Procedure for Calculating Empirical
Formulas of Compounds^©

By: Patrick Williford

Definition:

The empirical formula of a chemical substance is the substance stated with its elements noted as the smallest ratio of atoms (or lowest common subscripts).

For example: The formula for benzene is C₆H₆, but the smallest ratio of atoms for benzene is CH, so the empirical formula of benzene is CH. For lactic acid the formula is CH₃CH(OH)CO₂H => C₃H₆O₃, the empirical formula is CH₃O. However, several other chemicals have this same formula (as opposed to structure) including 1,3-dihydroxy acetone (HOCH₂)₂CO => C₃H₆O₃ which also has the empirical formula CH₃O.

Information Needed:

The identity of all elements of the compound.
The percent composition, the theoretical mass composition, or the experimentally measured mass composition of the compound.

Procedure:

If the percent composition of the compound is available, change it to the mass composition, in grams, by assuming; that 100.0 grams of the compound exists. Otherwise, itemize the mass composition that is available.
Convert the mass of each element from grams to moles by using the appropriate atomic mass.
Divide each mole value obtained in (2) by the smallest mole value obtained. Then look for values that approximate a fractional value (e.g., 2.33 = 7/3 and 1.5 = 3/2).
If the results of (3) involves only whole numbers, then the whole numbers indicate the atom ratio and should be used as the appropriate subscripts for writing the empirical formula of the compound.
If one or more of the results of (3) approximate a decimal fractions, then rewrite the decimal fractions as simple fractions (e.g., 1.67 = 5/3), then clear fractions by multiplying every result obtained in (3) by the least common denominator (LCD). The whole numbers obtained are used as the subscripts for writing the empirical formula of the compound.

Reasoning:

Atoms in a compound are integer values, one cannot have a fraction of an atom, so one must derive the integer ratio of atoms by multiplying each fractional value obtained by the LCD.

Example:

Combustion analysis of 0.2000 g of a compound yields 0.2998 g CO₂ and 0.0819 g H₂O, and one wants to find the empirical formula and percent composition of the compound.

One must identity each element C, H, and O.

Step #1 (Calculation of the number of moles of each substance):

Moles of CO₂ = (0.2998 g CO₂) x (1 mol CO₂ / 44.01 g CO₂) = 0.006812 mol CO₂

Moles of C = (0.2998 g CO₂) x (1 mol CO₂ / 44.01 g CO₂) x (1 mol C / 1 mol CO₂) = 0.006812 mol C

Moles of H = (0.0819 g H₂O) x (1 mol H₂O/ 18.02 g H₂O) x (2 mol H / 1 mol H₂O) = 0.00909 mol H

Since oxygen is present in excess (from the air) one needs to calculate the number of moles of it by using the calculated information already obtained. First one needs to know how many grams O there are in the compound, then one can calculate the number of moles O in the compound.

Grams of O = grams of unknown compound - grams of C - grams of H

Grams of C = (0.006812 mol C) x (12.01 g C / 1 mol C) = 0.08181 g C

Grams of H = (0.00909 mol H) x (1.008 g H / 1 mol H) = 0.00916 g H

Grams of O = 0.2000 g - 0.08181 g - 0.00916 g = 0.109 g O

Moles of O = (0.109 g O) x (1 mol O / 16.00 g O) = 0.00681 mol O

Step #2 (Calculation of the Empirical Formula):

One now has the beginning empirical formula C_0.006812H_0.00909O_0.00681.

Divide all values by the smallest value 0.00681.

The next empirical formula looks like C₁H_1.33O₁.

The last empirical formula is found by multiplying each subscript by the LCD, '3', to give C₃H₄O₃.

Step #3 (Calculation of percent mass of each element):

%C = (0.08181 g C / 0.2000 g substance) x 100% = 40.90%

%H = (0.00916 g H / 0.2000 g substance) x 100% = 4.58%

%O = (0.109 g O / 0.2000 g substance) x 100% = 54.5%

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Last Updated 23 January 1999
Last Accessed - - - Copyright 1997