Problem 83 - Chemistry The Central Science In Si Units

Question

A \(26.00-g\) sample of water containing tritium, \({ }_{1}^{3} \mathrm{H},\) emits \(1.50 \times 10^{3}\) beta particles per second. Tritium is a weak beta emitter with a half-life of 12.3 yr. What fraction of all the hydrogen in the water sample is tritium?

Step-by-Step Solution

Answer

The fraction of hydrogen that is tritium is approximately \(4.83 \times 10^{-13}\).

Step 1: Calculate the Number of Moles of Water

First, calculate the number of moles of water in the sample. The molar mass of water (H₂O) is approximately 18.015 g/mol. Use the formula for the number of moles \( n = \frac{m}{M} \), where \( m = 26.00 \) g and \( M = 18.015 \) g/mol.\[ n = \frac{26.00\, \text{g}}{18.015\, \text{g/mol}} = 1.443\, \text{mol} \]

Step 2: Calculate the Total Number of Hydrogen Atoms in Water Sample

Each molecule of water contains 2 hydrogen atoms. Therefore, to find the number of hydrogen atoms, multiply the number of moles of water by Avogadro's number \(6.022 \times 10^{23}\) molecules/mol and then by 2 hydrogen atoms per molecule.\[ 1.443 \times 6.022 \times 10^{23} \times 2 = 1.738 \times 10^{24}\, \text{hydrogen atoms} \]

Step 3: Calculate the Number of Tritium Atoms

The decay rate (activity \( A \)) given is \(1.50 \times 10^3\) decays per second. Use the decay formula \( A = \lambda N \), where \( \lambda = \frac{0.693}{t_{1/2}} \). Here, \( t_{1/2} = 12.3 \times 365 \times 24 \times 3600 \) seconds. Calculate \( \lambda \) and solve for \( N \), the number of tritium atoms.\[ t_{1/2} = 12.3 \times 365 \times 24 \times 3600 = 3.878 \times 10^8 \text{ s} \]\[ \lambda = \frac{0.693}{3.878 \times 10^8} \approx 1.787 \times 10^{-9} \text{ s}^{-1} \]\[ N = \frac{1.50 \times 10^3}{1.787 \times 10^{-9}} = 8.396 \times 10^{11} \text{ tritium atoms} \]

Step 4: Calculate the Fraction of Tritium

Calculate the fraction of hydrogen atoms that are tritium by dividing the number of tritium atoms by the total number of hydrogen atoms in the sample.\[ \text{Fraction} = \frac{8.396 \times 10^{11}}{1.738 \times 10^{24}} \approx 4.83 \times 10^{-13} \]

Key Concepts

Mole Calculations

Mole calculations are fundamental to chemistry as they allow us to relate the mass of a substance to the amount of particles it contains. The concept of the mole is centered around the number of particles (atoms, molecules, ions) present in a sample, typically expressed in terms of Avogadro's number (approximately \(6.022 \times 10^{23}\) particles/mole). For example, in the problem, we used the molar mass of water to determine how many moles of water were present in a sample. Using this formula,

• \( n = \frac{m}{M} \)

we calculated that the sample contains \(1.443\) moles of water.Next, knowing that each molecule of water includes two hydrogen atoms, we used the mole calculation to find the total number of hydrogen atoms present by multiplying the number of moles of water by Avogadro's number and then by 2.These calculations help us understand the scale of atomic and molecular reactions by linking measurable quantities of substances to the atomic scale through the number of moles.

Half-Life

The half-life of a radioactive isotope is the time it takes for half of the material to decay, a constant property that helps predict how a substance behaves over time. Each radioactive material, like tritium, has a characteristic half-life; for tritium, it is 12.3 years.The half-life is crucial because it allows us to calculate decay rates using the formula for activity:

• \( \lambda = \frac{0.693}{t_{1/2}} \)

In the problem, we converted years into seconds to gain an accurate value for the decay constant \(\lambda\). Once calculated, \(\lambda\) helps determine the activity or rate at which the substance is decaying. Understanding the half-life concept is essential as it plays a vital role in radioactivity and nuclear chemistry, helping us assess the longevity and stability of isotopes like tritium in the sample.

Beta Decay

Beta decay is a type of radioactive decay where a beta particle (which is a high-speed electron or positron) is emitted from the nucleus of an atom. Tritium, for example, undergoes beta decay, which is why it releases beta particles detectable over time.During beta decay, the neutron in the tritium atoms transforms into a proton, emitting a beta particle and an antineutrino. This transformation allows the atom to become more stable. Here’s the reaction equation for tritium:

• \( {}_{1}^{3} \text{H} \rightarrow {}_{2}^{3} \text{He} + \beta^- + \bar{u}_e \)

This decay process results in a different element, helium in this case, showing how radioactive decay can change one element into another. Beta decay impacts both the physical properties and behavior of elements, influencing the decay rate observed in the original exercise.

Avogadro's Number

Avogadro's number \(6.022 \times 10^{23}\) is a fundamental constant in chemistry, defining the number of particles in one mole of substance. It provides a bridge between the atomic scale and the macroscopic scale we experience in the lab.When we talk about a "mole" of a substance, we mean \(6.022 \times 10^{23}\) atoms, molecules, or ions of that substance. This concept is pivotal in converting between the mass of a material and the number of atoms, as seen in the exercise where the number of hydrogen atoms is calculated by multiplying the moles of water by Avogadro's number.Understanding Avogadro's number is crucial for comprehending the quantities involved in chemical reactions, as it allows scientists and students to efficiently convert between atomic-level descriptions and measurable quantities seen in experiments and everyday chemical applications.