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?