Formulas from Combustion Analysis
The reaction of the organic compound with excess oxygen
produces carbon dioxide and water.
Conservation of mass requires that all the carbon from the
organic material become carbon dioxide. Therefore the mass of
carbon in the carbon dioxide is the same as the mass of carbon in
the organic compound.
Similarly, the mass of hydrogen in the water must be the
mass of the hydrogen in the organic compound.
Conservation of mass also requires that the mass of the
compound be the sum of the mass of all the elements in the
compound. Therefore, if the organic compound contains oxygen,
the mass of oxygen can be determined by subtracting the mass of
hydrogen and carbon from the mass of the entire compound.
Example Problem:
What is the empirical formula of a hydrocarbon that produces
2.703 g CO2 and 1.108 g H2O when combusted?
Note: Since this is a hydrocarbon, the only elements it contains are
carbon and hydrogen.
Solution:
The empirical formula will be the simplest mol ratio of these
elements.
nC = 2.703 g CO2 x
nH
1 mol CO2 x 1 mol C
= 0.06142 mol C
44.01 g
1 mol CO2
= 1.108 g H2O x 1 mol H2O x 2 mol H = 0.1230 mol H
18.01 g
1 mol H2O
To get the simplest ratio, divide the moles of each substance by the
lowest mol value.
0.06142 mol C = 1
0.06142
0.1230 mol H = 2.003
0.06142
Round to the nearest whole number to use in the formula.
Therefore the empirical formula for this compound is CH2.
Question:
What is the empirical formula of a substance containing
carbon, hydrogen, and oxygen if 1.000 g of substance produces
1.467 g CO2 and 0.6003 g H2O upon combustion?
Solution strategy:
The mass of carbon dioxide and water can be used to find the
masses and moles of carbon and hydrogen. The mass of oxygen
can be determined from the mass of the original substance.
Because of conservation of mass:
Mass (g) substance = mass carbon + mass hydrogen + mass oxygen