DECEMBER 3rd

Do you remember the CH compounds or the compounds containing Carbon?
They are referred to as the ORGANIC COMPOUND--
The Question is...
How do you calculate the Empirical formula of Organic Compounds?







For example, given:

What's the empirical formula of a compound when a 5.00 gram sample is burned producing 15.0 grams of CO2 and 8.18 g of H2O?

Let the empirical formula of the compound be CxHy.

CxHy + zO2 -> xCO2 + y/2 H2O

1. Find the moles of C and H in the compound.

CO2: 15.0 g x 1mol/44.0 g = 0.341 mol

H2O: 8.18 x 1mol/18.02g = 0.454 mol

2.  0.341 moles of C                      0.341/0.341 = 1 x 3 =3

     0.454 x 2 = 0.908moles of H    0.908/0.341 = 2.66 x 3 = 8

3. Ratio of C:H = 3:8

 therefore, the empirical formula is C3H8.

4. Now, check the masses:

0.341 mol of C x 12g/1mol = 4.092 g of C

0.908 mol of C x 1.0g.1mol = 0.908 g of H

Total mass                           = 5.00 g

Therefore, there is no Oxygen atom in the compound.

*Here is a challenging question for an extra thinking~

Adipic acid is an organic compound composed of 49.31% C, 43.79% O, and the rest hydrogen. If the molar mass of adipic acid is 146.1g/mol, what are the empirical and molecular formulas for adipic acid?

if there were 100 g of the sample, it would contain

49.31 g C
43.79 g O
6.90 g H

convert everything to moles

49.31 g C------>4.11 mol C
43.79 g O------>2.74 mol O
6.90 g H -------->6.83 mol H

now, we have the molar ratio

C:O:H = 4.11 mol C : 2.74 mol O: 6.83 mol H

reduce the ratio to form in which the smallest term is 1
do this by dividing each term by the smallest value (2.74)

4.11 mol C : 2.74 mol O: 6.83 mol H = 1.5 mol C : 1 mol O :2.5 mol H

to eliminate fractions, multiply by the LCD (2 in this case)

1.5 mol C : 1 mol O :2.5 mol H = 3 mol C : 2 mol O :5 mol H

thus, the empirical formula is C3H5O2
======================================…
to find the molecular formula, we have to know how many times the empirical weight is the molecular weight
we do this by dividing the molecular weight by th empirical weight

Empirical Weight (EW)
=(12 g/mol C)(3 mol C) + (1.01 g/mol H)(5 mol H) + (16 g/mol O)(2 mol O)
= 73.05 g

Molecular Weight (MW) = 146.1 g

MW / EW = 146.1 g / 73.05 g = 2

multiply this by the subscripts of the elements in the empirical formula
===============

thus, the molecular formula is C6H10O4
thus, the empirical formula is C3H5O2

Did you get it??

Here is a summary of the steps

1. Find # of moles of the compounds containing C and H

2. Find the # of moles of atom of C, H

3. Determine the ratio of C:H based on the # of moles (divide by the smallest number)

4. IF there are decimals, use either rounding or multiply to get a whole number

5. FINALLY, PLEASE check the masses of each atom of C and H to see if it equals the mass (if given) in the question. If it does not, then there must be a O atom in the compound. Find the mass of O atom and convert to moles in order to determine the C:H:O ratio!