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Empirical formula

This is simply used to show a formula ratio between different elements. If you know the mass of each element in a compound you can calculate its empirical formula. Remember from the amount of substance post that we learnt about an important equation:

n = m / M    
                   n = Amount of substance
                   m = Mass
                   M = Molar mass

Let works with an example to show the empirical formula:

Analysis shows that 0.6075g of magnesium combines with 3.995g of bromine to form a compound.

Ar is the mass number of a singular atom, in this case it would be:

Mg = 24.3
Br = 79.9

                                                                       Mg   :   Br
 
This is where the ratio begins, starting with the mass divided by the molar mass. So I will do this separately first:
 
Mg = 0.6075 / 24.3
 
Resulting in 0.025
 
Br = 3.995 / 79.9
 
Resulting in 0.050

Ratio values do not have units so you don't need to stress about units here!
 
The weight of the element is given in the question and the relative atomic mass is found in any standardized periodic table of elements. From this we have simply divided the weight of the given element by the relative atomic mass! Using this we can now expand further with our next step of the ratio:
 
Mg    :    Br
0.6075/24.3   :    3.995/79.9
0.025    :   0.050
 
The last stage of this is to divide the smallest output number (0.025) into each of their ratio values. This is because ratios should always be whole numbers and not left as a decimal.
 
0.025 / 0.025 = 1
0.050 / 0.025 = 2
 
Making the ratio:

                                                                         Mg    :    Br
                                                       0.6075/24.3    :    3.995/79.9       weight / relative atomic massࣸ ⮯
                                                                     0.025    :    0.050      divide by 0.025 (smallest number)⮯      
1    :    2
 
 
This simply makes the compound MgBr₂ due to the output from the ratio. This is because for every one magnesium there is two bromine and this is simply plugged into the symbol form of this compound.
 
This isn't quite over yet as this kind of formula can be shown in a different way in question formats. Lets run through this one as well step by step:
 
Analysis of a compound showed that the following percentage composition for mass:
 
Na: 74.9%
O: 25.81%
 
Now once again we find the relative atomic mass of these individual elements:
 
Na - 23
O - 16
 
So to make this simple always assume you are using 100g of said compound as considering percentages in 100g is easiest as 1g is equal to 1% of the substance. As such the distribution would be:
 
Na - 74.9g
O - 25.81g
 
Then it's just a repeat of before:
 
Na = 74.19 / 23 = 3.226
O = 25.81 / 16 = 1.613
 
 
Na    :    O
                                                         74.19/23    :    25.81/16     weight / relative atomic massࣸ ⮯
                                                                   3.226    :    1.613     divide by 0.025 (smallest number)⮯   
 2    :    1
 
 
Making the compound Na₂O.
 
Simples, right?
 
Did you find this helpful? Feel free to drop a comment and let me know what you liked and if I have missed something!





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