# Dissociation Energy for CO2 per molecule

I'm trying to calculate the energy required to dissociate a single $\ce{CO2}$ molecule into its respective elements, and I'm given that it takes $373.6 \times 10^{3} \, \, \mathrm{\frac{joules}{gram}}$ where $44.01$ grams is the mass of $6.022 \times 10^{23}$ molecules of $\ce{CO2}$.

By dimensional analysis:

$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$

$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$

or,

$170.43\,\, \mathrm{eV/molecule}$

Is this derivation and the logical assumptions made herein correct?

• Your calculations are right, but the starting numbers are not. This just can't be real. Suppose you make such-and-such calculations and found that the density of certain compound is 170 kg/L, or that the height of certain human is 170 feet. What would you think then? – Ivan Neretin Sep 12 '15 at 17:11

$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$
$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$
$170.43\,\, \mathrm{eV/molecule}$