Question

An element has four isotopes, each differing by an atomic mass of 1. The ratio of their abundance from the lightest to heaviest isotope is 1:2:2:3, respectively. Determine the average atomic mass of the element, if the mass of the heaviest isotope is​

Answer 1

Answer:

A+⅛

Step-by-step explanation:

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Answer 2

Question,

An element has four isotopes, each differing by an atomic mass of 1. The ratio of their abundance from the lightest to heaviest isotope is 1:2:2:3, respectively. Determine the average atomic mass of the element, if the mass of the heaviest isotope is​ A.

Answer,

The average atomic mass of the element is A - $\frac{9}{8}$

Given,

An element has 4 isotopes, each contrasting by an atomic mass of 1.

The ratio of their abundance from the lightest to heaviest isotope → 1:2:2:3

Mass of the heaviest isotope →​ A

To Find,

The average atomic mass of the element

Solution,

The Average atomic mass of the element is given by -

Average atomic mass = M₁f₁ + M₂f₂ + M₃f₃ + M₄f₄ + .....

M → Mass of Isotope

f → Fraction of Abundance of the isotope

The Mass of the heaviest isotope = A = M₄

Therefore,

M₃ = A - 1

M₂ = A - 2

M₁ = A - 3

Now, the Fraction of Abundance of the isotope is given as

f = $\frac{Abundance of isotope}{Total abundance}$

f = $\frac{Abundance of isotope}{1+2+2+3} = \frac{Abundance of isotope}{8}$

f₁ = $\frac{1}{8}$

f₂ = $\frac{2}{8}$

f₃ = $\frac{2}{8}$

f₄ = $\frac{3}{8}$

Average atomic mass = M₁f₁ + M₂f₂ + M₃f₃ + M₄f₄

Average atomic mass = (A - 3)($\frac{1}{8}$) + (A - 2)($\frac{2}{8}$) + (A - 1)($\frac{2}{8}$) + (A)($\frac{3}{8}$)

Average atomic mass = $\frac{A - 3+2A-4+2A-2+3A}{8}$

Average atomic mass = $\frac{8A-9}{8}$ = A - $\frac{9}{8}$

Thus, the average atomic mass of the element is A - $\frac{9}{8}$

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