Answer
$$0.265 \mathrm{\ g}
$$
Work Step by Step
from Eq. $42-15$ and Eq. $42-18$ and from our knowledge that mass is proportional to the number of atoms), the amount decayed is
$$|\Delta m| =\left.m\right|_{t=16.0 \mathrm{h}}-\left.m\right|_{t_{f}=140 \mathrm{h}}=m_{0}\left(1-e^{-t_{t} \ln 2 / T_{I / 2}}\right)-m_{0}\left(1-e^{-t_{f} \ln 2 / T_{I / 2}}\right) \\ =m_{0}\left(e^{-t_{f}\left(\mathrm{h} 2 / T_{I / 2}\right.}-e^{-t_{t} \ln 2 / T_{I / 2}}\right)=(5.50 \mathrm{g})\left[e^{-(16.0 h / 12.7 \mathrm{h}) \ln 2}-e^{-(1.40 \mathrm{h} / 12.7) \ln 2}\right] \\ =0.265 \mathrm{g}
$$
