An element X with relative atomic mass 16.2 contains two isotopes168X with relative abundance of 90% and v8X with relative abundance of 10%. What is the value of m?

 

An element X with relative atomic mass 16.2 contains two isotopes¹⁶₈X with relative abundance of 90% and ^v₈X with relative abundance of 10%. What is the value of m?

Solution

Using the formula:

                                    Relative atomic mass = (M₁ X a₁) + (M₂ X a₂)

                                                                                     100                100

Where

M = Isotopic mass

a = percentage abundance

 

Relative atomic mass = 16.2

M₁ = 16

A₁ = 90%

M_{2 =} v

A₂ = 10%

 

16.2 = (16 X 90) + (m X 10)

                  100              100

 

16.2 = 14.4 + 0.1m

Collect like terms

16.2 – 14.4 = 0.1m

1.8 = 0.1m

Divide both sides by 0.1

1.8 = 0.1m

0.1     0.1       m = 18