# 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).

Remember 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.

There are fewer moles of oxygen than any other element, so we will assume
one mole of oxygen to establish the ratios.

(3.336 mol O/ 3.336) = *1 mol O*
(4.028 mol C/ 3.336) = *1.2 mol C*

(8.056 mol H/ 3.336) = *2.4 mol H*

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

**The mole ratio did not turn out to be whole numbers. Since the we cannot
have partial atoms in the empirical formula, a multiplication factor must
be applied to get whole numbers. In this case, 5 is the factor we need. **
**(1 mol O) (5) = 5 mol O**

**(1.2 mol C) (5) = 6 mol C**

**(2.4 mol H) (5) = 12 mol H**

**Now, the ratios are whole numbers, and we can write the empirical formula:**

**C**_{6}**H**_{12}**O**_{5}