Problem 114 - Chemistry Matter and Change

Question

Apply Indium has two naturally occurring isotopes and an atomic mass of 114.818 amu. In-11 3 has a mass of 112.904 amu and an abundance of 4.3\(\% .\) What is the identity and percent abundance of indium's other isotope?

Step-by-Step Solution

Answer

The other naturally occurring isotope of Indium is In-116, with a percent abundance of 95.7%.

Step 1: Write the formula for average atomic mass

The formula to calculate the average atomic mass of an element based on its isotopes is given by: Average atomic mass = (mass of isotope 1 × abundance of isotope 1) + (mass of isotope 2 × abundance of isotope 2) Here, the average atomic mass of Indium is 114.818 amu, and we have information about one of its isotopes, In-113, which has a mass of 112.904 amu and an abundance of 4.3%. We'll use "x" to represent the mass number of the other isotope and "y" to represent its percent abundance.

Step 2: Set up the equation

We can now set up the equation for the average atomic mass of Indium using the given information: 114.818 = (112.904 × 0.043) + (x × y) Since there are only two naturally occurring isotopes of Indium, the sum of their percent abundances must equal 100%. Therefore, we can also write an equation for the unknown isotope's percent abundance: y = 1 - 0.043

Step 3: Solve for y

We can now solve for the percent abundance of the unknown isotope (y): y = 1 - 0.043 y = 0.957 So, the percent abundance of the unknown isotope is 95.7%.

Step 4: Solve for x

Now, we can substitute y back into the atomic mass equation and solve for x: 114.818 = (112.904 × 0.043) + (x × 0.957) First, let's calculate the first term on the right side of the equation: 112.904 × 0.043 = 4.8548 Now, we can rewrite the equation as: 114.818 = 4.8548 + 0.957x Subtract 4.8548 from both sides: 110.9632 = 0.957x Now, divide both sides by 0.957: x ≈ 115.901 The mass number, x, is approximately 115.901, but since it needs to be a whole number, we can round it to the nearest integer to get 116.

Step 5: Identify the other isotope

Based on the calculated mass number and percent abundance, the other naturally occurring isotope of Indium is In-116, with a percent abundance of 95.7%.

Key Concepts

Atomic Mass

The concept of atomic mass is pivotal in understanding the composition of elements, especially when dealing with isotopes. Atomic mass, generally expressed in atomic mass units (amu), is the average mass of an element's isotopes. This average takes into account both the mass and the relative abundance of each isotope.
In the case of Indium, its atomic mass is 114.818 amu. This value is not just a simple average of its isotopic masses but rather a weighted average.
Weighted averages are essential because isotopes have different abundances in nature. So, rather than adding their masses and dividing by the number of isotopes, each mass is multiplied by its abundance first.

• This provides a more accurate representation of the element's mass in a natural context.
• This calculation method reflects the natural diversity and distribution of isotopes on Earth.

Percentage Abundance

Percentage abundance refers to the relative proportion of each isotope present in a sample of an element. This measure is expressed as a percentage and is crucial for calculating atomic mass.
When we say Indium-113 has a 4.3% abundance, it means that out of a hundred atoms of Indium, 4.3 are In-113.
In the textbook solution, the percentage abundance formula is applied to find the missing information of Indium's isotopes.

• The sum of the percentage abundances for all isotopes of an element always equals 100%.
• In the case of Indium, since we know the abundance of one isotope, we can infer the other by subtraction against 100%.

The example demonstrates how even one known abundance allows us to determine the other when only two isotopes exist. This is valuable in accurately determining the weighted atomic mass by plugging the percentages and isotopic masses into the average atomic mass formula.

Indium

Indium (symbol In, atomic number 49) is a fascinating element primarily known for its two isotopes: In-113 and In-115, though their atomic arrangement gives different mass numbers.
Its presence in nature is rare compared to elements like oxygen or nitrogen, making understanding its properties significant in fields like manufacturing and chemistry.
This element's atomic mass ties directly to its isotopic composition and abundance.

• With a given atomic mass of 114.818 amu, we explore how the isotopic distribution directly influences its overall atomic weight.
• Understanding the isotopes provides insights into its utility, especially In-115's prevalence, which is a result of its calculated abundance and atomic configuration.

In the problem, we see how mathematical approaches allow us to decipher Indium's elemental makeup, employing formulas that weigh isotopic mass against abundance. The calculation ultimately reveals the dominant In-115 isotope by mass, rounding vital isotopic figures to integers that align with atomic structure principles.