Assuming atomic masses of N-14 and N-15 to be exactly 14 and 15 amu respectively,
14x + 15 (1 - x) = 14.007,
where x*100% is the percent abundance of N-14, and (1 - x)*100% is the percent abundance of N-15.
From the equation above, x = 0.993 => percent abundance of N-14 is 99.3%, and the percent abundance of N-15 is 0.7%.
However, the actual masses of N-14 and N-15 isotopes rounded to three decimal places are 14.003 and 15.000 amu respectively. In this case
14.003x + 15.000 (1 - x) = 14.007;
x = 0.996.
=> percent abundance of N-14 is 99.6%, and the percent abundance of N-15 is 0.4%.
Answer: N-14 - 99.3%, N-15 - 0.7% assuming isotope masses to be 14 and 15 respectively;
N-14 - 99.6%, N-15 - 0.4% considering their real (actual) masses.
Comments
Dear Anna Patrice B. Azarse, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!
This is very helpful for my class assignments
Leave a comment