A chemist wishing to do an experiment requiring \(^{47} \mathrm{Ca}^{2+}\) (half-life \(=4.5\) days) needs \(5.0 \mu \mathrm{g}\) of the nuclide. What mass of \(^{47} \mathrm{CaCO}_{3}\) must be ordered if it takes \(48 \mathrm{h}\) for delivery from the supplier? Assume that the atomic mass of \(^{47} \mathrm{Ca}\) is \(47.0 \mathrm{u}\)

To find the mass of \(^{47}\mathrm{CaCO}_3\) that must be ordered, first calculate the initial amount of \(^{47}\mathrm{Ca}^{2+}\) needed considering its half-life and the delivery time: \(N_0 = \frac{5.0\mu\mathrm{g}}{(1/2) ^{\frac{48\mathrm{h}}{4.5\mathrm{days} \times 24\mathrm{h/day}}}}\). Next, calculate the moles of desired \(^{47}\mathrm{Ca}^{2+}\) as \(\frac{N_0}{47.0\mathrm{u}}\). The moles of \(^{47}\mathrm{CaCO}_3\) are the same as the moles of \(^{47}\mathrm{Ca}^{2+}\). Finally, calculate the mass of \(^{47}\mathrm{CaCO}_3\) using its molar mass: (mass of \(^{47}\mathrm{CaCO}_3\)) = (moles of \(^{47}\mathrm{CaCO}_3\)) × (\(47.0\mathrm{u} + 12.0\mathrm{u} + 3 \times 16.0\mathrm{u}\)). Don't forget to convert the mass into the appropriate unit (e.g., micrograms, milligrams, or grams).