Empirical Formula
Class Notes
Introduction:
There are several ways of showing the formula of a compound:
- Displayed or Structural formula: Shows all atoms and bonds between them
- Condensed formula: Show the atoms and which elements are bonded to each but does not show the bonds
- Molecular formula: Shows only the atoms present in a compound
- Empirical formula: Minimum ratio of the elements in a molecular formula. It is a simplification of the molecular formula.
Empirical Formula is the minimum whole ratio between the atoms in a molecule expressed in Moles.
Having the % composition as data or the amount of each element in grams in the compound, it is easy to work out the empirical formula:
Steps to calculate the empirical formula:
- Find the number of grams of each element, present in the compound.
- Convert the mass of each element to moles using the molar mass from the periodic table. (mass /Ar in grams = # of moles)
- Divide each mole value by the smallest number of moles calculated in 2 to get whole number ratios.
Example 1:
Calculate the empirical formula for a compound that has 19.60g of Iron and 5.60 g of Oxygen
19.60 g Fe ÷ 56 g Fe/mol = 0.35 /0.35 = 1 ==> Fe1
5.60 g O ÷ 16 g O/mol = 0.35 / 0.35 = 1 ==> O1
Empirical Formula: FeO
Example 2:
Calculate the empirical formula for a compound that has 69.94% of Iron and 30.06% of Oxygen.
69.94 g Fe ÷ 56 g Fe/mol = 1.24893 / 1.24893 = 1
30.06 g O ÷ 16 g O/mol = 1.87875 / 1.24893 = 1.5
we cannot round numbers that can be expressed by a fraction. In this case, we need to multiply BOTH NUMBERS by two to get two WHOLE NUMBERS
Fe = 1 x 2 = 2
O = 1.5 x 2 = 3
Example 2:
Calculate the empirical formula for a compound that has 69.94% of Iron and 30.06% of Oxygen.
69.94 g Fe ÷ 56 g Fe/mol = 1.24893 / 1.24893 = 1
30.06 g O ÷ 16 g O/mol = 1.87875 / 1.24893 = 1.5
we cannot round numbers that can be expressed by a fraction. In this case, we need to multiply BOTH NUMBERS by two to get two WHOLE NUMBERS
Fe = 1 x 2 = 2
O = 1.5 x 2 = 3