Question

Combustion of 19.81 mg of an industrial acid produces 41.98 mg CO₂, and 6.45 mg H₂O. If 0.250 mol of the acid has a mass of 41.5 g, determine the empirical and molecular formula of the acid.

Answer 1

OK, here's how you do these:

First, you need to find the amount of carbon and hydrogen that were in the sample:

41.98mg x (12.01/44.01) = 11.46 mg C
6.45mg x (2.02/18.02) = 0.723 mg H

Now, we get the weight of the oxygen by difference:

19.81 mg - (11.46 + 0.723) = 7.627 mg O

The reason you need to get oxygen by difference rather than just calculating the weight of the oxygen in the CO2 and H2O formed, is because you don't know how much of that oxygen came from the compound and how much came from the air during combustion.

Now we divide each weight by the respective atomic weight and we will get a set of numbers which will be proportional to the empirical formula: (Yes, I know I'm dividing mg by g, but we are treating them all the same and all we need here is the relative ratios)

11.46/12.01 = 0.954
0.723/1.01 = 0.716
7.627/16.00 = 0.477

Now, we want to get these into whole numbers, we will try dividing all of them by the smallest:

0.954/0.477 = 2
0.716/0.477 = 1.5
0.477/0.477 = 1.0

Now, to get all in whole numbers we need to multiply all of them by 2, so the empirical formula is:

C4H3O2

Remember, the empirical formula may or may not match the actual formula. We are going to now find the actual formula. To do this, we need to get the molar weight of terephthalic acid:

41.5g/0.250mole = 166g/mole

Now, we need to see how many empirical formulas fit into the actual molecular weight:

The weight of the empirical formula is:
[(4 x 12.01) + (3 x 1.01) + (2 x 16.00)] = 83.07

Now we divide the actual formula weight by the empirical formula weight: 166/83.07 = 1.998 = 2

So, the actual formula is 2 x the empirical formula or C8H6O4.