Question

1. pure nitrogen is found as a mixture of 2 isotopes, n - 14 and n - 15. nitrogens atomic mass on the periodic table is 14.01 amu. what does this tell you about percent abundance of n - 14 and n - 15 in any nitrogen sample?

Explanation:

Step1: Set up the equation

Let the percent - abundance of N - 14 be $x$ (expressed as a decimal), then the percent - abundance of N - 15 is $1 - x$. The atomic mass of N - 14 is 14 amu and of N - 15 is 15 amu. The average atomic mass formula is $Average\ atomic\ mass=(Mass\ of\ isotope1\times Abundance\ of\ isotope1)+(Mass\ of\ isotope2\times Abundance\ of\ isotope2)$. So, $14.01 = 14x+15(1 - x)$.

Step2: Expand and solve the equation

Expand the right - hand side: $14.01=14x + 15-15x$. Combine like terms: $14.01=15 - x$. Rearrange to solve for $x$: $x = 15 - 14.01=0.99$.

Step3: Convert to percentage

The percent - abundance of N - 14 is $0.99\times100 = 99\%$, and the percent - abundance of N - 15 is $(1 - 0.99)\times100 = 1\%$.

Answer:

The percent abundance of N - 14 is 99% and the percent abundance of N - 15 is 1%.