Question

A fictitious element Z has an average atomic mass of 223.06 u.223.06 u. Element Z has two naturally occuring isotopes. The more abundant isotope has an exact mass of 223.95 u223.95 u and a relative abundance of 65.51%.65.51%. Calculate the exact mass of the second isotope.

Answer 1

Answer:

221.37 u

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

For first isotope:

% = 65.51 %

Mass = 223.95 u

For second isotope:

Since the element has only 2 isotopes, so the percentage of second is 100 - first percentage.

% = 100 %  - 65.51 %  = 34.49 %

Let, Mass = x u

Given, Average Mass = 223.06 u

Thus,  

$223.06=\frac {65.51}{100}\times {223.95}+\frac {34.49}{100}\times {x}$

Solving for x, we get that:

x = 221.37 u

Thus mass of second isotope = 221.37 u