Saturday, December 4, 2010

Calculating the Empirical Formula of Organic Compounds

First we need to know what an organic compound is!
-         An organic compound is any substance containing carbon
Example: Methane is one of the simplest organic compounds

The empirical formula of an organic compound can be found by
1)      Burning the compound (this involves reacting it with 02)
2)      Collecting and weighing the products
3)      From the mass of the products, the moles of each element in the  original organic compound can be calculated.

Lets do an example together!

What is the empirical formula of a compound that burns to produce 16.9 g of CO2 and 3.46 g of H2O?

Step 1:  CxHy  +     z02   à    xCO2   +  y/2  H2O

This is the balanced chemical equation. From the equation we can see that all of the C and H in CxHy went into making xCO2 and y/2  H2O.
Therefore the moles of C and H are equal to the moles of C in CO2 and the moles of H in H2O.

Step 2:  Convert grams to moles

16.9 g of CO2  x   1 mole                    =   0.384 moles of CO2
                          44g of CO2
3.46 g of H2O      x   1 mole                =  0.192 moles of H2O
                               18g of H2O

Now we need to find the moles of C and H

Mole of C      0.384 moles   x     1 mol C                = 0.384 moles of C
                                                1 mol of CO2

Mole of H   0.192 moles   x  2 mol of H                 =  0.384 moles of H
                                             1 mol of H2O   

Step 3: Not needed in this problem.

Answer: CH

Lets try another problem!

A 7.30 gram sample of hydrocarbon is burned to give 23.8 grams of CO2 and 7.30 grams of H2O. What is the empirical formula?

Mol of CO2    23.8 g of CO2  x  1 mole                  =  0.541 moles of CO2
                                                  44g of CO2

Mol of H2O   7.30 g of H2O  x   1 mole              = 0.405 moles of H2O
                                                   18g of H2O

Mol of C     0.541 moles of CO2   x  1 mol of C     =  0.541 moles of C
                                                        1 mol of CO2

Mol of H    0.405 moles of H2O   x   2 mol of H     = 0.811 moles of H
                                                          1 mol H2O

Step 3: Divide by the smallest molar amount (0.541)

Carbon =  0.541       = 1          X         2     = 2

Hydrogen =  0.811   =  1.5      X         2     = 3

Change ratio to a whole number ration by multiplying.

Answer: C2H3

*Check Mass

0.541 mol C  x 12.0g C    =  6.49g C
                        1 mol C

0.811 mol H  x 1.01g H    =  0.819g H
                        1 mol H

Total Mass = 6.49g + 0.819g = 7.31g
No other elements are present in this sample.

Note: If mass numbers do not add up to the original mass, another element is present.

By Candace

No comments:

Post a Comment