Empirical formula

From a given chemical formula, we can generate a list of percent compositions of the elements that make up the compound.

We will now carry out the opposite task. Given the percent compositions of the elements that make up the compound, we will derive the chemical formula of the compound.

Example

An analysis of a substance shows that it is made up of carbon, hydrogen, and oxygen. The percentage composition of the elements in the compound are determined to be:

* 39.10 % carbon,
* 8.77 % hydrogen, and
* 52.13 % oxygen. 

From the above data derive the ratio of carbon, hydrogen and oxygen in the substance. In other words, find the formula in the form, C_xH_yO_z, where x, y and z are the simplest whole-number ratio of carbon, hydrogen to oxygen of the substance. The derived chemical formula is called the empirical formula of the substance.

• Step 1: Understand what the percentages mean.

When the elements of a compound are given as percent composition of the compound, it is always mass percentages. Then you can assume that if there is a 100.0 g sample of the compound:

* 39.10 g of the 100.0 g is contributed by the mass of carbon,
* 8.77 g of the 100.0 g is contributed by the mass of hydrogen,
* 52.13 g of the 100.0 g is contributed by the mass of oxygen. 

In other words,

* 39.10 g of carbon is equivalent to x' moles of carbon.
* 8.77 g of hydrogen is equivalent to y' moles of hydrogen.
* 52.13 g of oxygen is equivalent to z' moles of oxygen.

• Step 2: Convert grams of each element to moles.

To convert

divide by the atomic mass of the atom.

Element	Conversion to Moles	Moles of each Element
C:	(39.10 g C) / 12.011 g/mole C	x' = 3.258 mole C
H:	(8.77 g H) / 1.008 g/mole H	y' = 8.683 mole H
O:	(52.13 g O) / 15.999 g/mole O	z' = 3.258 mole O

The last column, x' (pronounced x prime), y' (pronounced y prime) and z' (pronounced z prime) gives the mole ratios of the elements.

• Step 3: Find the simplest mole ratio of C, H and O.

To derive the empirical formula of the substance, we need to find the smallest whole number mole ratio of C, H and O. This is carried out by dividing the number of moles of each element by the smallest of the mole ratios calculated.

Here, we divide through by 3.258 moles.

C: 	3.258 mole / 3.258 mole = 1.00

H: 	8.683 mole / 3.258 mole = 2.67

O: 	3.258 mole / 3.258 mole = 1.00

At this point, we manage to find the simplest mole ratio of C, H and O. If the ratio is in whole-numbers, then our job is complete. But we notice that the number ratio at this point is not whole-number, so we need to proceed to step 4.

Step 4: Convert the mole ratio obtained in Step 3 to the simplest whole-number mole ratio.

Here, by inspection, (or by trial and error), we find that it is necessary to multiply through by 3.

C: 	1.00 x 3 = 3

H: 	2.67 x 3 = 8

O: 	1.00 x 3 = 3

The empirical formula, which has the form _xH_yO_z, has x=3, y=8, and z=3.

The empirical formula is C₃H₈O₃.

Find out how the empirical formula relates to the molecular formula, the chemical formula of the compound.

Content suitability

BCIT courses: CHEM 0011