[EX5TASY]

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 0.246 mol of terephthalic acid has a mass of 40.8 g, determine the molecular formula for terphthalic acid.


Here is how I went about the problem:

1) First I converted the mg to grams. Therefore,
Terephthalic is 0.01981g
CO₂ 0.04198g
H₂ 0.00645g

2) Then I find the mole of each grams of elements:

0.04198g CO₂ x 1 mol CO₂ / 44.01g CO₂ x 1 molC / 1 mol CO₂ =  9.53e-4 mol C

I do these for the others as well and find,
7.16e-4 mol H
0.00226 mol O

Then, shouldn't I just be able to find the empirical formula by dividing each mol by the smallest mole? However, I get weird values.

9.53e-4 / 0.00226 =  0.42

Am I doing this right? Please *delete me*!! 

[Dan]

Do not calculate moles of O in this way, you will get a higher value since combustion is the reaction with oxygen.

Your moles of C and H are correct.

Convert to mass of C and mass of H.

Mass(C) + Mass(H) + Mass(O) = Mass(total)

mass(O) is the only unknown in this equation, so solve for mass(O), convert this mass to moles(O) and then find the molar ratio C:H:O to find the empirical formula

[EX5TASY]

OK, so I figured out the masses of each element to be:

C = 0.01144g
H = 7.217e-4g
O = 0.0362g

So how exactly can I convert grams of O to moles, since there are 2 compounds that combined, have that much grams of O?

[Dan]

O = 0.0362g

So how exactly can I convert grams of O to moles, since there are 2 compounds that combined, have that much grams of O?

No, use total mass of the starting material, not the products. The idea is that the C in the products must all come from the starting compound, so the amount od C in the products is the same as in the starting compound. Same is true for H, so we can work out the C and H content, which will then imply the O content, given we know the starting mass, form the equation I posted. Total mass is given in the question as 19.81mg, use that in the equation I posted earlier.

Because we are using the mass of original sample, doing it this way only considers the oxygen in the original compound, and not the oxygen added in the combustion reaction.

Once you have a mass of O, convert it to moles in the usual way (using the atomic mass).

[EX5TASY]

Sweet, I get it now. Thanks Dan!