Charcoal samples from Stonehenge in England were burned in \(\mathrm{O}_{2},\) and the resultant \(\mathrm{CO}_{2}\) gas bubbled into a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (limewater), resulting in the precipitation of \(\mathrm{CaCO}_{3}\). The \(\mathrm{CaCO}_{3}\) was removed by filtration and dried. A 788 -mg sample of the \(\mathrm{CaCO}_{3}\) had a radioactivity of $1.5 \times 10^{-2}$ Bq due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half-life of carbon-14, 5700 yr, calculate the age of the charcoal sample.

First, we need to find the current radioactivity of the sample in disintegrations per minute per gram (DPM/g). The sample has a radioactivity of \(1.5 \times 10^{-2}\) Bq, which is equivalent to \(1.5 \times 10^{-2}\) disintegrations per second. To convert this to disintegrations per minute, multiply by 60 seconds per minute: \[ \frac{1.5 \times 10^{-2}\,\text{disintegrations/s}}{1\,\text{min}} \times \frac{60\,\text{s}}{1\,\text{min}} = 0.9\,\text{disintegrations/min} \] Next, we need to find the mass of carbon in the sample, which can be found from the mass of the precipitated \(\mathrm{CaCO}_{3}\). The 788-mg sample of \(\mathrm{CaCO}_{3}\) contains carbon, and the mass ratio of carbon to \(\mathrm{CaCO}_{3}\) is: \[ \frac{\text{mass of carbon}}{\text{mass of }\mathrm{CaCO}_{3}} = \frac{\text{molar mass of carbon}}{\text{molar mass of }\mathrm{CaCO}_{3}} \] The molar mass of carbon is 12.01 g/mol, and the molar mass of \(\mathrm{CaCO}_{3}\) is 100.09 g/mol, so the mass ratio is: \[ \frac{\text{mass of carbon}}{788\,\text{mg}} = \frac{12.01\,\text{g/mol}}{100.09\,\text{g/mol}} \] Now we can find the mass of carbon in the sample: \[ \text{mass of carbon} = \frac{12.01\,\text{g/mol}}{100.09\,\text{g/mol}} \times 788\,\text{mg} = 94.2\,\text{mg} \] Now, we can find the radioactivity of the sample in disintegrations per minute per gram (DPM/g): \[ \frac{0.9\,\text{disintegrations/min}}{94.2\,\text{mg}} = 9.56 \times 10^{-3}\,\text{DPM/mg} \] Then, we will find the ratio of the current radioactivity to the initial radioactivity of living organisms: \[ \frac{9.56 \times 10^{-3}\,\text{DPM/mg}}{15.3\,\text{DPM/mg}} = 6.25 \times 10^{-4} \]

Now that we have the ratio of the current radioactivity to the initial radioactivity, we can use the half-life formula to calculate the age of the charcoal sample. The formula is: \[ t = \frac{\ln(\frac{\text{Current radioactivity}}{\text{Initial radioactivity}})}{-\lambda} \] Where \(t\) is the age of the sample, and \(\lambda\) is the decay constant, which can be calculated from the half-life using the formula: \[ \lambda = \frac{\ln(2)}{\text{half-life}} \] Using the given half-life of carbon-14, 5700 years, we can find the decay constant \(\lambda\): \[ \lambda = \frac{\ln(2)}{5700\,\text{yr}} = 1.21 \times 10^{-4}\,\text{yr}^{-1} \] Now we can substitute the values of the ratio of radioactivity and decay constant into the age formula to find the age of the sample: \[ t = \frac{\ln(6.25 \times 10^{-4})}{-1.21 \times 10^{-4}\,\text{yr}^{-1}} \approx 3450\,\text{yr} \] The age of the charcoal sample is approximately 3450 years.