Atomic Structure and Chemical Bonding — Question 2

Let,

the percentage of ¹⁶₈A be X and

percentage of ¹⁸₈A be 100-X

Average atomic mass = $\dfrac{(16 \times \text{X})+[18 \times \text{(100-X)}]}{100}$ = 16.2

⇒ 16X + 1800 - 18X = 1620

⇒ -2X + 1800 = 1620

⇒ -2X = 1620 - 1800

⇒ 2X = 180

⇒ X = $\dfrac{180}{2}$ = 90

∴ 100 - X = 100 - 90 = 10

Hence, percentage of the two isotopes in this sample are 90% of ¹⁶₈A and 10% of ¹⁸₈A