Experiment #7
Virtual Chemistry Laboratory
(STOICHIOMETRY) ELEMENTAL ANALYSIS
I.
PURPOSE OF THE EXPERIMENT
The elemental analysis of a compound is particularly useful in
determining the empirical formula of the compound.
II.
INTRODUCTION
Composition by Mass: The composition of a compound is often
expressed in terms of the weight percent of each element in the compound.
For example, consider Ethanol (Ethyl Alcohol) has a mass of 46.07g, which has
the formula C2H6O.
1 Mole of C2H6O contains
2 moles of “C”
6 moles of “H”
1 mole of “O” atom
%of" C" 
2moles" C" (12.01g / mol" C" )
x100%  52.14%
46.07g..Ethanol
%of" H" 
6moles" H" (1.008g / mol" H" )
x100%  13.13%
46.07g..Ethanol
%of" O" 
1moles" O" (16.01g / mol" O" )
x100%  34.73%
46.07g..Ethanol
Notice: that the sum of the weight percents of all the elements in a
compound must equal 100%.
Empirical Formula: The formula that contains the smallest set integer
ratios for the elements in the compound that gives the correct elemental
composition by mass.
Example: Octane , whose molecular formula is C8H18.
8: 18
Divided by 8
The smallest ratio of integers is
Therefore Empirical Formula is
1: 2.25 x 2(or)3(or)4
4: 9 (1:2.25 x 4)
[C4 H9]
Note: Molecular formula is an integer (n) multiple of the Empirical formula.
For Octane E.F = C4H9
M.F = C4nH9n, In this case n = 2, which produces C8H18.
Be aware that the elemental analysis is not perfectly accurate. The
experimental error will generally produces atom ratios that are not perfect
integers but are close to integers.
A separate measurement of the molecular mass is required to determine the
molecular formula .(Mass Spectrometry)
III. EXPERIMENTAL
3.1 Chemicals:
3.2 Equipment:
3.3 Procedure: Refer
(http://www.chm.davidson.edu/chemistryapplets/stoichiometry/CH.html)
IV. RESULTS and DISCUSSIONS
V. CONCLUSION