Problem 23 - Chemistry The Molecular Science

Question

The atomic weight of boron is 10.811 . The natural abundance of \({ }^{10} \mathrm{~B}\) is \(19.91 \%\). Determine the atomic weight of the only other natural isotope of boron.

Step-by-Step Solution

Answer

The atomic weight of \(^{11}\text{B}\) is approximately 11.00 amu.

Step 1: Express the Problem Mathematically

The atomic weight of a chemical element is the weighted average of the masses of its isotopes, based on their natural abundance. For boron, with two natural isotopes \[ ^{10}\text{B} \text{ and } ^{11}\text{B}, \]this can be expressed as: \[\bar{A} = (\text{mass of } ^{10}\text{B})(\%\text{ abundance of } ^{10}\text{B}) + (\text{mass of } ^{11}\text{B})(\%\text{ abundance of } ^{11}\text{B}).\] Substitute the known abundance, converting 19.91% to its decimal form 0.1991, and the atomic weight \( \bar{A}=10.811 \, \text{amu} \).

Step 2: Set Up the Equation with Known Values

Let's denote the atomic mass of \(^{11}\text{B}\) as \(x\). The abundance of \(^{11}\text{B}\) is \(1 - 0.1991 = 0.8009\). Now set up the equation: \[10.811 = (10)(0.1991) + (x)(0.8009)\] to calculate \(x\), the atomic mass of \(^{11}\text{B}\).

Step 3: Simplify and Solve the Equation

Calculate the first term: \((10)(0.1991) = 1.991\). Substitute this value into the equation:\[10.811 = 1.991 + (x)(0.8009)\].Next, isolate \(x\):\[10.811 - 1.991 = (x)(0.8009)\], which simplifies to:\[8.82 = (x)(0.8009)\].Solve for \(x\):\[x = \frac{8.82}{0.8009}\].

Step 4: Calculate the Atomic Mass of the Isotope

Perform the calculation: \[x = \frac{8.82}{0.8009} \approx 11.00\]. Thus, the atomic mass of \(^{11}\text{B}\) is approximately 11.00 amu.

Key Concepts

Isotopes

Isotopes are different forms of the same chemical element. While they have the same number of protons, they differ in the number of neutrons. This neutron variation leads to different atomic masses. For example, boron has two naturally occurring isotopes:

• ¹⁰B, with 5 protons and 5 neutrons.
• ¹¹B, with 5 protons and 6 neutrons.

Isotopes contribute to the diversity in atomic masses within the same element. This diversity is captured in the concept of the atomic weight of an element, which considers the isotopic distribution. Each isotope contributes to the atomic weight based on its abundance in nature. Understanding isotopes is fundamental for calculating the atomic weight of elements, especially those with multiple isotopes like boron.

Natural abundance

Natural abundance refers to the relative percentages of an element's isotopes as they occur in nature. It explains how much of each isotope exists compared to the total amount of the element. This is crucial for calculating the element's average atomic weight.
For boron, the isotopes ¹⁰B and ¹¹B have natural abundances of 19.91% and 80.09%, respectively. These percentages are used to calculate boron's atomic weight as 10.811 amu (atomic mass units). When determining an element's atomic weight, you multiply the mass of each isotope by its natural abundance. Then, sum these products for all isotopes.
Natural abundance plays a significant role in geochemistry and physics, where understanding isotope distribution can provide insights into processes such as climate change or geological time scales.

Boron

Boron is a fascinating element with diverse applications in everyday life and industry. Positioned as element number 5 in the periodic table, boron is a metalloid, bridging metals and nonmetals. Because of its unique properties, it proves useful in:

• Strengthening glass and ceramics.
• Serving as a micro-nutrient in plants.
• Acting as a fuel additive for improving combustion efficiency.

Boron occurs in nature primarily as its two isotopes: ¹⁰B and ¹¹B. The latter is more prevalent due to its higher natural abundance. This natural mixture results in boron's atomic weight of approximately 10.811 amu. The element's variety of isotopes and their natural abundancies make boron particularly interesting for both scientific study and practical applications. Such isotopic considerations are essential when calculating atomic weights, understanding chemical behaviors, and exploring elemental properties.