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Empirical Formulas from Percent Composition

Deriving Empirical Formulas from Percent Composition

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.

Determining an Empirical Formula from Percent Composition

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (see image below). What is the empirical formula for this gas?


An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. Image credit: “Dual Freq”/Wikimedia Commons


Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage.

This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

\(27.29\%\text{C} = \frac{27.29\text{ g C}}{100\text{ g compound}}\)

\(72.71\%\text{O} = \frac{72.71\text{ g O}}{100\text{ g compound}}\)

The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

\(27.29\text{ g C} \left (\frac{\text{mol C}}{12.01\text{ g}} \right ) = 2.272\text{ mol C}\)

\(72.71\text{ g O} \left (\frac{\text{mol O}}{16.00\text{ g}} \right ) = 4.544\text{ mol O}\)

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

\(\frac{2.272\text{ mol C}}{2.272} = 1\)

\(\frac{4.544\text{ mol O}}{2.272} = 2\)

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO2.

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