20 minutes.
The sample would lose one half the quantity of francium in each half-life.
[tex]1/16 = 1/2^{4} = 1/2 \times1/2 \times1/2 \times1/2[/tex]
Thus a mass decrease by a factor of 16 would correspond to a period of four half-lives. It took 80 minutes for the sample to lose all these francium, therefore
[tex]t_{1/2} = 80/4 = 20[/tex] minutes.
