to give the molecular
formula.
(C5H8)3 = C15H24
Tro's "Introductory Chemistry",
Chapter 6
56
Practice—Benzopyrene has a Molar Mass of 252 g
and an Empirical Formula of C5H3. What is its
Molecular Formula? (C = 12.01, H=1.01)
Answer: C20H12
57
Example 6.14—Determine the Molecular
Formula of Nicotine, which has a Molar Mass
of 162 g and is 74.0% C, 8.7% H, and the
Rest N.
(C=12.01, H=1.01, N=14.01)
Tro's "Introductory Chemistry",
Chapter 6
58
Example 6.14—Determine the Molecular Formula of
Nicotine, which has a Molar Mass of 162 g and is
74.0% C, 8.7% H, and the Rest N, Continued
Given: 74.0% C, 8.7% H, {100 – (74.0+8.7)} = 17.3% N 
in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N.
Find: CxHyNz
Conversion Factors:
1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 g
Solution Map:
gC
mol C
gH
mol H
gN
mol N
whole
mole number
pseudo- ratio ratio empirical
formula
formula
59
Example 6.14—Determine the Molecular Formula of
Nicotine, which has a Molar Mass of 162 g and is
74.0% C, 8.7% H, and the Rest N, Continued.
Apply solution map:
1 mol C
C5 = 5(12.01 g) = 60.05 g
74.0 g C 
 6.16 mol C
12.01 g
N1 = 1(14.01 g) = 14.01 g
1 mol H
H7 = 7(1.01 g) =
7.07 g
8.7 g H 
 8.6 mol H
C5H7N
=
81.13 g
1.01 g
1 mol N
17.3 g N 
 1.23 mol N
14.01 g
mol. mass nicotine
162 g

2
C6.16H8.6N1.23
mol. mass emp. form. 81.13 g
C 6.16 H 8.6 N1.23  C5H 7 N
1.23
1.23
{C5H7N} x 2 = C10H14N2
1.23
60
Recommended Study Problems Chapter 6
NB: Study problems are used to check the student’s understanding
of the lecture material. Students are EXPECTED TO BE ABLE
TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS.
If you encounter any problems, please talk to your professor or seek
help at the HACC-Gettysburg learning center.
Questions from text book Chapter 6, p 189
3, 7, 11, 17, 19, 25, 27, 37, 51, 57, 59, 69, 71, 73, 79, 85, 87, 89,
97, 102, 117, 116
ANSWERS
-The answers to the odd-numbered study problems are found at
the back of your textbook
61