Empirical Formula
This is the simplest whole number ratio of number of atoms of each element in a substance.
You normally encounter this during calculations involving %mass.
Let’s look at such a calculation in reverse to see where the information comes from. Consider the molecule ETHANOIC ACID CH3COOH It is written this way to highlight some aspects of the structure but its molecular formula could just as easily be written as C2H4O2.
From this it is clear that its empirical formula is CH2O because this shows the ratio of the numbers of atoms as being 1:2:1.
If we think about the mass of each of these atoms we know that the mass of each carbon atom is 12 x the mass of a hydrogen atom. Each oxygen atom is 1 x the mass of the hydrogen.
So the ratio of the masses of each kind of atom (remembering that the empirical formula contains two hydrogen atoms for every carbon and oxygen) is going to be 12:2:16. We can write this as a percentage of the total, (12/30) x100% for carbon (2/30) x100% for hydrogen and (16/30) x100% for oxygen.
This gives the percentage by mass to be 40.0% C, 6.7% H, 53.3% O.
If we had been given this information, could we have deduced the empirical formula? Yes, just do the calculation in reverse. First we divide each percentage mass by the appropriate ram. C 40/12 to give 3.33 H 6.7/1 to give 6.7 O 53.3/16 to give 3.33
Then divide each value by the smallest value (3.33 in this case) C 1, H 2, O 1
How could we work out the molecular formula? To do this we would have to be told the relative molecular mass. For ethanoic acid, it is 60. The CH2O unit has a formula mass of 30 (12 + 2 + 16). So we need two of the CH2O units. This means that the molecular formula is C2H4O2 .
How could we work out the structure? To do this we would need much more information. Perhaps we could be told some nuclear magnetic resonance data or some infra red data. Even the fact that the molecule reacts with sodium carbonate solution would be a start!
QUESTIONS
Calculate the empirical formula for each of the following substances.
You should use the following values for relative atomic mass:
H = 1 N = 14 O = 16 P = 31 S = 32 Cu = 64
1. What is the empirical formula of the phosphorus oxide that has 43.7% by mass of phosphorus and 28.4% by mass oxygen?
2. An oxide of nitrogen has 69.6% by mass of oxygen. What is the empirical formula?
3. What is the empirical formula of this compound of copper. It has the following percentages by mass. Cu 25.9% S 12.7% O 57.4% H 4%
Answers
1. P2O5
2. NO2
3. CuSO9H10
Formula from Mass Composition
Figuring out the empirical formula from a molecules mass composition
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