The empirical formula of a compound expresses a ratio between the numbers of atoms of different elements present in a molecule of the compound. This ratio is a mole ratio as well as a ratio between numbers of atoms. From the formula it is possible to calculate the percent composition of a compound. Going in the opposite direction from the composition of a compound, it is possible to calculate its empirical formula. Consider the compound chloroform. The percent composition of chloroform is 10.06% carbon, 0.85% hydrogen, and 89.09% chlorine. We know then that 100 g chloroform contain 10.06 g carbon, 0.85 g hydrogen, and 89.09 g chlorine. This weight relationship can be converted to a mole ratio by the following calculations:
| carbon | 10.06 g C | X | 1 mol C 12.01 g C |
= | 0.8376 mol C |
| hydrogen | 0.85 g H | X | 1 mol H 1.008 g H |
= | 0.84 mol H |
| chlorine | 89.09 g Cl | X | 1 mol Cl 35.45 g Cl |
= | 2.513 mol Cl |
These calculations show that the mole ratio between the elements in chloroform is 0.84 mol C to 0.84 mol H to 2.51 mol Cl. This ratio can be expressed by the formula:
C0.8376H0.84Cl2.513
However, formulas by definition can contain only whole numbers of atoms. The ratio can be changed to whole numbers by dividing each subscript by the smallest subscript, giving the formula of chloroform as:
C0.8376/0.8376H0.84/0.8376Cl2.513/0.8376 or CHCl3
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Example Calculate the empirical formula of a compound that contains 36.8% nitrogen and 63.25% oxygen Solution 1. Assume that you have 100 g of the compound, which will contain 36.8 g nitrogen and 63.25 g oxygen. 2. Convert these weights to moles.
This calculation gives the formula: 3. Change this ratio to whole numbers: But the ratio is not yet whole numbers as wanted. If each subscript is multiplied by 2, the subscripts then become whole numbers giving
the correct empirical formula for the compound. |
This example presents a situation often encountered in calculating empirical
formulas. When the result of a calculation of an empirical formula contains
a subscript more than 0.1 away from a whole number, it must not be rounded off.
Rather, the whole formula must be multiplied by a factor that will make that
subscript a whole number. In general, when the subscript is 1.5, multiply by
2. When the subscript is 1.3 or 1.7, multiply by 3.
We have calculated the formulas of compounds from their percent composition. The following example shows how to determine the formula of a compound when its composition is given not in percent but in grams.
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Example Analysis of 3.23 g of a compound shows that it contains 0.728 g of phosphorus and 2.50 g chlorine. What is the empirical formula of the compound? Solution 1. Calculate the number of moles of each element in 3.23 g of the compound. 2. Convert these weights to moles.
This calculation gives the formula: 3. Change this ratio to whole numbers:
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