Math Problem Statement

An atom has three stable isotopes. The first has a mass of 35.9675 u and an abundance of 0.33%, the second has a mass of 37.9627 u and an abundance of 0.063%, and the third has a mass of 39.9624 u and an abundance of 99.60%. Determine the atomic mass of the element.

Solution

To determine the atomic mass of the element, we will use the weighted average formula for atomic mass based on the mass and relative abundance of each isotope.

The formula is:

Atomic Mass=(Mass of Isotope 1×Abundance of Isotope 1100)+(Mass of Isotope 2×Abundance of Isotope 2100)+(Mass of Isotope 3×Abundance of Isotope 3100)\text{Atomic Mass} = \left( \frac{\text{Mass of Isotope 1} \times \text{Abundance of Isotope 1}}{100} \right) + \left( \frac{\text{Mass of Isotope 2} \times \text{Abundance of Isotope 2}}{100} \right) + \left( \frac{\text{Mass of Isotope 3} \times \text{Abundance of Isotope 3}}{100} \right)

Given dataTo determine the atomic mass of the element, we will use the weighted average formula for atomic mass based on the mass and relative abundance of each isotope.

The formula is:

Atomic Mass=(Mass of Isotope 1×Abundance of Isotope 1100)+(Mass of Isotope 2×Abundance of Isotope 2100)+(Mass of Isotope 3×Abundance of Isotope 3100)\text{Atomic Mass} = \left( \frac{\text{Mass of Isotope 1} \times \text{Abundance of Isotope 1}}{100} \right) + \left( \frac{\text{Mass of Isotope 2} \times \text{Abundance of Isotope 2}}{100} \right) + \left( \frac{\text{Mass of Isotope 3} \times \text{Abundance of Isotope 3}}{100} \right)

Given

  1. Mass of Isotope 1 = 35.9675u35.9675 \, \text{u}, Abundance = 0.33%0.33\%
  2. Mass of Isotope 2 = 37.9627u37.9627 \, \text{u}, Abundance = 0.063%0.063\%
  3. Mass of Isotope 3 = 39.9624u39.9624 \, \text{u}, Abundance = 99.60%99.60\%

Calculation:

Let's calculate the atomic mass step by step.

Atomic Mass=(35.9675×0.33100)+(37.9627×0.063100)+(39.9624×99.60100)\text{Atomic Mass} = \left( \frac{35.9675 \times 0.33}{100} \right) + \left( \frac{37.9627 \times 0.063}{100} \right) + \left( \frac{39.9624 \times 99.60}{100} \right)

Now, let's compute this.It seems there was an issue with the calculation tool. Let me proceed with manual steps to complete the computation.

Step 1: Contribution from Isotope 1

35.9675×0.33100=0.11869375u\frac{35.9675 \times 0.33}{100} = 0.11869375 \, \text{u}

Step 2: Contribution from Isotope 2

37.9627×0.063100=0.023916501u\frac{37.9627 \times 0.063}{100} = 0.023916501 \, \text{u}

Step 3: Contribution from Isotope 3

39.9624×99.60100=39.8029056u\frac{39.9624 \times 99.60}{100} = 39.8029056 \, \text{u}

Adding all contributions:

Atomic Mass=0.11869375+0.023916501+39.8029056=39.945515851u\text{Atomic Mass} = 0.11869375 + 0.023916501 + 39.8029056 = 39.945515851 \, \text{u}

Therefore, the atomic mass of the element is approximately 39.95 u.

Would you like more details or have any questions?

Related questions:

  1. What is the concept of isotopic abundance?
  2. How does the atomic mass of an element relate to its isotopes?
  3. Can isotopic mass and atomic mass ever be equal?
  4. How does the number of neutrons influence isotopic mass?
  5. What is the significance of atomic mass in the periodic table?

Tip:

The atomic mass reflects the average mass of an element’s isotopes, weighted by their natural abundance—crucial for chemical calculations and understanding element behavior.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Weighted Average
Isotopic Abundance
Atomic Mass Calculation

Formulas

Atomic Mass = (Mass of Isotope 1 × Abundance of Isotope 1 / 100) + (Mass of Isotope 2 × Abundance of Isotope 2 / 100) + (Mass of Isotope 3 × Abundance of Isotope 3 / 100)

Theorems

-

Suitable Grade Level

Grades 10-12