Question

An undiscovered element has three naturally occurring isotopes of X-55, X-57, and X-58. Isotope X-55 has an abundance of 27.80 % and isotope X-57 has an abundance of 44.39 %. What is the average mass of this element in amu?

Answer 1

Answer: The average atomic mass of this elements is 56.7221 amu.

Explanation: The average atomic mass is the sum of the masses of its isotopes each multiplied by their natural abundances.

`\text{average atomic mass}=\sum_(i=1)^(n)(mass)_i(\text{Fractional abundance})_i` .....(1)

`\text{Fractional abundance}=\frac{\%\text{ abundance}}{100}`

We are given 3 isotopes of an element.

For Isotope
`X^(55)`,

Mass = 55 amu

Fractional abundance = 0.2780

For isotope
`X^(57)`,

Mass = 57 amu

Fractional abundance = 0.4439

Total Fractional abundance = 1

For isotope
`X^(58)`,

Mass = 58 amu

Fractional abundance = Total abundance - abundances of the other isotopes

Fractional abundance = 1 - 0.7219

= 0.2781

Now, putting all the values in equation 1, we get

`\text{Average atomic mass}= (55 amu* 0.2780)+(57 amu* 0.4439)+(58 amu* 0.2781)`

Average atomic mass = 56.7221 amu.