Silver (ag) has two stable isotopes: 107ag, 106.90 amu, and 109ag, 108.90 amu. if the average atomic mass of silver is 107.87 amu,
what is the natural abundance of each isotope? (expressed as a percentage)

Explanation:

The average atomic mass of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we use the second equation to express A1 in terms of A2:

A1 = 1.0 - A2

We replace A1 in the first equation:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We use the value of A2 to calculate A1 in the second equation:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

I hope you won't be solid on this answer, But I researched it with my colleagues, and we all agreed it's the amount of atoms in the compound. (I hope this was you question.)