Chapter 6: Chemical Composition
+
4 Wood Boards
+
6 Nails
1 Fence Panel
For a 12 panels fence…
4 Dozen
Wood Boards
+
6 Dozen
Nails
1 Dozen
Fence Panels
+
4C
+
3H2
C4H6
For a mole of 1,3-Butadiene
…
4 moles of
C
+
3 moles of
H2
1 mole of
C4H6
1 Dozen = 12 items
• The mole is defined (since 1960) as the amount of substance of a system that contains as
many entities as there are atoms in 12 g of carbon-12.
• Symbol: mol.
• Coined by Wilhelm Ostwald in 1893
1 mol = 12 g of carbon-12
• A mole of carbon contains 6.0221415×1023 atoms of carbon, but the same is true for any
other element or molecule; in general:
1 mole = 6.0221415×1023 items
# of Molecules = # of moles X 6.022×1023
• 6.0221415×1023 is the Avogadro’s Number
m = number of nails
X mass of 1 nail
Example:
500g = 100 nails X 5g
m = number of moles X mass of 1 mole of the substance
m = n X M.M.
Where:
m = mass (g)
n = number of moles (mol)
M.M. = Molecular Mass (g/mol) = mass of 1 mole of the substance
Sometime the Molecular Mass is measured in Daltons (1Da = 1g/mol)
The molar mass of an element is numerically the same as the atomic
weight of the element. They are not however the same; they have
different units:
• The atomic weight is defined as one twelfth of the mass of an isolated
atom of carbon-12 and is therefore dimensionless
• The molar mass is measured in g/mol.
The molar mass of a compound is given by the sum of the atomic
weights of the atoms which form the compound.
Example: molar mass of Ca(NO3)2
Atoms
Atomic Weights
1 Ca
1 X 40.078
40.078
2N
2 X 14.007
28.014
6O
6 X 15.999
95.994
Total: 164.086 g/mol
Example:
A silver ring contains 1.1
x 1022 silver atoms. How
many moles of silver are
in the ring?
Information
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Conv. Fact.: 1 mole = 6.022 x 1023
• Write a Solution Map for converting the units :
atoms Ag
moles Ag
1 mole Ag
6.022 10 23 Ag atoms
Example:
A silver ring contains 1.1
x 1022 silver atoms. How
many moles of silver are
in the ring?
Information
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Conv. Fact.: 1 mole = 6.022 x 1023
Sol’n Map: atoms  mole
• Check the Solution:
1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag
The units of the answer, moles, are correct.
The magnitude of the answer makes sense
since 1.1 x 1022 is less than 1 mole.
Example:
Calculate the mass (in
grams) of 1.75 mol of
water
Information
Given: 1.75 mol H2O
Find: ? g H2O
C F: 1 mole H2O = 18.02 g H2O
• Write a Solution Map for converting the units :
mol H2O
g H2O
18.02 g H 2 O
1 mol H 2 O
Example:
Calculate the mass (in
grams) of 1.75 mol of
water
Information
Given: 1.75 mol H2O
Find: ? g H2O
C F: 1 mole H2O = 18.02 g H2O
Sol’n Map:
mol  g
• Check the Solution:
1.75 mol H2O = 31.5 g H2O
The units of the answer, g, are correct.
The magnitude of the answer makes sense
since 31.5 g is more than 1 mole.
Chemical Formulas as Conversion
Factors
• 1 spider  8 legs
• 1 chair  4 legs
• 1 H2O molecule  2 H atoms  1 O atom
Mole Relationships in
Chemical Formulas
• since we count atoms and molecules in mole
units, we can find the number of moles of a
constituent element if we know the number
of moles of the compound
Moles of Compound
1 mol NaCl
1 mol H2O
1 mol CaCO3
Moles of Constituents
1 mole Na, 1 mole Cl
2 mol H, 1 mole O
1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6
6 mol C, 12 mol H, 6 mol O
Example:
• Carvone, (C10H14O), is the main component in spearmint
oil. It has a pleasant odor and mint flavor. It is often
added to chewing gum, liquers, soaps and perfumes.
Find the mass of carbon in 55.4 g of carvone.
Example:
Find the mass of carbon in
55.4 g of carvone,
(C10H14O).
Information
Given: 55.4 g C10H14O
Find: g C
CF: 1 mol C10H14O = 150.2 g
1 mol C10H14O  10 mol C
1 mol C = 12.01 g
• Write a Solution Map for converting the units :
g
C10H14O
mol
C10H14O
1 mol C10 H14O
150.2 g C10 H14O
10 mol C
1 mol C10 H14O
mol
C
g
C
12.01 g C
1 mol C
Example:
Find the mass of carbon in
55.4 g of carvone,
(C10H14O).
Information
Given: 55.4 g C10H14O
Find: g C
CF: 1 mol C10H14O = 150.2 g
1 mol C10H14O  10 mol C
1 mol C = 12.01 g
SM: g C10H14O  mol C10H14O 
mol C  g C
• Check the Solution:
55.4 g C10H14O = 44.3 g C
The units of the answer, g C, are correct.
The magnitude of the answer makes sense since
the amount of C is less than the amount of C10H14O.
•
Percent Composition
Percentage of each element in a compound
– By mass
• Can be determined from
1. the formula of the compound
2. the experimental mass analysis of the
compound
• The percentages may not always total to 100%
due to rounding
part
Percentage 
 100%
whole
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 
100 g NaCl
 g NaCl
39 g Na
Example - Percent Composition from the
Formula C2H5OH
1. Determine the mass of each element in 1
mole of the compound
2 moles C = 2(12.01 g) = 24.02 g
6 moles H = 6(1.008 g) = 6.048 g
1 mol O = 1(16.00 g) = 16.00 g
2. Determine the molar mass of the
compound by adding the masses of the
elements
1 mole C2H5OH = 46.07 g
Sample - Percent Composition from the
Formula C2H5OH
3. Divide the mass of each element by the
molar mass of the compound and
multiply by 100%
24.02g
 100%  52.14%C
46.07g
6.048g
 100%  13.13%H
46.07g
16.00g
 100%  34.73%O
46.07g
Empirical Formulas
Hydrogen Peroxide
Molecular Formula = H2O2
Empirical Formula = HO
Benzene
Molecular Formula = C6H6
Empirical Formula = CH
Glucose
Molecular Formula = C6H12O6
Empirical Formula = CH2O
Finding an Empirical Formula
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
a) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply
all by 3, etc.
b) skip if already whole numbers
All these molecules have the same
Empirical Formula. How are the
molecules different?
Name
glyceraldehyde
Molecular
Formula
C3H6O3
Empirical
Formula
CH2O
erythrose
C4H8O4
CH2O
arabinose
C5H10O5
CH2O
glucose
C6H12O6
CH2O
All these molecules have the same
Empirical Formula. How are the
molecules different?
Name
glyceraldehyde
Molecular
Formula
C3H6O3
Empirical
Formula
CH2O
Molar
Mass, g
90
erythrose
C4H8O3
CH2O
120
arabinose
C5H10O5
CH2O
150
glucose
C6H12O6
CH2O
180
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