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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)

Respuesta :

maacastrobr

Answer:

  • The abundance of 107Ag is 51.5%.
  • The abundance of 109Ag is 48.5%.

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%.

mickymike92

The relative abundances of Ag-107 and Ag-109 are 43.5% and 56.5% respectively.

average atomic mass of elements

Average atomic mass = sum of mass of isotope * relative abundance

  • Average atomic mass of silver is 107.87 amu

Sum of relative abundances equals 100% or 1

let the relative abundance of Ag-107 be a,

let relative abundance of Ag-109 be b

  • a + b = 1
  • a = 1 - b

calculating the relative abundances:

107.87 = 107 * a + 109 * b

substitute a = 1 -b in the equation above

107.87 = 107 ( 1 - b) + 109b

107.87 = 107 -107b + 109b

107.87 - 107 = 2b

0.87 = 2b

b = 0.435

Then; a = 1 - 0.435

a = 0.565

Therefore, the relative abundances of Ag-107 and Ag-109 are 43.5% and 56.5% respectively.

Learn more about relative abundance and isotopes at: https://brainly.com/question/7988172

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