Using a capacitor to keep circuit powered for 1s after battery disconnected?
The device can work from 3.6 to 3.0 V according to the datasheet. I assume you will use a 3.6 V supply. Also I simplify the device as an impedance of 150 Ohms. So, the problem can be translated into that: In a RC circuit, what is the capacitance needed for a 150 Ohm resistance, having 3.6 V on t=0, having 3.0 V on t=1.

v(t) = V(0) * e^(-t/(R*C))

v(1) = v(0) * e^(-1/(150*C))

3 = 3.6 * e^(-1/(150*C))

C = 0.0365F = 36500 uF

It is bulky for a conventional electrolytic cap. Use a supercapacitor.

note: This formula I used is a solution of differential equation for natural response of RC circuit. For more information look at: https://en.wikipedia.org/wiki/RC_circuit

note2: It may not be bulky for electrolytic if you find a capacitor a bit higher than 3.6V volts, but I guess it is difficult to find.