evaluate the chemical formula

Question

Fatty acids have a general formula of CxHyOz. A technician wants to determine the type of pure fatty acid isolated from a metabolite mixture. As 16.71g of the pure fatty acid is fully combusted under oxygen (o2), it decomposes according to the below chemical equation.

CHO(l)+O₂(g)-->CO₂(g)+H₂O(l)

It is found that only 46.33g carbon dioxide (CO2) and 16.43g water (h2o) were obtained as the products. It is also known that the molecular weight of the fatty acid is 238 g/mol. (a) Please evaluate the chemical formula of the fatty acid.

Answer 1

First, we will find the empirical formula, and then we can find the molecular formula.

To find the empirical formula, we will find the moles of C, H, and O and reduce them to the simplest whole number.

moles C: 46.33 g CO2 x 1 mol CO2/44.01 g x 1 mol C/ mol CO2 = 1.0527 moles C

moles H: 16.43 g H2O x 1 mol H2O/18.02 g x 2 mol H / mol H2O = 1.8235 moles H

To find the moles of O, we will find grams of C and H and subtract that from the original weight of 16.71 g. That will give us grams of O which we will then convert to moles of O.

grams C = 1.0527 mols C x 12.01 g / mol = 12.64 g C

grams H = 1.8235 mols H x 1.01 g / mol = 1.84 g H

grams O = 16.71 g - 12.6 g - 1.84 g = 2.23 g O

moles O = 2.23 g O x 1 mol / 16 g = 0.1394 moles O

SUMMARY:

1.0527 moles C

1.8235 moles H

0.1394 moles O

Divide all by 0.1394 to try to get whole numbers:

1.0527 / 0.1394 = 7.55 moles C

1.8235 / 0.1394 = 13.08 moles H

0.1394 / 0.1394 = 1.00 moles O

Still not whole numbers, so we will multiply all by 2 to get..

15 moles C

26 moles H

2 moles O

Empirical formula = C₁₅H₂₆O₂

The molar mass of this would be (15 x 12) + (26 x 1) + (2 x 16) = 180 + 26 + 32 = 238 g/mol

Since this is the same as the molecular weight of the unknown acid, the molecular formula of the unknown is also C₁₅H₂₆O₂

Answer 2

Hello, Sakura,

I don't believe I have a correct answer, but I'll describe the process and you might spot the error. The equation for combustion isn't balanced, since we don't know the structure of the fatty acid. The problem is further complicated since we do not know how much of the oxygen is supplied by O2 and that which comes from the fatty acid. So we need to work with the facts we do have. I decided to calculate the moles of H and C that were formed from the H2O and CO2, since we have their masses and molar masses. We have to ignore the oxygen, since we don't know it's source. But if we have the moles, and masses of the C and H, we know they came from the fatty acid. The difference, in grams, is the oxygen that came from the fatty acid.

Moles CO2: 46.33g/44.01 g/mole = 1.053 moles CO2

Moles H2O: 16.3 g/18 g/mole = 0.9056 moles H2O

From these two we can obtain the grams of both C and H that came from the fatty acid.

1. 1.053 moles CO2 = 1.053 moles C, at 12g/mole means 12.63 grams C was from the fatty acid
2. 0.9056 moles H2O = 0.9056 moles H₂ atoms from the fatty acid, at 2g/mole means 1.81 grams of H was from the fatty acid.

So now we know:

12.63 grams C

1.81 grams of H

Unknown g of O

16.71 g of the fatty acid

This means (16.71-(12.63+1.81)) = 2.27 grams O from the fatty acid.

Convert these masses from the fatty acid into moles:

(12.63 grams C) /12g/mole = 1.053 moles C

(1.81 grams H) / 1 g/mole = 1.81 moles H

(13.27 grams O) / 16 g/mole = 0.142 moles O

Now we need to find a ratio amoung these value to make them whole numbers. This is where I had a problem - I don't see an easy way to make them all whole. The closest I get is:

C₁₅H₃₀O₂

That gives a molar mass of 242, close, but not the reported value of 238 g/mole. However, I do find pentadecanoic acid, a long chain fatty acid with a molecular weight of 242, so I'll blame the discrepancy on my lab partner.

I hope this helps, and please let me know if you find an answer that works for the 238 g/mole number. We could kick out 4 hydrogens and replace them with double bonded carbons in two locations, but stll, the data doesn't support that.

Bob