Empirical Formula 1

(Calculation of the empirical formula from mass data.)

Empirical Formula: The simplest ratio of the atoms present in a molecule.

Problem:

Find the empirical formula for the oxide that contains 42.05 g of nitrogen and 95.95 g of oxygen.

Strategy:

As with most stoichiometry problems, it is necessary to work in moles. The ratio of the moles of each element will provide the ratio of the atoms of each element.

  1. Convert the mass of each element to moles of each element using the atomic masses.

    (42.05 g N) (1 mol/ 14.01 g N) = 3.001 mol N

    (95.95 g O) (1 mol/ 16.00 g O) = 5.997 mol O

  2. Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles.

    There are fewest moles of nitrogen, so assume one mole of nitrogen in finding the ratio.

    (3.001 mol N/ 3.001) = 1 mol N

    (5.997 mol O/ 3.001) = 1.998 mol O

  3. Use the mole ratio to write the empirical formula.

    Since there cannot be partial atoms in the empirical formula, the mole ratios must be whole numbers. 1.998 is sufficiently close to 2 that we can round. Thus the empirical formula is:

    NO2

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