# Oxidation of malonic acid

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?

• 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. – Martin - マーチン Nov 29 '16 at 10:02
• @Martin It is kind of an online riddle and my answer isn't accepted – Ctx Nov 29 '16 at 10:16
• 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. – Zhe Jun 30 '17 at 13:27

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: $m(\ce{CO2}) \approx \pu{42.3 g}$.