Question

An element has only two naturally occurring isotopes. You know that isotope 1 has a mass of 44.000 amu and isotope 2 has a mass of 47.000 amu. If the average atomic mass is 45.873 amu, what are the natural abundances in percent for the two isotopes? 14.00m

Answer 1

Answer:

isotope 1: 37.6%

isotope 2: 62.4%

Explanation:

As the element has only two naturally occurring isotopes, we can name them as:

x = mass of isotope 2; so x = 47.000amu

(1-x) = mass of isotope 1; so (1-x) = 44.000amu

Then the average atomic mass of the element must be:

$45.873=(47.000x)+(44.000(1-x))$

Solving for x, we have:

$45.873=47.000x+44.000-44.000x$

$45.873=44.000+3.000x$

$45.873-44.000=3.000x$

$1.873=3.000x$

$x=\frac{1.873}{3.000}$

$x=0.624$

So, with x we can find (1-x):

$1-x=1-0.624$

$1-x=0.376$

So, the abundance of isotope 1 is 37.6% and isotope 2 has an abundance of 62.4%.