% Abundace Calc Error

[BrianC]

here's the prob and my work can anyone help me figure out where i'm going wrong? thanks

The element silver has two naturally occurring isotopes with masses of 106.91 amu and 108.90 amu. What would be the percent abundance of the first isotope, if the atomic weight were 108.0429 amu?

108.0429 = 106.91*X + 108.90*Y (and since Y=1-X)
108.0429 = 106.91*X + 108.90 - 108.90*X
108.0429 - 108.90 = 106.91*X + 108.90*X
-(-1.99 = -0.8571*X) = 1.99 = 0.8571*X
X = 0.4307
Y= 0.5693

i did that and the online thing i enter the homework into says it's incorrect :confused:

[BrianC]

little typo in there the third line of my work should be
108.0429 - 108.90 = 106.91*X - 108.90*X

[BrianC]

-(-1.99 = -0.8571*X) = 1.99 = 0.8571*X
should be -(-1.99X = -0.8571) = (1.99X = 0.8571)

[Sirus]

Try converting your answers to percentage values instead of decimals.