Chromium has an atomic mass of 51.9961 u and consists of four isotopes, cr50, cr52, cr53, and cr54. the cr52 isotope has a natural abundance of 83.79% and an atomic mass of 51.9405 u. the cr54 isotope has a natural abundance of 2.37% and an atomic mass of 53.9389 u. the natural abundances of the cr50 and cr53 isotopes exist in a ratio of 1: 0.4579, and the cr50 isotope has an atomic mass of 49.9460 u. determine the atomic mass of the cr53 isotope.

56.4459u

Explanation:

Given parameters:

Atomic mass of Cr = 51.9961u

Isotopes listed = Cr50, Cr52, Cr53 and Cr4

Natural abundance of Cr52 = 83.79%    Atomic mass = 51.9405u

Natural abundance of Cr54 = 2.37%      Atomic mass = 53.9389u

Natural abundance of Cr50:Cr53 = 1:0.4579 Atomic mass of Cr50 = 49.9460u

Unknown:

Atomic mass of Cr53 = ?

Solution:

Isotopes of an element have the same electronic configuration. They differ only in their masses. Isotopes have been studied extensively using the mass spectrometry method. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occur in nature.

The proportion is called the geonormal abundance or simply abundance. From this proportion, the relative atomic mass(RAM) of an element can be evaluated. The RAM is the average mass of all isotopes of an element.

RAM = m₁α₁ + m₂α₂ + m₃α₃ + ........... + mₙαₙ

where m is the mass of a given isotope

           α is the abundance of the isotope

To solve the problem we need to establish the relationship between the given atomic masses, their relative atomic mass and the abundances of the isotope:

  RAM = m₅₂α₅₂ + m₅₀α₅₀ + m₅₃α₅₃ + m₅₄α₅₄

The unknown in the expression here is the m₅₃, atomic mass of Cr53

We need to derive the abundances α₅₃ and α₅₀.

Given that their isotopic natural abundances is 1:0.4579;

The sum of natural abundances when expressed as percentage is equal to 100:

Therefore;

natural abundance of α₅₃ and α₅₀ = 100 - (83.79+2.37) = 13.84%

Since they occur in ratio 1:0.4579, we can determine their proportions.

sum of the ratio = 1+0.4579 = 1.4579

Proportion of α₅₃ = x 13.84  = 4.3469%

Proportion of α₅₀ =   x 13.84u = 9.4931%

Since we now know the abundances of the remaining isotopes, we can forge ahead to find the unknown atomic mass of Cr53.

Evaluating

RAM = m₅₂α₅₂ + m₅₀α₅₀ + m₅₃α₅₃ + m₅₄α₅₄

51.9961 =( 51.9405 x ) + ( 49.9460 x + (m₅₃ x ) + (53.9389 x )

51.9961 = 43.5209 + 4.7414 + 0.0435m₅₃  + 1.2784

0.0435m₅₃ = 51.9961 - 49.5407

0.0435m₅₃ = 2.4554

           m₅₃ = = 56.4459u

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.