CHAPTER 3-C
Percent Composition,
Molecular Formulae,
Combustion Analysis
CHEMISTRY
THIRD EDITION
Gilbert | Kirss | Foster | Davies
© 2012 by W. W. Norton & Company
Percentage Composition
•
Percent Composition: Identifies the elements
present in a compound as a mass percent of the
total compound mass.
•
The mass percent is obtained by dividing the mass
of each element by the total mass of a compound
and converting to percentage.
2
Empirical Formula
•
The empirical formula gives the ratio of the number
of atoms of each element in a compound.
Compound
Formula
Hydrogen peroxide H2O2
Benzene
C6H6
Ethylene
C2H4
Propane
C3H8
Empirical Formula
OH
CH
CH2
C3H8
3
Percentage Composition
•
Glucose has the molecular formula C6H12O6. What
is its empirical formula, and what is the percentage
composition of glucose?
Empirical Formula = smallest whole number ratio
CH2O
4
Percentage Composition
CH2O
Total mass = 12.01 + 2.02 + 16.00 = 30.03
%C = 12.01/30.03 x 100% = 39.99%
%H = 2.02/30.03 x 100% = 6.73%
%O = 16.00/30.03 x 100% = 53.28%
5
Percentage Composition
Saccharin has the molecular formula C7H5NO3S.
What is its empirical formula, and what is the
percentage composition of saccharin?
Empirical Formula is same as molecular formula
MW = 183.19 g/mole
%C = (7 x 12.011)/183.19 x 100% = 45.89% etc.
6
Empirical Formula
•
A compound’s empirical
formula can be determined
from its percent composition.
•
A compound’s molecular
formula is determined from
the molar mass and empirical
formula.
7
Empirical Formula
•
A compound was analyzed to be 82.67% carbon and 17.33%
hydrogen by mass. What is the empirical formula for the
compound?
•
Assume 100 g of sample, then 82.67 g are C and 17.33 g are
H
8
Empirical Formula
•
Convert masses to moles:
82.67 g C x mole/12.011 g = 6.88 moles C
17.33 g H x mole/1.008 g = 17.19 mole H
Find relative # of moles (divide by smallest number)
9
Empirical Formula
Convert moles to ratios:
6.88/6.88 = 1 C
17.19/6.88 = 2.50 H
Or 2 carbons for every 5 hydrogens
C2H5
10
Empirical Formula
Empirical Formula is: C2H5
Formula weight is: 29.06 g/mole
If the molecular weight is known to be 58.12 g/mole
Then the molecular formula must be:
C4H10
11
Empirical Formula
•
Combustion analysis is one of the
most common methods for
determining empirical formulas.
•
A weighed compound is burned in
oxygen and its products analyzed
by a gas chromatogram.
•
It is particularly useful for analysis
of hydrocarbons.
12
Combustion Analysis
Combustion Analysis: the technique of finding the mass
composition of an unknown sample (X) by examining the
products of its combustion.
X + O2 → CO2 + H2O
0.250 g of compound X produces:
0.686 g CO2 and 0.562 g H2O
13
Combustion Analysis
X + O2 → CO2 + H2O
1. Find the mass of C & H that must have been present in X
(multiply masses of products by percent composition of the
products).
C: 0.686 g x 12.01 g/44.01 g = 0.187 g C
H: 0.562 g x (2 x 1.008 g)/18.02 g = 0.063 g H
14
Combustion Analysis
X + O2 → CO2 + H2O
0.187 g C + 0.063 g H = 0.250 g total
So compound X must contain only C and H!!
2. Find the number of moles of C and H
C: 0.187 g x mole/12.01 g = 0.0156 moles C
H: 0.063 g x mole/1.008 g = 0.063 moles H
15
Combustion Analysis
X + O2 → CO2 + H2O
3. Find the RELATIVE number of moles of C and H in whole
number units (divide by smallest number of moles)
C: 0.0156/0.0156 = 1
H: 0.063/0.0156 = 4
If these numbers are fractions, multiply each by the same
whole number.
16
Combustion Analysis
X + O2 → CO2 + H2O
3. Write the Empirical Formula (use the relative numbers as
subscripts)
CH4
17
Combustion Analysis
Summary
1.
Find the mass of C and H in the sample.
2.
Find the actual number of moles of C and H in the
sample.
3.
Find the relative number of moles of C and H in whole
numbers.
4.
Write the empirical formula for the unknown compound.
18
Combustion Analysis
NOTE!
In step # 1 always check to see if the total mass of C and H adds
up to the total mass of X combusted. If the combined mass of
C and H is less than the mass of X, then the remainder is an
unknown element (unless instructed otherwise).
If a third element is known, calculate the mass of that element by
subtraction (at the end of step 1), and include the element in
the remaining steps.
19
Combustion Analysis
Combustion Analysis provides the Empirical Formula. If a
second technique provides the molecular weight, then the
molecular formula may be deduced.
1.
Calculate the empirical formula weight.
2.
Find the number of “formula units” by dividing the known
molecular weight by the formula weight.
3.
Multiply the number of atoms in the empirical formula by
the number of formula units.
20
Combustion Analysis
The molecular weight of glucose is 180 g/mole and its
empirical formula is CH2O. Deduce the molecular
formula.
1.
Formula weight for CH2O is 30.03 g/mole
2.
# of formula units = 180/30.03 = 6
3.
Molecular formula = C6H12O6
21