Help with capacitors and project
Simplified model of your load as a constant current source for a conservative bound. In that case, if your capacitor of capacitance \$C\$ is charged to \$V_c\$, and your constant current draw is \$I\$, then:

$$V(t) = V_c - I \cdot t / C$$

What is t when your voltage reaches zero? 

$$t = V_c \cdot C / I$$

Let's say your current draw is 100mA, \$V_c\$ is 6.3V and \$C\$ is 1F, then \$t\$ is 6.3 * 1 / 0.1 = 63 seconds. 

Problem with the constant current source model is your load actually stops drawing power when your voltage drops below some level (or has some more complex dependence on voltage), which significantly reduced the "run time" of your capacitor. That is to say, your real load is probably not a constant current source.

Also capacitors have leakage, but I'd expect that to be a secondary effect. Also my maths should be community checked. 

If instead of a constant current model, you imagine a purely resistive load, then the (slower) voltage decay follows the perhaps more familiar form:

$$V(t) = V_c \cdot exp^{-t/(RC)}$$