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C1 0.150 O 0.525 = Cl 0.150 0.150 O 0.525 0.150 = ClO 3.5

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:

  1. Deriving the number of moles of each element from its mass
  2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
  3. 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.

A flow chart is shown that is composed of six boxes, two of which are connected together by a right facing arrow and located above two more that are also connected by a right-facing arrow. These two rows of boxes are connected vertically by a line that leads to a right-facing arrow and the last two boxes, connected by a final right facing arrow. The first two upper boxes have the phrases, “Mass of A atoms” and “Moles of A atoms” respectively, while the arrow that connects them has the phrase, “Divide by molar mass,” written below it. The second two bottom boxes have the phrases, “Mass of X atoms” and “Moles of X atoms” respectively, while the arrow that connects them has the phrase, “Divide by molar mass” written below it. The arrow that connects the upper and lower boxes to the last two boxes has the phrase “Divide by lowest number of moles” written below it. The last two boxes have the phrases, “A to X mole ratio” and “Empirical formula” respectively, while the arrow that connects them has the phrase, “Convert ratio to lowest whole numbers” written below it.
The empirical formula of a compound can be derived from the masses of all elements in the sample.

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?

Two rounded, smooth black stones are shown.
Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)


For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

34.97 g Fe ( mol Fe 55.85 g ) = 0.6261 mol Fe 15.03 g O ( mol O 16.00 g ) = 0.9394 mol O

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

0.6261 0.6261 = 1.000 mol Fe 0.9394 0.6261 = 1.500 mol O

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:

2 ( Fe 1 O 1.5 ) = Fe 2 O 3

The empirical formula is Fe 2 O 3 .

Check your learning

What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?


N 2 O 5

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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?

A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.
An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (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 % C = 27.29 g C 100 g compound 72.71 % O = 72.71 g O 100 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 g C ( mol C 12.01 g ) = 2.272 mol C 72.71 g O ( mol O 16.00 g ) = 4.544 mol O

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

2.272 mol C 2.272 = 1 4.544 mol O 2.272 = 2

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO 2 .

Check your learning

What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?


CH 2 O

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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