Empirical Formula 2
Empirical Formula: The simplest ratio of the atoms present in a molecule.
Problem:
Find the empirical formula of a compound that is 48.38% carbon, 8.12% hydrogen,
and 53.5% oxygen by mass.
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.
- Get the mass of each element by assuming a certain overall mass for the
sample (100 g is a good mass to assume when working with percentages).
Remeber that percentages are a ratio multiplied by 100. You must convert
percentages back to their decimal value before working with them.
(.4838) (100 g) = 48.38 g C
(.0812 ) (100 g) = 8.12 g H
(.5350) (100 g) = 53.38 g O
- Convert the mass of each element to moles of each element using the atomic
masses.
(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C
(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H
(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O
- Find the ratio or the moles of each element by dividing the number of moles
of each by the smallest number of moles.
- Use the mole ratio to write the empirical formula.
