Silver has two naturally occurring isotopes. Ag- 107 with a mass of 106.905 amu and a natural abundance of 51.84 % and Ag-109. How do you use the atomic mass of silver listed in the periodic table to determine the mass of Ag-109?

Md Ariful Islam

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Md Ariful Islam

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Using my (not very precise) periodic table, I get the mass of Ag as 107.87.

This must be a combination of these two . So the % of the other isotope must be 48.16%.
To work with these percentages in an equation I'll use them as decimals.

So now I can start putting them into an equation:
$107.87 = (106.905 * 0.5184) + (x * 0.4816)$
Because both of the isotope masses, multiplied by their percentage, would give us the overall average mass.
$107.87 = 55.419552 * (x * 0.4816)$
Worked out the brackets that we can work out, and then rearrange to give:
$107.87 - 55.419552 = (x * 0.4816)$
Again rearrange:
$52.450448 = (x * 0.4816)$
And final rearranging:
$x = 52.450448 / 0.4816$
$x = 108.909$ (using same sig figs as given in question)

You probably want to work through that again using your periodic table value for Ag.

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