Quantitative
Composition of Compounds
Chapter 7
Hein and Arena
Version 2.0
12th Edition
Eugene Passer
Chemistry Department
Bronx Community
1 College
© John Wiley and Sons, Inc
Chapter Outline
7.1 The Mole
7.2 Molar Mass of Compounds
7.3 Percent Composition of
Compounds
7.4 Empirical Formula versus
Molecular Formula
7.5 Calculating Empirical
Formulas
7.6 Calculating the Molecular
Formula from the Empirical
Formula
2
7.1
The Mole
3
The mass of a single atom is too small to
measure on a balance.
mass of hydrogen atom = 1.673 x 10-24 g
4
This is an
infinitesimal
1.673 x 10-24 g
mass
5
• Chemists require a unit for counting
which can express large numbers of
atoms using simple numbers.
• Chemists have chosen a unit for
counting atoms.
• That unit is the
6
1 mole = 6.022 x
23
10 objects
7
6.022 x
23
10
is a very
LARGE
number
8
6.022 x
23
10
is
Avogadro’s Number
number
9
If 10,000 people started to count
Avogadro’s number and counted at the
rate of 100 numbers per minute each
minute of the day, it would take over 1
trillion years to count the total number.
10
1 mole of any element contains
6.022 x 1023
particles of that substance.
11
The atomic mass in grams
of any element23
contains 1 mole of atoms.
12
This is the same number of particles
6.022 x 1023
as there are in exactly 12 grams of
12
6
C
13
Examples
14
Species
Quantity
Number of H
atoms
H
1 mole
6.022 x
23
10
15
Species
Quantity
Number of
H2 molecules
H2
1 mole
6.022 x
23
10
16
Species
Quantity
Number of
Na atoms
Na
1 mole
6.022 x
23
10
17
Species
Quantity
Number of
Fe atoms
Fe
1 mole
6.022 x
23
10
18
Species C6H6
Quantity 1 mole
Number of
23
6.022 x 10
C6H6 molecules
19
1 mol of atoms = 6.022 x 1023 atoms
1 mol of molecules = 6.022 x 1023 molecules
1 mol of ions = 6.022 x 1023 ions
20
• The molar mass of an element is its
atomic mass in grams.
• It contains 6.022 x 1023 atoms
(Avogadro’s number) of the element.
21
H
Atomic
mass
1.008 amu
Mg
24.31 amu
24.31 g
6.022 x 1023
Na
22.99 amu
22.99 g
6.022 x 1023
Element
Number of
Molar mass
atoms
1.008 g
6.022 x 1023
22
Problems
23
How many moles of iron does 25.0 g of iron
represent?
Atomic mass iron = 55.85
Conversion sequence: grams Fe → moles Fe
Set up the calculation using a conversion factor
between moles and grams.
 1 mol Fe 
(grams Fe) 

 55.85 g Fe 
 1 mol Fe 
(25.0 g Fe) 
 0.448 mol Fe

 55.85 g Fe 
24
How many iron atoms are contained in 25.0 grams of
iron?
Atomic mass iron = 55.85
Conversion sequence: grams Fe → atoms Fe
Set up the calculation using a conversion factor
between atoms and grams.
 6.022 x 1023 atoms Fe 
(grams Fe) 

55.85
g
Fe


 6.022 x 1023 atoms Fe 
23
(25.0 g Fe) 

2.70
x
10
atoms Fe

55.85 g Fe


25
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
Molar mass Na = 22.99 g
Conversion sequence: atoms Na → grams Na
Set up the calculation using a conversion factor
between grams and atoms.
22.99 g Na


(atoms Na) 

23
 6.022 x 10 atoms Na 
22.99 g Na


(3.01 x 10 atoms Na) 
 11.5 g Na
23
 6.022 x 10 atoms Na 
26
23
What is the mass of 0.365 moles of tin?
Atomic mass tin = 118.7
Conversion sequence: moles Sn → grams Sn
Set up the calculation using a conversion factor
between grams and atoms.
1 molar mass Sn 

(moles Sn) 

 1 mole Sn 
118.7 g Sn 

(0.365 moles Sn) 
  43.3 g Sn
 1 mole Sn 
27
How many oxygen atoms are present in 2.00 mol of
oxygen molecules?
Two conversion factors are needed:
 6.022 x 1023 molecules O 2 


