Math Problem Statement

The atomic mass of calcium is 40.078 amu. Calcium has 2 common isotopes. One of the isotopes has a mass of 40.962 amu and occurs 46.323% of the time. (a) What is the % abundance of the other isotope? (b) What is the mass of the other isotope?

Solution

The problem provided involves the atomic mass of calcium, which is 40.078 amu. Calcium has two common isotopes, one of which has a mass of 40.962 amu and occurs 46.323% of the time. The question asks:

  1. a) What is the percentage abundance of the other isotope?
  2. b) What is the mass of the other isotope?

a) Find the percentage abundance of the other isotope:

The total percentage abundance for all isotopes must add up to 100%. Since one isotope has an abundance of 46.323%, the other isotope's abundance is:

Abundance of other isotope=100%46.323%=53.677%\text{Abundance of other isotope} = 100\% - 46.323\% = 53.677\%

b) Find the mass of the other isotope:

The average atomic mass of an element can be calculated by taking the weighted average of the masses of its isotopes. This is given by the equation:

Average atomic mass=(f1×m1)+(f2×m2)\text{Average atomic mass} = (f_1 \times m_1) + (f_2 \times m_2)

Where:

  • f1f_1 and f2f_2 are the fractional abundances of the two isotopes.
  • m1m_1 and m2m_2 are the masses of the two isotopes.

We are given:

  • Average atomic mass of calcium = 40.078 amu
  • Isotope 1: m1=40.962amum_1 = 40.962 \, \text{amu} and f1=46.323100=0.46323f_1 = \frac{46.323}{100} = 0.46323
  • Isotope 2: f2=53.677100=0.53677f_2 = \frac{53.677}{100} = 0.53677
  • We need to find m2m_2, the mass of the second isotope.

Now, substituting into the equation:

40.078=(0.46323×40.962)+(0.53677×m2)40.078 = (0.46323 \times 40.962) + (0.53677 \times m_2)

First, calculate the first term:

0.46323×40.962=18.9700.46323 \times 40.962 = 18.970

Now substitute this back into the equation:

40.078=18.970+(0.53677×m2)40.078 = 18.970 + (0.53677 \times m_2)

Subtract 18.970 from both sides:

40.07818.970=0.53677×m240.078 - 18.970 = 0.53677 \times m_2

21.108=0.53677×m221.108 = 0.53677 \times m_2

Now solve for m2m_2:

m2=21.1080.53677=39.32amum_2 = \frac{21.108}{0.53677} = 39.32 \, \text{amu}

Final Answers:

  • The percentage abundance of the other isotope is 53.677%.
  • The mass of the other isotope is approximately 39.32 amu.

Would you like further explanation on any of these steps? Here are a few related questions to deepen understanding:

  1. What would happen if the abundances of the isotopes were switched?
  2. How is the concept of atomic mass connected to the mole in chemistry?
  3. Why do elements have isotopes with different masses?
  4. What are the effects of isotopic abundance on chemical reactions?
  5. How does the mass spectrometer measure isotope abundance?

Tip: Remember, when dealing with isotopic abundances, always ensure that the sum of the percentage abundances equals 100%.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Weighted Average
Algebra

Formulas

Average atomic mass = (f1 * m1) + (f2 * m2)
Abundance of other isotope = 100% - abundance of first isotope

Theorems

-

Suitable Grade Level

Grades 10-12