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A fictitious element that has two isotopes has an average atomic mass of 117.3. If 65.43% of the atoms of the element have an atomic mass of 119.8, what is the atomic mass of the other isotope?

1 Answer

Sep 6, 2017

Approx... $113 \cdot amu$ $113*"amu"$

Explanation:

The weighted average of the isotopes is equal to $117.3 \cdot amu$ $117.3*"amu"$ .

And thus.........with units of $amu$ $"amu"$

$[65.43 % \times 119.8 + 34.57 % \times χ] = 117.3$ $[65.43%xx119.8+34.57%xxchi] =117.3$

Where $χ \equiv mass of the other isotope$ $chi-="mass of the other isotope"$ ....

And so we solve for $χ$ $chi$ ....

$χ = \frac{117.3 - 0.6543 \times 119.8}{0.3457} = 112.6 \cdot amu$ $chi=(117.3-0.6543xx119.8)/(0.3457)=112.6*"amu"$

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