Representation of Compounds

The formula for a chemical compound gives us information about the relative proportions of the different elements that constitute it. Conversely, knowledge of the composition of a compound enables us to determine its (empirical) formula. Knowing how to represent chemical compounds, and knowing how to determine a compound’s formula, is very important for the SAT Subject Test: Chemistry.

Law of Constant Composition

The law of constant composition states that any sample of a given compound will contain the same elements in the identical mass ratio. For instance, every sample of H2O will contain two atoms of hydrogen for every atom of oxygen, or, in other words, one gram of hydrogen for every eight grams of oxygen. This is hardly surprising since we already know that atoms prefer an octet structure and would combine with other atoms in predictable ways to achieve this.

Empirical and Molecular Formulas

There are two ways to express a formula for a compound. The empirical formula gives the simplest whole number ratio of the elements in the compound. The molecular formula gives the exact number of atoms of each element in a molecule of the compound, and is a multiple of the empirical formula (including a multiple of 1—that is, same as the empirical formula). For example, benzene is a molecule where six carbon atoms are joined together in a ring, with a hydrogen atom attached to each of them. Its molecular formula is therefore C6H6, but its empirical formula is just CH. For some compounds, the empirical and molecular formulas are the same, as in the case of H2O. An ionic compound, such as NaCl or CaCO3, will have only an empirical formula since there are no real molecules in the solid state in these cases, as discussed above.

Given a molecular formula, you can always write the empirical formula just by looking to see whether the numbers of atoms are already in the smallest whole number ratio. If not, you can factor out the common factor among them. C2H4 is not an empirical formula because you can factor out a two from the subscripts to get CH2. CH4, on the other hand, is already an empirical formula. If you are given an empirical formula, however, you need to know the molecular weight (or molar mass) of the compound to find out the actual molecular formula.

  1. Example: A compound with the empirical formula CH2O has a weight of 180 g/mol. What is the molecular formula?
  2. Solution: Let us first find what the formula weight is from the empirical formula: 1 × mass of carbon atom + 2 × mass of hydrogen atom + 1 × mass of oxygen atom = (1 × 12 + 2 × 1 + 1 × 16) g/mol = 30 g/mol
    The actual molecular weight is 6 times this; therefore the molecular formula must be 6 times the empirical formula: C6H12O6.

Percent Composition

The percent composition by mass of an element is the weight percent of the element in a specific compound. To determine the percent composition of an element X in a compound, the following formula is used:

The percent composition of an element may be determined using either the empirical or molecular formula.

  1. Example: What is the percent composition of chromium in K2Cr2O7?
  2. Solution: The formula weight of K2Cr2O7 is:
    2(39 g/mol) + 2(52 g/mol) + 7(16 g/mol) = 294 g/mol
    % composition of
  1. Example: What are the empirical and molecular formulas of a compound that contains 40.9% carbon, 4.58% hydrogen, 54.52% oxygen, and has a molecular weight of 264 g/mol?
  2. Solution: First, assume that we have a sample that weighs 100 g total. The percentage then translates directly into the weight of that element in the sample (e.g., 40.9% by weight means 40.9 g in a 100-g sample). Then convert grams to moles by dividing the weight of each element by its molar atomic mass:
  1. Next, find the simplest whole number ratio of the elements by dividing the number of moles by the smallest number obtained in the previous step.
  1. Finally, the empirical formula is obtained by converting the numbers obtained into whole numbers (multiplying them by an integer value). In this case, we want to turn 1.33 into an integer; the smallest number we can multiply it by to make it an integer is 3:
1.33 × 3 = 4
  1. The empirical formula is therefore 3 × C1H1.33O1 = C3H4O3.

This method gives the empirical formula because the elements are always in their smallest whole number ratio. A molecular formula of C6H8O6, which is a multiple of two of the empirical formula, would be entirely consistent with the percent compositions given above: You cannot distinguish between the two, or any multiple of the empirical formula, just by percent compositions alone. Incidentally, this is how the term empirical formula gets its name: The word empirical means experimental, and the values of percent compositions, obtained experimentally through simple analytical techniques, only allow us to determine the empirical formula. (Of course, nowadays with modern technology, we are not limited to experimental techniques that would only give us percent compositions.)

For the second part of the question on the molecular formula, we can use the same approach discussed earlier: Divide the molecular weight by the weight represented by the empirical formula. The resultant value is the number of empirical formula units in the molecular formula. The empirical formula weight of C3H4O3 is:

3(12 g/mol) + 4(1 g/mol) + 3(16 g/mol) = 88 g/mol

The molecular weight is given to be 264 g/mol. Therefore:

C3H4O3 × 3 = C9H12O9 is the molecular formula.