Oxidation of malonic acid

Question

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?

Answer 1

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.

Answer 2

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.