Rubidium has two naturally occurring isotopes with masses 84.91 amu and 86.91 amu and has an atomic mass of 85.47 amu. part a which mass spectrum is most likely to correspond to a naturally occurring sample of rubidium?

Answer is: mass spectrum of isotope with mass 84.91 amu.

Ar₁ (Rb) = 84.91 amu; the average atomic mass of isotope 1.

Ar₂ (Rb) = 86.91; the average atomic mass of isotope 2.

ω₁ (Rb) = ?.

ω₂ (Rb) = ?.

ω (Rb) = ω₁ + ω₂.

Ar (Rb) = 85.47 amu; average atomic mass.

Ar (Rb) = Ar₁ (Rb) · ω₁ (Rb) + Ar₂ (Rb) · ω₂ (Rb).

85.47 amu = 84.91 amu · ω ₁ + 86.91 amu · ω₂.

ω ₁ = 100% - ω₂.

85.47 amu = 84.91 amu · (1 - ω₂) + 86.91 amu · ω₂.

2ω₂ = 0.56.

ω₂ = 0.28 = 28%.

ω₁ = 72%.

84.91 amu isotope will correspond to naturally occurring sample of Rubidium.

Explanation:

Let me explain, what is the question and what is actually required from us to calculate.

In this question, we have been asked to calculate percentage relative abundance of both the naturally occurring isotopes and check which one is most likely to correspond to a naturally occurring sample of Rubidium with an atomic mass of 85.47 amu.

So, the isotope with greater percentage relative abundance will surely correspond to sample of Rubidium. Let's calculate the percentage abundances of both the isotopes.

Note: Atomic masses that we see on periodic table are actually average atomic weight. And our answer lies in the formula for average atomic weight.

Formula:

Average atomic weight = (percentage abundance x Mass of 1st isotope) + (percentage abundance x Mass of 2nd Isotope).

Now we know following quantities already:

Average atomic weight of Rubidium = 85.47

mass of 1st isotope = 84.91

mass of 2nd isotope = 86.91

Total Percentage abundance will be = 100%

So, let's name the 1st percentage abundance as M and other one as 1-M.

Average atomic weight = (percentage abundance x Mass of 1st isotope) + (percentage abundance x Mass of 2nd Isotope).

85.47 = (M x 84.91) + ((1-M) x 86.91)

85.47 = (M x 84.91) + (86.91 - 86.91M)

85.47 = 84.91M + 86.91 - 86.91M

85.47 - 86.91 = M (84.91-86.91)

-1.44 = - 2 M

M = - 1.44 / - 2

M = 0.72 x 100%

M = 72% So, percentage of abundance of 1st isotope of mass 84.91 amu is 72%

Whereas, 1-M = 100% - 72% = 28%, this is the percentage abundance of 2nd Isotope of mass 86.91 amu.

It can be clearly seen that, percentage abundance of 1st isotope of mass 84.91 amu is greater than 2nd isotope of mass 86.91 amu. Hence, 1st isotope of mass 84.91 amu will most likely correspond to naturally occurring sample of Rubidium.