1.24 calculate empirical and molecular formulae from experimental data

This is in three sections

Empirical formulae show the simplest ratio of atoms in a compound. Molecular formulae show the actual formulae of compounds. Often the two formulae are, in fact, the same.

1 - Empirical Formulae from Masses

e.g.      1.15 g of sodium and 0.40 g of oxygen react to give sodium oxide. What is the empirical formula of sodium oxide?

Step 1 – Draw a table like this:

	Na	O
Mass	1.15	0.4

Step 2 – Divide the masses by the relative atomic masses for each elements (giving moles):

	Na	O
Mass Moles (Mass÷Ar)	1.15 1.15/23 = 0.050	0.4 0.4/16 = 0.025

Step 3 – Work out the ratio of the above numbers. If you can’t see it, divide each number by the smallest number:

	Na	O
Mass	1.15	0.4
Moles	0.050	0.025
Moles ÷ Smallest Number	0.050/0.025 = 2	0.025 / 0.025 = 1

Step 4 – Use the ratio to give the empirical formula:

Ratio = 2 Na : 1 O

Therefore, empirical formula = Na2O

2 - Empirical Formula from Percentages

e.g.     A compound is made up of 40% Ca, 12% C and the rest is oxygen. Calculate the empirical formula.

Step 1 – Draw a table like this:

	Ca	C	O
%	40	12

Step 2 – Work out the missing percentage (sometimes you won’t need to do this):

	Ca	C	O
%	40	12	100 - 40 - 12 = 48

Step 3 – Divide the % values by the RAMs:

	Ca	C	O
% Moles (≡ % ÷ Ar)	40 40/40 = 1	12 12/12 = 1	48 48/16 = 3

Step 4 –  Work out the ratios and hence the empirical formula. Here is it is obvious. If not, divide each by the smallest number as shown in Example 1.

Ratio = 1 Ca : 1 C : 3 O

Therefore, empirical formula = CaCO3

3 - Molecular Formulae vs Empirical Formulae

•        The Empirical Formula is the simplest whole number ratio of atoms of each element in a compound. (This may or may not be the same as the Molecular Formula.)

•        The Molecular Formula is the ACTUAL number of atoms of each element in a molecule of the compound.

We can work out the Molecular Formula if we know the Empirical Formula (see Examples 1 and 2) and the Relative Formula Mass.

e.g.      A compound has the empirical formula CH2.  It has a Relative Formula Mass of 28.  What is its Molecular Formula?

Step 1  Work out the molar mass of the Empirical Formula unit:                  CH2 = 1 x 12 + 2 x 1 = 14

Step 2   Divide the Relative Formula Mass by the answer to Step 1: 28 ÷ 14 = 2

Step 3  Multiply the Empirical Formula by the answer to Step 2:                     CH2 x 2 = 2 C and 4 H

Therefore, the Molecular Formula = C2H4

You may find the answer to Step 2 is one. This simply means the Empirical Formula and the Molecular Formula are the same. There is nothing unusual about this, so don’t be surprised by it in the exam.

If you are given mass or percentage data for a compound, and are asked to work out its Molecular Formula, carry out Example 1 or Example 2 as appropriate, and then finish off with the steps shown in Example 3.

Calculations require practice! Show working clearly and use units