Empirical & Molecular Formulas + Percent Composition!

Welcome back Beryllium Chemist followers! Today we will be talking about a few things. Firstly, the difference between empirical and molecular formulas. Then, we will move on to percent composition! (That's a lot!) I hope you'll survive through this blog!

In chemistry, there are two types of formulas that you MUST know: molecular and empirical

Empirical Formula

• shows the RATIO of the atoms or moles in LOWEST-TERMS 
• when you form ionic compounds on the periodic table, they are in empirical formulas since they're in lowest terms
• to get to empirical formula, divide molecular formula by its greatest factor
• you can calculate the empirical formula from percent composition and by mass
  
  eg. NaCl is an empirical formula but Na2Cl2 is not

Molecular Formula

• shows the total atoms each element are present in the molecule of the compound
• it is a MULTIPLE and not in lowest terms
• formula of the ionic and covalent compounds
  
  for example, C3H6 is a molecular formula because it's simplest form is CH2

Here is the key to converting:

Empirical Formula x whole number = Molecular Formula
Empirical Formula MASS x whole number = Molecular Formula MASS
Empirical Formula Mass (GRAMS) x whole number = mass of ONE MOLE of the compound


So if carbon and hydrogen are present in a compound in a ratio of 1:2, then the EF would be CH2. What is the molecular formula if 28.0g is the mass of one mole?

Step 1: Find the empirical formula mass.
12.0 + 2(1.0) = 14.0g/mol

Step 2: Divide the molar mass with the empirical formula mass.
(remember molecular formula mass = EF mass x whole number)

28.0 / 14.0 = whole number = 2

Step 3: Multiply whole number by the empirical formula.
Recall: Molecular formula = empirical formula x whole number

2(CH2) = C2H4


Now lets try some questions where we try to determine the EF with a given mass!
If you are given the question: Determine the empirical formula with the given 30.4g Nitrogen and 69.6g Oxygen. Follow through with these steps.

Step 1: Convert grams to moles.
Nitrogen: 30.4g x 1mole/14g N = 2.1714 mol
Oxygen: 69.6g x 1mole/16g = 4.35 mol


Step 2: Divide each molar mass by the smallest molar amount.
Nitrogen: 2.1714/2.1714 = 1
Oxygen: 4.35/2.1714 = 2.003
And voila! Your final empirical formula is NO2.


BUT, if after step 2 your answers don't round easily to a whole number, you have to multiply them by a whole number.
For example: If you have 1.7344 Carbon and 4.2332 Nitrogen

Multiply by 2, and keeping going till you get a number close enough to a whole number
Carbon: 1.7344 x 2 = 3.4688. 1.7344 x 3 = 5.2032. 1.7344 x 4 = 6.9376
Nitrogen: 4.2332 x 2 = 8.4664. 4.2332 x 4 = 12.6996. 4.2332 x 4 = 16.9328Therefore your empirical formula will be C7N17


Hope this helps guys:)
-Kimberly