Empirical Formula of Organic Compounds By Sally Chen

- An organic compound is a covalent compound containing carbon.
- Process:
     -find CO2 and H2O (mole)
     -find C and H
     -C:H
     -CxHx
     -check the answer
     -if the totals don't balance, there is another element in the compound (usually O or N)
              -difference of totals = mass of the other element
     -CxHx or CxHxUx
- Example
     An organic compound weighed 99.99 g was burnt and in the product there is 191.29 g carbon dioxide and 117.36 g water.  (there IS the element O in the compound)
           - CO2:  191.29 g x 1 mol / 44.0 g = 4.3475 mol
             H2O:  117.36 g x 1 mol / 18.0 g = 6.52 mol
           - C:  4.3475 mol
             H:  13.04 mol
           - C:H = 1:2.999424957 = 1:3
           - Check:  C:  4.3475 mol x 12.0 g / 1 mol = 52.17 g
                       H:  13.04 mol x 1.0 g / 1 mol = 13.04 g
                       52.17 g + 13.04 g = 65.21 g < 99.99 g
           - O:  99.99 g - 65.21 g = 34.78 g
             34.78 g x 1 mol / 16.0 g = 2.17375 mol
           - C:H:O = 1:2.999424957:2.17375 = 1:3:2.17375 = 2:6:1
           - C2H6O

Chapter 4 Empirical & Molecular Formula & Percentage Composition -- By Nemo Jin

Chapter 4 

--Empirical & Molecular Formula & Percentage Composition

Percentage Composition

Definition:

--Percentage Composition is the percentage of the mass of a certain element in a compound 

Calculation:

--Calculate the molar mass of the compound

--Calculate the mass of each element in the compound

--The %composition of X element = mass of X element / total mass of the compound * 100%

Ex.

In ionic compound FeO

The percentage composition of Fe = (55.8 / 55.8 + 16.0)*100% = 77.7%

Applied skills:

(Connect Empirical Formula with Percentage Composition)

--Use percentage composition to find out the empirical formula of a certain compound.

Question: 

A compound is made of 72.8% Oxygen and 27.2% Carbon. Find out the empirical formula of that compound.

--Assume there is 100g compound 

--so that the mass of oxygen is 72.8g and the mass of carbon is 27.2g

--find out how many moles by doing “72.8g/16.0g=4.55 moles” and “27.2g/12.0g=2.27 moles” 

--the biggest number is divided by the smallest “4.55/2.27=2”

--write out the empirical formula, gives you the formula of Carbon Dioxide CO2


Empirical Formula And Molecular Formula 


--Empirical Formula of a compound gives you the lowest-term of ration of atoms in the formula, for example C2H3. 2 and 3 are both not divisible by any number that is larger than 1.


--Molecular Formula of a compound gives you all atoms exist in that compound. For example C4H6. 4 and 6 are both divisible by 2.


Conversion.
--Convert empirical formula to molecular formula you just need to multiply every atoms by n
--Convert molecular formula to empirical formula you just need to divide every number by the same number n.