Problem 87 - Chemistry The Central Science In Si Units

Question

A 0.53-g sample of fludeoxyglucose \(\left(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{FO}_{5}\right)\) contains radioactive fluorine- 18 (whose atomic mass is \(18.0 \mathrm{u}\) ). If \(68.3 \%\) of the fluorine atoms in the sample are fluorine-18 and the remainder are naturally occurring nonradioactive fluorine-19 atoms, how many disintegrations per second are produced by this sample? The half-life of fluorine- 18 is 110 min.

Step-by-Step Solution

Answer

Approximately \(1.263 \times 10^{17}\) disintegrations per second.

Step 1: Calculate Moles of Fludeoxyglucose

The molar mass of fludeoxyglucose is calculated from its molecular formula \(\text{C}_{6} \text{H}_{11} \text{FO}_{5}\) by adding the atomic masses of its elements: \(6(12.01) + 11(1.008) + 1(18.998) + 5(16.00) = 181.159\,\text{u}\). To find the moles in the sample, divide the mass by the molar mass: \(\frac{0.53\,\text{g}}{181.159\,\text{g/mol}} \approx 0.002927\,\text{mol}\).

Step 2: Determine Number of Molecules

To find the number of molecules of fludeoxyglucose, multiply the moles calculated in Step 1 by Avogadro's number \(6.022 \times 10^{23}\,\text{mol}^{-1}\): \(0.002927 \times 6.022 \times 10^{23} \approx 1.762 \times 10^{21}\).

Step 3: Calculate Number of Fluorine Atoms

Each molecule of fludeoxyglucose contains 1 fluorine atom. Thus, the number of fluorine atoms is equal to the number of molecules: \(1.762 \times 10^{21}\).

Step 4: Find Number of Fluorine-18 Atoms

With 68.3% of fluorine atoms being fluorine-18, calculate the number of these atoms: \(0.683 \times 1.762 \times 10^{21} \approx 1.203 \times 10^{21}\,\text{Fluorine-18 atoms}\).

Step 5: Determine Activity in Disintegrations per Second

Use the decay equation \(N(t) = N_0 e^{-\lambda t}\) where \(\lambda = \frac{\ln(2)}{T_{1/2}}\) to find decay constant \(\lambda\). With \(T_{1/2} = 110\,\text{min} = 6600\,\text{s}\), \(\lambda = \frac{\ln(2)}{6600}\,\text{s}^{-1}\). Calculate activity \(A = \lambda N\) where \(N\) is the number of fluorine-18 atoms. Thus, \( \lambda \approx 1.050 \times 10^{-4}\,\text{s}^{-1}\) and \(A \approx 1.050 \times 10^{-4} \times 1.203 \times 10^{21} \approx 1.263 \times 10^{17}\) disintegrations per second.

Key Concepts

Molar Mass Calculation

Calculating molar mass is essential for converting between mass and number of moles, enabling better understanding of chemical reactions and properties. For any molecule, the molar mass is calculated by summing the atomic masses of all the constituent atoms. In the case of fludeoxyglucose

• The molecule is composed of 6 carbon ( C), 11 hydrogen ( H), 1 fluorine ( F), and 5 oxygen ( O) atoms.
• The atomic mass of carbon is roughly 12.01 amu, hydrogen is 1.008 amu, fluorine is 18.998 amu, and oxygen is 16.00 amu.
• Add these values for each type of atom: \(6(12.01) + 11(1.008) + 1(18.998) + 5(16.00) = 181.159 \; \mathrm{u}\).

To find how many moles are in a given mass, such as a 0.53-g sample, divide the mass by the molar mass: \(\frac{0.53}{181.159} \approx 0.002927 \; \mathrm{mol}\). This calculation is useful for determining how many molecules are present, aiding in further analysis of chemical behavior.

Fluorine Isotopes

Fluorine, a reactive and lightweight element, has different isotopes that play distinct roles in practical applications. Fluorine-18 and fluorine-19 are the commonly found isotopes. While most fluorine exists as fluorine-19, fluorine-18, an isotope used for medical imaging, is radioactive and decays over time.

• Fluorine-19 is stable and the most abundant isotope; it accounts for almost all naturally occurring fluorine.
• Fluorine-18, on the other hand, is not stable. It has an atomic mass of 18.0 amu and is artificially produced.
• It decays by positron emission, making it valuable in PET scans for its ability to highlight activity in the body.

In our scenario, 68.3% of fluorine atoms in fludeoxyglucose are fluorine-18, giving critical context for further calculations such as half-life and disintegration rates.

Half-Life Calculation

Half-life is a crucial concept in understanding radioactive decay. It's the time required for half the atoms in a sample of a radioactive isotope to decay. For fluorine-18, the half-life is 110 minutes, or 6600 seconds.To calculate the decay constant ( \lambda ), which represents the probability per unit time that an atom will decay, use the formula: \[\lambda = \frac{\ln(2)}{T_{1/2}}\]Where \(T_{1/2}\) is the half-life. So for fluorine-18:

• With \(T_{1/2} = 6600 \; \text{s}\), \(\lambda \approx 1.050 \times 10^{-4} \; \text{s}^{-1}\).

Next, determining the activity of a sample, which is the rate of decay or disintegrations per second, can be done using:

• Activity \(A = \lambda N\), where \(N\) is the number of fluorine-18 atoms in the sample.
• For our purposes, \(N = 1.203 \times 10^{21}\) atoms, and \(A \approx 1.263 \times 10^{17}\) disintegrations per second.

This measure is invaluable in both predicting behavior of a sample and in medical or industrial applications where radioactive isotopes are used.