It is frequently useful to determine the chemical formula of an unknown compound. One method is combustion analysis. If the unknown compound is known to contain at least carbon and hydrogen, heating a measured amount in the presence of excess oxygen, a process known as combustion, produces carbon dioxide and water, which can be collected and weighed individually. All of the carbon atoms are contained in the carbon dioxide, and all of the hydrogen atoms are contained in the water. Since 1 mole of carbon dioxide is equivalent to 1 mole of carbon atoms, the number of moles of carbon and hydrogen in the unknown compound can be determined. If oxygen is also present in the unknown, it is distributed over the carbon dioxide and water products; its mass can be found by difference and then converted to moles of oxygen in the original or unknown compound. The mass or moles of other elements, such as nitrogen and sulfur, which form known compounds with oxygen, can be determined in the same fashion.

Another method is known as elemental analysis. A measured amount of the unknown substance is heated until it decomposes into its constituent elements, which are collected and analyzed individually. This method is called pyrolysis which is the intense heating of a compound or mixture in the absence of oxygen. The percent composition by mass of the compound is obtained. Example 1.7 illustrates how percent composition data can be used to deduce the chemical formula and ultimately the identity of an unknown substance.

Example 1.7

The percent composition by mass of an unknown compound is 40.9% carbon, 4.57% hydrogen, and 54.5% oxygen. By a separate analysis, its molecular weight or molar mass is found to be 176 g/mol. Find the molecular formula for the unknown compound:

A. the empirical formula

B. the molecular formula


A. Recall that the empirical formula represents the smallest group or combination of atoms, in the proper ratio, of which the molecule is composed. Assume a 100-g sample to work with. Then the given percentages can be translated directly into grams as follows:

Next, convert these masses to moles by dividing each one by its respective atomic mass:

At this point, the chemistry is done, and one could write the empirical formula in principle, using the calculated numbers as the subscripts: C3.41H4.53O3.41-

Chemical formula rules stipulate, however, that these subscripts must not only be in the proper ratio but also must be whole numbers (integers)- The problem now is to find a mathematical technique to maintain the ratio but change the subscripts into integers- One way is to divide through by the smallest number:

C3.41/3.41H4.53/3.41O3.41/3.41 = C1.00H1.33O1.00-

Only the 1.33 subscript is not an integer. This can be corrected by multiplying through by the factor 3:

C3.00H4 .00O3.00

or simply written as C3H4O3

This, then, is the empirical formula.

B. To find the molecular formula, divide the molar mass by the empirical mass. This ratio should always be a whole number (to three significant figures). Then multiply each of the subscripts in the empirical formula by this factor to obtain the molecular formula.

The empirical mass of the above chemical formula (C3H4O3) is calculated in the same way as a molar mass would be and equals 88.0 g/mol. Thus, the factor is:

Therefore, the correct molecular formula is СбН80б-

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