Question

Challenge Nitrogen has two naturally occurring isotopes, N-14 and N-15. Its atomic mass is 14.007. Which isotope is more abundant? Explain your answer.

Short answer

The more abundant isotope is N-14, as its relative abundance is approximately 0.993, compared to N-15's relative abundance of approximately 0.007. This conclusion is based on the given atomic mass and solving for the relative abundances of both isotopes.

Step-by-step solution

Understand the concept of atomic mass and isotopes

Atomic mass is the weighted average mass of the naturally occurring isotopes of an element. It takes into account the mass of each isotope and their relative abundance. An isotope is a variant of an element with the same number of protons (atomic number) but a different number of neutrons, resulting in different atomic masses.

Set up the equation for atomic mass

To find the atomic mass, we can use the equation: Atomic mass = Sum of (Isotope mass × Relative abundance) Since there are two naturally occurring isotopes, we can represent their relative abundances as x and (1-x), with x representing the relative abundance for N-14. So our equation becomes: 14.007 = (14 × x) + (15 × (1-x))

Solve the equation to find the relative abundance of N-14

Now, we can solve the equation for x, which is the relative abundance of N-14: 14.007 = 14x + 15 - 15x Combine the terms with x: x = 1 - 14.007 x ≈ 0.993

Compare the relative abundance of both isotopes

Since we now know the relative abundance of N-14, we can easily find the relative abundance of N-15 by subtracting x from 1: Relative abundance of N-15 = 1 - x ≈ 1 - 0.993 ≈ 0.007

Determine the more abundant isotope

Comparing the relative abundances, we can see that N-14 is more abundant than N-15: Relative abundance of N-14 ≈ 0.993 Relative abundance of N-15 ≈ 0.007 Considering these values, the more abundant isotope is N-14.

Key concepts

Atomic Mass

Atomic mass, sometimes referred to as atomic weight, represents the average mass of an element's isotopes as they occur naturally. Each isotope of an element has a distinct mass based on the number of neutrons present in its nucleus. However, because elements consist of a mixture of isotopes, atomic mass is calculated as a weighted average and thus includes contributions from each isotope and their respective natural abundances.

In the example of nitrogen, which has two naturally occurring isotopes (N-14 and N-15), its atomic mass is 14.007. This value indicates that the average mass per nitrogen atom—based on Earth's natural isotopic distribution—leans much closer to 14 than to 15, suggesting N-14's dominance. Atomic mass is an essential concept in chemistry, providing insight into the composition and behavior of elements in the environment.

Relative Abundance

Relative abundance is key in determining the atomic mass of an element. It refers to the proportion, or percentage, of each isotope of an element relative to the total amount of the element present. In simple terms, it tells us how much of each isotope exists compared to each other.

For nitrogen, knowing its isotopes N-14 and N-15, we calculate atomic mass based on their relative abundances: how frequently each isotope occurs naturally. The relative abundance can be expressed as a fraction or decimal, with the total always equating to 1 or 100%. For instance, if 99.3% of nitrogen atoms are N-14 and 0.7% are N-15, these proportions reflect their relative abundances.

• These percentages directly affect the calculated atomic mass.
• A higher relative abundance means greater contribution to atomic mass.

This concept is vital, especially when determining which isotope of an element is more prevalent, like with the more common N-14 in nitrogen.

Isotope Mass Calculation

To calculate isotope mass, we utilize a formula that reflects both the mass of each isotope and their relative abundances. The equation for computing atomic mass can be expressed as:

\[ \text{Atomic mass} = \sum (\text{Isotope mass} \times \text{Relative abundance}) \]

For nitrogen with isotopes N-14 and N-15, the formula morphs into:

• Let the abundance of N-14 be expressed as \( x \).
• Then, N-15's abundance is \( 1 - x \).

Substituting these values, the equation becomes:

\[ 14.007 = (14 \cdot x) + (15 \cdot (1-x)) \]

Solving this helps find \( x \), the abundance of N-14. Through algebraic manipulation, you find that \( x \approx 0.993 \), indicating N-14's overwhelming presence.

This method is fundamental in chemistry for assessing isotope masses and understanding how isotopic compositions influence the natural characteristics and applications of elements.