Question

Given that natural sample of iron has isotopes fe 54 fe 56 fe57 in the ratio of 5%,90%,5% respectively .what will be the average atomic mass of fe?

Answer 1

Explanation:

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Answer 2

Given,

A natural sample of iron has isotopes Fe 54, Fe 56, Fe 57 in the ratio of 5%, 90%, 5%, respectively.

To find,

The average atomic mass of Fe.

Solution,

We can simply solve this numerical problem by using the following process:

Let us assume that a sample contains 100 random atoms of Fe.

Mathematically,

The average atomic mass of any element

= (sum of the products of the individual isotopic atomic masses with their respective number)/(total number of atoms of all the isotopes)

According to the question;

Number of Fe 54 atoms in the sample = 5

Number of Fe 56 atoms in the sample = 90

Number of Fe 57 atoms in the sample = 5

Now,

The average atomic mass of Fe

= (sum of the products of the individual isotopic atomic masses with their respective number)/(total number of atoms of all the isotopes)

= {(54×5) + (56×90)+ (57×5)}/(5+90+5)

= {(270) + (5040)+ (285)}/(5+90+5)

= 5595/100 = 55.95

Hence, the average atomic mass of Fe is equal to 55.95.