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>>
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Formulas From Combustion Analysis
>> Parts of this equation/concept include:
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.
The empirical formula is the simplest mole ratio of elements in
the compound. So determine the moles of each element in the compound
and divide by the lowest value to get the simplest ratio. The procedure
is basically the same as that described in Chapter 4.
>> Example 1
What is the empirical formula of a hydrocarbon that produces
2.703 g CO2 and 1.108 g H2O when combusted?
Solution:
Since this is a hydrocarbon, the only elements it contains are
carbon and hydrogen. The empirical formula will be the simplest
mole ratio of these elements.
| mol carbon |
= |
2.703 g CO2 |
 |
|
|
|
= |
0.06142 mol C |
| mol hydrogen |
= |
1.108 g H2O |
|
|
|
|
= |
0.1230 mol H |
To get the simplest ratio, divide the moles of each substance
by the lowest mol value.
Round to the nearest whole number or simple fraction to use in
the formula. Therefore the empirical formula for this compound
is CH2.
>> Example 2
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:
The mass of carbon dioxide and water can be used to find the
moles of carbon and hydrogen. The mass of oxygen can be determined
from the mass of the original substance. Because of conservation
of mass
g substance = g carbon + g hydrogen + g oxygen
This requires that in addition to determining moles of carbon,
hydrogen, and oxygen, the mass of carbon and hydrogen needs to
be determined.
| 1.467 g CO2 |
|
|
|
|
= |
0.03333 mol C |
| 0.03333 mol C |
|
|
= |
0.4003 g C |
| 0.600 g H2O |
|
|
|
|
= |
0.0666 mol H |
| 0.0666 mol H |
|
|
= |
0.0672 g H |
| grams oxygen |
= |
g substance – g carbon
– g hydrogen |
| |
= |
1.000 – 0.4003 –
0.0672 |
| |
= |
0.5325 g |
| 0.5325 g O |
|
|
= |
0.03328 mol O |
To obtain the simplest ratio, divide by the lowest value of moles.
In this example, that is the moles of carbon
Round to the nearest whole number or simple fraction for use
in the formula. Thus the empirical formula is CH2O.
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To obtain the molecular formula, first determine the empirical
formula, as in the preceding examples. The additional information
of molar mass is required to complete these problems.
To obtain the molecular formula from the empirical formula, first
determine the apparent molar mass, based on the empirical formula.
Then divide the actual molar mass (given in the problem) by the
apparent molar mass and round to the whole number. Multiply each
subscript in the empirical formula by the factor obtained in this
manner to obtain the molecular formula.
>> Example 3
The molar mass of the substance in Example 2 is 120 g/mol. What
is its molecular formula?
Solution:
From Example 2, the empirical formula of the substance is CH2O.
Its apparent molar mass is
apparent molar mass = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
| multiplication factor |
= |
|
= |
3.996 |
= |
4 |
empirical formula = 4(CH2O) = C4H8O4
>> Example 4
What is the molecular formula of a substance containing carbon,
hydrogen, and oxygen if it has a molar mass of 234 g/mol and 0.360
g of substance produces 0.406 g CO2 and 0.250 g H2O
upon combustion?
Solution:
First, determine both the mass and moles of carbon and hydrogen
from carbon dioxide and water, respectively.
| mol carbon |
= |
0.406 g CO2 |
|
|
|
|
= |
0.00922 mol C |
| mass carbon |
= |
0.00922 mol C |
|
|
= |
0.111 g C |
| mol hydrogen |
= |
0.250 g H2O |
|
|
|
|
= |
0.0278 mol H |
| mass hydrogen |
= |
0.0278 mol H |
|
|
= |
0.0279 g H |
The mass of oxygen is determined from the mass of the substance:
| g oxygen |
= |
g substance – g carbon
– g hydrogen |
| |
= |
0.360 – 0.111 – 0.0279 |
| |
= |
0.221 g O |
The moles of oxygen is determined from mass of oxygen:
| mol oxygen |
= |
0.221 g O |
|
|
= |
0.0138 mol O |
Determine the empirical formula by dividing by the smallest mole
value. In this example, that is the moles of carbon.
Round to the nearest whole number or simple fraction. This gives
a ratio of C:H:O of 1:3:1 1/2 . Since formulas require whole numbers,
multiply the ratio by the denominator of the fraction. Then the
ratio is 2:6:3 and the empirical formula is C2H6O3.
To determine the molecular formula, compare the apparent molar
mass to the actual molar mass. The problem states that the actual
molar mass is 234 g/mol.
apparent molar mass = 2(12.01) + 6(1.008) + 3(16.00) = 78.07
g/mol
| multiplication factor |
= |
|
= |
2.997 |
= |
3 |
Multiplying the empirical formula by 3, the answer is C6H18O9.
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