How long can I practically power an IC with a capacitor?
The voltage on a capacitor will drop at a rate of \$ \dfrac{dv}{dt} = \dfrac{-I}{C}\$, assuming a constant current \$I\$ and capacitance \$C\$. 

So if the IC is guaranteed to work down to 2.90V (you have determine this value from the datasheet), the capacitance required to get it to work for 'x' minutes with 10mA draw will be: 

$$C = \frac{I \cdot T}{\Delta V} = \frac{0.01A \cdot (60 sec/min) \cdot (x\  min)}{0.4V}
 = 1.5 Farads/min \cdot x min $$

So for 5 minutes, you'd need 7.5F of capacitance. 

While that's a prohibitive amount of capacitance for a conventional electrolytic capacitor, it's not impossible for a "double layer" capacitor such as these. 



The cost of such a part is prohibitive for many consumer electronic devices.