please help with this problem???? (Chemistry)?

Question

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO2 and 6.45 mg H2O. If 0.242 mol of terephthalic acid has a mass of 40.2 g, determine the molecular formula for terphthalic acid.


SoI tried to figure out if 19.81mg = 41.98 mg CO2 and = 6.45 mg H2O then it's 2.119131752 times the acid for CO2 and 0.3255931348 times for H2O then 1 mol = 166.1157025g
then CO2 = 30% C and 70% O
and H2O = 11% H and 88% O
Then it's C9H6O18
But that's not correct can you help me?
What is the answer? and can you explain how you got it please.

wow #_# that's so complicated! thanks!

Answer 1

These are very simple to do:

First, get the molecular weight of terephthalic acid from the given numbers (we'll need it later):

40.2g/0.242mol = 166 g/mol

Now, the amount of carbon from combusted sample. We just want the part of the weight of the evolved CO2 that is due to the carbon which came from the sample:

41.98mg CO2 x [12/(12 + 32)] = 11.45mg carbon

The amount of hydrogen from combusted sample: We just want the part of the weight of the evolved water that came from the hydrogen in the sample.

6.45mg H2O x 2/[(2 X1) + 16] = 0.717mg hydrogen

Amount of oxygen in sample (before combustion); this is the original weight of the sample minus the "C and H weight" Remember, they said it contained only C, H, and O, so the once we subtract C & H we only have the oxygen left.

19.81mg -[11.45 + 0.717]mg = 7.64mg

We have to get the oxygen by difference because when we combust the sample we introduce "extra" oxygen to do the combustion and we don't know how much of the oxygen in the isolated CO2 and H2O is from the original sample, and how much was "added" during combustion.

Now we have the CHO ratio by weight but we need to know by moles to get a formula! So we divide all the weights by the molar weight of the respective element: If we really wanted to be anal here we would carry all the units through but we really are only interested in the mole ratios at this point

11.45/12 = 0.954 C
0.717/1 = 0.717 H
7.64/16 = 0.4775 O

These numbers are in the proper ratios but, we need to get these into some kind of "normal" ratio (one where they are all whole numbers, or nearly whole numbers). First, try dividing by the reciprocal of the smallest number (that is usually a good starting point). Dividing all by 0.4775 they become:

1.998 C
1.50 H
1.00 O

Now we are getting somewhere; since we can't have half a hydrogen, we will now multiply everything by 2 to get all whole numbers: They now become:

3.996 C
3.00 H
2.00 O

Now we have the empirical fromula: C4H302 but we need to get the actual formula, this is fairly easy, we just determine the molecular weight of the empirical formula and divide it into the molecular weight we calculated back at the very beginning. If there are no mistakes, it should come out a whole number (or close to a whole number)

[(12 x 4) + (1 x 3) + (2 x 16)] = 83amu for empirical formula

166g/mol/83g/mol = 2

So the actual formula is "double" the empirical formula:

C8H6O4 Wasn't that simple!