A formula that gives the simplest whole number ration of atoms in a compound found from an experiment. Empirical Formula means ‘from experiment’.
Molecular formula is the actual ration for a molecule
For example for Benzene:
– Actual Formula: C6H6
– Empirical Formula: CH
Examples
- Determine the Empirical Formula of methane given that 6.0g of Methane can be decomposed into 4.5g of carbon and 1.5g of hydrogen. Work out the empirical formula.
|
Carbon |
Hydrogen |
|
|
Mass |
4.5g |
1.5g |
|
Ar |
12 |
1 |
|
Moles |
4.5g / 12 = 0.375 |
1.5g /1 = 11.5 |
|
Divide by smallest |
0.375 / 0.375 |
1.5 / 0.375 |
|
Ratio |
1 |
4 |
= CH4
Empirical Formula from percentages
Empirical formula can be used using percentages, just using the percentage instead of it’s mass.
Examples
- The composition of a compound is 40% sulphur and 60% oxygen by weight. What is its empirical formula?
|
Sulphur |
Oxygen |
|
|
Mass |
40% |
60% |
|
Ar |
32.1 |
16.0 |
|
Moles |
40 / 32.1 = 1.25 |
60% / 16.0 = 3.75 |
|
Divide by smallest |
1.25 / 1.25 |
3.25 / 1.25 |
|
Ratio |
1 |
3 |
= SO3
- Calculate the percentage of iron present in Fe2O3
2Fe = 2 x 55.8 x100 = 69.92%
Fe2O3 2 x 55.8 + 48
- Percentage composition of 81.8% carbon, 6.1% hydrogen and 12.1% oxygen.
|
Carbon |
Oxygen |
Hydrogen |
|
|
Mass |
81.8% |
12.1% |
6.1% |
|
Ar |
12.0 |
16.0 |
1 |
|
Moles |
81.8 / 12.0 = 6.16 |
12.1% / 16.0 = 0.75625 |
6.1 / 1 = 6.1 |
|
Divide by smallest |
6.16 / 0.75625 |
0.75625/ 0.75625 |
6.1/0.75625 |
|
Ratio |
9 |
1 |
8 |
= C9H8O
- Blue Sulphate crystals yield the following percentage composition, Cu – 25%, S – 12.8%, O – 25.6% and water crystallisation – 36.1%. Find the empirical formula of the crystals
|
Copper |
Sulphur |
Oxygen |
H2O |
|
|
Mass |
25.0% |
12.1% |
25.6% |
36.1% |
|
Ar |
63.5 |
32.1 |
16.0 |
18 |
|
Moles |
25.0 / 63.5 = 0.39 |
12.1% / 32.1 = 0.37 |
25.6 / 16.0 = 1.60 |
36.1 / 18 =2.01 |
|
Divide by smallest |
0.39 / 0.37 |
0.37 / 0.37 |
1.60 / 0.37 |
2.01/ 0.37 |
|
Ratio |
1 |
1 |
4 |
5 |
= CuSO4 • 5H2O
- A sugar solution contains 40% carbon, 6.7% hydrogen and 53.3% oxygen.
|
Carbon |
Oxygen |
Hydrogen |
|
|
Mass |
40.0% |
53.3% |
6.7% |
|
Ar |
12.0 |
16.0 |
1 |
|
Moles |
40.0 / 12.0 = 3.3 |
53.3% / 16.0 = 3.3 |
6.7 / 1 = 6.7 |
|
Divide by smallest |
3.3 / 3.3 |
3.3 / 3.3 |
3.3 / 3.3 |
|
Ratio |
1 |
1 |
2 |
= CH2O
- 24% Carbon and 76% fluorine
|
Carbon |
Fluorine |
|
|
Mass |
24% |
76% |
|
Ar |
12 |
19 |
|
Moles |
24 / 12 = 2 |
76% / 19 =4 |
|
Divide by smallest |
2 / 2 |
1.5 / 0.375 |
|
Ratio |
1 |
2 |
Calculating molecular formula from empirical formula
This can be deduced from the empirical formula if the Mr is known (as the Mr can come from mass spectrometry)
In mass spectrometry the molecular ion is the last peak on the graph. The Mr is this value.
21% in this example graph is the Molecular Ion.
A compound has an empirical formula of ClCH2 and a molecular weight of 98.96 g/mol. What is its molecular weight formula
Mass of empirical unit?
Mass of Chlorine + Mass of Carbon + Mass of Hydrogen2
35.5 + 12 + 2 = 49.50 g/mol
Empirical Formula = Mass of Molecular unit = 98.96
Mass of Empirical Formula 49.50
= 2.00 empirical units per molecular unit
It takes two empirical units to make a molecular unit, so the molecular formula is Cl2C2H4