Empirical and Molecular Formula

So today............we're going to be doing Empirical and Molecular Formula.



What is the empirical formula?
It is the expression of the ratio of atoms in a formula in its lowest terms. By this definition, we can realize that all ionic compounds are empirical formulas.


So, for example, let's say we have the formula C₁₀H₁₄N₂, which is Nicotine (the thing that makes you addicted to cigarettes).


Now, to get the empirical formula, let's reduce all the subscripts to the lowest terms.
We'd get C₅H₇N. Of course, this is a whole new compound, but we can say this is the empirical formula of nicotine.

Now, find the empirical formula for a compound consisting of 63% Mn and 37% O. 

First, we assume that we have 100g of this compound, just to make everything easier.

Now, we take the 63 g(now grams of Mn, because 63% of 100 is 63) and convert it into moles.

Since the molar mass of Mn is 54.9, we do 

63 g X 1 mole  = 1.1 moles Mn (in sig figs)

             54.9g        

Then, we take the 37 g of O and do the same.

37 g X 1 mole = 2.3 moles O 

             16.0g

Now, since we have both parts of the compound in moles, we now divide each by the smallest molar amount. If you're going like HUH? just wait for the next step.

So, we can all agree that the number 1.1 is less than 2.3 right? 

That will be what we are dividing by. 

We take the 1.1 moles of Mn and divide by 1.1, giving us the whole number 1.

Then, we take 2.3 moles of O and divide by 1.1, giving us 2.1, which is around 2. 

The final part is to scale the ratios into whole numbers. But good thing for us, the numbers in this example are already whole numbers, so we don't need to multiply again. (In other words, we don't get 1.5, which means we would have to multiply by 2 to get a whole number)

We will get 1 Mn, and 2 O.

Our final answer for this question is...... MnO2.

OK..Now let's do some molecular formula. This is a multiple of the empirical formula and it can show the actual number of atoms that are necessary to combine into a molecule.



For this, we have a special formula. That is, the molar mass of a compound divided by molar mass of the empirical formula, will give us the molecular formula.


Let's say


The simplest formula for vitamin C is C₃H₄O₃. Experimental data indicates that the molar mass of vitamin C is about 180 g/mol. What is the molecular formula of vitamin C?


First, we take the empirical formula, and add up the molar masses of each element to get the total molar mass. Get your periodic table for this - you'll need it.


The molar mass of C is 12.0g, and we have 3, so 3 X 12 = 36.0 
Then, we have 4 Hydrogen, with each hydrogen at 1.0g each, so 4 X 1 = 4.0
Finally, we have 3 Oxygen at 16.0 g each, so 3 X 16 = 48.0
36.0 + 4.0 + 48.0 = 88.0 g/mol


Now we have both molar mass of compound, and the molar mass of the empirical formula. The last step for us is to put it into the formula.


So 180 g/mol  = 2.045, which is around 2.
       88 g/mol


This means that we need to take that original formula and multiply it entirely by 2.
2 X C₃H₄O₃
This gives us C₆H₈O₆, which means the molecular formula of vitamin C is C₆H₈O<sub>6


Time for a video! :)</sub> 

If you're still a little bit unsure, be sure to check out these helpful links!
http://www.ausetute.com.au/empirical.html
http://galileo.stmarys-ca.edu/jsigman/Chem07/MolecularFormulaPractice.htm
http://dl.clackamas.edu/ch104-03/practice.htm

That's all. Bye