Congratulations! You have discovered a new element called Bhauzium (Bz). The average atomic mass of Bz was found to be 348.72 amu and its atomic number is 123.
a) if the atomic masses of three isotopes of Bhsuzium are 342.92 amu, 356.43 amu (6.52%), and 351.29 amu, what is the relative abundance (%) of each isotope?
Let X = fraction of 342.92
356.43 is 6.52% or 0.0652 as a fraction.
351.29 = (1-X-0.0652) = 0.9348-X
Sum those values = 348.72
342.92X + 356.43(0.0652) + 351.29(0.9348-x) = 348.72
Solve for x which is the fraction of the first isotope and 0.9348-x is the fraction of the third isotope listed. The percentages should add to 100.0%.
To find the relative abundance of each isotope, we need to use the atomic masses and the percent abundance of the isotopes. Here's how you can calculate it:
1. Assign variables to the atomic masses and percent abundance of each isotope.
Let's use the following variables:
- Atomic mass of isotope 1 (A1) = 342.92 amu
- Atomic mass of isotope 2 (A2) = 356.43 amu
- Atomic mass of isotope 3 (A3) = 351.29 amu
- Percent abundance of isotope 2 (P2) = 6.52% = 6.52/100 = 0.0652
2. Calculate the percent abundance of isotope 1 (P1):
The total percent abundance should be 100%, so we can calculate P1 as:
P1 = 100% - P2
P1 = 100% - 6.52% = 93.48% = 93.48/100 = 0.9348
3. Calculate the percent abundance of isotope 3 (P3):
Since the total percent abundance should be 100%, we can calculate P3 as the remaining percentage:
P3 = 100% - (P1 + P2)
P3 = 100% - (93.48% + 6.52%) = 100% - 100% = 0
Therefore, the relative abundance (%) of each isotope is approximately as follows:
- Isotope 1: 93.48%
- Isotope 2: 6.52%
- Isotope 3: 0%
Note: Since the percent abundance of isotope 3 is 0%, it means that this isotope does not exist naturally for this element.