In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl
_{2} O
_{7} as the final empirical formula.
In summary, empirical formulas are derived from experimentally measured element masses by:
Deriving the number of moles of each element from its mass
Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained
[link] outlines this procedure in flow chart fashion for a substance containing elements A and X.
Determining a compound’s empirical formula from the masses of its elements
A sample of the black mineral hematite (
[link] ), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?
Solution
For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:
The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe
_{1} O
_{1.5} ). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:
For additional worked examples illustrating the derivation of empirical formulas, watch the brief
video clip.
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 (
[link] ). What is the empirical formula for this gas?
Solution
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: