Percent Composition and Empirical Formulas for AP Chemistry

Practice problems for these concepts can be found at:

• Stoichiometry: Review Questions for AP Chemistry
• Stoichiometry: Free-Response Questions for AP Chemistry

If the formula of a compound is known, it is a fairly straightforward task to determine the percent composition of each element in the compound. For example, suppose you want to calculate the percentage hydrogen and oxygen in water, H₂O. First calculate the molecular mass of water:

1 mol H₂O = 2 mol H + 1 mol O

Substituting the masses involved:

1 mol H₂O = 2 (1.0079 g/mol) + 16.00 g/mol = 18.0158 g/mol

(intermediate calculation—don't worry about significant figures yet)

As a good check, add the percentages together. They should equal to 100% or be very close.

Determine the mass percent of each of the elements in C₆H₁₂O₆ Formula mass (FM) = 180.158 amu

Answer:

The total is a check. It should be very close to 100%.

In the problems above, the percentage data was calculated from the chemical formula, but the empirical formula can be determined if the percent compositions of the various elements are known. The empirical formula tells us what elements are present in the compound and the simplest whole-number ratio of elements. The data may be in terms of percentage, or mass, or even moles. But the procedure is still the same: convert each to moles, divide each by the smallest number, then use an appropriate multiplier if needed. The empirical formula mass can then be calculated. If the actual molecular mass is known, dividing the molecular mass by the empirical formula mass gives an integer (rounded if needed) that is used to multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells which elements are in the compound and the actual number of each.

For example, a sample of a gas was analyzed and found to contain 2.34 g of nitrogen and 5.34 g of oxygen. The molar mass of the gas was determined to be about 90 g/mol. What are the empirical and molecular formulas of this gas?

The molecular formula may be determined by dividing the actual molar mass of the compound by the empirical molar mass. In this case the empirical molar mass is 46 g/mol.

Thus which, to one significant figure, is 2. Therefore, the molecular formula is twice the empirical formula—N₂O₄.

Be sure to use as many significant digits as possible in the molar masses. Failure to do so may give you erroneous ratio and empirical formulas.

Practice problems for these concepts can be found at:

• Stoichiometry: Review Questions for AP Chemistry
• Stoichiometry: Free-Response Questions for AP Chemistry