Finding Empirical Formula of Propenoic Acid

[Drakkith]

Homework Statement

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Propenoic acid is a reactive organic liquid used in the manufacture of plastics, coatings, and adhesives. An unlabeled container is thought to contain this acid. A 0.2033-g sample is combusted in an apparatus such as that shown below. The gain in mass of the H2O absorber is 0.102 g, whereas that of the CO2 absorber is 0.374 g.
(Mass gained by each absorber corresponds to the mass of CO₂ or H₂O produced)

What is the empirical formula of propenoic acid?

Homework Equations

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The Attempt at a Solution

First and foremost, I don't even know what propenoic acid consists of at all, other than carbon and hydrogen (since neither comes from the air). I'm going under the assumption that it has oxygen as well. It has to consist of something else other than C and H, as there's 0.08983 grams of our sample left over. Since nothing else other than C, H, and O is produced as a product, it narrows it down to O.

Finding the mass of C and H:

Carbon's mass per mol is 12.01 g, so it makes up 27.289% of the mass of CO₂. So 0.374 g CO₂ contains 0.10206 g C.

Hydrogen's mass per mol is 2.016, so it makes up 11.19% of the mass of H₂O, giving us a mass of 0.01141 g H.

.2033 g of Sample - 0.10206 g C- 0.01141 g H= 0.08983 g O.

Mole ratio:

0.10206g C / 12.01 g/mol C = 0.0084979 mol C.

0.01141 g H / 2.016 g/mol H = 0.0056597 mol H.

0.08983 g O / 16 g/mol O = 0.0056144 mol O.

Dividing the mols of C, H, and O by number of moles of O gives me 1.5, 1, and 1 as subscripts. Multiply by 2 to get a whole number ratio and I get C₃H₂O₂. But this is incorrect according to the mastering chemistry website we use for our homework. I'm not sure what I've done wrong.

 

[Ygggdrasil]

0.01141 g H / 2.016 g/mol H = 0.0056597 mol H.

You have the wrong atomic mass of hydrogen.

 

[Drakkith]

You have the wrong atomic mass of hydrogen.

Sigh... I used the mass of H₂...

Okay. Using the correct atomic mass gives me C₃H₄O₂, which turns out to be correct. Thanks!