1
mol
O


2
 2 atoms O 
 1 mol O 

2 
Conversion sequence:
moles O2 → molecules O → atoms O
 6.022 x 1023 molecules O 2   2 atoms O 
(2.00 mol O 2 ) 



1 mol O 2

  1 molecule O 2 
= 2.41 x1024 atoms O
28
7.2
Molar Mass of
Compounds
29
The molar mass of a compound can be
determined by adding the molar masses
of all of the atoms in its formula.
30
Calculate the molar mass of C2H6O.
2 C = 2(12.01 g) = 24.02 g
6 H = 6(1.01 g) = 6.06 g
1 O = 1(16.00 g) = 16.00 g
46.08 g
31
Calculate the molar mass of LiClO4.
1 Li = 1(6.94 g) = 6.94 g
1 Cl = 1(35.45 g) = 35.45 g
4 O = 4(16.00 g) = 64.00 g
106.39 g
32
Calculate the molar mass of (NH4)3PO4 .
3 N = 3(14.01 g) = 42.03 g
12 H = 12(1.01 g) = 12.12 g
1 P = 1(30.97 g) = 30.97 g
4 O = 4(16.00 g) = 64.00 g
149.12 g
33
Avogadro’s
Number of
Particles
23
10
6x
Particles
1 MOLE
Molar Mass
34
Avogadro’s
Number of
Ca atoms
23
10
6x
Ca atoms
1 MOLE Ca
40.078 g Ca
35
Avogadro’s
Number of
H2O molecules
23
10
6x
H2O
molecules
1 MOLE H2O
18.02 g H2O
36
These relationships are present when
hydrogen combines with chlorine.
H
6.022 x 1023 H
atoms
Cl
HCl
6.022 x 1023 Cl 6.022 x 1023 HCl
atoms
molecules
1 mol H atoms 1 mol Cl atoms
1 mol HCl
molecules
1.008 g H
35.45 g Cl
36.46 g HCl
1 molar mass H
atoms
1 molar mass
Cl atoms
1 molar mass
37
HCl molecules
In dealing with diatomic elements (H2, O2,
N2, F2, Cl2, Br2, and I2), distinguish between
one mole of atoms and one mole of
molecules.
38
Calculate the molar mass of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the molar mass of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g
39
Problems
40
How many moles of benzene, C6H6, are present in
390.0 grams of benzene?
The molar mass of C6H6 is 78.12 g.
Conversion sequence: grams C6H6 → moles C6H6
78.12 grams C6 H 6
Use the conversion factor:
1 mole C6 H 6
 1 mole C6 H 6 
= 5.000 moles C6 H 6
(390.0 g C6 H 6 ) 

 78.12 g C6 H 6 
41
How many grams of (NH4)3PO4 are contained in 2.52
moles of (NH4)3PO4?
The molar mass of (NH4)3PO4 is 149.12 g.
Conversion sequence: moles (NH4)3PO4
→ grams (NH4)3PO4
149.12 grams (NH 4 )3PO4
Use the conversion factor:
1 mole (NH 4 )3PO4
 149.12 g (NH 4 )3PO4 
(2.52 mol (NH 4 )3 PO 4 ) 

 1 mol (NH 4 )3PO4 
= 376g (NH 4 )3 PO 442
56.04 g of N2 contains how many N2 molecules?
The molar mass of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors
1 mol N 2
28.02 g N 2
6.022 x 1023 molecules N 2
1 mol N 2
 1 mol N 2   6.022 x 10 molecules N 2 
(56.04 g N 2 ) 


1
mol
N
28.02
g
N


2
2 
23
= 1.204 x 1024 molecules43 N 2
56.04 g of N2 contains how many nitrogen atoms?
The molar mass of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
1 mol N 2
28.02 g N 2
Use the conversion factor
2 atoms N
6.022 x 1023 molecules N 2
1 molecule N 2
1 mol N 2
 1 mol N 2   6.022 x 10 molecules N 2 
(56.04 g N 2 ) 


1
mol
N
28.02
g
N


2
2 
23
 2 atoms N 
 1 molecule N 

2 
= 2.409 x 10
24
atoms44 N
7.3
Percent Composition
of Compounds
45
Percent composition of a compound is the
mass percent of each element in the compound.
H2O
11.19% H by mass
88.79% O by mass
46
Percent Composition
From Formula
47
If the formula of a compound is known,
a two-step process is needed to calculate
the percent composition.
Step 1 Calculate the molar mass of the
formula.
Step 2 Divide the total mass of each
element in the formula by the
molar mass and multiply by
100.
48
total mass of the element
x 100 = percent of the element
molar mass
49
Calculate the percent composition of hydrosulfuric
acid H2S.
Step 1 Calculate the molar mass of H2S.
2 H = 2 x 1.01g = 2.02 g
1 S = 1 x 32.07 g = 32.07 g
34.09 g
50
Calculate the percent composition of hydrosulfuric
acid H2S.
Step 2 Divide the mass of each element
by the molar mass and multiply
by 100.
 2.02 g H 
H: 
 (100) = 5.93%
 34.09 g 
 32.07 g S
S: 
 34.09 g

 (100)  94.07%

S
H
94.07% 5.93%
51
Percent Composition
From Experimental Data
52
Percent composition can be calculated
from experimental data without knowing
the composition of the compound.
Step 1 Calculate the mass of the
compound formed.
Step 2 Divide the mass of each element
by the total mass of the
compound and multiply by 100.
53
A compound containing nitrogen and oxygen is found
to contain 1.52 g of nitrogen and 3.47 g of oxygen.
Determine its percent composition.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g = total mass of product
54
A compound containing nitrogen and oxygen is found
to contain 1.52 g of nitrogen and 3.47 g of oxygen.
Determine its percent composition.
Step 2 Divide the mass of each element
by the total mass of the
compound formed.
 1.52 g N 

