Information about reacting masses is used to calculate empirical formulae. This is obtained from experiments.
A hydrocarbon contains 4.8 g of carbon and 1.0 g of hydrogen. Calculate its empirical formula. (Ar of C = 12, Ar of H = 1)
Ar refers to the relative atomic mass of an element.
|1||Write the element symbols||C||H|
|2||Write the masses||4.8 g||1.0 g|
|3||Write the Ar values||12||1|
|4||Divide masses by Ar||4.8 ÷ 12 = 0.4||1.0 ÷ 1 = 1|
|5||Divide by the smallest number||0.4 ÷ 0.4 = 1||1.0 ÷ 0.4 = 2.5|
|6||Multiply (if needed) to get whole numbers||1 × 2 = 2||2.5 × 2 = 5|
|7||Write the formula||C2H5||C2H5|
The action at step 5 usually gives the simplest whole number ratio straightaway. Sometimes it does not, so both numbers may need to be multiplied to get a whole number (step 6). For example, multiply by 2 if there is a .5, by 3 if the number has a .33, or by 4 if it has a .25 in a number.
3.2 g of sulfur reacts completely with oxygen to produce 6.4 g of an oxide of sulfur. Calculate the empirical formula of the oxide of sulfur. (Ar of S = 32, Ar of O = 16)
|1||Write the element symbols||S||O|
|2||Write the masses||3.2 g||6.4 g - 3.2 g = 3.2 g|
|3||Write the Ar values||32||16|
|4||Divide masses by Ar||3.2 ÷ 32 = 0.1||3.2 ÷ 16 = 0.2|
|5||Divide by the smallest number||0.1 ÷ 0.1 = 1||0.2 ÷ 0.1 = 2|
|6||Multiply (if needed) to get whole numbers||(not needed - already have whole numbers)||(not needed - already have whole numbers)|
|7||Write the formula||SO2||SO2|
The empirical formula can worked out from diagrams of molecules. Here is a diagram showing a molecule of butane:
The molecular formula of butane is C4H10. This is the actual number of atoms of each element in a molecule of butane. This formula does not show the simplest whole number ratio because each number can be divided by 2. This gives the empirical formula of butane - C2H5.
The molecular formula and empirical formula of some substances are the same. For example, both types of formula for carbon dioxide are CO2.