5
$\begingroup$

Question
How many grams $\ce{CO2}$ will result when burning $\pu{100 g}$ malonic acid? The formula for malonic acid is $\ce{C3H4O4}$ so 3 $\ce{CO2}$ molecules should result from oxidating one malonic acid molecule. $$\ce{C3H4O4 + 2O2 -> 3CO2 + 2H2O}$$

My attempt

$$44.01\cdot\frac{100}{104.0615}\cdot3 = \pu{126.88 g}$$

with $\pu{44.01 g}$ being the mass of a mol $\ce{CO2}$ and $\pu{104.0615 g}$ for a mole of malonic acid.

However, this doesn't seem to be right. Can anyone hint me for the correct solution?

Improve this question
$\endgroup$
3
  • $\begingroup$ Welcome! Could you comment why it doesn't seem right, please. Apart from a little inconsistent use of significant figures, I can't see anything wrong either. $\endgroup$ – Martin - マーチン♦ Nov 29 '16 at 10:02
  • $\begingroup$ @Martin It is kind of an online riddle and my answer isn't accepted $\endgroup$ – Ctx Nov 29 '16 at 10:16
  • $\begingroup$ Why can't it be right? You're converting $\ce{C3H4O4}$ to something that has the same equivalent weight as $\ce{C3O6}$, which is has a higher weight than malonic acid. $\endgroup$ – Zhe Jun 30 '17 at 13:27
4
$\begingroup$

You actually are correct, the value is greater than 100g because when you burn a hydrocarbon, it is actually an exothermic redox reaction with oxygen. Meaning that the carbon in the compound is combining with atmospheric oxygen. So even if you use 100g of malonic acid, you could end up with more than 100g of CO2. Your answer is correct.

Improve this answer
$\endgroup$
2
$\begingroup$

Using Bunsen burner for this reaction will most likely result in the reaction you've shown. But please keep in mind that even aqueous solution of malonic acid upon heating above $\pu{70 ^\circ C}$ easily undergoes decarboxylation:

$$\ce{HOOC-CH2-COOH ->[\Delta] CH3-COOH +CO2}$$

If your "burning conditions" are rather mild, then you get three times less carbon dioxide: $m(\ce{CO2}) \approx \pu{42.3 g}$.

Apart from that, your result look fine to me as well.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.