 (100) = 30.5%
 4.99 g 
 3.47 g O

 4.99 g

 (100) = 69.5%

O
69.5%
N
30.5%
55
7.4
Empirical Formula versus
Molecular Formula
56
• The empirical formula or simplest
formula gives the smallest wholenumber ratio of the atoms present in a
compound.
• The empirical formula gives the
relative number of atoms of each
element present in the compound.
57
• The molecular formula is the true
formula of a compound.
• The molecular formula represents the
total number of atoms of each element
present in one molecule of a compound.
58
Examples
59
Molecular Formula
C2H4
Empirical Formula
CH2
Smallest Whole
Number Ratio
C:H 1:2
60
Molecular Formula
C6H6
Empirical Formula
CH
Smallest Whole
Number Ratio
C:H 1:1
61
Molecular Formula
H2O2
Empirical Formula
HO
Smallest Whole
Number Ratio
H:O 1:1
62
63
Two compounds can have identical
empirical formulas and different molecular
formulas.
64
65
7.5
Calculating
Empirical Formulas
66
Step 1 Assume a definite starting
quantity (usually 100.0 g) of the
compound, if the actual amount
is not given, and express the
mass of each element in grams.
Step 2 Convert the grams of each
element into moles of each
element using each element’s
molar mass.
67
Step 3 Divide the moles of atoms of
each element by the moles of
atoms of the element that had
the smallest value.
– If the numbers obtained are whole
numbers, use them as subscripts
and write the empirical formula.
– If the numbers obtained are not
whole numbers, go on to step 4.
68
Step 4 Multiply the values obtained in
step 3 by the smallest numbers
that will convert them to whole
numbers
Use these whole numbers as the
subscripts in the empirical
formula.
FeO1.5
Fe1 x 2O1.5 x 2
Fe2O3
69
• The results of calculations may differ
from a whole number.
– If they differ ±0.1, round off to the next
nearest whole number.
2.9  3
– Deviations greater than 0.1 unit from a
whole number usually mean that the
calculated ratios have to be multiplied by
a whole number.
70
Some Common Fractions and Their Decimal Equivalents
Decimal
Resulting Whole
Equivalent
Number
Common Fraction
1
4
0.25
1
1
3
0.333…
1
0.666…
2
0.5
1
0.75
3
2
3
1
2
3
4
Multiply the
decimal equivalent
by the number in
the denominator of
the fraction to get
a whole number.
71
Problems
72
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
K = 56.58 g
C = 8.68 g
O = 34.73 g
73
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 2 Convert the grams of each element to moles.
 1 mol K atoms 
 1.447 mol K atoms
K:  56.58 g K  

 39.10 g K 
 1 mol C atoms 
C atoms
atoms
C: 8.68 g C  
 0.723 mol C

 12.01 g C  C has the smallest number
of moles
 1 mol O atoms 
 2.171 mol O atoms
O:  34.73 g O  

74
 16.00 g O 
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 3 Divide each number of moles by the smallest
value.
1.447 mol
0.723 mol
K=
= 2.00
C:
= 1.00
0.723 mol
0.723 mol
0.723 mol C atoms
2.171 mol
O=
= 3.00
C has the smallest number
0.723 mol
of moles
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3
75
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
N = 25.94 g
O = 74.06 g
76
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 2 Convert the grams of each element to moles.
 1 mol N atoms 
 1.852 mol N atoms
N:  25.94 g N  

 14.01 g N 
 1 mol O atoms 
O:  74.06 g O  
 4.629 mol C atoms

 16.00 g O 
77
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 3 Divide each number of moles by the smallest
value.
1.852 mol
N=
= 1.000
1.852 mol
4.629 mol
O:
= 2.500
1.852 mol
This is not a ratio of whole numbers.
78
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 4 Multiply each of the values by 2.
N: (1.000)2 = 2.000
O: (2.500)2 = 5.000
Empirical formula N2O5
79
7.6
Calculating the Molecular Formula
from the Empirical Formula
80
• The molecular formula can be calculated
from the empirical formula if the molar mass
is known.
• The molecular formula will be equal to the
empirical formula or some multiple, n, of it.
• To determine the molecular formula evaluate
n.
• n is the number of units of the empirical
formula contained in the molecular formula.
molar mass
n=
= number of empirical
formula units
mass of empirical formula
81
What is the molecular formula of a compound which
has an empirical formula of CH2 and a molar mass of
126.2 g?
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = 2.02g
14.03g
126.2 g
n
 9 (empirical formula units)
14.03 g
The molecular formula is (CH2)9 = C9H18
82
83