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Quantitative
Composition of Compounds
Chapter 7
Hein and Arena
Version 1.1
Eugene Passer
Chemistry Department
1 College
Bronx Community
© John Wiley and Sons, Inc.
The Mole
2
1 mole = 6.022 x
23
10 units
One mole of a substance is defined as
the amount of the substance that
contains as many elementary entities
as there are atoms in 12.000 g of 12C.
3
6.022 x
23
10
units/mol
is
Avogadro’s Number
4
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
5
Problems
6
How many moles of iron are equivalent to 25.0 g of
iron?
Atomic mass iron = 55.85 g/mol
Conversion sequence: grams Fe → moles Fe
Set up the calculation using a conversion factor
between moles and grams.
 1 mol Fe 
x mol. Fe  (grams Fe)  55.85 g Fe 


 1 mol Fe 
(25.0 g Fe) 
 0.448 mol Fe

 55.85 g Fe 
7
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
Molar mass Na = 22.99 g/mol
Conversion sequence: atoms Na → mols Na → grams Na
Set up the calculation using conversion factors
between atoms/mol and grams/mol.
x g Na
3.01x1023
atoms Na
1 mol Na
6.022x1023 atoms Na
22.99 g Na
= 11.5 g Na
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1 mol Na
What is the mass of 0.365 moles of tin?
Atomic mass tin = 118.7 g/mol
Conversion sequence: moles Sn → grams Sn
Set up the calculation using a conversion factor
between grams and moles.
118.7 g Sn  43.3 g Sn

x g Sn  (0.365 moles Sn) 

 1 mole Sn 
9
How many oxygen atoms are equivalent to 2.00 mol
of oxygen molecules?
Two conversion factors are needed:
 6.022 x 10 molecules O2 


1
mol
O


2
23
 2 atoms O 
 1 mol O 

2 
Conversion sequence:
moles O2 → molecules O → atoms O
 6.022 x 1023 molecules O2   2 atoms O 
(2.00 mol O2 ) 
  1 molecule O 
1
mol
O


2
2 
= 2.41 x1024 atoms O
10
Molar Mass of
Compounds
11
The molar mass of a compound can be
determined by adding the atomic masses
of all of the atoms of each element in its
formula.
12
Calculate the molar mass of C2H6O, ethanol.
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
13
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
14
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
15
In dealing with diatomic elements (H2, O2,
N2, F2, Cl2, Br2, and I2), we need to
distinguish between one mole of atoms and
one mole of molecules.
16
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
17
Molar Mass Problems
18
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 
=4.992 mole C6H6
(390.0 g C6 H 6 ) 

 78.12 g C6 H 6 
19
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 )3PO 4
Use the conversion factor:
1 mole (NH 4 )3 PO 4
 149.12 g (NH 4 )3 PO4 
(2.52 mol (NH 4 )3 PO4 ) 

 1 mol (NH 4 )3 PO 4 
= 376g (NH 4 )3 PO420
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 and 6.022 x 1023 molecules N 2
28.02 g 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 2

 28.02 g N 2  
23
= 1.204 x 1024 molecules21 N 2
56.04 g of N2 contains how many N atoms?
The molar mass of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
Use the conversion factors
1 mol N 2 6.022 x 1023 molecules N 2
2 atoms N
28.02 g 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 1024 atoms22 N
Percent Composition
of Compounds
23
Percent composition of a compound is the
mass percent of each element in the compound.
H2O
11.21% H by mass
88.79% O by mass
Recall the law of definite composition
24
Percent Composition From Formula
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.
25
total mass of the element
x 100 = percent of the element
molar mass
26
Percent Composition
Problems
27
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
28
1/2
Calculate the percent composition of hydrosulfuric
acid H2S.
Step 2 Divide the total 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%
29
Empirical and Molecular
Formula
30
• The empirical formula or simplest
formula gives the smallest wholenumber ratio of the atoms present in a
compound.
• 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 formula unit of a
compound.
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Examples
32
Molecular Formula
C2H4
Empirical Formula
CH2
Smallest Whole
Number Ratio
C:H 1:2
33
Molecular Formula
C6H6
Empirical Formula
CH
Smallest Whole
Number Ratio
C:H 1:1
34
Two compounds can have identical
empirical formulas and different molecular
formulas; recall the law of multiple
proportions.
35
36
Calculating
Empirical Formulas
Step by Step
37
Step 1 Assume a definite starting
quantity (100% = 100g).
Convert % of each element to
grams of each element. In NaCl,
39.34% Na = 39.34 g Na
60.66% Cl = 60.66 g Cl
Step 2 Convert the grams of each
element into moles of each
element using atomic weights.
38
Step 3 Calculate mole ratios by
dividing the moles of each
element by the smallest number
of moles present.
– 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.
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Step 4 Calculate formula ratio by
multiplying the mole ratios
obtained in step 3 by the smallest
whole number that will convert
the mole ratios into whole
numbers
Use these whole numbers as the
subscripts in the empirical
formula.
40
• 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.
41
Empirical Formula Problems
42
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 Convert % each element to grams. Assume
100% = 100 grams of compound.
Step 2 Convert grams to moles.
Step 3 Calculate mole ratios.
Smallest whole number (2.9 ~3.0)
Step 4 Use smallest whole numbers as subscript in
formula.
43
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
44
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.
Element
K
C
O
%
56.58
8.68
34.73
mass(g)
56.58
8.68
34.73
mol.
mol. Ratio
45
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  

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 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.
Element
K
C
O
%
56.58
8.68
34.73
mass(g)
56.58
8.68
34.73
mol.
1.45
0.723
2.17
mol. Ratio
47
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.
0.723 mol
1.447 mol
C:
K=
= 1.00
= 2.00
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
48
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.
Element
K
C
O
%
56.58
8.68
34.73
mass(g)
56.58
8.68
34.73
K2CO3
mol.
1.45
0.723
2.17
mol. Ratio
2
1
3
49
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
50
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 
51
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
The O mole ratio is not a whole number.
52
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
53
Calculating the Molecular Formula
from the Empirical Formula
54
• The molecular formula can be calculated
from the empirical formula if the molar mass
of the molecular formula 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 empirical formula units
contained in the molecular formula.
molar mass
n=
= number of empirical
formula units
mass of empirical formula
55
What is the empirical formula of a compound that is
14.40% H and 85.60% C?
Element
C
H
%
mass(g)
85.60 85.60
14.40 14.40
mol.
mol. Ratio
7.13 7.13/7.13 =1
14.26 14.26/7.13=2
The Empirical Formula is CH2
56
What is the molecular formula of a compound which
has the empirical formula 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
57
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