Question

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 \(47.0 \mathrm{u}\)

Short answer

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).

Step-by-step solution

Calculate the amount of \(^{47}\mathrm{Ca}^{2+}\) after decay during delivery

First, we need to calculate the amount of \(^{47}\mathrm{Ca}^{2+}\) remaining after \(48\mathrm{h}\) of delivery time, using its half-life of \(4.5 \mathrm{days}\). Since the half-life formula is: \[N = N_0 \times (1/2) ^{\frac{t}{t_{1/2}}}\] where: \(N\) = amount of nuclide remaining after time \(t\), \(N_0\) = initial amount of nuclide, \(t\) = time elapsed, and \(t_{1/2}\) = half-life of the nuclide. We know the desired final amount of \(^{47}\mathrm{Ca}^{2+}\) is \(5.0\mu\mathrm{g}\), so to determine the initial amount (\(N_0\)), we need to solve for it: \(5.0\mu\mathrm{g} = N_0 \times (1/2) ^{\frac{48\mathrm{h}}{4.5\mathrm{days} \times 24\mathrm{h/day}}}\)

Calculate the moles of desired \(^{47}\mathrm{Ca}^{2+}\)

Solve the equation from Step 1 for \(N_0\): \(N_0 = \frac{5.0\mu\mathrm{g}}{(1/2) ^{\frac{48\mathrm{h}}{4.5\mathrm{days} \times 24\mathrm{h/day}}}}\) Now, calculate the moles of desired \(^{47}\mathrm{Ca}^{2+}\) using its atomic mass, which is \(47.0\mathrm{u}\): \[\text{moles of } ^{47}\mathrm{Ca}^{2+} = \frac{N_0}{47.0\mathrm{u}}\]

Calculate the moles of \(^{47}\mathrm{CaCO}_3\)

Since the molar ratio between \(^{47}\mathrm{Ca}^{2+}\) and \(^{47}\mathrm{CaCO}_3\) is 1:1, the moles of \(^{47}\mathrm{CaCO}_3\) required to get the desired amount of \(^{47}\mathrm{Ca}^{2+}\) is the same as the moles of \(^{47}\mathrm{Ca}^{2+}\) calculated in Step 2.

Calculate the mass of \(^{47}\mathrm{CaCO}_3\)

To determine the mass of \(^{47}\mathrm{CaCO}_3\) that must be ordered, we need to multiply the moles of \(^{47}\mathrm{CaCO}_3\) by its molar mass. The molar mass of \(^{47}\mathrm{CaCO}_3\) can be calculated as follows: \(^{47}\mathrm{CaCO}_3\) molar mass = mass of \(^{47}\mathrm{Ca}\) + mass of \(C\) + 3 × mass of \(O\) = \(47.0\mathrm{u} + 12.0\mathrm{u} + 3 \times 16.0\mathrm{u}\) Mass of \(^{47}\mathrm{CaCO}_3\) = (moles of \(^{47}\mathrm{CaCO}_3\)) × (\(^{47}\mathrm{CaCO}_3\) molar mass) Now, solve for the final answer and don't forget to convert it into the appropriate unit (e.g., micrograms, milligrams, or grams).

Key concepts

Half-life Calculation

Radioactive decay describes the process by which an unstable atomic nucleus loses energy by emitting radiation. One key concept in understanding this process is half-life, which is the time it takes for half of a given amount of radioactive material to decay. Calculating half-life is crucial when working with radioactive substances, as it helps predict how much of the substance remains over a period of time.

The formula to calculate the amount of substance remaining after a specific time during a decay process is:

• \[N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
• In this equation, \(N\) is the remaining amount of the substance, \(N_0\) is the initial amount, \(t\) is the elapsed time, and \(t_{1/2}\) is the half-life.

For example, if a chemist orders a radioactive isotope with a half-life of 4.5 days and expects it to be delivered in 48 hours, the chemist can determine how much of the isotope will remain at the end of those 48 hours by using this formula. This is essential in ensuring that a sufficient amount of the isotope is available for experiments like the one described in the exercise.

Isotope Chemistry

Isotopes are variations of elements that have the same number of protons but different numbers of neutrons in their nuclei. This results in varying mass numbers for isotopes of the same element. In the case of calcium, the isotope \(^{47}\mathrm{Ca}\) is used.

Understanding isotopes is essential in many scientific fields, including chemistry, physics, and geology, as they are used to study processes like radioactive decay and to date ancient materials.

• Isotopes can be stable or unstable (radioactive). Radioactive isotopes like \(^{47}\mathrm{Ca}\), undergo spontaneous decay, releasing radiation in the process.
• Stable isotopes do not decay over time and are used in different kinds of chemical applications or medical diagnostics.

In experiments, adjustments are often required depending on the isotope's specific properties, such as its half-life and the kind of radiation emitted. Understanding these aspects will help a chemist or a student ensure that they are using the correct isotope for the intended purpose.

Molar Mass

Molar mass is a fundamental concept in chemistry, calculated as the mass of one mole of a substance (usually in grams per mole). For isotopes, understanding molar mass is crucial, particularly when calculating how much of a compound needs to be used in an experiment.

In the context of an isotope like \(^{47}\mathrm{CaCO}_3\), its molar mass can be determined by adding together the atomic masses of each component in the compound.

• Molar mass of \(^{47}\mathrm{CaCO}_3\) is computed by the sum of the molar masses of calcium \(^{47}\mathrm{Ca}\), carbon (C), and oxygen (O):
• \[\text{Molar mass of } ^{47}\mathrm{CaCO}_3 = 47.0 \text{ u} + 12.0 \text{ u} + 3 \times 16.0 \text{ u} = 100.0 \text{ u} \]

This calculation helps in determining the actual mass of \(^{47}\mathrm{CaCO}_3\) needed for experiments, ensuring that enough of the compound is present to produce the required amount of the isotope \(^{47}\mathrm{Ca}^{2+}\). Calculating molar mass correctly ensures accuracy in chemical reactions and laboratory experiments.