Atomic Mass Chemistry Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard

An element has two isotopes. Isotope A has mass

10.01\,\text{amu}

and isotope B has mass

11.01\,\text{amu}

. The atomic mass listed on the periodic table is

10.81\,\text{amu}

. Calculate the percent abundance of each isotope.

Solution

1
Let $x$ = fraction of isotope A. Then $(1-x)$ = fraction of isotope B.
2
Set up equation: $10.01x + 11.01(1-x) = 10.81$ . Simplify: $10.01x + 11.01 - 11.01x = 10.81$ .
3
$-1.00x = -0.20$ , so $x = 0.20$ . Isotope A: $20.0\%$ , Isotope B: $80.0\%$ . (This is boron: ${}^{10}\text{B}$ and ${}^{11}\text{B}$ .)

Answer

\text{Isotope A: } 20.0\%,\quad \text{Isotope B: } 80.0\%

Working backward from the atomic mass to find isotope abundances requires setting up a system of equations. Since the atomic mass is closer to 11.01, isotope B must be more abundant.

About Atomic Mass

The weighted average mass of all naturally occurring isotopes of an element, expressed in atomic mass units (amu), where each isotope's mass is weighted by.

Learn more about Atomic Mass →

More Atomic Mass Examples

Example 1 easy

Define atomic mass and explain why the atomic mass of carbon is listed as [formula] rather than exac

Example 2 medium

Silicon has three stable isotopes: [formula] (92.23%, mass [formula]), [formula] (4.67%, mass [formu

Example 3 medium

Chlorine has two stable isotopes: [formula] (mass [formula], 75.77%) and [formula] (mass [formula],

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Related Concepts

Isotope Molar Mass